# Science: asteroid vs. hero physics



## Janx (Aug 28, 2018)

I'm working on a short story due in a few weeks (for an anthology, not school).  I need all my physicsful friends ideas to adjust the plan.

The big climax is the hero has just bonded with her new super sci-fi space suit that will let her fly up and help protect the earth from asteroids and stuff.  She can fly, propel matter, and using pico-printer-gel*, disassemble and reassemble matter.  

In the current draft, she has about 15 minutes to fly up and prevent an asteroid that's going to splatter the earth.  This is the "last one", since the previous about a 100 years ago (which aligned with Asteroid Bennnu's description and timing).  Mankind can't survive another hit.

There's limits to how much mass she can move, we don't have numbers, but the asteroid is too big.  She's a sniper (back on earth) and tends to deal with problems distantly.  I decided she'd have to get close for this one (and thus risk her life).

Since she can't just superman shove the asteroid off course, I thought she'd try aikido.  Let's say she just got up to space and there's 5 minutes left.  The asteroid is on her left (relatively) and coming in to kill us all.  She flies right, around the planet, picking up speed (a lot of speed), gathering up matter to build a lattice that she can ram/cushion with.  She rejoins the asteroid from behind and pushes with all her super-pushy-power.  Thus adding speed along a vector the asteroid is mostly going, and thus, causing it to skirt past the earth instead of slaughtering the last of humanity.

I don't need exact math, but I think there's a hole in the idea.  To get around a planet of 24K miles circumference in under 5 minutes is a whole lot of speed and acceleration.  Assuming she can now handle higher g-forces, wouldn't her new speed be very much faster than the asteroid, causing her to splat, instead of line up to shove?

Is there a better solution for dealing with an asteroid in 15 minutes (5 assuming it takes 10 to get up there)?

I don't think Nukes has the right vibe and lots of articles disproved that as a solution.

Is the original plan workable if I adjust the effective starting point for the asteroid so it will be far enough away and have just the right speed for her to join it by orbiting the planet once?

I can play with the parameters a bit since it's just fiction, but I need a short timeline and personal risk.  Her flinging an object at it might work, but how would she herself be at risk to add tension?

This is the big finish to my post-Ragnarok coming of age biker cultist story about a fallen angel.

Thanks!


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## Umbran (Aug 28, 2018)

Hm.  This is a good one.  

The kinetic energy of a non-relativistic object is 1/2 M V^2.  So, there are two basic ways for it to have a lot of KE.  One is for it to be big, and the other is for it to be fast.  Note that the energy goes linearly with the mass, but goes like the square of the velocity.  Double the mass, you double the KE.  Double the velocity, and you *quadruple* the energy.

And, part of making this work, or at least be cinematically plausible - make the asteroid less small, but faster.  That, overall, makes it easier to push around.

Now, not, pushing "along a vector it is mostly going" doesn't help.  Any portion of her effort that moves along the line of the asteroid's current motion makes it reach Earth faster, and hit harder.  She needs to add velocity more or less perpendicular to its current motion.  

We can set up things to minimize how much she needs to do.  If the thing is aimed dead center, she needs to move it 4000 miles in order to be a clear miss.  If we set it up as an off-center, almost glancing blow, then she only needs to move the thing a few hundred miles.  Mind you, she needs to move the thing a few hundred miles in 5 minutes.  If it is, say, 400 miles in 5 minutes is like 4800 miles per hour.  She has to add 5000 mph *to the mass of an asteroid* and that's if she takes the asteroid from its current path to +4800 MPH instantly.  If she takes time to accelerate it, the first moments are slow, and the end speed must be even higher to succeed.

Note - this is basically saying that she, alone, can create an impact roughly equivalent to the asteroid's impact herself.  She is capable of ending civilization.  Is that the power level of your hero?


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## Nagol (Aug 28, 2018)

The problem with dealing with asteroids with a (very) short timeframe is the any asteroid that can harm the denizens of Earth has to have a lot of mass and since you have so little time to react, tremendous forces need to be brought to bear.  

If you want to keep the time pressure, perhaps you could move the asteroid further out and unless its vector is changed in the next 5 minutes, the impact will be unavoidable? Assuming the answer is no, let's do some back of the envelope math!

I'll assume the rock is about 100m in radius and has a density of about 3, that means it masses 12 million tonnes.

Because the timeframe is short, you have to adjust the vector by a substantial amount.  That mass is moving at quite a clip (~11 km/s).  If you have 300 seconds, you're intercepting the asteroid about 3500 km from Earth.  Earth has a radius just over 6,000 km.  Worst case is a centre strike which would mean you'd need to change its vector by 45 degrees.  But let's assume the impact is projected to be a grazing hit about 1000 km from the edge which means you'd need to shift its angle by about 20 degrees.  

Now that shift needs to be instantaneous.  Every second of delay/acceleration means the distance is closing and the angle needs to become more extreme.  Vectors at these speeds are mostly additive so you need to impart a velocity of about 3 km/s which would have the rock just skim the atmosphere -- assuming we ignore gravitational attraction.  Let's bump that up to 3.5 km/s for a basic safety margin.  That change is what you'd need if you could impart it instantly.  Let's assume you spend a whole minute of acceleration and thus need to achieve a higher velocity of about 6 km/s to simplify the math. A = v / t so 6km/60 seconds = 100 m/s-squared or about 10 g of acceleration.  F = ma, so 12 billion kg x 100 m / s-squared is 1.2 trillion Newtons of force which is the same as 1.2 terawatts needs to be applied for the whole 60 seconds.

Regardless of how our hero applies the force -- whether directly to the asteroid threat or through a passive intermediary, the amount of energy remains the same.  Since you want the hero to use an intermediary, there needs to be something that acts as a force multiplier.  You don't have sufficient time to use the Earth's gravity well so let's look elsewhere.

One simple force multiplier would be a secondary large rock -- say a half the radius of the threat that is moving in a different orbit that is closer to the preferential vector that can be adjusted to impact the threat.  With a mass of about 10% of the target rock, the energy required to change its vector is equally reduced.  If the asteroids are mostly metallic then you could even potentially set up an elastic collision.  In an perfect scenario, the larger mass would have its velocity changed by ~5% of the closing velocity between the rocks.


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## Janx (Aug 28, 2018)

thanks for getting back to me so quick.

To umbran's question about her power level: not sure.  The space suit was part of a fleet of these for defending the earth from ballistic attack (mass drivers pushing big rocks from Mars).  So using math for evil would probably mean yes, she could.  I want her constrained that she can't just fly up and superman nudge the threat away.

I was trying to come up with a different approach to solving the problem (instead of brute force), and aikido inspired me to have her use the rock's momentum/direction to her advantage. Sounds like for that to work, she needs to push perpendicular, not "with" the general direction of the rock.

I can change how long it takes to get from ground to space, and then from space to the rock.  Mostly, I wanted a short enough time to be urgent, she she had to leave right away, and be able to exposition what's going on to the folks back home over radio for the first part of it, because people have a totally wrong understanding of what happened.

the rock should be aimed off-center, it's a misfire anyway, and 100 years later, we've come back around to meet it.  I can change the 100 years to whatever sounds plausible.  The first strike was timed to align with Asteroid Bennu (a real supposed danger).  Since this was pushed toward earth, it can be smaller, with a higher velocity.

Other than space debris (dead satelites, etc), I'm not sure if there's another rock floating around for her to grab.  Ironically enough, a named asteroid like Bennu coming by at just this same time would make for opportunity to be used as ammo.  

Or she could just grab enough matter to make a space bullet and fly with it so she can maintain acceleration on it, thus killing or deflecting the asteroid with a bullet/herself.  It's heroic, and it's the kind of idea a sniper would come up with.

So, it sounds like she needs a bullet that is about half the mass of the threat.  Umbran suggests a smaller, faster rock.  Not sure how small would work, to justify how much mass she could feasibly gather (think all that wrecked stuff from Gravity).  Maybe there's 1-2 tons worth of movie-metal to grab in a hurry? Space is big and that's kinda bullcrap, but let's pretend that's possible for her to scoop up while zooming along the right path.  I'm guessing that should be the limit of her power to move/manipulate anyway.

Is it feasible to launch a 4 ton rock from the asteroid belt toward earth, miss (let's say somebody had second thoughts after the first launch and interfered) and have it eventually come around a century later and be a threat that our last Valkyrie can solve? Would that 4 ton rock be a threat?


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## Nagol (Aug 28, 2018)

Janx said:


> thanks for getting back to me so quick.
> 
> To umbran's question about her power level: not sure.  The space suit was part of a fleet of these for defending the earth from ballistic attack (mass drivers pushing big rocks from Mars).  So using math for evil would probably mean yes, she could.  I want her constrained that she can't just fly up and superman nudge the threat away.
> 
> ...




A 4 ton rock would need so much velocity to be a threat that if it missed on the first pass it would exit the solar system.


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## Nagol (Aug 28, 2018)

The only way an intermediary helps is if it already has energy that can be used against the target.  If you have to pick up the target and accelerate it, it is only helpful if you have extra time or if energy output of the suit is not the limiting factor for the acceleration.  In other words, if the suit has the force to accelerate a 100 kg human at 50 g, but humans go unconscious at 3 g then the hero could afford to pick up and accelerate about another 1500 kg and stay conscious.

Let's change the asteroid's composition to have density of 8 (solid iron) and set the orbit to a ~100 year elliptical.  The math for calculating the instantaneous velocity isn't hard, particularly, but I'm rusty enough not to do it on the back of an envelope.  Let's guess the velocity at 1 AU would be about 3 times the base velocity of a circular orbit with the same period or 20 km/s and further assume its going counter to the planetary motion so we can increase the closing velocity to 50 km/s.

To achieve the same kinetic energy of my original assumption 12 million tonnes at 11 km /s, we have ~600,000 tonnes at 50 km/s and it has about a 25 m radius.

Let's push the rock further out to say the hero intercepts it 1 hour from Earth (about 180,000 km out).  The necessary deflection is much smaller.  We still need to shift about 1,000 km, but we have 3,600 seconds to complete the shift or merely about 350 m /s. 

If she picks up 1,500 kg and accelerates at 2 g for the whole trip, it takes a bit over an hour to arrive (4200 seconds) and will be travelling a bit over 40 km/s compared to the Earth.  That gives a relative closing velocity on the target of about 80 km/s.  The 1,500 kg mass in an elastic collision would cause the target to gain about 400 m/s vector (and thus miss Earth) and the 1,500 kg slug would bounce back at nearly 80 km/s.  Since it unlikely the slug could survive those g-forces, the hero would have to evade the expanding shrapnel cloud.

Actually, this assumes a side strike appropriately perpendicular to the target's forward motion.  So toss in another hour or so of travel time to line the shot up.


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## tomBitonti (Aug 28, 2018)

To be accurate, be careful to include both energy and momentum in the physics of the collision.  If I have the right sense of it, a very fast small object hitting a larger relatively still object either, with an elastic collision, bounces off at high speed (transferring momentum and a little energy) or with an inelastic collision does a lot of local damage and transfers almost the same momentum.  (This is a classic physics test problem, with the key observation being whether the collision is elastic or not, which tells you what analysis to do.)

As what Umbran said: The push should be perpendicular to the line of motion of the asteroid.  A perpendicular push will create the greatest deviation.  There is a case of a skimming trajectory, where the straight line motion of the asteroid would miss, but gravity curves the asteroid path into the earth, where it seems that speeding up the asteroid should work.

Some other considerations: How far away the asteroid starts from the earth, and at what speed, compared with escape velocity (11.2 km/s) and orbital velocity for a circular orbit (about 7.6 km/s): If the asteroid is left with less than escape velocity, the asteroid may go into orbit, or may impact the earth at a later time because of the particulars of the changed orbit the asteroid.

Thx!
TomB


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## Janx (Aug 28, 2018)

the suit is made of pico-printer-gel (tinier than nanites) and it can modify/repair her such that she can probably handle twice the G.  I made that up of course, but thus far, it's in specification for what I'd already figured out.  Calling it a suit is really a simplification for our discussion. 

So that might mean that she can go twice as fast, turning that hour long flight into aprox. 30 minutes (or as you added an hour, an hour instead of two).

I can still make it tense and "guys I need to go now" because she really does need to get there while it's still farther away.  Add in radio fade out after the Q&A with the folks back home , a bit of time passes and as she lines up for the shot, bring the radio back in for final words.

I can even have the science guy explain it to the rest of the party, "of course, she needs to achieve a speed of 80km/s and impact at a perpendicular angle to blah-blah" or some such.  

I don't want to cite too many numbers, but I wanted to have the strategy be interesting, heroic, and plausible if one of y'all crunched numbers.

I think I can make this work and be closer to plausible, since it is a sci-fi story.

Glad I asked y'all.  Let me know if there's any holes with my recap of what I think you said, or a different way to approach the problem (besides let everybody die, which I already dealt with


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## Janx (Aug 28, 2018)

tomBitonti said:


> To be accurate, be careful to include both energy and momentum in the physics of the collision.  If I have the right sense of it, a very fast small object hitting a larger relatively still object either, with an elastic collision, bounces off at high speed (transferring momentum and a little energy) or with an inelastic collision does a lot of local damage and transfers almost the same momentum.  (This is a classic physics test problem, with the key observation being whether the collision is elastic or not, which tells you what analysis to do.)
> 
> As what Umbran said: The push should be perpendicular to the line of motion of the asteroid.  A perpendicular push will create the greatest deviation.  There is a case of a skimming trajectory, where the straight line motion of the asteroid would miss, but gravity curves the asteroid path into the earth, where it seems that speeding up the asteroid should work.
> 
> ...




so are you saying, speeding up the asteroid might actually work?

And you raise the other elephant in the room, if the asteroid doesn't hit the earth (she wins), it's still out there, ready to be a problem again...

Any other solutions to that, besides let it hit the earth?  Or is there a way to hit the earth safely?  Or build a parachute for it? or wings to land it like a shuttle?

I don't think she can drill it fast enough or disolve it with her pico-printer-gel (think my version of nano-tech, but smaller)


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## Eltab (Aug 28, 2018)

Flying up to the Really Big Rock and doing a Superman trick sounds like a bit of a cliché.  (Let her best friend tell her so.)  Let's make her smart and clever as well as tough.

She wants to plant a motor on the asteroid and let THAT push it off collision course.  This could be a solar sail (very fragile thing provides very low-G push over a very long time) or a Saturn V moon rocket (tough robust structure provides high-G push over a short time).  Either way, she does the math and realizes she must go NOW.  Her friends, the parts suppliers, &c all think there's plenty of time no hurry.  If she proposes the solar sail and shows the physics but everybody blows her off because they watched _Armageddon_ as kids and now they think a moon rocket full of nukes will do the job, we set up that everybody acknowledges the seriousness of the situation but not the urgency.  That allows some people to change their minds and help her out once they think it through.

"Victory" for Our Heroine may be that she breaks up the Really Big Rock into several smaller parts that mere mortals can handle, plus a lot of gravel and fragments that make a very spectacular meteor shower as they burn up on re-entry.  This may look too much like that old movie, though...

As the writer you can align everything as you want.  The asteroid is arriving from a direction such that, if you push it to one side enough, the Moon's gravity does some more of the work for her.  (Hmmm, her computer simulations account for the Moon's gravity but the official simulation doesn't; she finds out the Asteroid is going to hit THIS time, not NEXT time, it crosses Earth's orbit.  Must hurry!)

Or … the incoming Really Big Rock is actually a Really Big Iceball (comet).  If you can make only one side melt, the natural jets (which normally create the tail and halo) will push it off its original course.  The solar sail can perform double duty IF it is in place early enough and can be installed in the right position.

Our Heroine may have to hack NASA's supercomputers along the way, to figure out how to place the sail so all the physics tricks can work together.


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## Tonguez (Aug 28, 2018)

So could the pico-gel be used to collect up all the space debris in orbit and use that as her bullet?

Of course there’s the power of narrativium which would allow cool effects like having the asteroid graze the atmosphere until her suit channels te increase energy to expel the asteroid in  fiery aurora explosion the lights up the sky around the world (ie every thing works through the power of cool)


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## tomBitonti (Aug 28, 2018)

Janx said:


> so are you saying, speeding up the asteroid might actually work?
> 
> And you raise the other elephant in the room, if the asteroid doesn't hit the earth (she wins), it's still out there, ready to be a problem again...
> 
> ...




Yes to speeding up the asteroid ... in very specific initial conditions.  (Think of a putt that grazes the cup going too fast.)

Yup, it's not just enough to perturb the initial trajectory.  Unless the asteroid is booted out of orbit entirely, it either needs to get into a stable orbit, or it will hit earth anyways, say, after looping away for a short bit.

A solution might be to dissolve the asteroid into a dispersed mist (or maybe a very fine gravel?).  I'm not sure what will happen in those cases.  Say, have the pico-gel go van-neumann (start self replicating, by doing an ultra dangerous removal of safeguards against such,"But the manual said never every enable full self replication!") and have it burrow through the asteroid and break it into millions of fragments.

Thx!
TomB


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## Nagol (Aug 28, 2018)

tomBitonti said:


> Yes to speeding up the asteroid ... in very specific initial conditions.  (Think of a putt that grazes the cup going too fast.)
> 
> Yup, it's not just enough to perturb the initial trajectory.  Unless the asteroid is booted out of orbit entirely, it either needs to get into a stable orbit, or it will hit earth anyways, say, after looping away for a short bit.
> 
> ...




Considering the thing is in a century orbit, the perturbation is likely to be insufficient to yank it into a secondary pass.  It's more likely to fly away and become a threat again in a few thousand years (maybe).  If the hero is a particularly lucky math genius then the perturbation could either (a) reduce eccentricity and keep the asteroid outside Earth orbit or (b) have it hit the moon as it moves away from the Earth.

Converting the asteroid to dust would have potential if there was time for the dust to start to disperse.  Dust or a single solid lump, it contains the same kinetic energy -- dust would just tend to release it faster as it hit the atmosphere.  More *boom* less splat.


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## Ovinomancer (Aug 29, 2018)

Hmm.  Earth is moving pretty quickly in space, so, depending on approach vectors, speeding the asteroid up or slowing it down a bit should generate enough 'miss' if you're far enough out.  

For example, if the asteroid is crossing the orbit of Earth perpendicularly, the window to strike Earth is only 13,000 km / 30 km/s or about 435 seconds (about 7 and a quarter minutes).  Assuming the asteroid aimed precisely at the center of Earth, you'd need to speed up or slow down the approach by half that 7 minute time, or about 4 minutes to be safe (240 seconds).  A 4 minute delta in arrival time depends on the velocity of the asteroid and how long you take to start the acceleration.  Assuming a 50 km/s speed on the asteroid, 4 minutes at 1 hour distance is a ratio:  needed speed over current speed = needed time over original time.  Or needed speed = original speed x (new time/old time).  In this case, needed speed = 50km/s x (56 minutes/60 minutes) = 50 km/s x (.933) = 46.67 km/s.  You'd need to generate a delta-v of 3.33 km/s at exactly 1 hour out (or earlier) to cause a complete miss.  I think this is lower than the necessary delta-v to accelerate the mass sideways 6500 km over an hour (6500km/360 seconds = 18 km/s, so, yep, 6 times less).

Now, if your geometry has the asteroid coming in much closer to the orbital direction, this changes, as earth's relative movement is much less with respect to the path of the asteroid.  However, a rock launched from the asteroid belt some time in the past should have a high angle of intercept to Earth's orbit, as it would be on a steep elliptical around the Sun.  But, assuming a near head on collision, say at about a 20 degree approach, combined delta-v would be 30km/s (Earth) plus cosine(20)x50km/s or about 77 km/s.  Earth relative speed on the perpendicular would be sin(20)x30km/s or about 10km/s.  This puts a Earth on-target time window of 1300 seconds or 21.7 minutes.  Again assuming a center strike, that's a time window of 11 minutes you have to modify.  Needed speed up/slow down delta-v at 1 hour would be 14.7 km/s.  This is still less than the needed perpendicular delta-v to cause a miss, although not much.

Orbital mechanics are funny.  Often speeding up or slowing down along your path causes huge changes in orbit, while orthogonal thrusts change orbits more slowly.  For the purposes of your story, gathering up a bunch of debris and smacking the asteroid just enough to slow it down would be sufficient for a high angle of intercept.  This is the most likely scenario given the orbital geometries of a rock launched from the belt that's coming around again.  

A neat way to tell this story might be that the angle of impact is such that meeting the asteroid and speeding it up is the right call (this would be the case if the asteroid was going to hit on the forward half of the Earth).  Timing could be such that the push means the ateroid scrapes the atmo and skips, causing an awesome light show and placing the heroine at risk.


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## Janx (Aug 29, 2018)

Tonguez said:


> So could the pico-gel be used to collect up all the space debris in orbit and use that as her bullet?
> 
> Of course there’s the power of narrativium which would allow cool effects like having the asteroid graze the atmosphere until her suit channels te increase energy to expel the asteroid in  fiery aurora explosion the lights up the sky around the world (ie every thing works through the power of cool)




adding more details (ain't like y'all are gonna spoil it for readers).  The suit is a Vid Liv Kostym, a swedish acronym I made up with Google Translate to spell VLK which thus smells like Valkyrie.  It's got mass driver bracelet/anklets, solar panel wings bonded to the back and a happy coat of pico-printer-gel for general protection and utility.  Plus BalDR, the Ballistic Defense Ring, satellites hosting an AI that does all the heavy lifting, math-wise.

So the mass drivers can move matter around up to some range and weight limit.  They'd be how she scoops up space debris to make a shield or something.  The pico-printer-gel does the fine detail work of re-organizing that matter when it's brought into range to make electronics, repairs, heal injuries by re-organizing atoms (hence why I went tinier than nano-tech).

All of it limited by the power of narrativium as Tonquez surmises.  Basically if it sounds good or plausible within what I described above, but not too powerful.  Much like Umbran asking if she could kill the planet.  I guess, but I already got an asteroid on the way to do that so it ain't OP right now...


Tom's idea to replicate the pico-gel is good.  I had similar thoughts but ruled them out for purposes of just dissolving the asteroid to save the day was "too easy"  Presumably, something about the total mass vs. time isn't possible.

Now I could have it that Mars has been sending these every year or something, rather than looping around a century later.  Mars doesn't play directly into the story, it's just the big final reveal of how nobody understood what really happened to Earth.


So to recap the ideas:
perpendicular attack to deflect
speedup to skip might work
dissolve it - not allowed per original restriction of too easy unless we think of some complication
put a rocket on it and push (hauling a rocket up might be too heavy or slow)
build a mass-driver on it and move itself (variant of rocket idea)


Did I get everything? There's a lot of posts since I got back, sorry if I missed something.  I'll be re-reading these after I finish my night-shift of work...

I'm hoping for one that is risky to her, and being non-obvious. Extra points for if its something a sniper would think of OR if it required her to get close and personal (the opposite of a sniper's mentality).  Also, in the current draft, I only have 300 words left in my 6K budget.

BTW, thanks to all of you for participating.  Good ideas from everybody.  I had a hunch that whatever I thought of might have a hole that y'all would find, which would ruin the story for some. Plus, hopefully this is fun.


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## Umbran (Aug 29, 2018)

Ovinomancer said:


> Orbital mechanics are funny.




Yes, they are.  But in this situation, those are not so telling, unless you have much longer timescales to work on.  The lightsail idea mentioned upthread can work on these premises - much lower thrust applied over days, months, or even years, to subtly change orbits.

Over the course of minutes as given in this scenario it is somewhat easier (and more understandable to an audience that isn't made of rocket scientists) to take this from the point of view of the Earth as the center of the coordinate system.  In that, we simply have a target (the Earth) and a projectile (the asteroid).


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## Ovinomancer (Aug 29, 2018)

Umbran said:


> Yes, they are.  But in this situation, those are not so telling, unless you have much longer timescales to work on.  The lightsail idea mentioned upthread can work on these premises - much lower thrust applied over days, months, or even years, to subtly change orbits.
> 
> Over the course of minutes as given in this scenario it is somewhat easier (and more understandable to an audience that isn't made of rocket scientists) to take this from the point of view of the Earth as the center of the coordinate system.  In that, we simply have a target (the Earth) and a projectile (the asteroid).



Treating Earth as a fixed point completely ignores the geometry of the problem.  The asteroid is essentially a bullet fired leading Earth, on a rendevous to meet it, not a bullet fired directly at a stationary Earth.

There's a 6x difference in needed delta-v 1 hour out by treating the problem as an orbital one rather than a fixed Earth one (3.3 km/s vs 18 km/s).  This was calculated as instantaneous delta-v at one hour out, so I have no idea where light sails come into that.  Heck, at 5 mins out, ypu need to slow the object by 40km/s (80% of it's 50km/s speed!) to cause a miss, but you'd need 108km/s (216%!!) to generate the same miss by perpendicular thrust.  Ignoring Earth's 30km/s speed and the fact this is a rendezvous problem not a fixed target problem is the oversight, here.

The acceleration needed will vary by object mass, but the delta-v doesn't care.  For the 1 hour out case, and assuming a 1EE14 kg mass at 50km/s (roughly the same as the Chicxulbub impact), let's look at a 10 minute burn application centered at 1 hour out.  The needed delta-v to slow or speed the stroid is 3.3km/s.  Over 10 minutes, the a is 5.5m/(s x s).  The F needed is 5.5EE14 N.  The lateral shove needs an a of 30m/(ss), and an F of 3EE15 N.  An order of magnitude more force!

So, no, no lightsails or slow accelerations here.


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## Janx (Aug 29, 2018)

Ovinomancer said:


> Treating Earth as a fixed point completely ignores the geometry of the problem.  The asteroid is essentially a bullet fired leading Earth, on a rendevous to meet it, not a bullet fired directly at a stationary Earth.
> 
> There's a 6x difference in needed delta-v 1 hour out by treating the problem as an orbital one rather than a fixed Earth one (3.3 km/s vs 18 km/s).  This was calculated as instantaneous delta-v at one hour out, so I have no idea where light sails come into that.  Heck, at 5 mins out, ypu need to slow the object by 40km/s (80% of it's 50km/s speed!) to cause a miss, but you'd need 108km/s (216%!!) to generate the same miss by perpendicular thrust.  Ignoring Earth's 30km/s speed and the fact this is a rendezvous problem not a fixed target problem is the oversight, here.
> 
> ...




so to recap this (which tests my rudimentary understanding as I digest and keep it simple in the story):

the planet-killer asteroid  has a mass of 1EE14 kg and is traveling at 50km/s (same size as Chicxulb)

by realizing that the rock is leading the moving target of Earth (a sniper thing to do), messing up that timing will save the day

If she can get behind it and speed the rock up, it takes less force than a pushing from the side, which is something she might be able to do

I can make it scrape the atmosphere for a lightshow and heat up, since she's personally pushing this thing, thus creating the danger through art of wordsmithing. To solve the repeat-performance problem, I can have her set the pico-printer-gel to chewing away at it, which in another hundred years, probably wouldn't be a problem. I'll come up with something to explain how she got so much force, but it's easy to justify in words that this is more feasible than pushing from the side once I throw in "order of magnitude"

I don't know how much techno-babble I'll put in, but y'all have given enough detail that if this version of the plan is plausible, I can take it from here.

Thanks!


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## Eltab (Aug 29, 2018)

Larry Niven's novel _The Smoke Ring_ has a poem (6 lines or so) about what direction a thing in orbit will really move when you push on it, compared to the things around it.  It might help you visualize the changing situation as Our Heroine is working on the problem.

Once in a while, borrowing everything I read from the library has a downside: I can't just copy the poem here.


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## Janx (Aug 29, 2018)

another wrinkle in the heroics is catching up to this thing and matching speed to then get behind it to push.

that seems like it could be action packed, but also complicated.  unlike superman, where smashing headlong into it would be easy, if she flies toward it, she's got momentum in the wrong direction once she gets there.  I assume she could zip out to some midpoint, then reverse direction and bring herself back up to speed as the asteroid catches up

I'm sure the mass-driver system she uses defies conventional physics, but...


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## Ovinomancer (Aug 29, 2018)

Janx said:


> another wrinkle in the heroics is catching up to this thing and matching speed to then get behind it to push.
> 
> that seems like it could be action packed, but also complicated.  unlike superman, where smashing headlong into it would be easy, if she flies toward it, she's got momentum in the wrong direction once she gets there.  I assume she could zip out to some midpoint, then reverse direction and bring herself back up to speed as the asteroid catches up
> 
> I'm sure the mass-driver system she uses defies conventional physics, but...



Space is BIG.  This asteroid moving at 50km/s still takes two hours to go between the moon's orbit and Earth.  You could slingshot the moon to come up behind a la Armegeddom but flip the expectation by shoving it past the Earth rather than blowing it up.

By the by, the close pass to Earth would likely fling the asterood out of a passing orbit, so no need to destroy it before it comes back -- it most likely won't.  An easy, and not unlikely, out is that it's thrown into the Sun after the near miss.


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## MarkB (Aug 29, 2018)

If you want to play up the sniper angle, maybe make the asteroid particularly elongated, maybe even barbell-shaped so it has a weak point in the middle.

The way it's oriented coming in, it's going to hit the atmosphere end-on and carve right through, like a bullet punching through armour, losing only a little mass to friction before striking the ground at almost full orbital velocity, causing catastrophic damage.

But if, instead of trying to divert it, she re-orients it so that it slams into the atmosphere side-on, maybe with some additional tumble, the stress of atmospheric entry will shatter it. Some of the mass will still make it to the ground, but at a shallower angle, and the majority of the asteroid will be broken down into small enough chunks that it burns up in the atmosphere.

Not a clean win, but enough to reduce it from a dinosaur-killer to, say, a really bad day for several hundred square miles plus a side-serving of nuclear winter.

But the only way to give it plenty of spin and still ensure that it hits atmosphere at the right orientation is for her to ride it all they way down to the edge of space, making last-minute adjustments as she goes - thus the extra peril.


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## Mustrum_Ridcully (Aug 30, 2018)

I am not sure the distances could possibly work out with the 5 minute time frame, but maybe instead of trying to deflect it to avoid hitting Earth, you just need to deflect it so it hits the Moon?


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## Eltab (Aug 30, 2018)

Mustrum_Ridcully said:


> I am not sure the distances could possibly work out with the 5 minute time frame, but maybe instead of trying to deflect it to avoid hitting Earth, you just need to deflect it so it hits the Moon?




Our Heroine adopts this plan, begins pushing.  She tells herself that she'll get out of the way when the heat of re-entry makes her uncomfortable.  Only a few minutes later does she realize: the Moon has no atmosphere so there won't BE any 'heat of re-entry' … and there's almost no room or time to get out from underneath the onrushing Really Big Rock.


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## Janx (Aug 30, 2018)

Mustrum_Ridcully said:


> I am not sure the distances could possibly work out with the 5 minute time frame, but maybe instead of trying to deflect it to avoid hitting Earth, you just need to deflect it so it hits the Moon?




I might need to expand the time frame to an hour or so, depends on how close to math correctness I need for verisimilitude.

Technically, I bet it's less distance from ground to "space" than it is from "space" to rock, so even in the original 15 minutes, I bet I can shorten how much the initial flight is to get a few more minutes on the secondary stage.

Mostly, I chose a time that seemed urgent and could be divided into stages for purposes of story telling.  This is all act 3 of a 5700 word story, so it's got to move somewhat quickly and keep the action pace up.  It's not quite like the usual asteroid movies as the first two acts are about other wrong views of why the world is like it is (post impact 100 years ago). Pacing wise, I've got it where I want it, but I'd like the science to be closer to right so if y'all read it, I wouldn't look like lousy science fiction writer


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## freyar (Aug 31, 2018)

I haven't bothered with the numbers, since I think those have probably been covered.  But assuming a "glancing blow" scenario, hitting the asteroid perpendicularly is the way to go.  What really matters is change in momentum, not really energy, though your heroine will need a lot of energy to have enough momentum to transfer to the rock.



Ovinomancer said:


> Treating Earth as a fixed point completely ignores the geometry of the problem.  The asteroid is essentially a bullet fired leading Earth, on a rendevous to meet it, not a bullet fired directly at a stationary Earth.




I did want to address this point.  Over the timescales we're talking about (say an hour, even up to a day or two), the earth is moving at close to a constant velocity.  So it is actually perfectly fine and even easier to think about everything from the earth's point of view (ie, "rest frame of the earth"), where the earth is sitting still and the asteroid is coming straight at it.  This is in fact the way the characters in the story, coming from earth, would think about it.  There's no need to worry about the earth's orbit.  Think about the asteroid like a bullet shot at a stationary earth all you like.


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## Ovinomancer (Aug 31, 2018)

freyar said:


> I haven't bothered with the numbers, since I think those have probably been covered.  But assuming a "glancing blow" scenario, hitting the asteroid perpendicularly is the way to go.  What really matters is change in momentum, not really energy, though your heroine will need a lot of energy to have enough momentum to transfer to the rock.
> 
> 
> 
> I did want to address this point.  Over the timescales we're talking about (say an hour, even up to a day or two), the earth is moving at close to a constant velocity.  So it is actually perfectly fine and even easier to think about everything from the earth's point of view (ie, "rest frame of the earth"), where the earth is sitting still and the asteroid is coming straight at it.  This is in fact the way the characters in the story, coming from earth, would think about it.  There's no need to worry about the earth's orbit.  Think about the asteroid like a bullet shot at a stationary earth all you like.




This is... not even wrong.  To attempt to explain, again, both the Earth and the asteroid are orbiting the Sun.  The Earth is in constant motion along it's orbital path, as is the asteroid.  The point of impact is where those two lines cross.  In other words, the asteroid isn't moving towards the Earth, it's moving towards _where the Earth is going to be_.

Imagine trying to hit a moving car with a baseball from the field next to the road the car is travelling along.  If you throw the baseball at the car, you will miss because by the time the baseball arrives at the road, the car will have already traveled past that point.  Instead, you throw the baseball ahead of the car so that when the baseball gets to the road, it _meets_ the car there. 

In this scenario, the hero is like a wind pushing the baseball after you throw it.  The car is moving so fast that it takes less wind to slow the thrown ball so that the car zooms past the rendezvous than it does to push the ball to the side so that it gets there at the same time but far enough off to one side.  Largely this is because, for reference, the car is huge and moving really fast so that a small change in speed leverages the speed of the car to cause the miss rather than having to push the baseball off target by half the length of the car from zero lateral speed.

If you insist on treating Earth as the center point, then the asteroid is going to appear to move under constantly changing acceleration (because it has to include Earth's orbital movement, which is elliptical, alongside it's own highly elliptical orbit, which is going to do weird things -- other planets appear to occasionally do loop-de-loops in the sky, for instance).  You're essentially recreating the problem of predicting planetary motions in an Earth-centric universe, a field of rather complicated mathematical modelling.  There's a reason everything got a lot easier (but not exactly easy) to predict when we moved to a heliocentric model.  Don't ignore Kepler!


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## tglassy (Aug 31, 2018)

So, didn't bother reading all 27 posts, so forgive me if this has been said.

How big is your asteroid?  Cause if you're looking at something, say, the size of Texas (like in Armageddon), then what you have done is create a scenario that you are unable to win.  You are constraining your hero to be unable to simply push it, and simply "flying real fast around the earth and hitting it perpendicularly so it goes off course" wouldn't work.  She's too small.  Even a few tons of mass would simply be too small.  It wouldn't push something massive like that off course, it would puncture a hole through it and go through the other side.  

If Superman really tried catching a falling plane by grabbing the nose, he wouldn't be able to.  The amount of strength needed to catch the plane would be FOCUSED on his FINGERS, focusing all that power into minute points, which would simply rip through the plane's skin like paper, not affecting the rest of the plane at all.  The plane isn't strong enough, structurally, to be able to withstand that level of force.  

In the same way, depending on the make up of the asteroid, I don't believe it would simply impact on the surface, diverting the asteroid.  It would be like a bullet.  She'd rip right through the thing.

Now, if it was smaller, and going faster, I still don't see her diverting it, but potentially shattering it.  If she hit the most dense portion of the asteroid, punching a hole through it, it could potentially disrupt the structure enough to make it break apart.  

Also hitting the larger asteroid with something of equal size, or of enough size to do anything, wouldn't push it off course.  The sheer mass and energy of both objects colliding would cause both objects to shatter, sending most of that debris to earth.  Or, well, it might actually give Earth a ring.  That would be cool.  

If the big asteroid would NOT shatter at something near its own mass slamming into it at those speeds, then the thing is so damn solid that the earth probably wouldn't even stop it.  It would punch through atmosphere, earth and mantle and rip straight through the core and out the other side without even slowing down.  

It's possible I'm not as scientifically accurate, but perhaps the answer lies in what the asteroid is made of.  If its mostly rock, granite and the like, have her analyze its structural weak point and shatter it with pinpoint precision.  If it's mostly metal... I don't know what to tell you, there.


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## tomBitonti (Aug 31, 2018)

Ovinomancer said:


> This is... not even wrong.  To attempt to explain, again, both the Earth and the asteroid are orbiting the Sun.  The Earth is in constant motion along it's orbital path, as is the asteroid.  The point of impact is where those two lines cross.  In other words, the asteroid isn't moving towards the Earth, it's moving towards _where the Earth is going to be_.
> 
> Imagine trying to hit a moving car with a baseball from the field next to the road the car is travelling along.  If you throw the baseball at the car, you will miss because by the time the baseball arrives at the road, the car will have already traveled past that point.  Instead, you throw the baseball ahead of the car so that when the baseball gets to the road, it _meets_ the car there.
> 
> ...




Changing from the earth’s Frame of reference and a sun centered frame of reference, a vector perpendicular to the velocity of the moon in the earth frame is no longer perpendicular in a solar frame. The shift is from 30-60 degrees, with the earth moving perpendicular to the moon in the solar frame, and varying the relative speeds from 1/2 to 2.

Over the course of a day, the earth’s motion shifts by one degree.  That doesn’t seem to change the situation much, in the sense of how you would want to adjust the asteroid’s movement.

In all of this, is the deflection caused by the earth’s gravity big enough to matter?  The effect in the earth frame puts the asteroid on an arc instead of a line, with (I’m thinking) more curvature closer to the earth.

Thx!
TomB


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## tomBitonti (Aug 31, 2018)

tglassy said:


> So, didn't bother reading all 27 posts, so forgive me if this has been said.
> 
> How big is your asteroid?  Cause if you're looking at something, say, the size of Texas (like in Armageddon), then what you have done is create a scenario that you are unable to win.  You are constraining your hero to be unable to simply push it, and simply "flying real fast around the earth and hitting it perpendicularly so it goes off course" wouldn't work.  She's too small.  Even a few tons of mass would simply be too small.  It wouldn't push something massive like that off course, it would puncture a hole through it and go through the other side.
> 
> ...




Yeah, Superman would simple travel straight on through.  But, a high speed impact of normal matter would turn into an explosion.

Thx!
TomB


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## Ovinomancer (Aug 31, 2018)

tomBitonti said:


> Changing from the earth’s Frame of reference and a sun centered frame of reference, a vector perpendicular to the velocity of the moon in the earth frame is no longer perpendicular in a solar frame. The shift is from 30-60 degrees, with the earth moving perpendicular to the moon in the solar frame, and varying the relative speeds from 1/2 to 2.



Huh?  I can't make heads or tails of that.  You seem to be talking about a soecific scenario, but haven't presented it.  I'm completely lost on your relative speed thing, though.


> Over the course of a day, the earth’s motion shifts by one degree.  That doesn’t seem to change the situation much, in the sense of how you would want to adjust the asteroid’s movement.



Well, that 1 degree is about 2.5 million km (per day), so you tell me if that seems like a lot.


> In all of this, is the deflection caused by the earth’s gravity big enough to matter?  The effect in the earth frame puts the asteroid on an arc instead of a line, with (I’m thinking) more curvature closer to the earth.
> 
> Thx!
> TomB




Not to the impact, actually.  At the speeds involved, the acceleration due to Earth's gravity is mostly going to be seen in the outgoing trajectory.  Near Earth Orbital speed is low, comparitively.  The ISS is going about 7 km/s, so the asteroid is an order of magnitude faster.  It will be bent, but the fastest acceleration will come when it's passing Earth, so it matters little for the intercept.

Go slower, and it matters more.


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## Janx (Aug 31, 2018)

tglassy said:


> So, didn't bother reading all 27 posts, so forgive me if this has been said.
> 
> How big is your asteroid?  Cause if you're looking at something, say, the size of Texas (like in Armageddon), then what you have done is create a scenario that you are unable to win.  You are constraining your hero to be unable to simply push it, and simply "flying real fast around the earth and hitting it perpendicularly so it goes off course" wouldn't work.  She's too small.  Even a few tons of mass would simply be too small.  It wouldn't push something massive like that off course, it would puncture a hole through it and go through the other side.




I think we're going with these stats:
the planet-killer asteroid has a mass of 1EE14 kg and is traveling at 50km/s (same size as Chicxulb)

I don't know if it could be smaller. I don't know how big that is, it's just a number, and it killed a lot of good dinosaurs.

Oviniomancer says it's less energy (still alot) to push it faster, past its intercept point, than to hit it perpendicular (another order of magnitude of fig newtons needed and in a post-disaster world, fig newtons are rare).

The hero can scoop up and reconfigure matter in a localized area (let's say withing 30 feet of her suit).  So she can break a number of physics rules, up to a point.  Let's say she can gather and push up to 2 tons of matter (could lift a car). It's more like like telekinesis, but yeah.

You raise a good question, that might enable other tactics.  What's a typical asteroid that big made of?  Can a 2 ton bullet driven by her, pierce and shatter the asteroid.

In other fun facts, her name is Pierce. So I kinda had that in mind way back when I started the story...


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## Nagol (Aug 31, 2018)

Janx said:


> I think we're going with these stats:
> the planet-killer asteroid has a mass of 1EE14 kg and is traveling at 50km/s (same size as Chicxulb)
> 
> I don't know if it could be smaller. I don't know how big that is, it's just a number, and it killed a lot of good dinosaurs.
> ...




1EE14 is 100,000,000,000,000 kg or 100,000,000,000 tonnes or 100,000 mega tonnes.  You don't need that much speed with something that size: typical asteroid collision speeds of 11 km/s is more than sufficient. 

A 2 tonne impactor isn't going to cause it to wobble let alone impart enough velocity to cause it to miss.  Momentum transfer is necessary and momentum is proportional to both mass and velocity.  My original calc of 1,500 kg vs. 600,000 tonnes was off by 3 orders of magnitude (I swapped kg for tonnes when I did the math).

Shattering the asteroid is only helpful if there is sufficient time for the debris to scatter and actually miss the Earth.  The energy released from the pieces striking is the same as if the object was one piece -- whether it is released in the atmosphere or on impact.  Air release may be less damaging, but at these scales it'll still be devastating.

There are 3 main types of asteroid: nickel-iron, rock, and more rarely carbonaceous.  Chicxulb was probably the latter.  Since this impactor is actually launched by an alien force, nickel-iron probably a better option -- the volume is smaller and they can use electromagnetic forces to change its orbit.  100,000,000,000 tonnes is about 12,000,000,000 cubic metres or a cube 2.3 km on a side.


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## tomBitonti (Aug 31, 2018)

Re: what angle to push the asteroid.  If the earth has motion perpendicular to the asteroid, the angle appears different if you examine it the frame of reference which is moving with the earth vs the angle in a sun centered frame of reference.

Put it like this:  Earth towards the bottom, moving upwards.  Asteroid to the right, moving to the left.  Paths intersect in the top left.  That’s the solar view.  The view moving with the earth simply has the earth to the left and the asteroid to the right, with the asteroid moving directly leftwards towards the earth.

In the earth centric view, a perpendicular push on the asteroid points directly down.  Translating that into the sun centric view, the push is no longer perpendicular to the motion of the moon.  If the earth is moving at the same speed as the asteroid, the angle in a solar frame seems to be 45 degrees.  If the earth is moving at half the speed, 30 degrees, and if at twice the speed, 60 degrees.  As a check, if the earth is still, the angle is the same, and if the earth is moving a lot (10x) faster the angle gets close to 90 degrees.  In which case you really are mostly speeding up or slowing down the asteroid — just in the sun centric view.

This is meant to unify the two perspective: Both are correct.

Thx!
TomB


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## Ovinomancer (Sep 1, 2018)

tomBitonti said:


> Re: what angle to push the asteroid.  If the earth has motion perpendicular to the asteroid, the angle appears different if you examine it the frame of reference which is moving with the earth vs the angle in a sun centered frame of reference.
> 
> Put it like this:  Earth towards the bottom, moving upwards.  Asteroid to the right, moving to the left.  Paths intersect in the top left.  That’s the solar view.  The view moving with the earth simply has the earth to the left and the asteroid to the right, with the asteroid moving directly leftwards towards the earth.
> 
> ...



Took me a few readings, but I follow you here, I just don't see the point.  Your example here shows that treating this from the Earth's perspective and applying a perpendicular push to the apparent motion of the asteroid means that you're being very inefficient because you aren't fully slowing the asteroid nor are you committing fully to pushing it to the side.  It'll work, but you're borrowing the worst of both for no benefit.  Also, Earth doesn't change speeds, so... maybe you meant the asteroid?  Yes, if the asteroid approaches infinite speed, the inefficient push approaches the line of motion, but that's not helpful anywhere.

This whole translation doesn't illuminate the issues, so I'm confused as to what your purpose is?


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## freyar (Sep 1, 2018)

Ovinomancer said:


> This is... not even wrong.  To attempt to explain, again, both the Earth and the asteroid are orbiting the Sun.  The Earth is in constant motion along it's orbital path, as is the asteroid.  The point of impact is where those two lines cross.  In other words, the asteroid isn't moving towards the Earth, it's moving towards _where the Earth is going to be_.




You missed what I said about being in the rest frame of the earth *over a short time*.  For someone sitting on the earth, in the last little bit before the asteroid hits, it's going to look like the asteroid is coming straight at the earth.  Over longer periods of time, yes, you're right that you have to take the earth's acceleration around the sun into account. But not over the course of a few hours. Looking at the numbers, the difference between the earth's actual orbit and a constant velocity path is less than the earth's radius for about a 12-hour period, so that's about how long we can neglect the acceleration (sorry, I was just guessing when I said a day or two).



> Imagine trying to hit a moving car with a baseball from the field next to the road the car is travelling along.  If you throw the baseball at the car, you will miss because by the time the baseball arrives at the road, the car will have already traveled past that point.  Instead, you throw the baseball ahead of the car so that when the baseball gets to the road, it _meets_ the car there.



That's in the rest frame of the baseball field.  In the rest frame of the car --- assuming no acceleration, which would change your scenario as well --- the baseball has to come straight at the car, or it will miss.  You and the baseball field are, however, moving compared to the car.  This is a simple example of change of reference frame from introductory physics.




> In this scenario, the hero is like a wind pushing the baseball after you throw it.  The car is moving so fast that it takes less wind to slow the thrown ball so that the car zooms past the rendezvous than it does to push the ball to the side so that it gets there at the same time but far enough off to one side.  Largely this is because, for reference, the car is huge and moving really fast so that a small change in speed leverages the speed of the car to cause the miss rather than having to push the baseball off target by half the length of the car from zero lateral speed.




Depends on what frame you're measuring the wind's speed from.  In the earth's frame (the natural one for our hero), slowing down the asteroid won't stop it from hitting the earth.  It will just make it hit later.



> If you insist on treating Earth as the center point, then the asteroid is going to appear to move under constantly changing acceleration (because it has to include Earth's orbital movement, which is elliptical, alongside it's own highly elliptical orbit, which is going to do weird things -- other planets appear to occasionally do loop-de-loops in the sky, for instance).  You're essentially recreating the problem of predicting planetary motions in an Earth-centric universe, a field of rather complicated mathematical modelling.  There's a reason everything got a lot easier (but not exactly easy) to predict when we moved to a heliocentric model.  Don't ignore Kepler!




That's only if you're talking about longer time periods than this problem.  The OP was talking about a 5-minute time frame, which seems to have been extended maybe to a few hours.  If you want to talk about a day, you have to worry about this a little.  More than a day or two, yes, I agree you have to worry about the full orbital mechanics.  Kepler is certainly important over the course of a year, but one thing we learn in physics is how to know when different effects are actually important.


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## freyar (Sep 1, 2018)

Ovinomancer said:


> Took me a few readings, but I follow you here, I just don't see the point.  Your example here shows that treating this from the Earth's perspective and applying a perpendicular push to the apparent motion of the asteroid means that you're being very inefficient because you aren't fully slowing the asteroid nor are you committing fully to pushing it to the side.  It'll work, but you're borrowing the worst of both for no benefit.  Also, Earth doesn't change speeds, so... maybe you meant the asteroid?  Yes, if the asteroid approaches infinite speed, the inefficient push approaches the line of motion, but that's not helpful anywhere.
> 
> This whole translation doesn't illuminate the issues, so I'm confused as to what your purpose is?




The point is that the push needs to be perpendicular to the asteroid's motion in the earth's rest frame (again assuming short time periods).  In the heliocentric frame, it will be angled differently, depending on the orientation of the asteroid's motion with respect to the earth's.


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## Ovinomancer (Sep 1, 2018)

Ovinomancer said:


> Hmm.  Earth is moving pretty quickly in space, so, depending on approach vectors, speeding the asteroid up or slowing it down a bit should generate enough 'miss' if you're far enough out.
> 
> For example, if the asteroid is crossing the orbit of Earth perpendicularly, the window to strike Earth is only 13,000 km / 30 km/s or about 435 seconds (about 7 and a quarter minutes).  Assuming the asteroid aimed precisely at the center of Earth, you'd need to speed up or slow down the approach by half that 7 minute time, or about 4 minutes to be safe (240 seconds).  A 4 minute delta in arrival time depends on the velocity of the asteroid and how long you take to start the acceleration.  Assuming a 50 km/s speed on the asteroid, 4 minutes at 1 hour distance is a ratio:  needed speed over current speed = needed time over original time.  Or needed speed = original speed x (new time/old time).  In this case, needed speed = 50km/s x (56 minutes/60 minutes) = 50 km/s x (.933) = 46.67 km/s.  You'd need to generate a delta-v of 3.33 km/s at exactly 1 hour out (or earlier) to cause a complete miss.  I think this is lower than the necessary delta-v to accelerate the mass sideways 6500 km over an hour (6500km/360 seconds = 18 km/s, so, yep, 6 times less).
> 
> ...




I reviewed this math, and I made an error that changes the analysis.  I made a magnitude error in the step where I calculated the deflection speed at 1 hour when I used 360 seconds instead of 3600 seconds. This error was propogated.  At 50km/s, it's better to generate velocity lateral to the movement of the asteroid in the solar frame, not slow it.  There is a breakpoint in speed where slowing is better, but it's dependent on time.  At 1 hour, that breakpoint is slightly less than 30km/s.

Mea culpa, [MENTION=8835]Janx[/MENTION].


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## Ovinomancer (Sep 1, 2018)

freyar said:


> The point is that the push needs to be perpendicular to the asteroid's motion in the earth's rest frame (again assuming short time periods).  In the heliocentric frame, it will be angled differently, depending on the orientation of the asteroid's motion with respect to the earth's.




I disagree.  The problem here is the same one you make about -- when you shifted the frame from the road/field to the car, you DON'T shift the vector of the wind -- it remains the same.  Ergo, if you go with an Earth centric reference frame, the asteroid's motion is actually comprised of it's vector and the Earth's speed vector, creating the apparent straight line movement towards Earth.  If you restart the Earth, you don't change the Asteroid's original vector, you just recover the Earth's vector back to Earth.  The push that was perpendicular in the Earth frame is still in the same direction, which is now diagonal to the asteroid's path.  This means you're both slowing the asteroid AND pushing it to the side.  At 5 minutes out, the breakpoint speed where slowing becomes more efficient than pushing it to the side is:


Speed of Earth = 30 km/s; radius of Earth (with atmo at a bit of slop): 6500km
Time needed for Earth to move out of the way: 6500km/30km/s = 216.7 seconds.  This is the time you need to generate by slowing the asteroid for Earth to get out of the way.

S1 * T1 = S2 * T2.  Rearranging, we find S2 = S1 * T1/T2.  T2 = T1 + 216.7s.  T1 = 300 s (five minutes).  So, simplifying, S2 = S1 *(300s/516.7s) -> S2 = 0.581*S1.

21.667 km/s - speed needed to push perpendicular to the asteroid's path to cause a miss (6500km/300 seconds <-- right this time!).  Note the similarity in the time for Earth to move and this.  Just a fun fact, no importance.

Setting S2 equal to 21.667 km/s, we get an S1 of 37.3 km/s.

So long as the asteroid's speed isn't exactly 37.3 km/s, your perpendicular push in the Earth reference frame will be less efficient than a different angle at generating a miss.

Here's a graphic showing the Earth centric frame of three scenarios:  a 20km/s asteroid 5 minutes out in green, a 30km/s asteroid 5 minutes out in purple, and a 50km/s asteroid 5 minutes out in yellow-orange.  The big things here is that the graphic is _to scale._  The Earth is the proper size, and the path of the asteroids is what they will appear to travel over the 5 minutes.


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## tglassy (Sep 1, 2018)

So....I’m going to say go with the Rule of Cool. 

Find the solution that sounds the coolest, and go with that. Go watch One Punch Man. He’s awesome. Completely invulnerable, can jump to the moon and back, and can, literally, kill anything with one punch. He’s freaking amazing. Nobody cares about the science behind it. 

Now, it’s great to have some science to make things make sense. That way, you can create one or two “fictional” things, like a super suit, and use the science to know how it would interact with the rest of the world. 

But if you stick with the Rule of Cool, you can’t go wrong.


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## Tonguez (Sep 1, 2018)

Would a cascade of multiple pick bullets be better than one punch through? She collects multiple debris bullets drags them in orbit around the planet and sets them all to collide with the asteroid ....


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## Nagol (Sep 1, 2018)

Tonguez said:


> Would a cascade of multiple pick bullets be better than one punch through? She collects multiple debris bullets drags them in orbit around the planet and sets them all to collide with the asteroid ....




Momentum is effectively mass * velocity.  Multiple impacts aren't "better" than a single impact in the sense that you need at *least* as much momentum transfer from the multiple impacts as you do from a single impact.  I say at least because the velocity you need to impart grows the closer the target gets to Earth so time to impact matters a lot.  If you can set up a single impact to hit at 1 minute, and multiple impacts, the last of which hits at one minute, then the multiple impact option might be better if you have an easier time arranging multiples, say for example, your suit can only accelerate a few tonnes at a time.  If you only have an hour to disaster then if time to final impact is 10 minutes, you will probably need to accelerate substantially more mass than a single hit at one minute.  The other issue is you'll have to set up the shots guessing about how the earlier shots will affect the target.  It's size isn't very large in the scheme of things and small uncertainties in its mass could easily lead to complete misses of your later shots.

With a much longer time horizon, multiple impacts can be much more useful in the sense that you get small changes with each impact and those changes could lead to having more time available.  Multiple impacts might also allow you time to reevaluate and to recover from a mishap more easily. You also reduce the possibility of secondary unfortunate consequence -- like say breaking the object into two and needing to prevent both parts from striking or responding to a miss/unexpected result.


----------



## tomBitonti (Sep 1, 2018)

> I disagree. The problem here is the same one you make about -- when you shifted the frame from the road/field to the car, you DON'T shift the vector of the wind -- it remains the same. Ergo, if you go with an Earth centric reference frame, the asteroid's motion is actually comprised of it's vector and the Earth's speed vector, creating the apparent straight line movement towards Earth. If you restart the Earth, you don't change the Asteroid's original vector, you just recover the Earth's vector back to Earth. The push that was perpendicular in the Earth frame is still in the same direction, which is now diagonal to the asteroid's path. This means you're both slowing the asteroid AND pushing it to the side. At 5 minutes out, the breakpoint speed where slowing becomes more efficient than pushing it to the side is:




This seems to be both agreeing and disagreeing: In the earth centric frame, the push is perpendicular to the motion of the asteroid in that frame.  That same push, when transormed into a sun centered frame, can be as you say, mostly speeding up or slowing down the asteroid, and just a little to the side, depending on the relative motions of the earth and of the asteroid.

The reason for using an earth centered frame of reference is that the problem and the resulting maths are much simpler.

A theme here which is important is that there is *no* special (“distinguished”) frame of reference.  Just more and less natural and more and less useful ones.

In this problem, in the Earth frame of reference, is seems much simpler to find the most efficient push to give to the asteroid.  Once that most efficient push is determined, it’s simple to transform to other frames of interest, say, a sun centered frame, or a frame that moves with the asteroid.

Thx!
TomB


----------



## tomBitonti (Sep 1, 2018)

Tonguez said:


> Would a cascade of multiple pick bullets be better than one punch through? She collects multiple debris bullets drags them in orbit around the planet and sets them all to collide with the asteroid ....




Lots of small pushes instead of one big one.  That seems safer in not causing the asteroid to shatter.

But, it would still very likely be overwhelmed by the hugeness of the asteroid.  I wonder - over how much area must the push be spread to stay within the strength of material of the asteroid?

Thx!
TomB


----------



## Janx (Sep 1, 2018)

tomBitonti said:


> Lots of small pushes instead of one big one.  That seems safer in not causing the asteroid to shatter.
> 
> But, it would still very likely be overwhelmed by the hugeness of the asteroid.  I wonder - over how much area must the push be spread to stay within the strength of material of the asteroid?
> 
> ...




plus I have to consider how much mass is floating up in orbit and how much time it takes to scoop up.  Scooping up a shredded ISS might be feasible.

It sounds like we're back to the perpendicular push per Ovinomancer's corrected math.

If she can scoop up enough material and turn it into a big mass driver(TM) mounted on the side of the asteroid, that solves the fiction vs. physics of big numbers.  We could make the risk be that she has to power it, and it's gonna be close with her back to the earth as it scrapes by.


----------



## tomBitonti (Sep 1, 2018)

Janx said:


> plus I have to consider how much mass is floating up in orbit and how much time it takes to scoop up.  Scooping up a shredded ISS might be feasible.
> 
> It sounds like we're back to the perpendicular push per Ovinomancer's corrected math.
> 
> If she can scoop up enough material and turn it into a big mass driver(TM) mounted on the side of the asteroid, that solves the fiction vs. physics of big numbers.  We could make the risk be that she has to power it, and it's gonna be close with her back to the earth as it scrapes by.




With a mass driver you could scoop up stuff from the asteroid itself!

That would work for a distant asteroid, but doesn’t seem to work for an an asteroid which is at most a day away.

Thx!
TomB


----------



## MarkB (Sep 1, 2018)

So, this molecular 3D-printing stuff, how sophisticated and how quick is it? One option would be to use it to break down matter on the asteroid, turn it into pico-printer-gel manufacturers, and then use those in turn to break down more of it, in an exponential replication loop. Once the whole thing is turned into printers, they then turn each other into propulsion units and fly themselves away.


----------



## tomBitonti (Sep 1, 2018)

Starting with:



			
				https://en.wikipedia.org/wiki/Chicxulub_impactor said:
			
		

> the Chicxulub asteroid, was an asteroid or other celestial body some
> 10 to 15 kilometres (6 to 9 mi) in diameter




That doesn't have enough detail.  Further basic results are also lacking in deail.

Finally:

Assessments of the energy, mass and size of the Chicxulub Impactor 
Hector Javier Durand-Manterola and Guadalupe Cordero-Tercero
Departamento de Ciencias Espaciales, Instituto de Geofísica, Universidad Nacional 
Autonoma de México
Arxiv March 19, 2014



			
				https://arxiv.org/ftp/arxiv/papers/1403/1403.6391.pdf said:
			
		

> ... the aim of this study is to estimate the most relevant features of
> this one such as the size, mass and kinetic energy. We found that the
> kinetic energy of the impactor is in the range from 1.3x10^24 J to
> 5.8x10^25 J. The mass is in the range of 1.0x10^15 kg to 4.6x10^17
> ...




Since our hero was having difficulty with a much smaller rock having a
diameter of 1km, the following uses the lowest values from the ranges
of the estimate:

Diameter          
1.06 10^4 m    

Mass:
1.0 10^15 kg

Kinetic Energy (KE):
1.3 10^24 J

Then:

Volume:        
6.2 10^11 m^3  

Density:
1.6 10^3 kg/m^3

Speed:
5.1 10^4 m/s (51 km/s)

In support of these values, _Assessments_ writes:



> In the second model, to calculate the kinetic energy of the impactor,
> we needed crater diameter, density of the projectile, density of the
> target, earth's gravity and impactor velocity.  We considered the
> density of the projectile as 1650 kg/m^3 for comets (Greenberg, 1998),
> ...




Motion of 51 km/s is 1.84x10^5 km/hr, or 4.4x10^6 km/day.

(In comparison, Earths average orbital speed is about 30 km/s.)

Working from the equations of motion under constant acceleration,
and that relate force, mass, and acceleration:

Under constant acceleration:
    D = 1/2 A T^2
Or:
    A = 2 D / T^2

Definition of force:
    F = M A
Then:
    A = F / M

From which:
    F / M = 2 D / T^2
Or:
    F = 2 D M / T^2

Putting in:

The radius of the Earth:
    D = 6.4x10^6 m

The mass of the asteroid:
    M = 1.0 10^15 kg

The available time (one hour):
    T = 3.6 10^3

Results in:
    F = 1.0 10^15 kg m / s^2

Total _impulse_ is:
    I = F T
    I = 3.6 10^18 kg m / s

From: http://www.b14643.de/Spacerockets_2/United_States_1/Saturn-5/Design/SaturnV.htm

The total impulse of the Saturn V for Apollo 17 (launched 12-Jul-1972) was:
    I(SV) = 8.7x10^9 kg m / s

That is, pushing the asteroid out of the way would require the total
impulse of 2.4x10^8 (240,000,000, or 240 million) Saturn V's.

Notes:

The impulse requirements do not vary with the speed of the asteroid!
What matters is how long one has to push the asteroid out of the way.
That this should be the case can be seen by adopting an asteroid-centric
frame of reference, in which case the asteroid is still and
the earth which is in motion.  When looked at from this frame of
reference, all that matters is how long until the Earth and the
asteroid collide, not how fast the earth is moving.

From the force and total impulse equations:

    F = 2 D M / T^2

    I = F T = 2 D M / T

The force requirement increases linearly with the distance and mass,
and decreases in portion to the square of the available time.  But,
the total impulse requirement decreases linearly with the available
time.  (That is, as the available time increases, while the necessary
force is reduced, the duration of application of that force increases,
moderating a square factor to a linear one.)

Requirements for an asteroid having a similar composition but having a
different radius vary according to the change of radius of the
asteroid.  Changing from 10km to 1km reduces the force requirement by
a factor of 1000 (10^3).  Note that a "small" 1 km asteroid, still
needs the total impulse of 240,000 Saturn V's.

Thx!
TomB


----------



## freyar (Sep 2, 2018)

Ovinomancer said:


> I disagree.  The problem here is the same one you make about -- when you shifted the frame from the road/field to the car, you DON'T shift the vector of the wind -- it remains the same.  Ergo, if you go with an Earth centric reference frame, the asteroid's motion is actually comprised of it's vector and the Earth's speed vector, creating the apparent straight line movement towards Earth.  If you restart the Earth, you don't change the Asteroid's original vector, you just recover the Earth's vector back to Earth.  The push that was perpendicular in the Earth frame is still in the same direction, which is now diagonal to the asteroid's path.  This means you're both slowing the asteroid AND pushing it to the side. <snip>




You certainly do shift the vector of the wind.  But the main point I wanted to make is that you certainly can work in the earth's frame of reference without worrying about pseudo-forces over short time periods, and you now seem to be doing that.  I think you have some assumptions hidden in your calculations, so I can't really comment on those.

In the earth's frame, you need some component of the push perpendicular to the asteroid's travel, but you're right that the optimal push isn't quite perpendicular.  I don't have time to type out my work now, but I think the max deflection for an instantaneous impulse of magnitude dp on an asteroid of momentum p has sin(x)=dp/p, where x is the angle of the push from perpendicular to the asteroid velocity.


----------



## Ovinomancer (Sep 2, 2018)

freyar said:


> You certainly do shift the vector of the wind.  But the main point I wanted to make is that you certainly can work in the earth's frame of reference without worrying about pseudo-forces over short time periods, and you now seem to be doing that.  I think you have some assumptions hidden in your calculations, so I can't really comment on those.



You _don't _shift the wind.  Unless and until you rotate the diagram, in which case you shift the wind, but rotation doesn't work well in this case because Earth is so big and the impact point, while appearing to be in line with Earth's center in one frame isn't in another.  Here's a to scale diagram showing Earth as the circle on the left of a diameter including 150km of atmo (so it rounds up to a nice 13,000km diameter) and three asteroids (20km/s, 30km/s, 50km/s) at one hour from impact.  This shows the travel of Earth in that time and the travel of the asteroids in that time in both the Solar frame (Earth moving up, asteroids moving left) and the Earth frame (diagonal asteroid paths).  The grey arrow is the direction of an example push so that it's apparent that the shift in frame doesn't rotate the push.  As you can see, you can't just rotate the Earth frame because the asteroids aren't moving towards Earth's center but instead a point on the circumference, and that circumference is so large you can't ignore it for the purposes of the example.





> In the earth's frame, you need some component of the push perpendicular to the asteroid's travel, but you're right that the optimal push isn't quite perpendicular.  I don't have time to type out my work now, but I think the max deflection for an instantaneous impulse of magnitude dp on an asteroid of momentum p has sin(x)=dp/p, where x is the angle of the push from perpendicular to the asteroid velocity.



You must mean angle from the vector of the asteroid's travel, because otherwise a 90 degree push would be sin(0)=0.

That accounted for, the issue here is obvious when you consider it.  The maximum dp occurs at sin(+/-90), but it would obviously require a much, much harder push if you went past 90 to generate a miss if you pushed from any angle behind the asteroid (ie, speeding it up while you generate a lateral velocity) but your formula has dp dropping off, not increasing.  Also, it's possible to generate a miss by accelerating the asteroid "ahead" of the Earth, but your formula says needed dp there is 0.

I agree, there's an angle that minimizes pushed needed, but your assumption doesn't do it.  Momentum is the wrong frame of thinking, as the asteroid's mass is constant so it falls out of the equations and you're just dealing with velocities.  This is how most orbital problems work -- you deal in delta-v not momentum change.  The challenge here is the point of impact being on the surface of a large circle instead of the center in the Earth frame, so the math gets messy.  In the solar frame, it's two quickly moving objects.  The math doesn't resolve into a simple sine equation.


----------



## Eltab (Sep 3, 2018)

MarkB said:


> One option would be to use it to break down matter on the asteroid, turn it into pico-printer-gel manufacturers, and then use those in turn to break down more of it, in an exponential replication loop. Once the whole thing is turned into printers, they then turn each other into propulsion units and fly themselves away.




The last third of _2010_.  It makes a good plotline.  

But the author wants Our Heroine to be in peril at the climax, not watching safely from a short distance.
Maybe her suit gets infected with the 'make motors' nanite-virus, too?  Or would that distract from the big picture?


----------



## Janx (Sep 3, 2018)

tomBitonti said:


> With a mass driver you could scoop up stuff from the asteroid itself!
> 
> That would work for a distant asteroid, but doesn’t seem to work for an an asteroid which is at most a day away.
> 
> ...




Good idea.  I figure to limit the pico-printer-gel to arranging atoms/molecues.  So I need carbon and hydrogen to make hydrocarbons, frex.

Did we ever figure out the composition of the asteroid?  Making rockets or mass drivers (hers are bracelet/anklets, so bigger=better).  Seems a modest upscale could be feasible. and justify the # fig newtons needed to entice this beast to move.

Do we know what asteroids are made of before we get there?  Presumably kind of yes, but  how much do we know?


----------



## MarkB (Sep 3, 2018)

Eltab said:


> The last third of _2010_.  It makes a good plotline.
> 
> But the author wants Our Heroine to be in peril at the climax, not watching safely from a short distance.
> Maybe her suit gets infected with the 'make motors' nanite-virus, too?  Or would that distract from the big picture?




Maybe distance isn't an option. The pico-gel printers can relay commands between each other, but her suit is the command node for them, and needs to be in the loop, so she has to remain physically present during the whole process.


----------



## tomBitonti (Sep 3, 2018)

Janx said:


> Good idea.  I figure to limit the pico-printer-gel to arranging atoms/molecues.  So I need carbon and hydrogen to make hydrocarbons, frex.
> 
> Did we ever figure out the composition of the asteroid?  Making rockets or mass drivers (hers are bracelet/anklets, so bigger=better).  Seems a modest upscale could be feasible. and justify the # fig newtons needed to entice this beast to move.
> 
> Do we know what asteroids are made of before we get there?  Presumably kind of yes, but  how much do we know?




Re: Asteroid composition.  Then short answer is that it varies, according to distance from the sun:



			
				https://www.esa.int/Our_Activities/Space_Science/Asteroids_Structure_and_composition_of_asteroids said:
			
		

> Beyond their shape, there are a number of ways to classify asteroids. The first is by composition. The typical composition of an asteroid depends on its distance from the Sun. At the outer edges of the asteroid belt, that is between three and three and a half times further from the Sun than the Earth, over eighty percent of the asteroids are known as C-type.
> 
> The C stands for carbon and the surfaces of these asteroids are almost coal-black. These asteroids contain large quantities of carbon molecules as well as the more usual rocks and metals. They are very similar in composition to the carbonaceous chondrite meteorites that sometimes fall on Earth. It is thought that these meteorites are chippings, smashed off during collisions between asteroids.
> 
> Closer to the Sun, at just over twice the Earth's orbital distance, the proportion of C-type asteroids is only about 40 percent. Here, the majority of asteroids are grey, without the carbon material and principally made of silicate compound rock.




For more details, I tried _Mineralogy and Surface Composition of Asteroids_, Vishnu Reddy, Et. Al.
https://arxiv.org/ftp/arxiv/papers/1502/1502.05008.pdf
But it is far too dense, and doesn't present composition results in any simple representation.

Further searches found:



			
				https://nssdc.gsfc.nasa.gov/planetary/text/asteroids.txt said:
			
		

> The majority of asteroids fall into the following three categories:
> 
> C-type (carbonaceous):  Includes more than 75 percent of known asteroids. Very dark with an albedo of 0.03-0.09. Composition is thought to be similar to the Sun, depleted in hydrogen, helium, and other volatiles. C-type asteroids inhabit the main belt's outer regions.
> 
> ...




And:



			
				http://www.astronomysource.com/tag/c-type-asteroids/ said:
			
		

> In broad terms there are three classifications of asteroid based on their composition:
> 
> C-type, which are the most common, are carbonaceous, and consist of clay and silicate rocks.  They exist furthest from the Sun, and so have been least altered by heat, meaning that they are the most ancient. Due to the fact that some have never even reached temperatures above 50°C, it is estimated they can contain up to 22% water.
> 
> ...




What I've read (albeit, mostly science fiction) has mass driver payloads jacketed in iron, because a magnetic material is needed for the mass driver.

Thx!
TomB


----------



## Eltab (Sep 4, 2018)

Janx said:


> Do we know what asteroids are made of before we get there?



Somebody mentioned upthread that this is not a natural asteroid, it is a thrown projectile.  The throwers, if they want to maximum damage, would send a metallic slug down on the target.  Next best would be a stone composition.  Iceballs will not melt as much as you would think in fiction, but still are brittle and could be shattered / broken into chunks /melted with the 'magnifying glass' trick before arrival.

Want to throw the readers for a loop?  The thing used to be a carbonaceous chrondite, and inside a magnetic jacket is mostly oily goop, frozen or molasses-like or quicksand-y due to vacuum and ambient light / heat.


----------



## Janx (Sep 4, 2018)

Eltab said:


> Somebody mentioned upthread that this is not a natural asteroid, it is a thrown projectile.  The throwers, if they want to maximum damage, would send a metallic slug down on the target.  Next best would be a stone composition.  Iceballs will not melt as much as you would think in fiction, but still are brittle and could be shattered / broken into chunks /melted with the 'magnifying glass' trick before arrival.
> 
> Want to throw the readers for a loop?  The thing used to be a carbonaceous chrondite, and inside a magnetic jacket is mostly oily goop, frozen or molasses-like or quicksand-y due to vacuum and ambient light / heat.




Actually, Marsies flew over and grabbed some rocks and flung them at earth because they were tired of taxation without representation.


----------



## Umbran (Sep 4, 2018)

Janx said:


> Actually, Marsies flew over and grabbed some rocks and flung them at earth because they were tired of taxation without representation.




"We are tired of taxation without representation.  So, we will murder billions!"

Yeah, that's some ethical high ground there. :/


----------



## Tonguez (Sep 4, 2018)

Maybe they don’t have tea on Mars


----------



## Janx (Sep 4, 2018)

Umbran said:


> "We are tired of taxation without representation.  So, we will murder billions!"
> 
> Yeah, that's some ethical high ground there. :/




Allegedly they did feel bad about it.  Nobody's exactly gotten their side of the story other than they tried to stop the second launch.

Who knows how it really went down...

I just figured it was enough of a twist on the previous two perspectives I present.


----------



## Umbran (Sep 4, 2018)

Janx said:


> Allegedly they did feel bad about it.




"Yeah, we feel really bad about destroying the birthplace of humanity.  But there was serious cash involved...." 



> Nobody's exactly gotten their side of the story other than they tried to stop the second launch.
> 
> Who knows how it really went down...




The only really relevant point is that it didn't go down *accidentally*.  There is no, "Whoops!  We threw a mountain-sized rock across interplanetary distances, and *just happened* to hit the Earth.  So sorry!"  Whoever did it, meant it.  Consider what would be necessary for that to place those responsible on the right side of history.


----------



## Nagol (Sep 4, 2018)

Umbran said:


> "Yeah, we feel really bad about destroying the birthplace of humanity.  But there was serious cash involved...."
> 
> 
> 
> The only really relevant point is that it didn't go down *accidentally*.  There is no, "Whoops!  We threw a mountain-sized rock across interplanetary distances, and *just happened* to hit the Earth.  So sorry!"  Whoever did it, meant it.  Consider what would be necessary for that to place those responsible on the right side of history.




They needed to win?  Then they're on the right side of history much like Rome was on the right side compared to Carthage.  Unfortunately, they missed and their descendants and the descendants of their enemies now have to deal with the leftover threat.


----------



## Janx (Sep 4, 2018)

Umbran said:


> "Yeah, we feel really bad about destroying the birthplace of humanity.  But there was serious cash involved...."
> 
> 
> 
> The only really relevant point is that it didn't go down *accidentally*.  There is no, "Whoops!  We threw a mountain-sized rock across interplanetary distances, and *just happened* to hit the Earth.  So sorry!"  Whoever did it, meant it.  Consider what would be necessary for that to place those responsible on the right side of history.




and that is the makings for another story. 


I haven't decided on the details, as it doesn't matter for the current story, from Earth's perspective long after the event and knowledge has been lost.

But in my mind, a number of elements exist to work with:
that was 100 years ago, what does being angry at Mars now matter?
Why do we assume Mars was a unified people in agreement and responsible for the act?
What if you dropped the first bomb over Hiroshima and as it's falling, regretted it?
Who says the people responsible weren't punished already?
How is Mars doing these days?

Except for the first point, none of these factors play into the current story about people who've scavenged the heck out of things to survive and don't know somebody was mad at them 100 years ago, let alone have the means to retaliate.  To what end?


----------



## Janx (Sep 4, 2018)

Nagol said:


> They needed to win?  Then they're on the right side of history much like Rome was on the right side compared to Carthage.  Unfortunately, they missed and their descendants and the descendants of their enemies now have to deal with the leftover threat.




Could be.  I was being cheeky when I said taxation without representation.

Consider what Babylon 5 or The Expanse shows us of Earth's treatment toward settlements/colonies.


----------



## tomBitonti (Sep 4, 2018)

This is the problem when you have independent folks out in the belt with the capability to adjust orbits (as in _The Expanse_).  Individuals piloting asteroid tenders would have the capacity to do tremendous harm.  That seems unstable, with a terrible disaster just a matter of time.

This could be the basis for a decent set of stories -- the decentralization of control from individuals as a necessary step for surviving into a solar system spanning civilization.  I can't say I've read stories that use this as a basic idea.

This actually seems to be where we are headed with just an Earthbound civilization: Moving control over high energy means -- large vessels and large airplanes, then to smaller vessels and trucks, and eventually to cars -- from people to automation.  We are building that technology right now, and I am thinking that one of the motivations is exactly to remove control from non-approved actors.

Thx!
TomB


----------



## Tonguez (Sep 5, 2018)

Umbran said:


> "Yeah, we feel really bad about destroying the birthplace of humanity.  But there was serious cash involved...."
> 
> 
> 
> The only really relevant point is that it didn't go down *accidentally*.  There is no, "Whoops!  We threw a mountain-sized rock across interplanetary distances, and *just happened* to hit the Earth.  So sorry!"  Whoever did it, meant it.  Consider what would be necessary for that to place those responsible on the right side of history.




Humans have this great capacity to rationalise things in hindsight, so I'm sure that should a society with the capacity to launch asteroids be established on Mars then relations between them and Earth will balance out eventually


----------



## freyar (Sep 5, 2018)

Ovinomancer said:


> You _don't _shift the wind.  Unless and until you rotate the diagram, in which case you shift the wind, but rotation doesn't work well in this case because Earth is so big and the impact point, while appearing to be in line with Earth's center in one frame isn't in another.  Here's a to scale diagram showing Earth as the circle on the left of a diameter including 150km of atmo (so it rounds up to a nice 13,000km diameter) and three asteroids (20km/s, 30km/s, 50km/s) at one hour from impact.  This shows the travel of Earth in that time and the travel of the asteroids in that time in both the Solar frame (Earth moving up, asteroids moving left) and the Earth frame (diagonal asteroid paths).  The grey arrow is the direction of an example push so that it's apparent that the shift in frame doesn't rotate the push.  As you can see, you can't just rotate the Earth frame because the asteroids aren't moving towards Earth's center but instead a point on the circumference, and that circumference is so large you can't ignore it for the purposes of the example.




We _must_ somehow be talking about different things.  Let me just give an example on the earth.  Start in the rest frame of some city streets laid out on a rectangular grid with the wind blowing directly toward the right on the grid.  Suppose there is a car driving "up" on the grid, perpendicular to the wind in the frame of the streets.  In the rest frame of the car, the grid is moving "down," and the wind is moving at the diagonal down and right.  It has clearly shifted.  The same is true in this case: Pierce's velocity will be an apparently different vector when measured in the earth or sun rest frames.  You agree, right?



> You must mean angle from the vector of the asteroid's travel, because otherwise a 90 degree push would be sin(0)=0.



No, I mean the angle from the perpendicular.  So if the asteroid is infinitely massive and therefore has infinite momentum p for finite momentum transfer dp, you push perpendicular to the asteroid's travel to get the maximum change in the direction of the trajectory.



> That accounted for, the issue here is obvious when you consider it.  The maximum dp occurs at sin(+/-90), but it would obviously require a much, much harder push if you went past 90 to generate a miss if you pushed from any angle behind the asteroid (ie, speeding it up while you generate a lateral velocity) but your formula has dp dropping off, not increasing.  Also, it's possible to generate a miss by accelerating the asteroid "ahead" of the Earth, but your formula says needed dp there is 0.



I must not have made my assumptions clear.  (1) I am working in the earth's rest frame. (2) I am assuming that Pierce can transfer a maximum total amount of momentum dp to the asteroid; this fits with the idea of Pierce ramming into the asteroid once, but it also works with a continuous push over some time (EDIT: to be clear, the continuous push would need to end sufficiently before the asteroid passes the earth that it's on a straight-line path again as it passes earth, so this really is best for a single big push).  Therefore, in vector terms, the initial asteroid momentum is vec(p), and the final asteroid momentum is vec(p)+vec(dp).  (3) In the earth's frame, assuming Pierce can't ram the asteroid enough to turn it around, we are concerned about the change in direction of the asteroid's path, which we can give by the angle y between the initial and final asteroid momenta.  (5) If vec(p) is oriented at angle x compared to the perpendicular to vec(p) (and with parallel component oriented opposite vec(p) in order to slow down the asteroid), the deflection angle y is given by tan = dp cos(x)/ (p-dp sin(x)). (6) Because tangent is monotonic, y is maximized by maximizing the tangent.  We have assumed dp is fixed, so we are maximizing with respect to x.  Differentiating, we find maximization occurs when 
-dp sin(x)/ (p-dp sin(x))+dp^2 cos(x)^2/(p-dp sin(x))^2=0, which is equivalent to cos(x)=0 (which minimizes the deflection angle at y=0) or
-sin(x) (p-dp sin(x))+dp cos (x)^2 =0, which solves to sin(x) =dp/p (remembering cos^2+sin^2=1).
(7) All this ignore's the earth's gravity, but I think we've all agreed that the asteroid is probably moving fast enough that gravitational attraction is unimportant as a first approximation.



> I agree, there's an angle that minimizes pushed needed, but your assumption doesn't do it.  Momentum is the wrong frame of thinking, as the asteroid's mass is constant so it falls out of the equations and you're just dealing with velocities.  This is how most orbital problems work -- you deal in delta-v not momentum change.  The challenge here is the point of impact being on the surface of a large circle instead of the center in the Earth frame, so the math gets messy.  In the solar frame, it's two quickly moving objects.  The math doesn't resolve into a simple sine equation.



Since the asteroid's mass is constant, the change in its momentum and its velocity tell us the same thing.  But what's important is that we know Pierce has some max momentum she can transfer to the asteroid, so that's what we have to work with.  The important thing is that, no matter how the asteroid is going to hit the earth if left alone, all that matters to whether it will hit the earth is how much Pierce can deflect it in the earth's frame.  This we can figure out using conservation of momentum.  What I've told you is, given how much momentum Pierce can gather and transfer to the asteroid, what's the best angle for her to take in order to deflect it the most.


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## Eltab (Sep 5, 2018)

Janx said:


> Consider what Babylon 5 or The Expanse shows us of Earth's treatment toward settlements/colonies.



"The War of Earth's Aggression" _is_ a sci-fi trope.  You are not required to invoke it for your Future History if you don't want to.
It would make your universe different from so many other authors', if you didn't.


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## Ovinomancer (Sep 6, 2018)

freyar said:


> We _must_ somehow be talking about different things.  Let me just give an example on the earth.  Start in the rest frame of some city streets laid out on a rectangular grid with the wind blowing directly toward the right on the grid.  Suppose there is a car driving "up" on the grid, perpendicular to the wind in the frame of the streets.  In the rest frame of the car, the grid is moving "down," and the wind is moving at the diagonal down and right.  It has clearly shifted.  The same is true in this case: Pierce's velocity will be an apparently different vector when measured in the earth or sun rest frames.  You agree, right?




Here’s a thought experiment for you:  in your city frame, there’s a flagpole ahead of the car with a flag blowing in the wind.  The wind blows left to right, so the flag blows left to right.  The car is approaching the stationary flagpole.  Now, let’s shift the frame to that of the car.  Now, it appears that the car is stationary and the flagpole is approaching the car.  But the flag – the flag is still blowing left to right with the wind.  It doesn’t swivel to blow diagonally when you change the frame. 

Here’s a second though experiment.  Let’s use your setup and agree that the wind shifts direction when you change frames.  Agreeing to that, can we also agree that the wind exerts a force on the car?  In the city frame, that force is left to right, ie, it pushes the car to the right.  Perhaps the car is steering to overcome or the friction of the tires to lateral movement is sufficient, but, in any case, the wind is a force acting to push the car to the right in the city frame, yes?  Now, shift to the car frame and assume, for the sake of the argument, that the wind is now blowing diagonally down and right towards the car.  There’s now some part of the force pushing the car backwards.  If this is true, the car must compensate against the wind in two directions rather than one, and that means it must have some forward force even in the car stationary frame or the rate that the city moves past must slow with the added force.   This is clearly not the case – the car isn’t dealing with a different force when you change the frame.

What you’ve done is a translation on velocity.  In this translation, you’ve subtracted the velocity of the car and also subtracted that velocity from the city, resulting in the car having a zero velocity and the city gaining a velocity exactly opposite the car’s original velocity.  But the wind is a force, ma, and ma+v doesn’t change the vector of the ma any more than moving a building would affect it’s velocity in the translation.  Any given particle of the field of particles comprising the wind will, indeed, adopt the car’s velocity when you shift frames, but the force applied is still left to right.

To make an even more clear example, let’s go back to the ball and the car.  In the field’s frame, the car has a velocity v(car) upwards and the ball has a velocity v(ball) leftwards.  There’s a wind applying a force to both blowing from right to left (ie, accelerating the ball leftwards).  The car, due to steering or friction, resists this leftward force of the wind.  Now, let’s switch frames.  The car’s velocity is subtracted so it now has a relative velocity of 0.  The car’s velocity is also subtracted from the ball, so it now has the velocity of the car downward and now appears to move left and down towards the car (in the lower right-hand corner).  The wind still blows.  If the wind’s direction doesn’t change, then the ball is being accelerated leftwards and will, increasingly, appear to veer to the left as viewed by the car.  But, if what you say is true, and the wind also shifts to blow left and down, then the ball merely accelerates towards the car in a straight line.  But, if we sum the forces, we have a problem.  In the field frame, the ball is experiencing a net force to the left and the car is experiencing the same force with a balancing force (steering or friction) to the right.  Now, if we shift and use the former (my) assumption that the wind does not shift, the forces on everything are the same – the car is still at 0 net force and the ball is still being accelerated to the leftwards.  If, however, we use your assumption that the wind shifts, the car is now experiencing a down and left force and must compensate by pushing back right (steering or friction) and forward (ie, using the motor) while the ball is experiencing a net acceleration down and right.  Your assumption means the car has to press the gas to counter the force of the wind when you shift from the field frame to the car frame.  The ball is there to illustrate that the real effect is that the force doesn’t change vectors and will still be accelerated leftwards and no left and down.

I may have just stumbled on a better way to say this:  slope doesn’t change if you move the intercept.  Y= mx+b.  m is independent of b.  Changing b is like the frame shifts we’re talking about – the force vector doesn’t change if you move around the velocities, just like the slope of a line doesn’t change if you move it’s y-intercept.  To put it in calculus terms, you’re changing the +c, which doesn’t affect the integral.



> No, I mean the angle from the perpendicular.  So if the asteroid is infinitely massive and therefore has infinite momentum p for finite momentum transfer dp, you push perpendicular to the asteroid's travel to get the maximum change in the direction of the trajectory.
> 
> I must not have made my assumptions clear.  (1) I am working in the earth's rest frame. (2) I am assuming that Pierce can transfer a maximum total amount of momentum dp to the asteroid; this fits with the idea of Pierce ramming into the asteroid once, but it also works with a continuous push over some time (EDIT: to be clear, the continuous push would need to end sufficiently before the asteroid passes the earth that it's on a straight-line path again as it passes earth, so this really is best for a single big push).  Therefore, in vector terms, the initial asteroid momentum is vec(p), and the final asteroid momentum is vec(p)+vec(dp).  (3) In the earth's frame, assuming Pierce can't ram the asteroid enough to turn it around, we are concerned about the change in direction of the asteroid's path, which we can give by the angle y between the initial and final asteroid momenta.  (5) If vec(p) is oriented at angle x compared to the perpendicular to vec(p) (and with parallel component oriented opposite vec(p) in order to slow down the asteroid), the deflection angle y is given by tan = dp cos(x)/ (p-dp sin(x)). (6) Because tangent is monotonic, y is maximized by maximizing the tangent.  We have assumed dp is fixed, so we are maximizing with respect to x.  Differentiating, we find maximization occurs when
> -dp sin(x)/ (p-dp sin(x))+dp^2 cos(x)^2/(p-dp sin(x))^2=0, which is equivalent to cos(x)=0 (which minimizes the deflection angle at y=0) or
> ...




The thing that jumps out at me here is that you’ve assumed that the translation to the Earth frame centers the line of approach on Earth’s center.  It doesn’t.  Look at the graphic I posted above.  The size of the Earth isn’t negligible in this calculation – it’s a sizable fraction of the distance traveled in 1 hour for all the cases discussed.  Further, the impact being on the circumference of a 6500km radius circle means that the approaching vectors will not be coming straight down for an observer at that point but will instead by steeply angled with respect to the vertical (again, reference the diagram).  This affects your formula which assumes a point mass for Earth such that a perpendicular to the asteroid’s path is parallel to the tangent line at the point of impact.  This isn’t so, and dramatically affects the outcomes.  For instance, your formula now has a ‘right’ and a ‘wrong’ side for pushing, where a hypothetical deflection from one side’s perpendicular causes a miss but the opposite force on the other perpendicular still results in a hit.  For that reason, your simplification fails a first approximation due to bad assumptions.

In the Solar frame, I can pretty easily figure what’s needed to generate a miss with either a lateral push or a slowdown push.  The formulas are dependent on the asteroid speed and the time to impact.

V(slow) = V(ast)(1-t(i)/(t(i)+217))  --  Slowdown velocity is equal to the velocity of the asteroid times 1 minus the time of impact divided by the time of impact plus 217 seconds.  The 217 comes from the time it takes Earth at 30km/s to clear it's own 6500km radius.  That doesn't change.

V(lateral) = r(E)/t(i) -- lateral velocity is equal to the radius of the Earth divided by the time to impact.

This approximation treats Earth as a flat disc.  For a sphere, you have to calculate a T(m) which is equal to r(E)/V(asteroid), or the time it takes the asteroid to travel a radius of the Earth.  This is the time to clear the widest part of Earth from the impact point (assuming impact at the closest point).  You add this to t(i) in the above.  It complicates things a bit for a bit more fidelity.  I'm going with the easy here.

So, for an asteroid going 20km/s 1 hour our, the V(s) is 20km/s(1-3600s/3817s) = 1.14 km/s.  The V(l) is 1.81 km/s.  Either of these will generate a miss. And optimal push would be at some angle between 0 and 90 degrees and would have a value of less than 1.14.  

I've run out of time at this point.  I might get back to the optimization in the Solar frame problem so that it could be checked against your formula.


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## Eltab (Sep 6, 2018)

Ovinomancer said:


> Here’s a thought experiment for you:  in your city frame, there’s a flagpole ahead of the car with a flag blowing in the wind.  The wind blows left to right, so the flag blows left to right.  The car is approaching the stationary flagpole.  Now, let’s shift the frame to that of the car.  Now, it appears that the car is stationary and the flagpole is approaching the car.  But the flag – the flag is still blowing left to right with the wind.  It doesn’t swivel to blow diagonally when you change the frame.



Presuming the wind lines up with the streets (which I think you are describing), then

Have the moving car make a right-hand turn, and the wind will be coming from behind it.
So from the car's point of view, when it goes into motion down the original perpendicular street, the wind IS coming from another direction (not from its due left any more).


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## Ovinomancer (Sep 6, 2018)

Eltab said:


> Presuming the wind lines up with the streets (which I think you are describing), then
> 
> Have the moving car make a right-hand turn, and the wind will be coming from behind it.
> So from the car's point of view, when it goes into motion down the original perpendicular street, the wind IS coming from another direction (not from its due left any more).



Yes, but then you're not translating to a different reference frame, you're rotating the coordinate frame.  Changing the coordinate frame is a different exercise.

 This is apparent if you set that coordinate frame prior to the turn of the car -- the wind won't change direction.  Obviously, if the wind is blowing in the +y direction but you then invert the y-axis, the wind will now be blowing in the -y direction.  You've switched your coordinates, not changed the direction of the wind.


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## MarkB (Sep 6, 2018)

Ovinomancer said:


> Here’s a thought experiment for you:  in your city frame, there’s a flagpole ahead of the car with a flag blowing in the wind.  The wind blows left to right, so the flag blows left to right.  The car is approaching the stationary flagpole.  Now, let’s shift the frame to that of the car.  Now, it appears that the car is stationary and the flagpole is approaching the car.  But the flag – the flag is still blowing left to right with the wind.  It doesn’t swivel to blow diagonally when you change the frame.




But in order for a _moving_ flagpole to still have its flag blowing straight out sideways, wouldn't the wind _have_ to be blowing diagonally? After all, with no prevailing wind at all, the natural propensity for a flag on a moving pole would be to stream straight backwards from it, opposite to the direction of motion.


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## Ovinomancer (Sep 6, 2018)

MarkB said:


> But in order for a _moving_ flagpole to still have its flag blowing straight out sideways, wouldn't the wind _have_ to be blowing diagonally? After all, with no prevailing wind at all, the natural propensity for a flag on a moving pole would be to stream straight backwards from it, opposite to the direction of motion.



Sum the forces.


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## tomBitonti (Sep 7, 2018)

So.   I sat down and worked through this again, and the 90’ answer is wrong.  (It _is_ correct asymptotically as the asteroid distance increases.)

In the Earth frame, with the Earth simplified to a disc, and having the adjustment a single delta-v, the adjustment changes the asteroid direction from being straight at the Earth to being tangential to the disc of the Earth.  Lots of vectors will do this.  The shortest is perpendicular to the adjusted velocity.

Very far away (many multiples of the Earth’s radius), the original and modified asteroid velocities are nearly parallel. 

Thx!
TomB


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## Ovinomancer (Sep 7, 2018)

tomBitonti said:


> So.   I sat down and worked through this again, and the 90’ answer is wrong.  (It _is_ correct asymptotically as the asteroid distance increases.)
> 
> In the Earth frame, with the Earth simplified to a disc, and having the adjustment a single delta-v, the adjustment changes the asteroid direction from being straight at the Earth to being tangential to the disc of the Earth.  Lots of vectors will do this.  The shortest is perpendicular to the adjusted velocity.
> 
> ...



In the Earth frame, do you have the asteroid approaching the disc at an angle or did you have it approach perpendicularly?  Because the latter isn't correct -- the angle of appoach is 45 degrees for a 30km/s asteroid. It makes a big difference in the deflection vector.


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## Janx (Sep 7, 2018)

Hey, thanks everybody for all the ideas and math and diagrams and discussion for my story.  Y'all put it a lot of brain time for me.  Thank you.

I miss seeing this level of liveliness in the Misc sub-forum.  It was fun discussing a weighty topic.

I know ovinomancer says there was a math mistake, but my new current draft is using the "push it faster" plan because it sounded non-intuitive to a layman, but lined up with what a sniper might think.  She builds more mass driver rings when she gets there to provide the huge amount of power it needs.

I tried writing a "push it from the side' version, and I just didn't like the flow (which is my fault either way).

I've got to write a poem for the beginning.  Which allows me to cut out a huge chunk of Norse exposition.  It also creates symmetry with the ending poem, a quoted High Flight (which is not copyrighted, I checked).  Seems to work.  Then it's edit like an editor time and I can wrap this up and submit it.

The science may be wrong, but if anybody calls me on it, I can say "you're right.  My physicist friends debated it and warned me, but I liked this version better"  And smile, because if I get to that point, it means somebody read my story.


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## Ovinomancer (Sep 8, 2018)

Janx said:


> Hey, thanks everybody for all the ideas and math and diagrams and discussion for my story.  Y'all put it a lot of brain time for me.  Thank you.
> 
> I miss seeing this level of liveliness in the Misc sub-forum.  It was fun discussing a weighty topic.
> 
> ...



"Push it faster" still works perfectly for an asteroid hitting "high".  The error was that it worked an order of magnitude better -- it's better for a large class of asterood speeds, but not an order of magnitude better.  Glad your story worked out!


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## Eltab (Sep 9, 2018)

Janx said:


> I've got to write a poem for the beginning.  Which allows me to cut out a huge chunk of Norse exposition.  It also creates symmetry with the ending poem, a quoted High Flight (which is not copyrighted, I checked).  Seems to work.  Then it's edit like an editor time and I can wrap this up and submit it.



The "Norse World" thread (other sub-forum) is full of cool stuff; maybe some material that will help out?.


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## Janx (Sep 10, 2018)

Eltab said:


> The "Norse World" thread (other sub-forum) is full of cool stuff; maybe some material that will help out?.




there's a norse world thread?


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## Eltab (Sep 10, 2018)

Janx said:


> there's a Norse World thread?



I hope this is a link:
http://www.enworld.org/forum/showthread.php?652482-Norse-World

In case the link doesn't work, as of the time I wrote this post, the thread was on page two of the D&D 5e sub-forum.  The author and primary contributor has a ton of cool stuff and information.  He may know where to find a poem or perhaps a snippet from one of the Sagas.


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## freyar (Sep 13, 2018)

Sorry I haven't been able to respond to this for a week.  But it's important enough to get things right for the public that I don't want to let it pass. There is a lot correct here, but it's missing enough that the overall message is incorrect.



Ovinomancer said:


> Here’s a thought experiment for you:  in your city frame, there’s a flagpole ahead of the car with a flag blowing in the wind.  The wind blows left to right, so the flag blows left to right.  The car is approaching the stationary flagpole.  Now, let’s shift the frame to that of the car.  Now, it appears that the car is stationary and the flagpole is approaching the car.  But the flag – the flag is still blowing left to right with the wind.  It doesn’t swivel to blow diagonally when you change the frame.
> 
> Here’s a second though experiment.  Let’s use your setup and agree that the wind shifts direction when you change frames.  Agreeing to that, can we also agree that the wind exerts a force on the car?  In the city frame, that force is left to right, ie, it pushes the car to the right.  Perhaps the car is steering to overcome or the friction of the tires to lateral movement is sufficient, but, in any case, the wind is a force acting to push the car to the right in the city frame, yes?  Now, shift to the car frame and assume, for the sake of the argument, that the wind is now blowing diagonally down and right towards the car.  There’s now some part of the force pushing the car backwards.  If this is true, the car must compensate against the wind in two directions rather than one, and that means it must have some forward force even in the car stationary frame or the rate that the city moves past must slow with the added force.   This is clearly not the case – the car isn’t dealing with a different force when you change the frame.




In both cases, the force of an object from the wind is in the direction of the air's velocity _relative to the object_.  Let me just address your second example, since the logic holds in the first example also.  You are right that the car should experience the same force in both the city frame and the car frame.  You're also right that the force in the car frame is "down-right" because the wind velocity is at an angle with respect to the grid in this frame.  Now look at the frame of the city.  You are right that the wind blows across the grid, exerting a force to the right on the car.  You have forgotten that the car is moving "up" in this frame and pushing through the air, which by Newton's 3rd law is pushing back -- "down."  So in this frame, the car still experiences a "down-right" force from the air.

The simplest example of this I can think of at the moment is the following.  Suppose you're standing outside with no wind, ie, the air is still with respect to the ground. Now you start running.  You will feel wind in your face.  This is because, in your frame, the air is moving "backward" and exerting a force (what you feel) on your skin.  In the ground's frame, the air exerts a force on you because you are pushing through it.




> What you’ve done is a translation on velocity.  In this translation, you’ve subtracted the velocity of the car and also subtracted that velocity from the city, resulting in the car having a zero velocity and the city gaining a velocity exactly opposite the car’s original velocity.  But the wind is a force, ma, and ma+v doesn’t change the vector of the ma any more than moving a building would affect it’s velocity in the translation.  Any given particle of the field of particles comprising the wind will, indeed, adopt the car’s velocity when you shift frames, but the force applied is still left to right.




Yes, that is what I've done. You have to change the velocity of _everything_ when changing frames, including the wind.  You're also right that the force experienced by an object is the same in both frames, and I've explained why that happens in your example.



> To make an even more clear example, let’s go back to the ball and the car.  In the field’s frame, the car has a velocity v(car) upwards and the ball has a velocity v(ball) leftwards.  There’s a wind applying a force to both blowing from right to left (ie, accelerating the ball leftwards).  The car, due to steering or friction, resists this leftward force of the wind.  Now, let’s switch frames.  The car’s velocity is subtracted so it now has a relative velocity of 0.  The car’s velocity is also subtracted from the ball, so it now has the velocity of the car downward and now appears to move left and down towards the car (in the lower right-hand corner).  The wind still blows.  If the wind’s direction doesn’t change, then the ball is being accelerated leftwards and will, increasingly, appear to veer to the left as viewed by the car.  But, if what you say is true, and the wind also shifts to blow left and down, then the ball merely accelerates towards the car in a straight line.  But, if we sum the forces, we have a problem.  In the field frame, the ball is experiencing a net force to the left and the car is experiencing the same force with a balancing force (steering or friction) to the right.  Now, if we shift and use the former (my) assumption that the wind does not shift, the forces on everything are the same – the car is still at 0 net force and the ball is still being accelerated to the leftwards.  If, however, we use your assumption that the wind shifts, the car is now experiencing a down and left force and must compensate by pushing back right (steering or friction) and forward (ie, using the motor) while the ball is experiencing a net acceleration down and right.  Your assumption means the car has to press the gas to counter the force of the wind when you shift from the field frame to the car frame.  The ball is there to illustrate that the real effect is that the force doesn’t change vectors and will still be accelerated leftwards and no left and down.




I'm afraid I don't quite follow what you're setting up here.  But you're right that the force doesn't change directions --- but the wind has to change velocities because the force of a wind on an object is in the direction of the _difference_ of the object's and wind's velocities.



> I may have just stumbled on a better way to say this:  slope doesn’t change if you move the intercept.  Y= mx+b.  m is independent of b.  Changing b is like the frame shifts we’re talking about – the force vector doesn’t change if you move around the velocities, just like the slope of a line doesn’t change if you move it’s y-intercept.  To put it in calculus terms, you’re changing the +c, which doesn’t affect the integral.



Again, you're right, but you've forgotten parts of the force, as I've said above, so your overall conclusions above are incorrect.  And I'm not entirely sure what you're talking about integrating in the last sentence here.



> The thing that jumps out at me here is that you’ve assumed that the translation to the Earth frame centers the line of approach on Earth’s center.  It doesn’t.  Look at the graphic I posted above.  The size of the Earth isn’t negligible in this calculation – it’s a sizable fraction of the distance traveled in 1 hour for all the cases discussed.  Further, the impact being on the circumference of a 6500km radius circle means that the approaching vectors will not be coming straight down for an observer at that point but will instead by steeply angled with respect to the vertical (again, reference the diagram).  This affects your formula which assumes a point mass for Earth such that a perpendicular to the asteroid’s path is parallel to the tangent line at the point of impact.  This isn’t so, and dramatically affects the outcomes.  For instance, your formula now has a ‘right’ and a ‘wrong’ side for pushing, where a hypothetical deflection from one side’s perpendicular causes a miss but the opposite force on the other perpendicular still results in a hit.  For that reason, your simplification fails a first approximation due to bad assumptions.



No, I have made no such assumption.  No matter the impact location on the earth, at a given time before impact, there is a minimum angle that the relative velocity (that is, the asteroid's velocity in the earth's frame) must be changed to avoid hitting the earth somewhere.  The calculation I gave found the direction of push that Pierce can make to maximize the deflection angle, assuming Pierce can impart a fixed (or maximum) magnitude of momentum to the asteroid.  To be fair, this calculation is axially symmetric about the relative velocity, so it doesn't tell you which direction around that circle to push in order to get clear of the closest edge of the earth, but the calculation does what I said it does.




> In the Solar frame, I can pretty easily figure what’s needed to generate a miss with either a lateral push or a slowdown push.  The formulas are dependent on the asteroid speed and the time to impact.
> 
> V(slow) = V(ast)(1-t(i)/(t(i)+217))  --  Slowdown velocity is equal to the velocity of the asteroid times 1 minus the time of impact divided by the time of impact plus 217 seconds.  The 217 comes from the time it takes Earth at 30km/s to clear it's own 6500km radius.  That doesn't change.
> 
> ...



As with the pictures you've uploaded previously, you seem to be making some assumption about the direction of the asteroid's approach in the solar frame.  For example, if the collision is head on, slowing down or speeding up the asteroid won't help (unless of course you can slow it enough that it just precedes the earth around its orbit).


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## freyar (Sep 13, 2018)

MarkB said:


> But in order for a _moving_ flagpole to still have its flag blowing straight out sideways, wouldn't the wind _have_ to be blowing diagonally? After all, with no prevailing wind at all, the natural propensity for a flag on a moving pole would be to stream straight backwards from it, opposite to the direction of motion.



Yes.


Ovinomancer said:


> Sum the forces.



He did.  See my last post.


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## freyar (Sep 13, 2018)

Ovinomancer said:


> In the Earth frame, do you have the asteroid approaching the disc at an angle or did you have it approach perpendicularly?  Because the latter isn't correct -- the angle of appoach is 45 degrees for a 30km/s asteroid. It makes a big difference in the deflection vector.




You're clearly making some assumption about the approach of the asteroid and the earth in the solar frame --- in genera, you can have an asteroid approach with any angle and any speed.  Did I miss something in the thread where Janx said how the asteroid is approaching?


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## tglassy (Sep 13, 2018)

Y’all know that nobody actually cares about this stuff, right?


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## Ovinomancer (Sep 13, 2018)

freyar said:


> Sorry I haven't been able to respond to this for a week.  But it's important enough to get things right for the public that I don't want to let it pass. There is a lot correct here, but it's missing enough that the overall message is incorrect.
> 
> 
> 
> ...



Yes, I see that you don't get what I've set up, and it goes directly to your complaints above.  Let's try again.

Let's set up the situation without the wind.  In city frame, the car's motor is applying force to the car to move it along the road.  The air, which is stationary, is applying resistance to that force in equal measure (the car is at constant velocity, so the forces are equal).  What you have is:

Up: force applied by car's motor = down: force applied by air resistance.

Now, let's switch this to the car's frame.  The car is not stationary and it appear that there's a strong headwind (air resistance) pushing the car backwards.  The car's motor is applying force to counter that headwind (air resistance) and keep the city moving past at a constant rate.  You have:

down: headwind (air resistance) = up: car's motor,

Good so far? Okay, let's add the wind.

Now there's a cross wind of some force F.  First the city frame.  The car's motor hasn't changed force, so the air resistance also hasn't changed force.  Up and down forces are still equal.  But, we've added a lateral force, the wind(F).  This will push the car in the direction of the wind unless countered.  Let's assume that the friction of the tires on the surface counter the wind, here, such that the lateral forces are balanced.  You not have:
Up: car's motor = down: air resistance.
Right: wind (F) = Left: tire friction.

All forces balance.  Now, let's shift to the car's frame, and follow your suggestion that the wind also shifts direction when we shift frames.  We run into a problem.  If the wind shifts by some angle x, the lateral force of that wind(F) is now cos(x)*F, and there's now a up/down componenet of sin(x)*F.  If we sum the forces now, we have:

Down: air resistance + sin(x)*F = up: car's motor
Right: cos(x)*F = Left: tire friction

So, you're arguing that the tires exert less lateral friction force in the car frame than in the city frame, and that the car's motor must push harder to combat the increase in downward force because some part of the wind is adding to the air resistance.  This doesn't add up, though, as the forces on the car DO NOT CHANGE with a frame shift -- they still must be the same forces as in any other frame.  

The error in your thinking above is that the up/down component of the air's movement was already accounted for in the frames by the declaration of constant velocity -- meaning that the force of the car's motor exactly counters the force of the air resistance/wind in the up/down direction.  You have the lateral and up/down forces changing in your frame change, and that just doesn't happen.




> Again, you're right, but you've forgotten parts of the force, as I've said above, so your overall conclusions above are incorrect.  And I'm not entirely sure what you're talking about integrating in the last sentence here.
> 
> 
> No, I have made no such assumption.  No matter the impact location on the earth, at a given time before impact, there is a minimum angle that the relative velocity (that is, the asteroid's velocity in the earth's frame) must be changed to avoid hitting the earth somewhere.  The calculation I gave found the direction of push that Pierce can make to maximize the deflection angle, assuming Pierce can impart a fixed (or maximum) magnitude of momentum to the asteroid.  To be fair, this calculation is axially symmetric about the relative velocity, so it doesn't tell you which direction around that circle to push in order to get clear of the closest edge of the earth, but the calculation does what I said it does.



Um, okay, let's look at your equation as the force of the push goes to zero.  When this happens the limit of angle x goes to 0 degrees, which is nonsensical.  IE, the less you push, the more perpendicular you should push.  Further, if you increase your push to the velocity of the asteroid, sinx goes to 90.  That kinda makes sense in that completely stopping the asteroid relative to Earth will cause a miss (although gravity now becomes a dominating factor).  But what happens if you exceed the push?  There exists some dp greater than p where your formula says that pushing the asteroid even faster directly at Earth generates a miss.

Your error here was assuming some dp and then finding a formula that created an x that seemed right to you.  You're not testing other dp to find if the formula works.  It doesn't.



> As with the pictures you've uploaded previously, you seem to be making some assumption about the direction of the asteroid's approach in the solar frame.  For example, if the collision is head on, slowing down or speeding up the asteroid won't help (unless of course you can slow it enough that it just precedes the earth around its orbit).



Yes, those assumptions were established well upthread as plausible given the launch point of the asteroid and how it should interact with Earth's orbit.  A head on example in the Solar frame requires an extra-solar origin for the asteroid or an orbital period of much greater than 100 years (ie, the ellipse of the asteroid's path would have to be at Earth's orbit close to perihelion, and the combined speeds would be YUGE.  Further, for your purposes, a head on in the Solar frame is exactly identical to the Earth frame, with the Earth's 30km/s transferred to the asteroid's velocity.  At that point, your formula breaks very badly, as it suggests that the weaker your push, the more towards the perpendicular you should push but the stronger you push the more towards opposing the asteroid's path your should push.  That doesn't work.

In the Earth frame, at a given distance, and assuming an asteroid path aimed straight at Earth, there exists a minimum angle that will cause a miss.  This angle doesn't change based on the speed of the asteroid because the extended path will still need to be outside that angle.  This angle is easy to determine if you know the distance (d) and Earth's radius (Re) as it's simply Tan-1 (Re/d).  At sufficient distance, this angle is small.  To generate a miss, you must create an instantaneous dp such that the resulting path exceeds this miss angle.  This generates some math, but it's not too bad.  Let's call the miss angle (m), the angle of applied dp (x), and the angle the asteroid takes as (a); all angles are measured clockwise from the path.  Tan (a) is going to equal the lateral portion of dp, or dpcos(x) divided by p minus the vertical portion of dp, or dpsin(x).  This is tan(a) = dpcos(x)/(p-dpsin(x)).  We can set Tan(a) equal to Tan(m) and get that dpcos(x)/(p-dpsin(x))=Re/d.  From this, we can see a few things.  One, dp cannot go to 0, and two, x also cannot go to 0 (or 180).  This eliminates a push in the direction of the asteroid from generating a miss, as I stated above.  Now, let's solve for a dp if x = 90, ie, a perfectly lateral push.  IN this case, the formula goes go dp/p=Re/d.  Assuming p is known, we have a dp(lat) that generates a miss.  Now, can this dp applied to any other angle x, generate a miss (ignoring 270)?  Doing some substitution from above, this would mean that dp(lat)cos(x)/(p-dp(lat)sin(x))=dp(lat)/p.  There is no solution for x for a known dp(lat) in this scenario where x can be anything other than 90.  For a dp that exceeds dp(lat), you can have angles that aren't 90, because the resultant angle of miss can still be greater than m.

If we apply your forumla to this, though, where your sin(x)=dp/p, a known dp(lat) would result in an angle that isn't 90.  Your construction doesn't work even if we use your assumption that the path of the asteroid is directly through Earth's center.  In the realistic case, where it is not, it also doesn't work.


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## Ovinomancer (Sep 13, 2018)

tglassy said:


> Y’all know that nobody actually cares about this stuff, right?




At least 2 people do, and what a strange argument for the random geek stuff forum of a forum dedicated to pretending to be elves.  I mean, if no one cares here....


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## Ovinomancer (Sep 13, 2018)

freyar said:


> Yes.
> 
> He did.  See my last post.




Nope!  Flagpole in city frame.  No up/down forces.  Lateral forces are wind - being attached to the flagpole = 0.

Car frame:  forces don't change.  The 'head wind' the car feels isn't present for the flag, which is moving precisely along with and at the same speed as this 'headwind'.  So, up/down forces are still zero.  Side forces are also the same, ie: wind - being attached to the flagpole = 0.  If the wind shifted to the down and right, the flagpole is now pushing up and left, but a change of frame doesn't affect the sum of the forces, so that's impossible.

Sum.  The.  Forces.


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## MarkB (Sep 13, 2018)

Ovinomancer said:


> Nope!  Flagpole in city frame.  No up/down forces.  Lateral forces are wind - being attached to the flagpole = 0.
> 
> Car frame:  forces don't change.  The 'head wind' the car feels isn't present for the flag, which is moving precisely along with and at the same speed as this 'headwind'.  So, up/down forces are still zero.  Side forces are also the same, ie: wind - being attached to the flagpole = 0.  If the wind shifted to the down and right, the flagpole is now pushing up and left, but a change of frame doesn't affect the sum of the forces, so that's impossible.
> 
> Sum.  The.  Forces.




The headwind does exist for the flagpole, it's just precisely countered by its movement. Being precisely balanced out doesn't make it non-existent. And since the headwind does exist for the car, when you add in the lateral wind those two factors combine into a diagonal wind. It is impossible that they would not.

Forget the flagpole for a moment, and consider instead a crisp packet blowing across the car's path, moved purely from the force of the wind. From the city's perspective, if the crisp packet starts blowing left-to-right across the car's path, it is moving precisely perpendicular to the car, while the car is moving forwards to intersect its path. Start them at the right moment, and they'll collide.

But switch to the car's perspective. The car is perfectly still, and the crisp packet is carried by the wind, starting forward and to the left of the car. If the wind is still directly perpendicular to the car's facing, and the car is stationary, then the car cannot possibly ever intersect with the crisp packet's path. The only way in which it can do so is if the packet is travelling diagonally, from a position forward and to the left of the car to a position behind and to the right of the car.


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## Ovinomancer (Sep 13, 2018)

MarkB said:


> The headwind does exist for the flagpole, it's just precisely countered by its movement. Being precisely balanced out doesn't make it non-existent. And since the headwind does exist for the car, when you add in the lateral wind those two factors combine into a diagonal wind. It is impossible that they would not.
> 
> Forget the flagpole for a moment, and consider instead a crisp packet blowing across the car's path, moved purely from the force of the wind. From the city's perspective, if the crisp packet starts blowing left-to-right across the car's path, it is moving precisely perpendicular to the car, while the car is moving forwards to intersect its path. Start them at the right moment, and they'll collide.
> 
> But switch to the car's perspective. The car is perfectly still, and the crisp packet is carried by the wind, starting forward and to the left of the car. If the wind is still directly perpendicular to the car's facing, and the car is stationary, then the car cannot possibly ever intersect with the crisp packet's path. The only way in which it can do so is if the packet is travelling diagonally, from a position forward and to the left of the car to a position behind and to the right of the car.




How does the headwind exist for the flagpole?  Let's look at a no wind situation.  The flag-pole in relation to the city frame has no movement.  There's no wind, not push, it's entirely stationary.  Now, shift that to the car's frame.  The flagpole now appears to be moving toward the car along with the city.  The air is also moving towards the car at the same speed.  The flagpole doesn't experience any forces from the headwind, as it's moving along with it and not opposing it in any way.

When you add the wind, the above doesn't change for the up/down direction.  The wind blows sideways against the flagpole exactly the same way in both frames.  You can even see this if you shift to the flagpole's frame (hint: it's the same as the city frame).

As for the crisp packet, what does it look like with no wind?  The crisp packet, alongside the road, is still moving 'downward' at the speed of the frame.  When the wind blows to the right, and pushes it in front of the car, the speed at which it's moving 'downward' _does not increase_.  IE, the wind acts to push the packet to the right, but doesn't increase it's speed 'downward'.  The packet is still moving with all of the air at the same speed 'downward' with or without the wind.  It's speed laterally increases with the wind, but the wind adds no speed 'downward'.


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## MarkB (Sep 13, 2018)

Ovinomancer said:


> How does the headwind exist for the flagpole?  Let's look at a no wind situation.  The flag-pole in relation to the city frame has no movement.  There's no wind, not push, it's entirely stationary.  Now, shift that to the car's frame.  The flagpole now appears to be moving toward the car along with the city.  The air is also moving towards the car at the same speed.  The flagpole doesn't experience any forces from the headwind, as it's moving along with it and not opposing it in any way.




Why does that matter? The car is experiencing the headwind, and we're viewing things from the car's frame of reference. If the headwind exists for the car, then it exists for every other object in the model. The other objects simply happen to be travelling at a velocity that matches it.


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## MarkB (Sep 13, 2018)

Ovinomancer said:


> As for the crisp packet, what does it look like with no wind?  The crisp packet, alongside the road, is still moving 'downward' at the speed of the frame.  When the wind blows to the right, and pushes it in front of the car, the speed at which it's moving 'downward' _does not increase_.  IE, the wind acts to push the packet to the right, but doesn't increase it's speed 'downward'.  The packet is still moving with all of the air at the same speed 'downward' with or without the wind.  It's speed laterally increases with the wind, but the wind adds no speed 'downward'.




Bear in mind that from the car's frame of reference, the car is stationary but the city around it is moving. For there to be no wind from the car's frame of reference, there would need to be an upward-blowing wind from the city's frame of reference, blowing at precisely the same velocity as the car.


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## Ovinomancer (Sep 13, 2018)

MarkB said:


> Why does that matter? The car is experiencing the headwind, and we're viewing things from the car's frame of reference. If the headwind exists for the car, then it exists for every other object in the model. The other objects simply happen to be travelling at a velocity that matches it.




Um, no, no it doesn't exist for everything else in the model.  That's a big conceptual error, there. The point is that the flagpole is _at rest with the city in all frames_.  That doesn't change.  The car is not at rest with the city frame, so when you change frames, either the car is moving relative to the city frame or the city is moving relative to the car frame.  But the flagpole remains at rest with the city regardless of the frame.  The headwind the car experiences in the car frame is also at rest with the city -- it's blowing exactly at the same speed the city is moving in any frame.  In the city frame, the air resistance is stationary with the city while the car moves against it.  In the car frame, the air moves against the car with the city while the car remains motionless.  But all of the air is moving with the city.  When you add the wind, it creates a field effect to the right for all of that air, which is still at rest with regards to the city in the up/down direction in all frames.


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## Ovinomancer (Sep 13, 2018)

MarkB said:


> Bear in mind that from the car's frame of reference, the car is stationary but the city around it is moving. For there to be no wind from the car's frame of reference, there would need to be an upward-blowing wind from the city's frame of reference, blowing at precisely the same velocity as the car.




Yes... you're almost there!


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## MarkB (Sep 13, 2018)

Ovinomancer said:


> Um, no, no it doesn't exist for everything else in the model.  That's a big conceptual error, there. The point is that the flagpole is _at rest with the city in all frames_.




But the city itself is not at rest in the car's frame. That's the part you're missing.


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## Ovinomancer (Sep 13, 2018)

MarkB said:


> But the city itself is not at rest in the car's frame. That's the part you're missing.




Where am I missing that?  I just said that in the post you editted down to the above!  Come on, now.  The flagpole is part of the city frame, as is the air.  It ALL is at rest with respect to each other and ALL in exactly the same motion with respect to the car.  What they do not do is change motion with respect to each other.

Look, the forces are the same no matter what frame you're looking in.  If you add up the forces in one frame, they are exactly the same in a different frame.  What you're confusing here is trying to account for things partially.  There are no forces operating on the flagpole in the up or down direction regardless of the frame.  In the city frame, the air isn't moving against the flagpole in the up or down direction.  In the car's frame, the air isn't moving against the flagpole in the up or down direction.  The flagpole AND the air appear to be moving in the down direction, at exactly the same speed, from the car's perspective.  But the actual forces on the flagpole do. not. change.  Sum the forces yourself and see.

A force does not change direction with a frame shift of perspective.  What you're proposing is that the car's motor would have to work harder if you change how you think of it's motion.  Clearly this is false, so the error has to be in how you're conceiving of this.  Make the jump.


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## MarkB (Sep 13, 2018)

Ovinomancer said:


> Where am I missing that?  I just said that in the post you editted down to the above!  Come on, now.  The flagpole is part of the city frame, as is the air.  It ALL is at rest with respect to each other and ALL in exactly the same motion with respect to the car.  What they do not do is change motion with respect to each other.
> 
> Look, the forces are the same no matter what frame you're looking in.  If you add up the forces in one frame, they are exactly the same in a different frame.  What you're confusing here is trying to account for things partially.  There are no forces operating on the flagpole in the up or down direction regardless of the frame.  In the city frame, the air isn't moving against the flagpole in the up or down direction.  In the car's frame, the air isn't moving against the flagpole in the up or down direction.  The flagpole AND the air appear to be moving in the down direction, at exactly the same speed, from the car's perspective.  But the actual forces on the flagpole do. not. change.  Sum the forces yourself and see.
> 
> A force does not change direction with a frame shift of perspective.  What you're proposing is that the car's motor would have to work harder if you change how you think of it's motion.  Clearly this is false, so the error has to be in how you're conceiving of this.  Make the jump.




You are seriously missing the point here. In one frame the city is stationary and the car is in motion. In the other the car is stationary and the city is in motion. Whether the car is working to move forwards across a stationary city, or to remain stationary as the city tries to carry it backwards, its motor exerts the same effort.

The error is in how you are conceiving of this. Regardless of the frame of reference, you are insisting upon picturing things only from the perspective of the city.


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## Ovinomancer (Sep 13, 2018)

MarkB said:


> You are seriously missing the point here. In one frame the city is stationary and the car is in motion. In the other the car is stationary and the city is in motion. Whether the car is working to move forwards across a stationary city, or to remain stationary as the city tries to carry it backwards, its motor exerts the same effort.



YES!!!!



> The error is in how you are conceiving of this. Regardless of the frame of reference, you are insisting upon picturing things only from the perspective of the city.



No.  Sigh.

Okay.  Again.  No wind.  Car moves through city in city frame at 20 kph.  It therefore feels a 20 kph headwind due to this movement, yes?  Okay.  Now, go to the car frame, the headwind is still 20 kph, right?  Agreed?

Good.

Now, add in a 5 kph crosswind in the city frame.  So, now car is experiencing a 20 kph headwind AND a 5 kph crosswind.  Combined, this feels like a 20.6 kph wind coming in at about 14 degrees counterclockwise from the direction of the car's travel.  Now, let's switch to the car's frame -- the wind is *still *20.6 kph from 14 degree off of dead ahead. The car doesn't feel a different combined wind when you change the frame.

Now, that 20.6 kph wind at 14 degrees?  It's still a 20 kph headwind and a 5 kph crosswind combining.  IE, _the crosswind doesn't change direction when you shift the frame._


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## MarkB (Sep 13, 2018)

Ovinomancer said:


> Okay.  Again.  No wind.  Car moves through city in city frame at 20 kph.  It therefore feels a 20 kph headwind due to this movement, yes?  Okay.  Now, go to the car frame, the headwind is still 20 kph, right?  Agreed?
> 
> Good.
> 
> ...




But we're not just talking about the crosswind. We're talking about the sum of the crosswind and the headwind - the latter of which only exists in the car's frame. So, as you've so elegantly demonstrated, the wind does change depending upon which frame you're in.


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## Ovinomancer (Sep 13, 2018)

MarkB said:


> But we're not just talking about the crosswind. We're talking about the sum of the crosswind and the headwind - the latter of which only exists in the car's frame. So, as you've so elegantly demonstrated, the wind does change depending upon which frame you're in.




Gah.  Okay, once more unto the breach.

NO CROSSWIND CASE:
City frame -- car drives at 20 kph.  Air resists.  Force of the car's motor is exactly cancelled by the force of the air's resistance.  
Car's frame -- headwind blows at 20 kph.  Car resists.  Force of the wind's push is exactly cancelled by the force of the car's motor.

CROSSWIND CASE:
5 kph crosswind at 90 degrees to the path of the car.
City frame -- car drives at 20 kph.  Air resists.  Force of the car's motor is exactly cancelled by the force of the air's resistance.  Wind blows from left.  Pushes car to right.  Force of push exactly cancelled by tire friction.
Car frame -- headwind blows at 20 kph.  Car resists.  Force of the wind's push is exactly cancelled by the force of the car's motor.  Wind blows from left.  Pushes car to right.  Force of push exactly cancelled by tire friction.

In both cases, the forces in the direction the car is pointing DO NOT CHANGE.  In the crosswind case, there's a new force pushing the car to the right.  This force doesn't change IN EITHER FRAME.

What you're doing is saying that the headwind is part of the crosswind.  It isn't.  It exists exactly the same whether the crosswind exists or not.  If the crosswind exists, it exists exactly the same in either frame.  The vector of adding the headwind to the crosswind in the car frame DOES NOT MEAN THE WIND CHANGES DIRECTION.  It means you can add them if you want, like all vectors.  But the forces to the front of the car and to the side of the car do not change when you change the frame.  The crosswind applies the exact same force to the side of the car in either frame.  It doesn't change direction.

Honestly, when I first saw this example, I was unexcited because of this exact problem -- the presence of air would lead to people making this exact mistake of conception with 'headwinds'.


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## freyar (Sep 13, 2018)

Ovinomancer said:


> Yes, I see that you don't get what I've set up, and it goes directly to your complaints above.  Let's try again.
> 
> Let's set up the situation without the wind.  In city frame, the car's motor is applying force to the car to move it along the road.  The air, which is stationary, is applying resistance to that force in equal measure (the car is at constant velocity, so the forces are equal).  What you have is:
> 
> ...




So far, you are right.  But you need to remember that the air resistance and the "horizontal" wind force you've labeled F are just two components of the force of the air on the car.  The problem you are having is below.


> All forces balance.  Now, let's shift to the car's frame, and follow your suggestion that the wind also shifts direction when we shift frames.  We run into a problem.  If the wind shifts by some angle x, the lateral force of that wind(F) is now cos(x)*F, and there's now a up/down componenet of sin(x)*F.  If we sum the forces now, we have:
> 
> Down: air resistance + sin(x)*F = up: car's motor
> Right: cos(x)*F = Left: tire friction
> ...




Actually, no.  I don't have the up/down or lateral forces changing.  The problem you're having is that you are double counting the air resistance and "vertical" component of the wind force.  You still have the same force of the air on the car --- magnitude F "horizontally" and the same magnitude "vertically" as in the city frame.  It's just now that you say both components are from the air moving, rather than saying in the city frame that the horizontal component is from the wind moving and the vertical component is from the car moving through the air.  Remember, the force of the air on the car is due only to the *relative velocity* of the air and the car, which is the same in any reference frame.  





> Um, okay, let's look at your equation as the force of the push goes to zero.  When this happens the limit of angle x goes to 0 degrees, which is nonsensical.  IE, the less you push, the more perpendicular you should push.  Further, if you increase your push to the velocity of the asteroid, sinx goes to 90.  That kinda makes sense in that completely stopping the asteroid relative to Earth will cause a miss (although gravity now becomes a dominating factor).  But what happens if you exceed the push?  There exists some dp greater than p where your formula says that pushing the asteroid even faster directly at Earth generates a miss.
> 
> Your error here was assuming some dp and then finding a formula that created an x that seemed right to you.  You're not testing other dp to find if the formula works.  It doesn't.




Yes, as dp goes to zero, if you want to deflect the asteroid at all, you need to be pushing it as much "to the side" of its initial motion as possible in order to change its direction of motion.

Yes, x goes to +90 degrees as dp matches the initial momentum of the asteroid.  This is pushing directly *against* the asteroid's motion, so the final asteroid momentum along its line of motion is p-dp->0 in this limit, meaning the asteroid stops.  That's a "total deflection," or the best you can do.  Note that pushing the asteroid forward is a negative value of x.  That *never* optimizes the deflection angle.

I may have forgotten to note that I've assumed dp<p (which should be clear from the formula) because otherwise Pierce could just easily turn the asteroid around, and we don't need any dramatic "just missing the earth" bit.





> Yes, those assumptions were established well upthread as plausible given the launch point of the asteroid and how it should interact with Earth's orbit.  A head on example in the Solar frame requires an extra-solar origin for the asteroid or an orbital period of much greater than 100 years (ie, the ellipse of the asteroid's path would have to be at Earth's orbit close to perihelion, and the combined speeds would be YUGE.  Further, for your purposes, a head on in the Solar frame is exactly identical to the Earth frame, with the Earth's 30km/s transferred to the asteroid's velocity.  At that point, your formula breaks very badly, as it suggests that the weaker your push, the more towards the perpendicular you should push but the stronger you push the more towards opposing the asteroid's path your should push.  That doesn't work.




You are giving very specific numbers, like saying "the angle of appoach is 45 degrees for a 30km/s asteroid" in a response to tomB above.  While I admit I didn't read the whole thread, that seems more specific than the range of "reasonable" launch points.  But it doesn't really matter.

Yes, head-on in solar and earth frames looks the same but with different asteroid speeds.  My formula doesn't break at all.  Look, line up the asteroid's initial velocity/momentum (remember, these are proportional) along the horizontal axis.  Pierce then pushes.  The asteroid now has a final momentum with a vertical component --- entirely due to Pierce's perpendicular push -- and a horizontal component, which is its initial momentum minus Pierce's parallel push slowing it down.  The slope of the new velocity is the vertical part divided by the horizontal part.  This slope can get larger if one of two things happens: the numerator gets bigger, or the denominator gets smaller.  But if Pierce's push has some fixed total magnitude, you reduce the amount you can shrink the horizontal component of the asteroid momentum by making the numerator bigger, so these are competing effects.  My calculation tells you how to balance them to get the biggest slope.

I will go through the next bit in more detail...



> In the Earth frame, at a given distance, and assuming an asteroid path aimed straight at Earth, there exists a minimum angle that will cause a miss.  This angle doesn't change based on the speed of the asteroid because the extended path will still need to be outside that angle.  This angle is easy to determine if you know the distance (d) and Earth's radius (Re) as it's simply Tan-1 (Re/d).  At sufficient distance, this angle is small.  To generate a miss, you must create an instantaneous dp such that the resulting path exceeds this miss angle.



Yes!


> This generates some math, but it's not too bad.  Let's call the miss angle (m), the angle of applied dp (x), and the angle the asteroid takes as (a); all angles are measured clockwise from the path.  Tan (a) is going to equal the lateral portion of dp, or dpcos(x) divided by p minus the vertical portion of dp, or dpsin(x).



You are getting your trigonometry wrong.  If x is the angle from the initial velocity, dp sin(x) is the amount perpendicular to the initial velocity, and dp cos(x) is the amount along the initial velocity, which you have to subtract from the initial momentum.  You are effectively choosing x to be measured from the perpendicular to the initial asteroid velocity, with the push parallel to the initial velocity directed backward against the asteroid's velocity.  That's what I did also, but this may be part of your confusion.



> This is tan(a) = dpcos(x)/(p-dpsin(x)).  We can set Tan(a) equal to Tan(m) and get that dpcos(x)/(p-dpsin(x))=Re/d.  From this, we can see a few things.  One, dp cannot go to 0, and two, x also cannot go to 0 (or 180).  This eliminates a push in the direction of the asteroid from generating a miss, as I stated above.  Now, let's solve for a dp if x = 90, ie, a perfectly lateral push.



A perfectly perpendicular push is x=0. Cos(90 degrees)=0, which means there is no deflection at all.



> IN this case, the formula goes go dp/p=Re/d.  Assuming p is known, we have a dp(lat) that generates a miss.  Now, can this dp applied to any other angle x, generate a miss (ignoring 270)?  Doing some substitution from above, this would mean that dp(lat)cos(x)/(p-dp(lat)sin(x))=dp(lat)/p.  There is no solution for x for a known dp(lat) in this scenario where x can be anything other than 90.



This is true until the last sentence.  You can plot this on some software like maple or mathematica and see that there is in fact another solution for x even when dp=dp(lat). You can also see that there is a range of x in which tan(a) is greater than dp(lat)/p.  The smaller dp(lat) is, however, the more that range is crammed toward x=0, which means you have to push more perpendicularly.



> For a dp that exceeds dp(lat), you can have angles that aren't 90, because the resultant angle of miss can still be greater than m.



This is true, but you can also have dp<dp(lat) which will get tan(a)>tan(m).  There is still some minimum necessary value of dp, however.



> If we apply your forumla to this, though, where your sin(x)=dp/p, a known dp(lat) would result in an angle that isn't 90.  Your construction doesn't work even if we use your assumption that the path of the asteroid is directly through Earth's center.  In the realistic case, where it is not, it also doesn't work.



You're missing the point of what I'm doing.  If you have any fixed dp (and p), this formula maximizes tan(a).  But I didn't look at all about the necessary deflection tan(m), so I didn't say anything about what you call dp(lat) until this post.  Furthermore, there is *nothing* in my calculation that talks about the asteroid hitting the center of the earth. 




Ovinomancer said:


> Nope!  Flagpole in city frame.  No up/down forces.  Lateral forces are wind - being attached to the flagpole = 0.
> 
> Car frame:  forces don't change.  The 'head wind' the car feels isn't present for the flag, which is moving precisely along with and at the same speed as this 'headwind'.  So, up/down forces are still zero.  Side forces are also the same, ie: wind - being attached to the flagpole = 0.  If the wind shifted to the down and right, the flagpole is now pushing up and left, but a change of frame doesn't affect the sum of the forces, so that's impossible.
> 
> Sum.  The.  Forces.



The wind in the car's frame is down and right.  However, the flag is also moving down, so it is pushing additionally through the air.  That means there is an additional force from the air on the flag that cancels the push "down" from the wind.  In other words, the total force from the air on the flag is to the right, just like the relative velocity of the air and flag.  You have forgotten about a force, so you are not summing the forces correctly.  

Look, I doubt I can convince you, so I am mostly talking to other readers, if anyone else still cares.  But I'm not sure it's productive continuing this discussion if you can't consider what I'm saying carefully.


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## freyar (Sep 13, 2018)

I just saw more discussion came up.  I will look at it later, but I doubt there's much point participating.  It seems like everyone else gets it.


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## Ovinomancer (Sep 14, 2018)

freyar said:


> So far, you are right.  But you need to remember that the air resistance and the "horizontal" wind force you've labeled F are just two components of the force of the air on the car.  The problem you are having is below.
> 
> 
> Actually, no.  I don't have the up/down or lateral forces changing.  The problem you're having is that you are double counting the air resistance and "vertical" component of the wind force.  You still have the same force of the air on the car --- magnitude F "horizontally" and the same magnitude "vertically" as in the city frame.  It's just now that you say both components are from the air moving, rather than saying in the city frame that the horizontal component is from the wind moving and the vertical component is from the car moving through the air.  Remember, the force of the air on the car is due only to the *relative velocity* of the air and the car, which is the same in any reference frame.



YES! Except for the weird statement about double counting, which I'm not doing.  Regardless of whether of not the frame is the city or the car, the force of the air is the same because of the relative velocity between the air and the car.  Which means, when we add the wind from the side, we're adding a strictly lateral component to the mix.  This doesn't increase the up/down velocity of the air, it only adds a lateral component.  Even if you want to sum the vectors in the car frame so you get the 20.6 at 14 degrees, the forward force from the air doesn't change with frame and neither does the lateral force from the wind.

IN a case where we have a no wind model and then add the wind, then the added force of the wind is the new factor -- none of the old factors change.  And the new force applied by the wind is ALWAYS lateral -- it never changes direction.



> Yes, as dp goes to zero, if you want to deflect the asteroid at all, you need to be pushing it as much "to the side" of its initial motion as possible in order to change its direction of motion.
> 
> Yes, x goes to +90 degrees as dp matches the initial momentum of the asteroid.  This is pushing directly *against* the asteroid's motion, so the final asteroid momentum along its line of motion is p-dp->0 in this limit, meaning the asteroid stops.  That's a "total deflection," or the best you can do.  Note that pushing the asteroid forward is a negative value of x.  That *never* optimizes the deflection angle.
> 
> I may have forgotten to note that I've assumed dp<p (which should be clear from the formula) because otherwise Pierce could just easily turn the asteroid around, and we don't need any dramatic "just missing the earth" bit.




You're again saying that the weakest push must be more lateral, but the stronger the push the force moves to slow the asteroid.  That doesn't make sense, like, at all.  Take your statement about optimizing x.  Your point is that there exists some dp where the application of said dp generates the maximum value of x.  But is that what you're actually solving for?  Consider that your formula entirely fails to account for angles greater than 90.  There's a whole quadrant where  the angle of x measured from one perpendicular is valid that you're ignoring.  Further, a dp much, much larger than p creates x's that are nonsensical.  Take a dp 3 times greater than p such that your formula becomes Sin(x) = 3.  That's invalid.  If your formula can only work (supposedly) at a value of 0<dp<p, it's not really valid, is it?

Let's take a test case.  Let's assume an asteroid of velocity 30km/s.  Let's assume a dp with a velocity of 15km/s.  Since p and dp are both momentums of the same object, we can solve in velocities.  So, sin(x)=15/30=1/2.  x equals 45 degrees.  Half of the velocity of the asteroid maximizes deflection at 45 degrees.  Now, let's put that into some situations.  First, 1 hour out.  The needed miss angle is tan-1( 6500km/(30km/s * 3600s)) or 3.4 degrees.  The highest angle of push that achieves this is going to be (from perpendicular) dpcos(x)=(6500km/3600s) -> cos(x)=.13541.  x equals 82.2 degrees.  ANY push at 15 km/s 1 hour out will result in a miss so long as x is equal to or less than 82.2 degrees.  Compare this to your 45 degrees optimization.  Optimum is anything less than 82.2 degrees.





> You are giving very specific numbers, like saying "the angle of appoach is 45 degrees for a 30km/s asteroid" in a response to tomB above.  While I admit I didn't read the whole thread, that seems more specific than the range of "reasonable" launch points.  But it doesn't really matter.



Then you shouldn't complain, yeah?  I mean, if you haven't read everything up til now, complaining that there are assumptions in place you didn't read is kinda on you.



> Yes, head-on in solar and earth frames looks the same but with different asteroid speeds.  My formula doesn't break at all.  Look, line up the asteroid's initial velocity/momentum (remember, these are proportional) along the horizontal axis.  Pierce then pushes.  The asteroid now has a final momentum with a vertical component --- entirely due to Pierce's perpendicular push -- and a horizontal component, which is its initial momentum minus Pierce's parallel push slowing it down.  The slope of the new velocity is the vertical part divided by the horizontal part.  This slope can get larger if one of two things happens: the numerator gets bigger, or the denominator gets smaller.  But if Pierce's push has some fixed total magnitude, you reduce the amount you can shrink the horizontal component of the asteroid momentum by making the numerator bigger, so these are competing effects.  My calculation tells you how to balance them to get the biggest slope.



See my above where I show, by actually using your formula in a real scenario, that it doesn't work.  Pick a scenario, let's run your numbers.



> I will go through the next bit in more detail...
> 
> 
> Yes!
> ...



Yup, I boneheaded there.  I first messed up my coordinate placement.  Total bonehead on my part.  What I get for working in degrees.  The point here was it was the perpendicular portion of dp/p- the parallel portion of dp.  That's what I was solving with, despite my botch here.  The perpendicular part of dp is cos(x) when you measure from the perpendicular.


> This is true until the last sentence.  You can plot this on some software like maple or mathematica and see that there is in fact another solution for x even when dp=dp(lat). You can also see that there is a range of x in which tan(a) is greater than dp(lat)/p.  The smaller dp(lat) is, however, the more that range is crammed toward x=0, which means you have to push more perpendicularly.



Are you assuming this, or did you do this?  If you did this, show, please, because I would be shocked for that to be true.  Let me do a scenario and see.  Let's do the asteroid, 30km/s approach, 1 hour out, path through the center of Earth (so, straight in).  

d = 30km/s * 3600 s = 108,000 km.
Re = 6500 km

Re/d = 0.0602
Re/d=dp/p
removing mass from p's we have dv/v or Re/d=dv/v
v = 30km/s
dv = 1.801 km/s (this aligns with my previous solves for this value, so good check).

Now to use the formula (using 0 degree as perpendicular to the path):

dp(lat)cos(x)/(p-dp(lat)sin(x))=dp(lat)/p

1.801cos(x)/(30-1.801sin(x)=0.0602

1.801cos(x)=0.0602(30-1.801sin(x))= 1.801-0.1084sin(x)

cos(x) = 1 - .0602 sin(x) or 1=cos(x)+.0602sin(x)

That doesn't solve anywhere but 0 degrees.  Even solving for very close to 0, say 0.001, it doesn't solve.  Its close, but it's not 1.  It gets worse with larger x, and we can agree that it certainly fails as you approach 90 degrees.





> This is true, but you can also have dp<dp(lat) which will get tan(a)>tan(m).  There is still some minimum necessary value of dp, however.



A dp less the dp(lat) will not cause a miss.  Run whatever model you want, doesn't happen with the straight in assumption.  dp(lat) IS the minimum necessary dp.



> You're missing the point of what I'm doing.  If you have any fixed dp (and p), this formula maximizes tan(a).  But I didn't look at all about the necessary deflection tan(m), so I didn't say anything about what you call dp(lat) until this post.  Furthermore, there is *nothing* in my calculation that talks about the asteroid hitting the center of the earth.



I show above how that doesn't work, though.  And what I said was the path of the asteroid goes through the center of the Earth, not that the asteroid *hits* the center of the Earth.  

But, again, I refer to the specific scenario I did above.  The minimum deflection angle is ~2 degrees.  The maximum push angle at dp = 1/2 p to achieve this is ~82 degree.  This doesn't line up with your optimization formula.  Further, in my later example, I solve for the minimum necessary dp to cause a miss, dp(lat) and show that this can ONLY be applied at 0 degrees (ie laterally) to cause the miss.  There is no smaller dp that will work, and the angle is 0.  Your formula generates an angle of sin(x)=1.801/30 or 3.44 degrees.  This will not cause a miss.  We can backsolve for this by applying the force generated with that angle.  The lateral movement of the asteroid after application is dp(lat)cos(3.44)=1.798km/s.  The parallel slowdown is p-dp(lat)sin(x) = 30-1.801sin(3.4) = 29.89km/s.  We can solve for to see if this generates a miss by calculating the new time to impact by the new lateral speed.  (Again, Earth as a flat disc for this simplification.)  The new time is 108,000km/29.89km/s= 3613 second.  Distance traveled laterally in that time is 3613s x 1.798km/s = 6496.174 km.  The needed distance was 6500km.  A miss was not achieved.

You can quibble about the 6500km distance, but changing it changes all of the computations throughout equally, so it won't make a difference.  You maximization formula for a given dp results in an angle of push that doesn't generate a miss, while a 0 degree angle (ie perpendicular) _does _generate a miss at that dp.  It's close, but there's not an award for close in this case.

And, to forestall, the path through the center of the Earth is only possible for a head-on collision, which is not indicated by the scenario in the story (a rock launched from Mars that missed and is coming around again).  A more likely scenario is going to be an angle of approach from perpendicular to Earth's surface of tan(approach)=v(asteroid)/v(Earth).



> The wind in the car's frame is down and right.  However, the flag is also moving down, so it is pushing additionally through the air.  That means there is an additional force from the air on the flag that cancels the push "down" from the wind.  In other words, the total force from the air on the flag is to the right, just like the relative velocity of the air and flag.  You have forgotten about a force, so you are not summing the forces correctly.



The wind FOR THE CAR is down and right.  The wind FOR THE FLAGPOLE is only right.  The city moving past the car with it's air is what generates the apparent headwind on the car, but the flagpole is moving with that air and so feels no headwind at all.



> Look, I doubt I can convince you, so I am mostly talking to other readers, if anyone else still cares.  But I'm not sure it's productive continuing this discussion if you can't consider what I'm saying carefully.



You can't convince me because you're making an easy error of conception.  The 'wind' problem is fraught with this.

Let's switch to an airless moonbase with a mooncar.  In the moonbase frame, the car is moving north at a speed.  In this case, the mooncar's motor is opposed by the rolling friction of the tires and the friction of the drivetrain (elements present in the car, but ignored for simplicity).  These balance out.  If we switch to the mooncar's frame, now the ground is applying rolling friction to the tires and drivetrain that needs to be opposed by the motor -- still at a constant velocity.

Let's now add a laser firing from the west such as to hit the car at a specific point in it's travel north.  In the moonbase frame, the car arrives at that point at the same time as the laser, which then, though the push of photons and the explosive vaporizing of the skin of the moonskin, pushes the mooncar to the west.  Let's say it's a weak laser, and the friction of the tires of the mooncar can oppose it such that the mooncar doesn't actually move.  Now, translate that to the mooncar's frame.  The moonbase is moving past until the laser arrives.  What direction is the laser going in relation to the mooncar?  If your answer isn't "west", we have a problem.


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## freyar (Sep 14, 2018)

MarkB said:


> But we're not just talking about the crosswind. We're talking about the sum of the crosswind and the headwind - the latter of which only exists in the car's frame. So, as you've so elegantly demonstrated, the wind does change depending upon which frame you're in.



Perfect.


Ovinomancer said:


> Gah.  Okay, once more unto the breach.
> 
> NO CROSSWIND CASE:
> City frame -- car drives at 20 kph.  Air resists.  Force of the car's motor is exactly cancelled by the force of the air's resistance.
> ...




I bolded your error in the quote.  MarkB and I are not saying that the forces change from frame to frame.  But the wind velocity, which is *by definition* the velocity of the air, changes exactly by this vector addition.  This is your problem, just a simple conceptual one.  The "left-right" crosswind component doesn't change, but the total does.


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## Ovinomancer (Sep 14, 2018)

freyar said:


> Perfect.
> 
> 
> I bolded your error in the quote.  MarkB and I are not saying that the forces change from frame to frame.  But the wind velocity, which is *by definition* the velocity of the air, changes exactly by this vector addition.  This is your problem, just a simple conceptual one.  The "left-right" crosswind component doesn't change, but the total does.




Sigh, no, you're saying that a force that exists in either frame changes it's nature because you can't get past the air.  In the city frame, the car experiences the same 'headwind' as it does in the car frame, it's just caused by the car's movement against the still air.  The same 'wind' is existing either way.  It's only because you suddenly see the whole mass of air moving that you want to say it's a new thing called wind.  It's not, it's the same thing, just from a different viewpoint.  Since it's the same thing, it doesn't change it's nature.  Here's a test -- measure the 'wind speed' from the car in both frames without a cross wind.  Does it change?  No, it doesn't.  So, then, why are you insisting that it change when you add a crosswind?  The 'headwind' strength is exactly the same in any frame with or without a crosswind.  The crosswind doesn't change direction -- you're confusing the shift in appearance of the 'headwind' for an entirely new thing when it's present in the city frame exactly the same way.  

Force diagrams:
City frame:
..............'headwind'
.....................|
.....................v
crosswind ->  car <- tire friction
.....................^
.....................|
...............car motor

Now, Car frame:

..............'headwind'
.....................|
.....................v
crosswind ->  car <- tire friction
.....................^
.....................|
...............car motor

At any instantaneous point, the forces are thus.  You can add these forces together, because they are vectors, but that doesn't change their direction, just the direction of your new summation.  For instance, you can add the 'headwind' to the tire friction and claim there's now a new force from the top left called headtirewindfriction, but it doesn't change the forces actually working on the car.  Same with the crosswind and the motor.  Just because you think of the 'headwind' and the crosswind as winds _does not make them the same thing_.  The 'headwind' is static air resisting being pushed out of the way in the city frame, and exactly the same thing in the car frame, only you're calling it a 'wind'.  It's not, it's the same thing.  The only "wind" in the scenario is the crosswind, and it's present exactly the same way in both frames.  You're change frame and confusing yourself as to what's what.  Again, I suspected this would happen when this scenario was first proposed, but had the crazy idea I could point it out and get past it.  Days later, you're still insisting that 'wind' also means the air resistance from the car's motion, which is the same force in any frame.

Gah.  You guys are killing me, here.  II really thought this would be obvious by now, but it's the same thing.  Any chance you'll look at the mooncar example so we might get past this wind thing you're stuck on?


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## freyar (Sep 14, 2018)

This will probably be my last post, but we're making progress here, so I think it's worth one last try.



Ovinomancer said:


> YES! Except for the weird statement about double counting, which I'm not doing.  Regardless of whether of not the frame is the city or the car, the force of the air is the same because of the relative velocity between the air and the car.  Which means, when we add the wind from the side, we're adding a strictly lateral component to the mix.  This doesn't increase the up/down velocity of the air, it only adds a lateral component.  Even if you want to sum the vectors in the car frame so you get the 20.6 at 14 degrees, the forward force from the air doesn't change with frame and neither does the lateral force from the wind.
> 
> IN a case where we have a no wind model and then add the wind, then the added force of the wind is the new factor -- none of the old factors change.  And the new force applied by the wind is ALWAYS lateral -- it never changes direction.




In your previous post, you listed both an "air resistance" and a "sin(x)*F" force.  That's double counting.  In the frame of the car, the air resistance is apparently due to a headwind, that is, a "vertical" component to the wind's total velocity.  The wind from the west or left or whatever is the same component in either frame.  It doesn't rotate.  You just have to add another component.  You seem to be hung up on a very strange definition of wind.  When I or apparently MarkB or any textbook on the subject that I've read (which is several) refers to wind velocity, including when specifically talking about changing frames, wind velocity is defined as the velocity of air molecules, which changes from one reference frame to another.





> You're again saying that the weakest push must be more lateral, but the stronger the push the force moves to slow the asteroid.  That doesn't make sense, like, at all.  Take your statement about optimizing x.  Your point is that there exists some dp where the application of said dp generates the maximum value of x.  But is that what you're actually solving for?  Consider that your formula entirely fails to account for angles greater than 90.  There's a whole quadrant where  the angle of x measured from one perpendicular is valid that you're ignoring.  Further, a dp much, much larger than p creates x's that are nonsensical.  Take a dp 3 times greater than p such that your formula becomes Sin(x) = 3.  That's invalid.  If your formula can only work (supposedly) at a value of 0<dp<p, it's not really valid, is it?



The formula is not "invalid," it just has a well-defined regime of validity.  At dp=p, the optimal push is opposite the motion of the asteroid, which will stop the asteroid perfectly.  You then move into a different regime of optimization since at larger dp, you can now move the asteroid straight backward.  If our superhero can do that, she doesn't need to worry about deflecting the asteroid to miss the earth just barely, she can just launch it back from whence it came.  That seems to run counter to the drama needed in the story.  But, incidentally, the way I've chosen angles, x runs between -90 degrees and +90 degrees.  It's just that a push from negative angles, which helps speed up the asteroid in the earth's frame, is never the optimal way to push.  Also, I don't "fail to account" for angles greater than 90 degrees.  They don't exist in my coordinate system.



> Let's take a test case.  Let's assume an asteroid of velocity 30km/s.  Let's assume a dp with a velocity of 15km/s.  Since p and dp are both momentums of the same object, we can solve in velocities.  So, sin(x)=15/30=1/2.  x equals 45 degrees.  Half of the velocity of the asteroid maximizes deflection at 45 degrees.  Now, let's put that into some situations.  First, 1 hour out.  The needed miss angle is tan-1( 6500km/(30km/s * 3600s)) or 3.4 degrees.  The highest angle of push that achieves this is going to be (from perpendicular) dpcos(x)=(6500km/3600s) -> cos(x)=.13541.  x equals 82.2 degrees.  ANY push at 15 km/s 1 hour out will result in a miss so long as x is equal to or less than 82.2 degrees.  Compare this to your 45 degrees optimization.  Optimum is anything less than 82.2 degrees.



First of all, the inverse sine of 1/2 is 30 degrees, not 45 degrees (source: the scientific calculator on my computer, so maybe you will believe that).  But perhaps the problem is that I used the word "optimal" or "optimum" in the technical sense.  The optimal angle x given a fixed dp is the angle which maximizes the deflection of the asteroid for that fixed dp.  It doesn't mean you can't get the asteroid to miss the earth for a different angle.  Or that, if dp is too small, that you can get the asteroid to miss the earth at all.  I hope that makes more sense now.



> Then you shouldn't complain, yeah?  I mean, if you haven't read everything up til now, complaining that there are assumptions in place you didn't read is kinda on you.



Yeah, sure.  But what you're giving is a lot more specific than what I've seen on a quick glance through from other people, that's all.



> See my above where I show, by actually using your formula in a real scenario, that it doesn't work.  Pick a scenario, let's run your numbers.



Sure, let's look at that example.  Take dp/p=1/2, so I say sin(x)=1/2 (which is x=30 degrees) maximizes the angle of deflection.  For that value of sine, cos(x)=sqrt(1-sin(x)^2)=srqt(3)/2.  We have agreed that the deflection angle is given by tan(a) = dp cos(x)/(p-dp sin(x)).  I therefore claim that the largest possible angle of deflection for this value of dp/p is 
tan(a) = (1/2)(sqrt(3)/2) /(1-(1/2)(1/2)) = (2/3)(sqrt(3)/2) = sqrt(3)/3.  
Can you find a larger angle of deflection?  If I get a chance, I'll do a plot later.



> Yup, I boneheaded there.  I first messed up my coordinate placement.  Total bonehead on my part.  What I get for working in degrees.  The point here was it was the perpendicular portion of dp/p- the parallel portion of dp.  That's what I was solving with, despite my botch here.  The perpendicular part of dp is cos(x) when you measure from the perpendicular.



No problem!



> Are you assuming this, or did you do this?  If you did this, show, please, because I would be shocked for that to be true.



I had my computer plot it using Maple symbolic mathematics software.  If I get time, I will attach that later for the example above.



> Let me do a scenario and see.  Let's do the asteroid, 30km/s approach, 1 hour out, path through the center of Earth (so, straight in).
> 
> d = 30km/s * 3600 s = 108,000 km.
> Re = 6500 km
> ...



OK, let's just look at the last equation
1=cos(x)+.0602sin(x).
We agree that it is solved for x=0, meaning the right-hand side is 1 when x=0.  Now think about the derivative of the right-hand side, which is -sin(x)+.0602 cos(x).  At x=0, sin(x)=0, cos(x)=1, so this derivative is positive at x=0.  That means, for values of x that are a little bit positive, the right-hand side is larger than 1.  But for x=90 degrees, cos(x)=0, sin(x)=1, so the right-hand side is 0.0602, which is less than 1.  So somewhere in between, the right-hand side must hit 1 again, meaning that there is another solution.  According to Maple software, this other solution is x=6.89... degrees (EDIT: I initially gave the answer in radians, which is x=0.1202548706...) (This is by the Intermediate Value Theorem of calculus.)  There is also a range of x between x=0 and the other solution, where the right-hand side is larger than 1.  



> A dp less the dp(lat) will not cause a miss.  Run whatever model you want, doesn't happen with the straight in assumption.  dp(lat) IS the minimum necessary dp.



Actually, based on the logic above, if you use dp=dp(lat), you can get a larger deflection with an angle x>0.  That means you can reduce dp and still find an angle to hit the asteroid at and still deflect it enough.  I don't have time right now to do the math, but it shouldn't be hard.  I will see if I have time when doing the plots I mentioned.



> I show above how that doesn't work, though.  And what I said was the path of the asteroid goes through the center of the Earth, not that the asteroid *hits* the center of the Earth.
> 
> But, again, I refer to the specific scenario I did above.  The minimum deflection angle is ~2 degrees.  The maximum push angle at dp = 1/2 p to achieve this is ~82 degree.  This doesn't line up with your optimization formula.  Further, in my later example, I solve for the minimum necessary dp to cause a miss, dp(lat) and show that this can ONLY be applied at 0 degrees (ie laterally) to cause the miss.  There is no smaller dp that will work, and the angle is 0.  Your formula generates an angle of sin(x)=1.801/30 or 3.44 degrees.  This will not cause a miss.  We can backsolve for this by applying the force generated with that angle.  The lateral movement of the asteroid after application is dp(lat)cos(3.44)=1.798km/s.  The parallel slowdown is p-dp(lat)sin(x) = 30-1.801sin(3.4) = 29.89km/s.  We can solve for to see if this generates a miss by calculating the new time to impact by the new lateral speed.  (Again, Earth as a flat disc for this simplification.)  The new time is 108,000km/29.89km/s= 3613 second.  Distance traveled laterally in that time is 3613s x 1.798km/s = 6496.174 km.  The needed distance was 6500km.  A miss was not achieved.



Don't have time right now, so I will see if I can check these later (though now I'm getting enough "homework" on this that it may have to wait until the weekend or even Monday).



> You can quibble about the 6500km distance, but changing it changes all of the computations throughout equally, so it won't make a difference.  You maximization formula for a given dp results in an angle of push that doesn't generate a miss, while a 0 degree angle (ie perpendicular) _does _generate a miss at that dp.  It's close, but there's not an award for close in this case.
> 
> And, to forestall, the path through the center of the Earth is only possible for a head-on collision, which is not indicated by the scenario in the story (a rock launched from Mars that missed and is coming around again).  A more likely scenario is going to be an angle of approach from perpendicular to Earth's surface of tan(approach)=v(asteroid)/v(Earth).



While I won't claim to be infallible, I'm pretty confident that, if you get a miss at x=0, you can get a miss at some other small angle x with the same dp.  And I don't care about whether it's a head-on collision or not.  All we need to know is the require angle of deflection to know if we hit the earth or not.




> The wind FOR THE CAR is down and right.  The wind FOR THE FLAGPOLE is only right.  The city moving past the car with it's air is what generates the apparent headwind on the car, but the flagpole is moving with that air and so feels no headwind at all.



This is your problem.  When you say "FOR THE CAR," you clearly mean "in the rest frame of the car" or "relative to the car."  However, in the rest frame of the car, the wind is down and right.  Period.  There is that wind for the flag, but the flag is also pushing through the air in a way that counteracts some of that wind.  When you say "FOR THE FLAGPOLE," you mean "in the rest frame of the flagpole" or "relative to the flagpole."  Which does mean that the flagpole "feels" no headwind because what the flag feels is the quantity as measured in it's rest frame.  So you are getting hung up in nomenclature.  



> You can't convince me because you're making an easy error of conception.  The 'wind' problem is fraught with this.
> 
> Let's switch to an airless moonbase with a mooncar.  In the moonbase frame, the car is moving north at a speed.  In this case, the mooncar's motor is opposed by the rolling friction of the tires and the friction of the drivetrain (elements present in the car, but ignored for simplicity).  These balance out.  If we switch to the mooncar's frame, now the ground is applying rolling friction to the tires and drivetrain that needs to be opposed by the motor -- still at a constant velocity.
> 
> Let's now add a laser firing from the west such as to hit the car at a specific point in it's travel north.  In the moonbase frame, the car arrives at that point at the same time as the laser, which then, though the push of photons and the explosive vaporizing of the skin of the moonskin, pushes the mooncar to the west.  Let's say it's a weak laser, and the friction of the tires of the mooncar can oppose it such that the mooncar doesn't actually move.  Now, translate that to the mooncar's frame.  The moonbase is moving past until the laser arrives.  What direction is the laser going in relation to the mooncar?  If your answer isn't "west", we have a problem.



You mean pushing the car to the east in the moonbase frame, since the laser is coming from the west, but no matter.  This is fine.  But changing to the mooncar frame is a tricky question, since you're involving light.  That means we have to consider relativistic corrections, which means the force does change from frame to frame IIRC.  Anyway, my answer is that the laser is coming from the north-west, not exactly the west.  If the mooncar is moving at nonrelativistic speeds, it's only very slightly from the north, but that's it.  But, like I say, we have to be a bit careful of relativistic effects even with a slow mooncar, since there's a laser involved.

It would be somewhat easier to consider a space ship flying past a space station with a laser, since then we can just look at momentum conservation and not worry about forces, actually.  Maybe I can do that for you if time allows.


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## MarkB (Sep 14, 2018)

Ovinomancer said:


> Here's a test -- measure the 'wind speed' from the car in both frames without a cross wind.  Does it change?  No, it doesn't.




If you're measuring the 'wind speed' from the car, you are explicity within the car frame - that's the entire basis of there being two different frames. If you're in the city frame, you're not measuring the wind speed from the car, you're only measuring it from the city.


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## Ovinomancer (Sep 14, 2018)

freyar said:


> This will probably be my last post, but we're making progress here, so I think it's worth one last try.
> 
> 
> 
> In your previous post, you listed both an "air resistance" and a "sin(x)*F" force.  That's double counting.  In the frame of the car, the air resistance is apparently due to a headwind, that is, a "vertical" component to the wind's total velocity.  The wind from the west or left or whatever is the same component in either frame.  It doesn't rotate.  You just have to add another component.  You seem to be hung up on a very strange definition of wind.  When I or apparently MarkB or any textbook on the subject that I've read (which is several) refers to wind velocity, including when specifically talking about changing frames, wind velocity is defined as the velocity of air molecules, which changes from one reference frame to another.



That was the same force presented in different ways, not added together.  No double counting involved.

The point of the car example was to explore the question of 'does a frame change from solar to Earth change the direction of a push on the asteroid.'  The crosswind in the example is supposed to be the push on the asteroid.  But, since there's now an added mass of static air that resists the motion of the car, and when the frame change to to car is made this force is remained to a 'headwind', the failure of conception is that this is an actual wind like the crosswind instead of an artifact of the frame change renaming forces.  As you point out, you can sum the forces, as they are vectors, and with the confusion on -wind add the 'head'wind to the 'cross'wind to get something you've called the wind, but the actual components of these forces haven't changed.  The 'crosswind' still only pushes the car laterally, even with the frame change (as you note above).  When you switch over to a scenario that lacks the confusion of the static air in the scenario, it should become entirely apparent that my point about forces not changing direction due to a frame change is completely correct.  The 'crosswind' only ever adds lateral movement in the same direction and same magnitude regardless of the frame.  You adding the renamed air resistance as a 'headwind' notwithstanding.






> The formula is not "invalid," it just has a well-defined regime of validity.  At dp=p, the optimal push is opposite the motion of the asteroid, which will stop the asteroid perfectly.  You then move into a different regime of optimization since at larger dp, you can now move the asteroid straight backward.  If our superhero can do that, she doesn't need to worry about deflecting the asteroid to miss the earth just barely, she can just launch it back from whence it came.  That seems to run counter to the drama needed in the story.  But, incidentally, the way I've chosen angles, x runs between -90 degrees and +90 degrees.  It's just that a push from negative angles, which helps speed up the asteroid in the earth's frame, is never the optimal way to push.  Also, I don't "fail to account" for angles greater than 90 degrees.  They don't exist in my coordinate system.



Well, it ignores the 90 to 180 side of things as well (ie, it breaks on that side).  I still disagree that your formula works at all, narrowed coordinate regime or no.



> First of all, the inverse sine of 1/2 is 30 degrees, not 45 degrees (source: the scientific calculator on my computer, so maybe you will believe that).  But perhaps the problem is that I used the word "optimal" or "optimum" in the technical sense.  The optimal angle x given a fixed dp is the angle which maximizes the deflection of the asteroid for that fixed dp.  It doesn't mean you can't get the asteroid to miss the earth for a different angle.  Or that, if dp is too small, that you can get the asteroid to miss the earth at all.  I hope that makes more sense now.



Yup, bonehead math fail, again.  Going too fast and not paying attention to the setup because I'm working on the rest.  Doesn't change my points, though.



> Yeah, sure.  But what you're giving is a lot more specific than what I've seen on a quick glance through from other people, that's all.



Again, not my problem that you have confusion on this because you didn't read the whole thread.  Dunno what to so, man.



> Sure, let's look at that example.  Take dp/p=1/2, so I say sin(x)=1/2 (which is x=30 degrees) maximizes the angle of deflection.  For that value of sine, cos(x)=sqrt(1-sin(x)^2)=srqt(3)/2.  We have agreed that the deflection angle is given by tan(a) = dp cos(x)/(p-dp sin(x)).  I therefore claim that the largest possible angle of deflection for this value of dp/p is
> tan(a) = (1/2)(sqrt(3)/2) /(1-(1/2)(1/2)) = (2/3)(sqrt(3)/2) = sqrt(3)/3.
> Can you find a larger angle of deflection?  If I get a chance, I'll do a plot later.
> 
> ...



Turns out we're both not fully right.  It appears there's multiple solutions to the formula -- I found one between 6.2 and 6.3 and another definite at almost 88 degrees and a number of (around 8 or so) where it may solve.  I'm using numerical substitution in excel rather than expensive plotting software, so...

81.6 - 81.7
75.3-75.4 and 75.5-75.6
and so one.  A number of points of crossing.

I guess it goes to show you shouldn't eyeball trig.  Good point looking at the slope -- nice catch.

I'm not sure what this says about either of our assertions.  The generated angle from sin(x)=dp(lat)/p still doesn't work, but it appears a number of other possible angles will with that dp(lat) value.  In thinking about it, there's a definite minimum miss angle along which the asteroid will miss the disc regardless, but that describes a infinite set of possible triangles comprised of the modified parallel and lateral vectors.  I'm back to my (unspoken) thinking that the optimization problem is a differential equation and can't be solved via trig.


Don't have time right now, so I will see if I can check these later (though now I'm getting enough "homework" on this that it may have to wait until the weekend or even Monday).


While I won't claim to be infallible, I'm pretty confident that, if you get a miss at x=0, you can get a miss at some other small angle x with the same dp.  And I don't care about whether it's a head-on collision or not.  All we need to know is the require angle of deflection to know if we hit the earth or not.




> This is your problem.  When you say "FOR THE CAR," you clearly mean "in the rest frame of the car" or "relative to the car."  However, in the rest frame of the car, the wind is down and right.  Period.  There is that wind for the flag, but the flag is also pushing through the air in a way that counteracts some of that wind.  When you say "FOR THE FLAGPOLE," you mean "in the rest frame of the flagpole" or "relative to the flagpole."  Which does mean that the flagpole "feels" no headwind because what the flag feels is the quantity as measured in it's rest frame.  So you are getting hung up in nomenclature.
> 
> 
> You mean pushing the car to the east in the moonbase frame, since the laser is coming from the west, but no matter.  This is fine.  But changing to the mooncar frame is a tricky question, since you're involving light.  That means we have to consider relativistic corrections, which means the force does change from frame to frame IIRC.  Anyway, my answer is that the laser is coming from the north-west, not exactly the west.  If the mooncar is moving at nonrelativistic speeds, it's only very slightly from the north, but that's it.  But, like I say, we have to be a bit careful of relativistic effects even with a slow mooncar, since there's a laser involved.
> ...




But, where is the force applied?  It's not from the WNW, but from due west.  That's the point -- the force applied doesn't change direction.  The approach of the force, whatever it may be, may appear to change direction, but the force does not.  The laser pushes the car eastward irrespective of the framing.


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## Janx (Sep 14, 2018)

Would y'all like another physics problem? It's a similar deflection issue, just different scale of objects.


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## Nagol (Sep 14, 2018)

Janx said:


> Would y'all like another physics problem? It's a similar deflection issue, just different scale of objects.




Are you throwing miniature blackholes at stars now?


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## freyar (Sep 14, 2018)

Ovinomancer said:


> That was the same force presented in different ways, not added together.  No double counting involved.
> 
> The point of the car example was to explore the question of 'does a frame change from solar to Earth change the direction of a push on the asteroid.'  The crosswind in the example is supposed to be the push on the asteroid.  But, since there's now an added mass of static air that resists the motion of the car, and when the frame change to to car is made this force is remained to a 'headwind', the failure of conception is that this is an actual wind like the crosswind instead of an artifact of the frame change renaming forces.  As you point out, you can sum the forces, as they are vectors, and with the confusion on -wind add the 'head'wind to the 'cross'wind to get something you've called the wind, but the actual components of these forces haven't changed.  The 'crosswind' still only pushes the car laterally, even with the frame change (as you note above).  When you switch over to a scenario that lacks the confusion of the static air in the scenario, it should become entirely apparent that my point about forces not changing direction due to a frame change is completely correct.  The 'crosswind' only ever adds lateral movement in the same direction and same magnitude regardless of the frame.  You adding the renamed air resistance as a 'headwind' notwithstanding.



Of course the forces don't change, I haven't said that they do.  But you said the "wind" doesn't change.  The standard definition of wind is the (bulk flow) velocity of air as measured in a given frame.  That's a velocity.  It certainly changes.  The issue which seems clear now after a long discussion is that you are using a non-standard definition of wind and are disagreeing with those of us who are using the standard definition.
The other confusion is that you seem to be changing what you're saying from post to post.  In post 68 of this thread, you said that if the wind changes direction from frame to frame, the force would change, ignoring the fact that in the city's frame there is a component of the force (of air on the car) from the motion of the car through the air.  Then in post 84, you put that component of force in both frames, but you rotated the "lateral" component of the force from the air (which is due to "wind" in both frames).  In post 98, you are making a distinction between "headwind" and "crosswind," which is fine in that a car has a preferred orientation but isn't really relevant to overall definition of the wind.  Wind velocity is a vector quantity, and how we choose axes to determine components of that vector doesn't change what the vector is.  
Anyway, I think I'll stop talking about wind now.  This isn't really getting anywhere.



> Well, it ignores the 90 to 180 side of things as well (ie, it breaks on that side).  I still disagree that your formula works at all, narrowed coordinate regime or no.



Going from negative 90 degrees to positive 90 degrees covers the full 180 degrees from back to front.  Or, if we can think about a sphere, it covers the full 180 degrees from the south pole to the north pole.  (The 360 degrees around the equator don't really matter for what I've been talking about.)

I have to get going, so I will have to respond to the rest later.  But my ENWorld time is usually a bit short, and I'd like to get back to monster conversions, so hopefully the plots I can post will be convincing.


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## Eltab (Sep 15, 2018)

Janx said:


> Would y'all like another physics problem? It's a similar deflection issue, just different scale of objects.



Figure out which force of nature, and at what magnitude, will be necessary to get our three professional arguers to notice your earlier post that you had enough physics to write the story.


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## Janx (Sep 15, 2018)

Nagol said:


> Are you throwing miniature blackholes at stars now?




Alas, I don't know how feasible it is to generate a blackhole of any size, let alone move it.


Hopefully this is all fun.  I see some of you have gone deeper into math with flagpoles and cars.


The new problem is actually old to me.  It seemed like a good idea and is the basis of a heroic scene. I've skimped on details.

The bad guy has an AR15 because of Reasons(TM).  He's going to shoot a lot of unsuspecting people over yonder.

The hero steps in the way, let's say 10 feet in front (give or take, I had to assume the turret isn't setup right at the edge).  This helps block LOS and sucks up some bullets if the real plan doesn't work.

The real plan is that there are high speed cameras, computers and a rapid firing railgun launching matter to deflect the bullets from some unknown position.

The tech can fire at the speed of plot.  It is not allowed to kill the bad guy, for Reasons(TM). The hero doesn't love this plan, and would appreciate a second opinion.

Is it plausible that the bullets can be shot out of the air such that nobody gets hurt, including the hero?

PS. It is simply coincidence that both of these are deflection problems.  I worked out this situation a few years ago, and am trudging through the rewrite of the novel to finally get to write this bullet scene.  The Asteroid story came up this year.


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## freyar (Sep 15, 2018)

Eltab said:


> Figure out which force of nature, and at what magnitude, will be necessary to get our three professional arguers to notice your earlier post that you had enough physics to write the story.




As a physicist by trade, discussion is a big part of the job.  Disagreement comes along with the territory sometimes. Education is also a big part of the job, so (like I said) I'm responding in as much detail as I am in case it helps anyone else reading along to learn something.


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## freyar (Sep 15, 2018)

Ovinomancer said:


> But, where is the force applied?  It's not from the WNW, but from due west.  That's the point -- the force applied doesn't change direction.  The approach of the force, whatever it may be, may appear to change direction, but the force does not.  The laser pushes the car eastward irrespective of the framing.




Let me go back and quote your original mooncar/laser scenario, just so we don't have to skip around the thread to read it.


> Let's switch to an airless moonbase with a mooncar. In the moonbase frame, the car is moving north at a speed. In this case, the mooncar's motor is opposed by the rolling friction of the tires and the friction of the drivetrain (elements present in the car, but ignored for simplicity). These balance out. If we switch to the mooncar's frame, now the ground is applying rolling friction to the tires and drivetrain that needs to be opposed by the motor -- still at a constant velocity.
> 
> Let's now add a laser firing from the west such as to hit the car at a specific point in it's travel north. In the moonbase frame, the car arrives at that point at the same time as the laser, which then, though the push of photons and the explosive vaporizing of the skin of the moonskin, pushes the mooncar to the west. Let's say it's a weak laser, and the friction of the tires of the mooncar can oppose it such that the mooncar doesn't actually move. Now, translate that to the mooncar's frame. The moonbase is moving past until the laser arrives. What direction is the laser going in relation to the mooncar? If your answer isn't "west", we have a problem.




To simplify the problem, I'm going to make it a spaceship coasting past a starbase with a laser.  I will lay down an arbitrary 2D N-S-E-W set of directions.

OK, in the starbase frame, the ship is moving north.  The laser is somewhere north-west of the ship.  In the base frame, it fires a laser beam due east (ie, west to east).  It hits the ship.  Let's forget about vaporizing part of the ship, since that introduces unknown amounts of chemical energy, unknown direction of the exhaust, etc.  The light in the laser itself can push the ship.  Let's also assume that the light is perfectly reflected, so we don't have to worry about heating the ship and what the thermal radiation coming off the ship does.  So then the laser hits the car (pushing it east) and reflects off to the west (also pushing the ship to the east).  So, in the end, the ship moves off toward the northeast (at least until the pilot corrects course).  In other words, as you say, the force is to the east.

In the frame of the ship, the base and laser are moving south.  While the laser is still north of the ship, it fires.  The laser beam moves southeast and hits the ship, pushing to the southeast.  The beam reflects off the ship in a southwest direction, pushing the ship to the northeast.  The north-south component of the imparted momentum cancels, so the ship now moves to the east.

I think we both agree on this.  If the ship is moving at a significant fraction of lightspeed, we have to modify the discussion a bit, particularly if you want to talk about forces and not just momentum conservation.  But the point is that the velocity of the laser beam (particularly its direction of travel) can change without changing the direction of the imparted momentum.

Incidentally, the aberration of light, the change of the direction of motion of a lightbeam in different frames of reference, is an important effect historically (first observed 1725-6) and was one of the observations that most influenced Einstein in the development of special relativity.


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## Janx (Sep 15, 2018)

freyar said:


> As a physicist by trade, discussion is a big part of the job.  Disagreement comes along with the territory sometimes. Education is also a big part of the job, so (like I said) I'm responding in as much detail as I am in case it helps anyone else reading along to learn something.




I really appreciate that effort.  I was just worried about abusing your generosity.


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## freyar (Sep 15, 2018)

Ovinomancer said:


> Turns out we're both not fully right.  It appears there's multiple solutions to the formula -- I found one between 6.2 and 6.3 and another definite at almost 88 degrees and a number of (around 8 or so) where it may solve.  I'm using numerical substitution in excel rather than expensive plotting software, so...
> 
> 81.6 - 81.7
> 75.3-75.4 and 75.5-75.6
> ...



Just a reminder, we are looking for solutions to the equation 1=cos(x)+0.0602 sin(x) between x=0 and 90 degrees.  I suspect there is some problem with your excel, and I'll see if I can attach a plot.  But you are not limited to excel.  Try Wolfram alpha (free!).  If you type cos(x)+0.0602 sin(x)=1 in the box, it will give you a plot showing both solutions in the 0 to 90 degree range as red dots.  It will also give you a formula for all solutions below that.  If you want numerical values, you can click the "approximate solution" button above the formula to get the answer in radians, which convert to degrees by multiplying with 180/pi.  If you want to have a zoom in of the plot, you can instead type "plot cos(x)+0.0602 sin(x) from 0 to 10 degrees" (or whatever range you want) in the text box.  But it should be clear from the first plot that the only solutions between 0 and 90 degrees are x=0 and the one at about 0.12 radians.

Here are my plots. The blue curve is our trig function, and the purplish line is 1.  The horizontal axis is the angle x in degrees.  You can see there is a solution at x=0 and 6.89 degrees.  Incidentally, I'm using Mathematica for plotting today, since I like its plotting features better.



This comes back to my original formula.  We agree that the angle of deflection is tan(a)=dp cos(x)/( p-dp sin(x))= (dp/p) cos(x)/(1-(dp/p)sin(x)).  We were talking about an example with dp/p = 1/2.  I claim that the maximum deflection --- where tan(a) is biggest --- occurs for sin(x)=1/2, which is x=30 degrees.  I also claim that maximum deflection is given by tan(a)=sqrt(3)/3.  Here is my plot.  The blue curve is tan(a) as a function of x (horizontal axis in degrees), the purplish line is sqrt(3)/3=0.57735..., and the red vertical line is at 30 degrees.  You can make your own judgment about whether my formula is correct.  You can also do a version of this plot on Wolfram alpha for yourself, but you'll probably have to use radians for x.


Editing because it posted when it was supposed to preview: Anyway, we can also figure out the minimum value of dp/p needed to deflect the asteroid by required angle given by tan(m).  We just use the optimal angle given by sin(x)=dp/p for a given dp.  Then the deflection angle (the best one for that dp) is 
tan(a) = (dp/p) sqrt(1-(dp/p)^2)/(1-(dp/p)^2) = (dp/p)/sqrt(1-(dp/p)^2).  Then you have to find when this tan(a)>=tan(m).  I believe that equality occurs at dp/p=sin(m), so that should be the minimal value of dp/p that will deflect the asteroid the required amount.

Once again, I've been assuming dp/p<1 because otherwise Pierce could just say, "this is simple," and just be a dumb brute and blast the asteroid back from whence it came without having to stress or think about it.


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## Eltab (Sep 16, 2018)

Janx said:


> Is it plausible that the bullets can be shot out of the air such that nobody gets hurt, including the hero?



Do you have a sketch of where everything is, relative to each other?  Important scenery features / objects we should know about?

It is plausible, but technologically on the bleeding edge, that the crowd at a distance can be protected.
It is not plausible to protect the hero in this matter.  It would be more plausible for the hero to hotwire the bullet-interceptor and shoot at the villain.

I imagine that the protective system would be similar to President Reagan's original "Star Wars" / Strategic Defense Initiative concept, on a smaller scale.  
You would want to deflect the bullets downward, into the ground, if outdoors.  If indoors, into walls, padded ceilings, &c that can be repaired easily.  
Your nightmare is that you hit a bullet and it shatters into a cone of still-fast-moving fragments that act like shotgun pellets.  
Perhaps the interceptor-bullets are magnetic and the whole thing sticks together and heads off in a different direction, sum of the vectors (plus is no longer aerodynamic).
The big problem I see is that your computer has to scan the area, detect the bullets, not have false positives (a dragonfly), calculate their flight trajectories, detect and adjust for wind, calculate intercept trajectories, turn your own machinegun to the correct direction, fire, repeat while the villain's bullets are still in mid-flight.

P.S.  I remember reading this in _Arm in Arm_, a book of almost-nonsense

BULLETIN -- BULLETIN -- BULLETIN
Shott shot at Nott.
Nott, not wanting to be shot, shot at Shott.
Nott's shot shot Shott's shot.
Luckily, neither Nott nor Shott was shot.


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## Nagol (Sep 16, 2018)

Janx said:


> Alas, I don't know how feasible it is to generate a blackhole of any size, let alone move it.
> 
> 
> Hopefully this is all fun.  I see some of you have gone deeper into math with flagpoles and cars.
> ...




There is no real reason you can't shoot a bullet out of the air with sufficient tech. It's sort-of how phalanx guns work to intercept missiles. It's more can the apparatus to do so be close enough to intercept before the bullet hits the hero.  That is going to be tricky.

The constraints for interception are how accurately can you gauge the target's trajectory, how quickly can you get something to intercept the bullet, and how much deviation do you need to impart to its momentum.

The further the bullet can be allowed to travel before launching a counter, the better your trajectory prediction can be.  The longer the travel time of your interceptor, the better your prediction needs to be.  The further the travel time of the bullet after the interceptor hits, the less deflection you need. 

The constraints on distance of interceptor device is how long does it take to develop a target solution, how long does it take to accelerate the interceptor, and what is the flight time to intercept.
Let's start with what we know.  an AR-15 bullet travels at about 1,100 metres per second.  It will reach the hero in about 3/1,000 of a second after clearing the barrel.

Acquiring a trajectory is likely to take at least two images.  A very high end commercial camera shoots about 16,000 frames per second (a truly astounding single-purpose camera can shoot a trillion frames per second but isn't fit for purpose). Assume we get something just above that.  So you lose the first 1/8000 of a second gaining the trajectory.  We'll assume the computer effectively takes zero time to calculate a response trajectory.

Let's accelerate the interceptor at 1000 g to a maximum speed of Mach 10 (4,000 metres per second).  That takes 4/10 of a second.   Oops.  That didn't help.  Assuming we have zero flight time, and needed to hit Mach 10, the necessary acceleration would be more than 1,500,000 g. ( t = (3/1000 - 1/8000), v = 4000, a = v/t) to strike before the bullet hits the hero.  Because we don't care overmuch, let us increase the acceleration to 15,000,000 g and simply stipulate the interceptor isn't destroyed of deformed by the forces.  That costs us about 1/4000 of a second.  That leaves us with 1/400 of a second to intercept.  At 4,000 m/s the interceptor would have to be no more than 10 metres away.  Which is probably too close.  So let's increase both the acceleration and final speed 10-fold.  The inceptor is now accelerated at 150,000,000 g to a maximum speed of 40,000 m/s (past the point of ridiculousness in both cases).  Because both acceleration and velocity increased by the same multiple, the time remains the same so we are left with 1/400 second transit time so we can move the inceptor to 100 m away.  Which is good because at those speeds, the interceptor is going to burn up in the atmosphere like a miniature meteor.

Let's assume for safety's sake we need to change the bullet's trajectory 45 degrees.  Call the bullet 5 grams so the forward momentum is 5 Newton seconds.  Since the interceptor is traveling at 40,000 m/s, it's mass must be at least 0.125 grams for a perpendicular hit.

This scenario suffers from the same problem as the meteor one -- the projectile is just too close to its target.  Interception works much better when there is time to respond.


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## Umbran (Sep 16, 2018)

Nagol said:


> There is no real reason you can't shoot a bullet out of the air with sufficient tech.




Yes.  But... why don't you just shoot the bad guy's gun, and get it over with?



> Acquiring a trajectory is likely to take at least two images.  A very high end commercial camera shoots about 16,000 frames per second (a truly astounding single-purpose camera can shoot a trillion frames per second but isn't fit for purpose).




You have tech enough to see, calculate, fire, and intercept in in 3 milliseconds.  You aren't using a camera.  You are using a laser or radar targeting system.




> This scenario suffers from the same problem as the meteor one -- the projectile is just too close to its target.  Interception works much better when there is time to respond.




Agreed.  But, there are variations - in this case, for a superhero game, it is plausible that you shoot the bullets with a laser powerful enough to vaporize them before they reach their target.

Or, you know, have the hero carry out a big metal plate to block the bullets.  Unless they are the rare hero than can accept taking some bullets as a backup plan, but doesn't have above-human strength.


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## Nagol (Sep 16, 2018)

Umbran said:


> Yes.  But... why don't you just shoot the bad guy's gun, and get it over with?




Because that wasn't what was asked?



> You have tech enough to see, calculate, fire, and intercept in in 3 milliseconds.  You aren't using a camera.  You are using a laser or radar targeting system.




I've seen apps that can predict which roulette slot a ball will drop from less than a second of video shot from a phone camera.  The problem with RADAR is the wavelength limits your accuracy +/- 15 cm is a huge problem when you're trying to hit something a cubic cm in volume. The problem with lasers is single line of resolution.  Great for hand-targeting cars less useful for bullets.  You could use a 100 GHz or higher frequency RADAR, I suppose.



> Agreed.  But, there are variations - in this case, for a superhero game, it is plausible that you shoot the bullets with a laser powerful enough to vaporize them before they reach their target.



 Not part of the initial parameters.



> Or, you know, have the hero carry out a big metal plate to block the bullets.  Unless they are the rare hero than can accept taking some bullets as a backup plan, but doesn't have above-human strength.



  Not part of the original parameters.


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## MarkB (Sep 16, 2018)

Janx said:


> The real plan is that there are high speed cameras, computers and a rapid firing railgun launching matter to deflect the bullets from some unknown position.
> 
> The tech can fire at the speed of plot.  It is not allowed to kill the bad guy, for Reasons(TM). The hero doesn't love this plan, and would appreciate a second opinion.
> 
> Is it plausible that the bullets can be shot out of the air such that nobody gets hurt, including the hero?




Do you at least have sight of the shooter, at a reasonably high resolution? A lot of time can be shaved off the interception if you don't worry about tracking the bullet itself, but instead calculate its trajectory and launch time based upon the shooter's aim and when they tighten their finger on the trigger. Launch your interceptor in the split second after they commit to the shot but before the bullet leaves the chamber, and you have time to launch a larger mass to intercept it - even a stream of projectiles to help ensure a hit.


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## Nagol (Sep 16, 2018)

MarkB said:


> Do you at least have sight of the shooter, at a reasonably high resolution? A lot of time can be shaved off the interception if you don't worry about tracking the bullet itself, but instead calculate its trajectory and launch time based upon the shooter's aim and when they tighten their finger on the trigger. Launch your interceptor in the split second after they commit to the shot but before the bullet leaves the chamber, and you have time to launch a larger mass to intercept it - even a stream of projectiles to help ensure a hit.




Wouldn't work.  Bullets have too much spread due to imperfections in the bullet and the rifling which acts as minor limit on accuracy, but matters much more when you are trying to hit a cubic centimeter target in the air not to mention the motion of the person firing is not smooth either.  If you are close enough to predict these things, you're probably on the ledge with the hero.

ETA:  Looked up at the AR-15 tests, its cone is a mere 1.1 inches over 100 yards so that part is probably fine for the distances were talking. The shooter's motion would still probably throw you off too much though.


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## Umbran (Sep 16, 2018)

Nagol said:


> Because that wasn't what was asked?




Rule #1 - when someone asks for help, they don't generally tell you their problem, they tell you their preferred solution.  That is not always the *best* solution, or even a good one.



> I've seen apps that can predict which roulette slot a ball will drop from less than a second of video shot from a phone camera.  The problem with RADAR is the wavelength limits your accuracy...




So, the word laser was in there for a reason....



> Not part of the initial parameters.




Dude, you're a RPG player, yes?  Since when is remaining within the initial parameters part of your operating system?


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## Nagol (Sep 16, 2018)

Umbran said:


> Rule #1 - when someone asks for help, they don't generally tell you their problem, they tell you their preferred solution.  That is not always the *best* solution, or even a good one.




I'll respond to parameters given.  Training people to give me problems instead of solutions was something I did for decades.  Thankfully, not any more.  Besides, it was pitched as an integral part of a story-scene.  



> So, the word laser was in there for a reason....



 and lasers won't work for other reasons as I pointed out.  LIDAR is great if your target is large and/or still.



> Dude, you're a RPG player, yes?  Since when is remaining within the initial parameters part of your operating system?




Since forever.  Whenever my gaming circle decides to speculate about player alignments, the consensus is I'm a strong LN.


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## Janx (Sep 17, 2018)

MarkB said:


> Do you at least have sight of the shooter, at a reasonably high resolution? A lot of time can be shaved off the interception if you don't worry about tracking the bullet itself, but instead calculate its trajectory and launch time based upon the shooter's aim and when they tighten their finger on the trigger. Launch your interceptor in the split second after they commit to the shot but before the bullet leaves the chamber, and you have time to launch a larger mass to intercept it - even a stream of projectiles to help ensure a hit.




I'm back from a busy day or so of busy stuff. So I'll try to clarify what I can for everybody.  Thanks to Nagol for reminding Umbran of the parameters.  Umbran's right to ask, it's just that i've got restrictions that I chose to work with, like not killing bad guys in the first book.

To Eltab, I don't have a diagram (it's a vision in my head based on a real place I saw from the crowd's perspective).  It is on a roof/near top floor of a building under construction across from an open roof sports place.

As this is for my first novel that I hope to get Rich and Famous(TM) from, I've got to be a little more circumspect with the details. That's a big if...

The actual gun is on a turret, controlled by a computer.  You can google how to make one (everything in the first novel is or almost is Mythbusters Plausible).  I may stretch that in places, but I try to base things on stuff I've seen.

The hero is not bullet proof but does have the equivalent of a bullet proof vest.  It's unlikely he could survive all 50 rounds for any variety of reasons.

The hero's tech isn't allowed to kill in this book, call it a design decision by the hero before the book started.  He might regret that line of thinking by the time this is over.  Also, the beta-readers seem convinced that he won't kill somebody. Not sure how they got that idea, but the guy as evolved on the page will have to find where that line is someday.

The tech should have LOS to the gun, recognize it as a threat, and project direction of shot based on  the barrel.  I am not sure if it could see the trigger being pulled, but I could rule that in.

We can move the turret a little farther away (perhaps another 10 feet to reach the power outlet).  The tech can be perched on something higher and closer (on the same block as the building where this confrontation takes place).

Yes, I just had the idea of the hero unplugging the turret before he strolls into the scene, or something.  And then having the bad guy fire it manually, which might be more gratifying for the bad guy (they have feelings too).

By high speed cameras, I mean like the ones Mythbusters uses to watch bullets hit things.  Allegedly, the tech exists.  The sci-fi part being that computers can handle the data and use it to shoot down bullets. 

It sounds like there are hints of this being plausible from a angle/mass thing, but the distances suck (which is why I can move the gun back and the tech closer). I don't have to deliver the math to an editor, but it'd be nice when one of y'all read the book, you'd say, "well, you know, if the conditions were such and such, it just might be possible..." in an internet argument.


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## Janx (Sep 17, 2018)

Umbran said:


> Rule #1 - when someone asks for help, they don't generally tell you their problem, they tell you their preferred solution.  That is not always the *best* solution, or even a good one.
> 
> 
> 
> ...




Just a note to both my EN friends  [MENTION=177]Umbran[/MENTION] and @Morrus;


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## Ovinomancer (Sep 17, 2018)

freyar said:


> Let me go back and quote your original mooncar/laser scenario, just so we don't have to skip around the thread to read it.
> 
> 
> To simplify the problem, I'm going to make it a spaceship coasting past a starbase with a laser.  I will lay down an arbitrary 2D N-S-E-W set of directions.
> ...




Right -- regardless of frame, the force (or dp, whichever your prefer) remains the same and in the same direction.  The apparent shift in direction of approach does not change this fact -- the force applied does not change direction regardless of frame.

To bring this back to the original point -- if you change the frame, you DO NOT change the direction of the push on the asteroid. The vector of the _applied_ force or the dp or whichever you prefer _remains the same._


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## Ovinomancer (Sep 17, 2018)

freyar said:


> Just a reminder, we are looking for solutions to the equation 1=cos(x)+0.0602 sin(x) between x=0 and 90 degrees.  I suspect there is some problem with your excel, and I'll see if I can attach a plot.  But you are not limited to excel.  Try Wolfram alpha (free!).  If you type cos(x)+0.0602 sin(x)=1 in the box, it will give you a plot showing both solutions in the 0 to 90 degree range as red dots.  It will also give you a formula for all solutions below that.  If you want numerical values, you can click the "approximate solution" button above the formula to get the answer in radians, which convert to degrees by multiplying with 180/pi.  If you want to have a zoom in of the plot, you can instead type "plot cos(x)+0.0602 sin(x) from 0 to 10 degrees" (or whatever range you want) in the text box.  But it should be clear from the first plot that the only solutions between 0 and 90 degrees are x=0 and the one at about 0.12 radians.
> 
> Here are my plots. The blue curve is our trig function, and the purplish line is 1.  The horizontal axis is the angle x in degrees.  You can see there is a solution at x=0 and 6.89 degrees.  Incidentally, I'm using Mathematica for plotting today, since I like its plotting features better.
> View attachment 101470



Yep, I missed that Excel was using radians.  It was late, I wasn't paying attention, and it should have jumped up and smacked me.  So, yes, between 6 and 7 degrees (I see you edited your original claim to include this update, which is, oddly again, one of the reasons I went to check, because ~.12 degrees doesn't solve as you initially claimed, seems we both had issues mistaking radians for degrees).  Sadly, I originally had this worked out in degrees using a calculator, which helped confuse me on the Excel sheet because, in radians, there's also a point between 6 and 7 radians were the value crosses 1.  



> This comes back to my original formula.  We agree that the angle of deflection is tan(a)=dp cos(x)/( p-dp sin(x))= (dp/p) cos(x)/(1-(dp/p)sin(x)).  We were talking about an example with dp/p = 1/2.  I claim that the maximum deflection --- where tan(a) is biggest --- occurs for sin(x)=1/2, which is x=30 degrees.  I also claim that maximum deflection is given by tan(a)=sqrt(3)/3.  Here is my plot.  The blue curve is tan(a) as a function of x (horizontal axis in degrees), the purplish line is sqrt(3)/3=0.57735..., and the red vertical line is at 30 degrees.  You can make your own judgment about whether my formula is correct.  You can also do a version of this plot on Wolfram alpha for yourself, but you'll probably have to use radians for x.
> View attachment 101472
> 
> Editing because it posted when it was supposed to preview: Anyway, we can also figure out the minimum value of dp/p needed to deflect the asteroid by required angle given by tan(m).  We just use the optimal angle given by sin(x)=dp/p for a given dp.  Then the deflection angle (the best one for that dp) is
> ...



Okay, here's the issue -- your formula does work, but not how you originally said it does -- in the Earth frame.  Your formula works in the _Solar frame_, and only works in the Earth frame in a case where the path of the asteroid goes through the center of Earth.  All the shifting of the scenarios made me miss this when we did shift to this assumption.  As I documented above, the assumption that the path of the Asteroid goes through the center of Earth is erroneous in all cases except a head-on collision.  Even when we make the flat disc assumption for simplification of the problem, the change in frame to the Earth frame means the asteroid is not approaching the disc from the perpendicular anymore, and so any calculated angle from the path of the asteroid in the Earth frame is incorrect (although it may cause a miss).  The Solar frame doesn't have this problem, so your formula works.  It also works in this problem, because we've changed the assumption to a head on collision where the path of the asteroid lies through the center of the Earth/approaches the disc of Earth from the perpendicular.

One of the reasons I was making the 'forces don't change direction on frame change' argument so strongly is because of this -- the apparent approach vector doesn't matter, it's the force applied that matters, and this becomes apparent if you actually don't use a point mass for Earth and have the asteroid impact on the surface.  When you shift frames, the impact point being the same means the asteroid's path doesn't lie through the center of Earth (it actually doesn't in the Solar frame, either, but the path does align with the center of Earth at the time of impact in the cases laid out above -- it doesn't have to, but that changes the math which is still easier to deal with in the Solar frame).


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## freyar (Sep 17, 2018)

Ovinomancer said:


> Right -- regardless of frame, the force (or dp, whichever your prefer) remains the same and in the same direction.  The apparent shift in direction of approach does not change this fact -- the force applied does not change direction regardless of frame.
> 
> To bring this back to the original point -- if you change the frame, you DO NOT change the direction of the push on the asteroid. The vector of the _applied_ force or the dp or whichever you prefer _remains the same._




Right, the direction of the *change in momentum* of the asteroid (or the force on the asteroid, assuming a sharp push) remains the same in all frames.  But the direction of the asteroid's initial motion is different in different frames.  The discussion has been about the angle between these two vectors, which is different in the different frames.

That is why it can be impossible to cause a miss in the earth's frame by pushing straight back on the asteroid and just slowing down (again, unless you can push hard enough to stop the asteroid completely) but possible to cause a miss by just slowing the asteroid but not changing its direction in the solar frame.  



Ovinomancer said:


> Yep, I missed that Excel was using radians.  It was late, I wasn't paying attention, and it should have jumped up and smacked me.  So, yes, between 6 and 7 degrees (I see you edited your original claim to include this update, which is, oddly again, one of the reasons I went to check, because ~.12 degrees doesn't solve as you initially claimed, seems we both had issues mistaking radians for degrees).  Sadly, I originally had this worked out in degrees using a calculator, which helped confuse me on the Excel sheet because, in radians, there's also a point between 6 and 7 radians were the value crosses 1.




Yes, it's hard to remember to use degrees, which are a weird unit.  But please note that I edited it about an hour and a quarter after the initial post.  

The solutions between 6 and 7 radians are just the two solutions we've been talking about shifted by 2 Pi because sine and cosine are periodic.




> Okay, here's the issue -- your formula does work, but not how you originally said it does -- in the Earth frame.  Your formula works in the _Solar frame_, and only works in the Earth frame in a case where the path of the asteroid goes through the center of Earth.  All the shifting of the scenarios made me miss this when we did shift to this assumption.  As I documented above, the assumption that the path of the Asteroid goes through the center of Earth is erroneous in all cases except a head-on collision.  Even when we make the flat disc assumption for simplification of the problem, the change in frame to the Earth frame means the asteroid is not approaching the disc from the perpendicular anymore, and so any calculated angle from the path of the asteroid in the Earth frame is incorrect (although it may cause a miss).  The Solar frame doesn't have this problem, so your formula works.  It also works in this problem, because we've changed the assumption to a head on collision where the path of the asteroid lies through the center of the Earth/approaches the disc of Earth from the perpendicular.




Let me explain what my formula is for again and how it might be useful for the asteroid problem Janx posed again, since I must not have made it clear enough.
1) Ignore our discussion about reference frames for a moment.  In any arbitrary frame, some object (in our case, the asteroid) is moving with some initial momentum vector p.  We are able to strike it and change its momentum by vector dp.  Now imagine a plane drawn perpendicular to the vector p (the equatorial plane, so to speak) and define x as the angle between vector dp and that plane where positive x corresponds to the component of dp parallel to p pushing against p (that is, slowing the object down some).  Then choosing x such that sin(x)=dp/p maximizes the deflection angle (the angle between the object's initial velocity and final velocity) in the frame we're working in for a fixed magnitude of dp.  In other words, if you can only push so hard, how do you get the most bang for your buck?

2) The problem we are considering is how to get an asteroid to miss the earth.  I would like to consider this problem in terms of the deflection angle of the asteroid.  What frame lets us do that?  *The earth frame*.  In the rest frame of the earth, the earth is sitting still.  The question is just how to get the asteroid not to be aimed at the earth, in other words, how to deflect the direction of the asteroid's motion.  That is *not* true in the solar frame.  *As you have told us* in fact, it is possible in the solar frame to get the asteroid to miss the earth by slowing it down or speeding it up without changing its direction.  This may or may not be the easiest thing to do, but it can work.  As stated before, all this rests on the assumption that the asteroid is no more than a couple of hours out, so we can safely ignore the fact that the earth is accelerating in its orbit.

3) *Nowhere* is it necessary to assume that the asteroid hits the earth at the perpendicular or that its path will go through the center of the earth.  However the asteroid is going to hit, if we're looking at this in the earth frame, there is a minimal deflection angle needed to cause a miss.  Now, if the asteroid's trajectory isn't straight down the center of the earth, that minimal deflection angle may not be the same in all directions.  So then you also have to choose the direction of vector dp in the equatorial plane.  It might help if you sketch this out. I'd really rather not have to take the time to draw pictures.



> One of the reasons I was making the 'forces don't change direction on frame change' argument so strongly is because of this -- the apparent approach vector doesn't matter, it's the force applied that matters, and this becomes apparent if you actually don't use a point mass for Earth and have the asteroid impact on the surface.  When you shift frames, the impact point being the same means the asteroid's path doesn't lie through the center of Earth (it actually doesn't in the Solar frame, either, but the path does align with the center of Earth at the time of impact in the cases laid out above -- it doesn't have to, but that changes the math which is still easier to deal with in the Solar frame).




If you want to know whether the asteroid hits the earth, the approach vector (initial asteroid velocity) always matters in any frame.  Think about it.  If the asteroid isn't initially aimed to hit some point of the earth, who cares?  (Extreme example, I know.)  Whether it hits the center of the earth is not really the point. I would also argue that the earth frame has simpler math.  How can it make the problem easier if the earth is moving?  That's an extra motion we have to keep track of.  And, even if we wanted the earth to move, why choose the solar frame and not, say, the rest frame of Jupiter or something?  As long as things are happening fast enough that we can ignore the acceleration of the earth's and asteroid's orbits to a reasonable first approximation, the sun doesn't affect the physics at all.  So let's not introduce an extra confusing factor. Reducing a calculation to only the necessary elements is an early lesson in physics, or at least it should be.

And that's all without getting into the psychology of it all.  Humans have a strong tendency to think of the earth as still, and Pierce probably isn't going to overcome that kind of visceral intuition in the short time she has to deflect this thing.  After all, how often do you think "the earth is moving past me at 60 mph" rather than "I'm driving at 60 mph" when you're out in your car?  Furthermore, Pierce is presumably leaving on her mission from somewhere on earth, so she is providing force/thrust to herself to give herself a velocity *relative to the earth*.  Why should she get out into space and then have to think to herself, "OK, now how fast am I going relative to the sun?"

I hope this clears things up.


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## Ovinomancer (Sep 17, 2018)

freyar said:


> Right, the direction of the *change in momentum* of the asteroid (or the force on the asteroid, assuming a sharp push) remains the same in all frames.  But the direction of the asteroid's initial motion is different in different frames.  The discussion has been about the angle between these two vectors, which is different in the different frames.



But it's not really, is it?  It's about a preferred coordinate system for measurement -- the angle is the same referenced to the same vectors in either frame.  Only if you add the asteroid's solar speed and Earth's solar speed when you change to Earth's frame does the apparent vector of the asteroid's momentum change, but the push is still in the same direction with regards to the asteroid.  Measuring that push from a new vector is changing the coordinate system, not the observer frame.  I brought up the difference between an observer frame shift and a coordinate shift long ago.  The car example, for instance, deals only with an observer frame shift, not a coordinate shift.  You're doing both now with the asteroid and claiming it's analogous -- it's not.


> That is why it can be impossible to cause a miss in the earth's frame by pushing straight back on the asteroid and just slowing down (again, unless you can push hard enough to stop the asteroid completely) but possible to cause a miss by just slowing the asteroid but not changing its direction in the solar frame.



Yes, but it's also impossible to cause a miss by pushing in that same direction in the Solar frame.  No one's arguing that there are bad angles of push.  Why this is a sticking point, I'm not sure.




> Yes, it's hard to remember to use degrees, which are a weird unit.  But please note that I edited it about an hour and a quarter after the initial post.
> 
> The solutions between 6 and 7 radians are just the two solutions we've been talking about shifted by 2 Pi because sine and cosine are periodic.



Um, yes?  I wasn't confused about that once I realized my error in radians vs degrees.



> Let me explain what my formula is for again and how it might be useful for the asteroid problem Janx posed again, since I must not have made it clear enough.
> 1) Ignore our discussion about reference frames for a moment.  In any arbitrary frame, some object (in our case, the asteroid) is moving with some initial momentum vector p.  We are able to strike it and change its momentum by vector dp.  Now imagine a plane drawn perpendicular to the vector p (the equatorial plane, so to speak) and define x as the angle between vector dp and that plane where positive x corresponds to the component of dp parallel to p pushing against p (that is, slowing the object down some).  Then choosing x such that sin(x)=dp/p maximizes the deflection angle (the angle between the object's initial velocity and final velocity) in the frame we're working in for a fixed magnitude of dp.  In other words, if you can only push so hard, how do you get the most bang for your buck?



I understand that.  I've challenged that it doesn't work except in the head-on case because Earth is off-center and it might not generate a miss depending on approach path.  You acknowledge this in point 3, and offer a solution of not measuring from the path of the asteroid, which is what I've been saying.



> 2) The problem we are considering is how to get an asteroid to miss the earth.  I would like to consider this problem in terms of the deflection angle of the asteroid.  What frame lets us do that?  *The earth frame*.  In the rest frame of the earth, the earth is sitting still.  The question is just how to get the asteroid not to be aimed at the earth, in other words, how to deflect the direction of the asteroid's motion.  That is *not* true in the solar frame.  *As you have told us* in fact, it is possible in the solar frame to get the asteroid to miss the earth by slowing it down or speeding it up without changing its direction.  This may or may not be the easiest thing to do, but it can work.  As stated before, all this rests on the assumption that the asteroid is no more than a couple of hours out, so we can safely ignore the fact that the earth is accelerating in its orbit.



Yeah, your formula doesn't create a miss in some geometries (it can fail in minimum dp cases for all but the head-on case, frex).  I'm definitely not confused that the objective is to cause a miss.  Explaining that to me is rather condescending.  Yes, I made a few blunders because I haven't dusted off my trig outside of narrow applications for about a decade, but I've picked up everything you've laid down and caught a number of my own errors on doublecheck.  I'm an electrical engineer by trade, so, yeah, I may be rusty but I don't need to be explained to that we're trying to get an asteroid to miss the Earth at this point.


> 3) *Nowhere* is it necessary to assume that the asteroid hits the earth at the perpendicular or that its path will go through the center of the earth.  However the asteroid is going to hit, if we're looking at this in the earth frame, there is a minimal deflection angle needed to cause a miss.  Now, if the asteroid's trajectory isn't straight down the center of the earth, that minimal deflection angle may not be the same in all directions.  So then you also have to choose the direction of vector dp in the equatorial plane.  It might help if you sketch this out. I'd really rather not have to take the time to draw pictures.



I'm not sure what you mean by 'equatorial plane.'  I'm going to assume you mean the 'East-West' in the non-rotated coordinate scheme (where Earth's movement relative to the Sun is 'up'), yeah?  Okay, I'll buy that for the exact reason that it's the same plane as the Solar frame case under discussion, so I know that works.  But, point in fact, this means that you're now agreeing with me that the optimum deflection angle is NOT from the perpendicular of the asteroid's path with sin(x)=dp/p, but instead from a different reference (excepting the head-on case)?  Hallelujah!  I'm confused, though, that you started this point with a refutation of this.




> If you want to know whether the asteroid hits the earth, the approach vector (initial asteroid velocity) always matters in any frame.  Think about it.  If the asteroid isn't initially aimed to hit some point of the earth, who cares?  (Extreme example, I know.)  Whether it hits the center of the earth is not really the point. I would also argue that the earth frame has simpler math.  How can it make the problem easier if the earth is moving?  That's an extra motion we have to keep track of.  And, even if we wanted the earth to move, why choose the solar frame and not, say, the rest frame of Jupiter or something?  As long as things are happening fast enough that we can ignore the acceleration of the earth's and asteroid's orbits to a reasonable first approximation, the sun doesn't affect the physics at all.  So let's not introduce an extra confusing factor. Reducing a calculation to only the necessary elements is an early lesson in physics, or at least it should be.



I disagree on the frame for the example case.  For others, sure.  In the solar frame, I can very easily calculate the forces that will cause a miss in both the lateral and the parallel directions.  Determining optimum angle of push with a force in between those two somewhat straightforward.  In the Earth frame, I now have a lopsided target and non-right angles to calculate the minimum miss angles above and below.  Let me present the drawing I did earlier:



As you see above, each speed (red was 20km/s, purple 30 km/s, and yellow 50 km/s lateral velocity) creates a different angle because of the displacement due to the size of Earth. Even if you use the disc Earth simplification, the angles of approach in the Earth frame don't change.  This  makes figuring the optimum angle harder -- it's not symmetrical and it's not sin(x)=dp/p measured from the perpendicular to the asteroid's path.

Let's take what we've already determined for a 30km/s asteroid in the above Solar frame (purple line).  There exists a dp such that it will cause a miss if the angle of the push is perpendicular to the path of the asteroid.  We solved for that, it was an instantaneously applied 1.81 km/s (downwards).  That force can be applied to 0 degrees from perpendicular or ~6.12 degrees counterclockwise from perpendicular.   We've agreed, here.

Now, if we switch to the Earth frame, the angle of the asteroids apparent path from the observer on the surface is 45 degrees.  We've also agreed that until we change coordinates, the force we figured out above is still in the same direction, or perpendicular to the original path.  This means that force is still applied downwards even with the new path.  If we rotate the coordinate frame to match the current path of the asteroid, we rotate our measurements 45 degrees counterclockwise.  Going from the vertical of the asteroid's path now, the original force is *clockwise *by 45 degrees -- ie, behind the perpendicular.  If we use your formula for the optimum angle of push now, sin(x)=dp/p where dp=1.81km/s and p=sqrt(30km/s^2+30km/s^2) = 42.4 km/s.  x = 2.43 degrees clockwise from perpendicular.  We don't have matching answers here.

If, instead, you use the 'equatorial' plane, then at least your optimum is between the two values for 1.81 (0 < 2.43 < 6.12).  I'm not sure how valid that is, though.  

To test the above, 1.81 km/s applied downwards means the down velocity component in the Earth frame is now 31.8 km/s and the 'left' component velocity is still 30km/s.  It takes 1 hour to approach (this was the time assumption for the above graph, and the one we've been using recently), so it will take 3600s to reach the disc.  At which point it will be 1.81 km/s * 3600 s = 6516 km further 'down', which is a miss.  Barely, as predicted.  If applied at 2.43 degrees, the left component of that is -0.0767 km/s and the 'down' part is 1.808.  Left speed is 29.9233, so it will take 3609 seconds to approach, which means it will be 6,526.4 km further down.  A bigger miss.  Your formula does optimize on dp for push angle when used from the 'equatorial' plane (given the asteroid approaches on this plane in the solar frame, not sure of other cases).  Notably, these angles work when mirrored across the horizontal as well (ie, 180 degrees and 177.57 degrees). 

If we use the angle generated from the path of the asteroid in the Earth frame, it doesn't miss.  I'm going to rotate it back to the initial coordinate plane, which means that push is now 47.43 degrees from the vertical in the 'equatorial' plane.  Left component is -1.3256 km/s and down component is 1.2244 km/s.  Asteroid approach time is 108,000 km/(30 km/s - 1.3256 km/s) = 3766.4 s.  Down distance is then 4611.58.  This is not a miss.



> And that's all without getting into the psychology of it all.  Humans have a strong tendency to think of the earth as still, and Pierce probably isn't going to overcome that kind of visceral intuition in the short time she has to deflect this thing.  After all, how often do you think "the earth is moving past me at 60 mph" rather than "I'm driving at 60 mph" when you're out in your car?  Furthermore, Pierce is presumably leaving on her mission from somewhere on earth, so she is providing force/thrust to herself to give herself a velocity *relative to the earth*.  Why should she get out into space and then have to think to herself, "OK, now how fast am I going relative to the sun?"



I don't have a problem with it.  But then, I like orbital mechanics stuff.  If you can go into space, you figure it out pretty quickly.  It's, heh, a different frame of reference.



> I hope this clears things up.



Not sure.  I'm still saying what I have been, and you've moved a bit in my direction.  I've agreed your optimization formula works, at least in some scenarios.  We're drifting towards agreement, but I think the big issue to overcome is the frame of reference issue -- forces don't change direction on a frame change, and this has pretty big repercussions.  Also, assuming sin(x)=dp/p maximizes x in all cases isn't warranted -- you have to be careful where you measure from.  The frame change to Earth needs to account for the non-perpendicular impact on Earth, which adds additional complexity in the Earth frame.  We're closer, but not there.


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## Nagol (Sep 17, 2018)

Janx said:


> I'm back from a busy day or so of busy stuff. So I'll try to clarify what I can for everybody.  Thanks to Nagol for reminding Umbran of the parameters.  Umbran's right to ask, it's just that i've got restrictions that I chose to work with, like not killing bad guys in the first book.
> 
> To Eltab, I don't have a diagram (it's a vision in my head based on a real place I saw from the crowd's perspective).  It is on a roof/near top floor of a building under construction across from an open roof sports place.
> 
> ...




You're somewhat stuck with the distance.  In effect, the first foot or so is lost figuring out where the shot is going to go and firing a counter.  Every additional foot the bullet needs to travel gives you between 1 (if the shot is as slow as the bullet) and 40 feet (if the shot is at really high meteoric speed) your counter projectile launcher can be placed from the point of collision.  The actual distance is in effect a ratio for the interceptor's velocity to the AR-15 bullet of 1,100 m/s.

Considering the amount of energy needed to launch the projectiles, why not simply magnetize the weapon?  It won't fire if the hammer can't be pulled back.


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## Janx (Sep 18, 2018)

Nagol said:


> You're somewhat stuck with the distance.  In effect, the first foot or so is lost figuring out where the shot is going to go and firing a counter.  Every additional foot the bullet needs to travel gives you between 1 (if the shot is as slow as the bullet) and 40 feet (if the shot is at really high meteoric speed) your counter projectile launcher can be placed from the point of collision.  The actual distance is in effect a ratio for the interceptor's velocity to the AR-15 bullet of 1,100 m/s.
> 
> Considering the amount of energy needed to launch the projectiles, why not simply magnetize the weapon?  It won't fire if the hammer can't be pulled back.




Oh no, you've fallen under Umbran's influence and gone off script 

I don't know if/how to magnetize the weapon (this hero is constrained to Mythbusters Plausible solutions, aka I saw a video or article on a piece of tech and assumed it can do more).  I don't think the hero will get to touch the gun or he'd turn the safety on or something clever   Also, the scene might end up looking like:
Hero walks past the gun turret, admiring the craftsmanship.
BBEG: you're too late, the countdown is nearly complete. Muah ah ah ah!
Hero: Yep.
Gun: 3..2..1..click.
BBEG: Drat!  Curse you and your meddling kind!
Hero: Yep.

Clearly, that's compelling drama right there, and the finest writing I've ever writ. 

Now if there's a way to project magnetic force, I'm keenly interested in knowing more.  One of the hardest things about my project is figuring out how to repurpose tech and science toward my setting.  The hardest part of that is finding non-lethal FX that I can use to solve problems, because killing the bad guy is surprisingly easy.


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## Nagol (Sep 18, 2018)

Janx said:


> Oh no, you've fallen under Umbran's influence and gone off script
> 
> I don't know if/how to magnetize the weapon (this hero is constrained to Mythbusters Plausible solutions, aka I saw a video or article on a piece of tech and assumed it can do more).  I don't think the hero will get to touch the gun or he'd turn the safety on or something clever   Also, the scene might end up looking like:
> Hero walks past the gun turret, admiring the craftsmanship.
> ...




You invited off-script.  Though I'll keep providing on-script replies too.
​Killing the bad guy typically requires less energy and considerably less finesse.

The power required to accelerate the rail gun / mass driver projectile is stupendous.  The magnetic field it needs to get the insane acceleration needed requires that.  One thing a powerful changing magnetic field does is establish eddy currents in nearby conductors like metallic objects.    Those currents create a reversed magnetic field which, considering the strength of the original field, will probably stop the mechanism assuming the rifle just doesn't get too hot to hold.  Radio/microwave EMF also cause eddy currents in metals though an emission strong enough to lock up the gun would probably cook the person holding it.

Another completely off the wall tactic might be to create a short-lived impermeable plasma.  Impermeable plasma acts like a impenetrable solid with respect to gas.  I have seen "claims" that that impermeability would extend to solids though I've never seen any serious literature to support that.  Even if it only extends to gas, it would be a great tactic to handle an explosion which is a pressure wave though gas.


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## Eltab (Sep 18, 2018)

Janx said:


> Hero walks past the gun turret, admiring the craftsmanship.
> BBEG: you're too late, the countdown is nearly complete. Muah ah ah ah!
> Hero: Yep.
> Gun: 3..2..1..click.
> ...



Unbeknownst to the hero or the villian, _Chuck Norris levelled a steely glare_ at the gun from across the plaza. It just about melted...

More seriously:
The villain and his gun are on top of a building under construction, shooting across the street (and down) into a baseball stadium?


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## Janx (Sep 19, 2018)

Eltab said:


> Unbeknownst to the hero or the villian, _Chuck Norris levelled a steely glare_ at the gun from across the plaza. It just about melted...
> 
> More seriously:
> The villain and his gun are on top of a building under construction, shooting across the street (and down) into a baseball stadium?




mostly, across.  We could see the building situation from the upper deck of the actual stadium, thus inspiring the original situation.

Ultimately, unless some cool and dramatic alternative idea pops out, like Umbran says, I'll likely go with the original plan, but tighten the parameters so its closer to plausible.

Discussing it with physics minded people tells me about future tech ideas like a plasma shield or heating something up.  And it confirms the physics/tech defying aspects (like rapid fire rail gun isn't here yet).  It's not a bad idea to know the aspects that might work, and the aspects that are total BS 

And y'all are going in the Acknowledgements when I get that far and hopefully get a book deal.


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## Eltab (Sep 19, 2018)

Janx said:


> And y'all are going in the Acknowledgements when I get that far and hopefully get a book deal.



I'm going to take that as a plural 'you', because I haven't contributed that much analysis to the discussion.

Thank you.


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## freyar (Sep 19, 2018)

Ovinomancer said:


> But it's not really, is it?  It's about a preferred coordinate system for measurement -- the angle is the same referenced to the same vectors in either frame.  Only if you add the asteroid's solar speed and Earth's solar speed when you change to Earth's frame does the apparent vector of the asteroid's momentum change, but the push is still in the same direction with regards to the asteroid.  Measuring that push from a new vector is changing the coordinate system, not the observer frame.  I brought up the difference between an observer frame shift and a coordinate shift long ago.  The car example, for instance, deals only with an observer frame shift, not a coordinate shift.  You're doing both now with the asteroid and claiming it's analogous -- it's not.




Shifting the asteroid's velocity (not speed) by the earth's velocity (both measured in the solar frame) is the mathematical representation of switching to the earth's frame.  That's what changing frames means.  If someone taught you otherwise, they did you a disservice.  The little formula I wrote down uses the angle between the asteroid's initial momentum and the change in momentum it experiences when Pierce hits it.  I have said all along that I have applied that in the earth frame.  You therefore can't just apply my argument in the solar frame without adjustment.  

This is exactly the problem you've been having with the car and the wind.  The velocity of object B as measured in object A's rest frame is by definition the velocity of B relative to A.  In a "lab frame" this relative velocity is vB-vA (subtraction done vectorially), and that's vB as measured in A's rest frame.  I don't know how to make this clearer given how much we've discussed it already.  If you can't understand or accept this fact, I don't see much point in continuing.  I'll answer the rest below, but this is a key point.




> Yes, but it's also impossible to cause a miss by pushing in that same direction in the Solar frame.  No one's arguing that there are bad angles of push.  Why this is a sticking point, I'm not sure.




As you pointed out all the way back in post 14 of this thread, in the solar frame, the asteroid is aiming where the earth will be, and the earth is moving. Therefore, you can slow the asteroid down without changing its direction and still get it to miss the earth without changing its direction.  In the solar rest frame, not the earth's rest frame. Are you changing your mind on this?  Because you were right then.  The only problem then was that you insisted that the earth frame is invalid.




> Um, yes?  I wasn't confused about that once I realized my error in radians vs degrees.



Right, and, as I've said before, I'm not just writing for you but for other readers who maybe aren't quite as familiar with the math.



> I understand that.  I've challenged that it doesn't work except in the head-on case because Earth is off-center and it might not generate a miss depending on approach path.  You acknowledge this in point 3, and offer a solution of not measuring from the path of the asteroid, which is what I've been saying.



I'm glad you understand.  However, you've clearly misunderstood my point 3.  I'll address that below.  



> Yeah, your formula doesn't create a miss in some geometries (it can fail in minimum dp cases for all but the head-on case, frex).  I'm definitely not confused that the objective is to cause a miss.  Explaining that to me is rather condescending.  Yes, I made a few blunders because I haven't dusted off my trig outside of narrow applications for about a decade, but I've picked up everything you've laid down and caught a number of my own errors on doublecheck.  I'm an electrical engineer by trade, so, yeah, I may be rusty but I don't need to be explained to that we're trying to get an asteroid to miss the Earth at this point.




The head-on, path through the center of the earth case actually requires a greater deflection angle and therefore harder push dp than most other cases.  

As for the explanation, I'm sorry you found it condescending since that wasn't the intent.  I was trying to be methodical and clear about why I am choosing to apply the results of point 1 in the earth frame.  You haven't responded to that part.  In any case, you clearly have a fine grasp of the math and basic physical laws. I just want to straighten out some confusion on working in reference frames.

But, if we're talking about tone of posts, please go back and look at yours, particularly post 27 in the thread where you called my prior post "not even wrong."  I am well aware of the origin of the quote and its less than complimentary use in physics discussions.



> I'm not sure what you mean by 'equatorial plane.'  I'm going to assume you mean the 'East-West' in the non-rotated coordinate scheme (where Earth's movement relative to the Sun is 'up'), yeah?  Okay, I'll buy that for the exact reason that it's the same plane as the Solar frame case under discussion, so I know that works.  But, point in fact, this means that you're now agreeing with me that the optimum deflection angle is NOT from the perpendicular of the asteroid's path with sin(x)=dp/p, but instead from a different reference (excepting the head-on case)?  Hallelujah!  I'm confused, though, that you started this point with a refutation of this.



No, I am not agreeing with you at all.  Once again, in the solar frame, you can get the asteroid to miss just by slowing it down, so deflection angle isn't the only thing to consider there.  Let me explain what I said in my third point:
a) First off, I am now working exclusively in the earth's reference frame, so everything is measured relative to the earth.
b) I defined the equatorial plane as the plane through the center of the asteroid perpendicular to its velocity vector (see my point 1).  If we break dp up into its components, it has one component along the asteroid's velocity and another component in the equatorial plane.  The magnitude of the asteroid's deflection angle --- we called this angle a --- is unaffected by the orientation of that second component (all that matters are the magnitudes of the components), but which direction that deflection goes in does depend on the orientation of the component in the equatorial plane.
c) If our hero can intercept (and push) the asteroid at a certain time before collision with the earth, she has to deflect it by a minimal angle, which we have called m. To determine this angle, we draw a line tangent to the earth through the point where she intercepts the asteroid.  The required minimal deflection angle is the angle between that tangent line and the asteroid's original path.  The tangent line is a grazing trajectory where the asteroid just skims the top of the earth (or the atmosphere, however you like).  Actually, there is a whole family of such tangent lines, but typically one will require a smaller deflection angle than the others. So the push should then be oriented properly in the asteroid's equatorial plane to take advantage of the smallest required deflection angle. 
d) Note that the earlier Pierce can intercept the asteroid, the smaller the minimal required deflection angle will be, which I think we all agree on already.

I'm not going to reproduce your example and go through the whole argument, since it doesn't get to what I want to explain and what you seem to be confused about.  In fact, you seem to be confusing what we've been calling angle x (the angle of the push relative to the asteroid's equatorial plane and which is equal to arcsin(dp/p) if we want to maximize deflection angle a given a fixed magnitude dp) with the minimal required deflection angle m.  I will instead draw everything that I am talking about:


The earth is the big hollow circle.  The filled one is the asteroid at the point where Pierce intercepts it.  The black line is the asteroid's initial trajectory, and it's continuation through the earth is dotted.  The green lines are tangents to the earth, indicating the minimal deflection needed to miss the earth (realistically, Pierce should give some leeway because we're ignoring the earth's gravity, etc).  The solid one is the one with the smallest deflection angle, which is labeled m.  The dashed one is another tangent which will cause the asteroid to miss but requires more deflection.  I've labeled dp as a red vector hitting the asteroid along with angle x.  On the other side of the asteroid is a red dashed vector indicating a push that could get the asteroid to deflect along the dashed green trajectory. None of this has numbers worked out, and it's not to scale, since that's not the point.  But we know from previous work that the minimum push Pierce has to exert has dp/p = sin(m) oriented such that sin(x)=dp/p, or x=m.  Please note that I very specifically did not use a head-on collision with the asteroid going through the center of the earth.  Please also be forgiving if not all the tangents are precisely tangent, etc, since I did this in a hurry.

OK, last thing.  I haven't wanted to use the "appeal to authority" fallacy because it's, well, a fallacy, but I would just like to ask you to consider that I know what I'm talking about even if you still find this confusing.  As you may know from upthread or other EN World discussions, I am a physicist professionally.  Not only that, I have taught a course on relativity, including a large section on changing reference frames in Newtonian mechanics every academic year for the past 7 years.  So I hope that clears things up.  I am going to try to get myself to stay out of this thread, since I can't really take credit for this at work.


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## Eltab (Sep 22, 2018)

freyar said:
			
		

> ... since I can't really take credit for this at work.



I too (and probably other people) wish I could find a way to get paid for RPG'ing.
​


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## Janx (Nov 22, 2019)

A couple months ago, the anthology featuring my story came out.  I'm told my story was pretty good.

Thanks to everybody who discussed the topic with me (and went above and beyond with math and diagrams).  I probably didn't choose a physics-correct solution in the story, but at least I know it


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## Eltab (Nov 23, 2019)

Janx said:


> A couple months ago, the anthology featuring my story came out.  I'm told my story was pretty good.



Congrats on being published, and praised.  Best wishes for your sales (and income).


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## Janx (Nov 24, 2019)

Eltab said:


> Congrats on being published, and praised.  Best wishes for your sales (and income).




Alas, this was for the local writers guild, so pro bono.  Though the praise reminded me "I knew I should have tried to sell it..."

As a side note, most short stories for profit are sold for $.03-$.08/word for First Rights, meaning they have to be the first to publish it, usually holding that for a year or so.  After that, I could find a market willing to take published stories.  Those are rarer.  So one has to be careful in giving away content because it restricts your opportunities.

I've found an opportunity that opens in January for "recovering" post-apocalyptic, which this story might qualify, but they want a disabled protagonist (where the disability hinders in the story). And they take pre-published stuff.  I don't quite want to retool this story, so I've got something else in the works.


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