# Science: asteroid vs. hero physics

#### Nagol

##### Unimportant
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

Staff member
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?

#### Nagol

##### Unimportant
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.

#### Janx

##### 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.

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.

#### Janx

##### Hero
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.

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

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

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

#### Ovinomancer

##### No flips for you!
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.

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.

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.

#### Ovinomancer

##### No flips for you!
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
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.
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).

#### freyar

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.

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.

#### Ovinomancer

##### No flips for you!
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.

#### Nagol

##### Unimportant
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.

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.

#### Janx

##### Hero
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.

#### Nagol

##### Unimportant
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.

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.

#### Eltab

##### Lord of the Hidden Layer
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.
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?

#### Janx

##### Hero
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.

#### Eltab

##### Lord of the Hidden Layer
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.

#### freyar

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.

#### Eltab

##### Lord of the Hidden Layer
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.

#### Janx

##### Hero
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

#### Eltab

##### Lord of the Hidden Layer
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).

#### Janx

##### Hero
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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