Science: asteroid vs. hero physics

[Janx]

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.

 

[freyar]

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.

 

[freyar]

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.

 

[Janx]

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.

 

[freyar]

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.

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.

 

[Eltab]

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.

 

[Nagol]

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.

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.

 

[Umbran]

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.

 

[Nagol]

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.

 

[MarkB]

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.