# Science: asteroid vs. hero physics

#### Ovinomancer

##### No flips for you!
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!

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.

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

#### freyar

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.
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'.

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.

#### Ovinomancer

##### No flips for you!
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:
.....................|
.....................v
crosswind -> car <- tire friction
.....................^
.....................|
...............car motor

Now, Car frame:

.....................|
.....................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?

#### freyar

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

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

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.
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

##### Legend
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.

#### Ovinomancer

##### No flips for you!
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.

No problem!

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

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.
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.

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.

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.

#### Janx

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

#### Nagol

##### Unimportant
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?

#### freyar

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.

#### Eltab

##### Lord of the Hidden Layer
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. #### Janx

##### Hero
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

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

##### Hero
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.

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#### Eltab

##### Lord of the Hidden Layer
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

##### Unimportant
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.

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#### Umbran

Staff member
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

##### Unimportant
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

##### Legend
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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