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Science: asteroid vs. hero physics
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<blockquote data-quote="freyar" data-source="post: 7493959" data-attributes="member: 40227"><p>This will probably be my last post, but we're making progress here, so I think it's worth one last try.</p><p></p><p></p><p></p><p>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.</p><p></p><p></p><p></p><p></p><p>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.</p><p></p><p></p><p>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.</p><p></p><p></p><p>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.</p><p></p><p></p><p>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 </p><p>tan(a) = (1/2)(sqrt(3)/2) /(1-(1/2)(1/2)) = (2/3)(sqrt(3)/2) = sqrt(3)/3. </p><p>Can you find a larger angle of deflection? If I get a chance, I'll do a plot later.</p><p></p><p></p><p>No problem!</p><p></p><p></p><p>I had my computer plot it using Maple symbolic mathematics software. If I get time, I will attach that later for the example above.</p><p></p><p></p><p>OK, let's just look at the last equation</p><p>1=cos(x)+.0602sin(x).</p><p>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. </p><p></p><p></p><p>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.</p><p></p><p></p><p>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).</p><p></p><p></p><p>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.</p><p></p><p></p><p></p><p>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. </p><p></p><p></p><p>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.</p><p></p><p>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.</p></blockquote><p></p>
[QUOTE="freyar, post: 7493959, member: 40227"] 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. 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. 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. 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. 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. 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. 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. [/QUOTE]
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