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Science: asteroid vs. hero physics
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<blockquote data-quote="Ovinomancer" data-source="post: 7493890" data-attributes="member: 16814"><p>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.</p><p></p><p>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.</p><p></p><p></p><p></p><p>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?</p><p></p><p>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.</p><p></p><p></p><p></p><p></p><p>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.</p><p></p><p></p><p>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.</p><p></p><p></p><p>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.</p><p></p><p>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). </p><p></p><p>d = 30km/s * 3600 s = 108,000 km.</p><p>Re = 6500 km</p><p></p><p>Re/d = 0.0602</p><p>Re/d=dp/p</p><p>removing mass from p's we have dv/v or Re/d=dv/v</p><p>v = 30km/s</p><p>dv = 1.801 km/s (this aligns with my previous solves for this value, so good check).</p><p></p><p>Now to use the formula (using 0 degree as perpendicular to the path):</p><p></p><p>dp(lat)cos(x)/(p-dp(lat)sin(x))=dp(lat)/p</p><p></p><p>1.801cos(x)/(30-1.801sin(x)=0.0602</p><p></p><p>1.801cos(x)=0.0602(30-1.801sin(x))= 1.801-0.1084sin(x)</p><p></p><p>cos(x) = 1 - .0602 sin(x) or 1=cos(x)+.0602sin(x)</p><p></p><p>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.</p><p></p><p></p><p> </p><p></p><p>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.</p><p></p><p></p><p>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 <strong>hits</strong> the center of the Earth. </p><p></p><p>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.</p><p></p><p>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) <em>does </em>generate a miss at that dp. It's close, but there's not an award for close in this case.</p><p></p><p>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).</p><p></p><p></p><p>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.</p><p></p><p></p><p>You can't convince me because you're making an easy error of conception. The 'wind' problem is fraught with this.</p><p></p><p>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.</p><p></p><p>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.</p></blockquote><p></p>
[QUOTE="Ovinomancer, post: 7493890, member: 16814"] 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. 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. 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. 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. 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. 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. 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. 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 [B]hits[/B] 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) [I]does [/I]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 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. 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. [/QUOTE]
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