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Message: <blockquote data-quote="Ovinomancer" data-source="post: 7495547" data-attributes="member: 16814">But it's not really, is it?  It's about a preferred coordinate system for measurement -- the angle is the same referenced to the same vectors in either frame.  Only if you add the asteroid's solar speed and Earth's solar speed when you change to Earth's frame does the apparent vector of the asteroid's momentum change, but the push is still in the same direction with regards to the asteroid.  Measuring that push from a new vector is changing the coordinate system, not the observer frame.  I brought up the difference between an observer frame shift and a coordinate shift long ago.  The car example, for instance, deals only with an observer frame shift, not a coordinate shift.  You're doing both now with the asteroid and claiming it's analogous -- it's not.Yes, but it's also impossible to cause a miss by pushing in that same direction in the Solar frame.  No one's arguing that there are bad angles of push.  Why this is a sticking point, I'm not sure.Um, yes?  I wasn't confused about that once I realized my error in radians vs degrees.I understand that.  I've challenged that it doesn't work except in the head-on case because Earth is off-center and it might not generate a miss depending on approach path.  You acknowledge this in point 3, and offer a solution of not measuring from the path of the asteroid, which is what I've been saying.Yeah, your formula doesn't create a miss in some geometries (it can fail in minimum dp cases for all but the head-on case, frex).  I'm definitely not confused that the objective is to cause a miss.  Explaining that to me is rather condescending.  Yes, I made a few blunders because I haven't dusted off my trig outside of narrow applications for about a decade, but I've picked up everything you've laid down and caught a number of my own errors on doublecheck.  I'm an electrical engineer by trade, so, yeah, I may be rusty but I don't need to be explained to that we're trying to get an asteroid to miss the Earth at this point.I'm not sure what you mean by 'equatorial plane.'  I'm going to assume you mean the 'East-West' in the non-rotated coordinate scheme (where Earth's movement relative to the Sun is 'up'), yeah?  Okay, I'll buy that for the exact reason that it's the same plane as the Solar frame case under discussion, so I know that works.  But, point in fact, this means that you're now agreeing with me that the optimum deflection angle is NOT from the perpendicular of the asteroid's path with sin(x)=dp/p, but instead from a different reference (excepting the head-on case)?  Hallelujah!  I'm confused, though, that you started this point with a refutation of this.I disagree on the frame for the example case.  For others, sure.  In the solar frame, I can very easily calculate the forces that will cause a miss in both the lateral and the parallel directions.  Determining optimum angle of push with a force in between those two somewhat straightforward.  In the Earth frame, I now have a lopsided target and non-right angles to calculate the minimum miss angles above and below.  Let me present the drawing I did earlier:[ATTACH]101513[/ATTACH]As you see above, each speed (red was 20km/s, purple 30 km/s, and yellow 50 km/s lateral velocity) creates a different angle because of the displacement due to the size of Earth. Even if you use the disc Earth simplification, the angles of approach in the Earth frame don't change.  This  makes figuring the optimum angle harder -- it's not symmetrical and it's not sin(x)=dp/p measured from the perpendicular to the asteroid's path.Let's take what we've already determined for a 30km/s asteroid in the above Solar frame (purple line).  There exists a dp such that it will cause a miss if the angle of the push is perpendicular to the path of the asteroid.  We solved for that, it was an instantaneously applied 1.81 km/s (downwards).  That force can be applied to 0 degrees from perpendicular or ~6.12 degrees counterclockwise from perpendicular.   We've agreed, here.Now, if we switch to the Earth frame, the angle of the asteroids apparent path from the observer on the surface is 45 degrees.  We've also agreed that until we change coordinates, the force we figured out above is still in the same direction, or perpendicular to the original path.  This means that force is still applied downwards even with the new path.  If we rotate the coordinate frame to match the current path of the asteroid, we rotate our measurements 45 degrees counterclockwise.  Going from the vertical of the asteroid's path now, the original force is clockwise by 45 degrees -- ie, behind the perpendicular.  If we use your formula for the optimum angle of push now, sin(x)=dp/p where dp=1.81km/s and p=sqrt(30km/s^2+30km/s^2) = 42.4 km/s.  x = 2.43 degrees clockwise from perpendicular.  We don't have matching answers here.If, instead, you use the 'equatorial' plane, then at least your optimum is between the two values for 1.81 (0 < 2.43 < 6.12).  I'm not sure how valid that is, though.  To test the above, 1.81 km/s applied downwards means the down velocity component in the Earth frame is now 31.8 km/s and the 'left' component velocity is still 30km/s.  It takes 1 hour to approach (this was the time assumption for the above graph, and the one we've been using recently), so it will take 3600s to reach the disc.  At which point it will be 1.81 km/s * 3600 s = 6516 km further 'down', which is a miss.  Barely, as predicted.  If applied at 2.43 degrees, the left component of that is -0.0767 km/s and the 'down' part is 1.808.  Left speed is 29.9233, so it will take 3609 seconds to approach, which means it will be 6,526.4 km further down.  A bigger miss.  Your formula does optimize on dp for push angle when used from the 'equatorial' plane (given the asteroid approaches on this plane in the solar frame, not sure of other cases).  Notably, these angles work when mirrored across the horizontal as well (ie, 180 degrees and 177.57 degrees). If we use the angle generated from the path of the asteroid in the Earth frame, it doesn't miss.  I'm going to rotate it back to the initial coordinate plane, which means that push is now 47.43 degrees from the vertical in the 'equatorial' plane.  Left component is -1.3256 km/s and down component is 1.2244 km/s.  Asteroid approach time is 108,000 km/(30 km/s - 1.3256 km/s) = 3766.4 s.  Down distance is then 4611.58.  This is not a miss.I don't have a problem with it.  But then, I like orbital mechanics stuff.  If you can go into space, you figure it out pretty quickly.  It's, heh, a different frame of reference.Not sure.  I'm still saying what I have been, and you've moved a bit in my direction.  I've agreed your optimization formula works, at least in some scenarios.  We're drifting towards agreement, but I think the big issue to overcome is the frame of reference issue -- forces don't change direction on a frame change, and this has pretty big repercussions.  Also, assuming sin(x)=dp/p maximizes x in all cases isn't warranted -- you have to be careful where you measure from.  The frame change to Earth needs to account for the non-perpendicular impact on Earth, which adds additional complexity in the Earth frame.  We're closer, but not there.</blockquote>

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