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Message: <blockquote data-quote="freyar" data-source="post: 7493845" data-attributes="member: 40227">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, 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 dpYou 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.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.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.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.  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.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. 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.  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.</blockquote>

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