Sorry I haven't been able to respond to this for a week. But it's important enough to get things right for the public that I don't want to let it pass. There is a lot correct here, but it's missing enough that the overall message is incorrect.

In both cases, the force of an object from the wind is in the direction of the air's velocity *relative to the object*. Let me just address your second example, since the logic holds in the first example also. You are right that the car should experience the same force in both the city frame and the car frame. You're also right that the force in the car frame is "down-right" because the wind velocity is at an angle with respect to the grid in this frame. Now look at the frame of the city. You are right that the wind blows across the grid, exerting a force to the right on the car. You have forgotten that the car is moving "up" in this frame and pushing through the air, which by Newton's 3rd law is pushing back -- "down." So in this frame, the car still experiences a "down-right" force from the air.

The simplest example of this I can think of at the moment is the following. Suppose you're standing outside with no wind, ie, the air is still with respect to the ground. Now you start running. You will feel wind in your face. This is because, in your frame, the air is moving "backward" and exerting a force (what you feel) on your skin. In the ground's frame, the air exerts a force on you because you are pushing through it.

Yes, that is what I've done. You have to change the velocity of *everything* when changing frames, including the wind. You're also right that the force experienced by an object is the same in both frames, and I've explained why that happens in your example.

I'm afraid I don't quite follow what you're setting up here. But you're right that the force doesn't change directions --- but the wind has to change velocities because the force of a wind on an object is in the direction of the *difference* of the object's and wind's velocities.

Yes, I see that you don't get what I've set up, and it goes directly to your complaints above. Let's try again.

Let's set up the situation without the wind. In city frame, the car's motor is applying force to the car to move it along the road. The air, which is stationary, is applying resistance to that force in equal measure (the car is at constant velocity, so the forces are equal). What you have is:

Up: force applied by car's motor = down: force applied by air resistance.

Now, let's switch this to the car's frame. The car is not stationary and it appear that there's a strong headwind (air resistance) pushing the car backwards. The car's motor is applying force to counter that headwind (air resistance) and keep the city moving past at a constant rate. You have:

down: headwind (air resistance) = up: car's motor,

Good so far? Okay, let's add the wind.

Now there's a cross wind of some force F. First the city frame. The car's motor hasn't changed force, so the air resistance also hasn't changed force. Up and down forces are still equal. But, we've added a lateral force, the wind(F). This will push the car in the direction of the wind unless countered. Let's assume that the friction of the tires on the surface counter the wind, here, such that the lateral forces are balanced. You not have:

Up: car's motor = down: air resistance.

Right: wind (F) = Left: tire friction.

All forces balance. Now, let's shift to the car's frame, and follow your suggestion that the wind also shifts direction when we shift frames. We run into a problem. If the wind shifts by some angle x, the lateral force of that wind(F) is now cos(x)*F, and there's now a up/down componenet of sin(x)*F. If we sum the forces now, we have:

Down: air resistance + sin(x)*F = up: car's motor

Right: cos(x)*F = Left: tire friction

So, you're arguing that the tires exert less lateral friction force in the car frame than in the city frame, and that the car's motor must push harder to combat the increase in downward force because some part of the wind is adding to the air resistance. This doesn't add up, though, as the forces on the car DO NOT CHANGE with a frame shift -- they still must be the same forces as in any other frame.

The error in your thinking above is that the up/down component of the air's movement was already accounted for in the frames by the declaration of constant velocity -- meaning that the force of the car's motor exactly counters the force of the air resistance/wind in the up/down direction. You have the lateral and up/down forces changing in your frame change, and that just doesn't happen.

Again, you're right, but you've forgotten parts of the force, as I've said above, so your overall conclusions above are incorrect. And I'm not entirely sure what you're talking about integrating in the last sentence here.

No, I have made no such assumption. No matter the impact location on the earth, at a given time before impact, there is a minimum angle that the relative velocity (that is, the asteroid's velocity in the earth's frame) must be changed to avoid hitting the earth somewhere. The calculation I gave found the direction of push that Pierce can make to maximize the deflection angle, assuming Pierce can impart a fixed (or maximum) magnitude of momentum to the asteroid. To be fair, this calculation is axially symmetric about the relative velocity, so it doesn't tell you which direction around that circle to push in order to get clear of the closest edge of the earth, but the calculation does what I said it does.

Um, okay, let's look at your equation as the force of the push goes to zero. When this happens the limit of angle x goes to 0 degrees, which is nonsensical. IE, the less you push, the more perpendicular you should push. Further, if you increase your push to the velocity of the asteroid, sinx goes to 90. That kinda makes sense in that completely stopping the asteroid relative to Earth will cause a miss (although gravity now becomes a dominating factor). But what happens if you exceed the push? There exists some dp greater than p where your formula says that pushing the asteroid even faster directly at Earth generates a miss.

Your error here was assuming some dp and then finding a formula that created an x that seemed right to you. You're not testing other dp to find if the formula works. It doesn't.

As with the pictures you've uploaded previously, you seem to be making some assumption about the direction of the asteroid's approach in the solar frame. For example, if the collision is head on, slowing down or speeding up the asteroid won't help (unless of course you can slow it enough that it just precedes the earth around its orbit).

Yes, those assumptions were established well upthread as plausible given the launch point of the asteroid and how it should interact with Earth's orbit. A head on example in the Solar frame requires an extra-solar origin for the asteroid or an orbital period of much greater than 100 years (ie, the ellipse of the asteroid's path would have to be at Earth's orbit close to perihelion, and the combined speeds would be YUGE. Further, for your purposes, a head on in the Solar frame is exactly identical to the Earth frame, with the Earth's 30km/s transferred to the asteroid's velocity. At that point, your formula breaks very badly, as it suggests that the weaker your push, the more towards the perpendicular you should push but the stronger you push the more towards opposing the asteroid's path your should push. That doesn't work.

In the Earth frame, at a given distance, and assuming an asteroid path aimed straight at Earth, there exists a minimum angle that will cause a miss. This angle doesn't change based on the speed of the asteroid because the extended path will still need to be outside that angle. This angle is easy to determine if you know the distance (d) and Earth's radius (Re) as it's simply Tan-1 (Re/d). At sufficient distance, this angle is small. To generate a miss, you must create an instantaneous dp such that the resulting path exceeds this miss angle. This generates some math, but it's not too bad. Let's call the miss angle (m), the angle of applied dp (x), and the angle the asteroid takes as (a); all angles are measured clockwise from the path. Tan (a) is going to equal the lateral portion of dp, or dpcos(x) divided by p minus the vertical portion of dp, or dpsin(x). This is tan(a) = dpcos(x)/(p-dpsin(x)). We can set Tan(a) equal to Tan(m) and get that dpcos(x)/(p-dpsin(x))=Re/d. From this, we can see a few things. One, dp cannot go to 0, and two, x also cannot go to 0 (or 180). This eliminates a push in the direction of the asteroid from generating a miss, as I stated above. Now, let's solve for a dp if x = 90, ie, a perfectly lateral push. IN this case, the formula goes go dp/p=Re/d. Assuming p is known, we have a dp(lat) that generates a miss. Now, can this dp applied to any other angle x, generate a miss (ignoring 270)? Doing some substitution from above, this would mean that dp(lat)cos(x)/(p-dp(lat)sin(x))=dp(lat)/p. There is no solution for x for a known dp(lat) in this scenario where x can be anything other than 90. For a dp that exceeds dp(lat), you can have angles that aren't 90, because the resultant angle of miss can still be greater than m.

If we apply your forumla to this, though, where your sin(x)=dp/p, a known dp(lat) would result in an angle that isn't 90. Your construction doesn't work even if we use your assumption that the path of the asteroid is directly through Earth's center. In the realistic case, where it is not, it also doesn't work.