#### Ovinomancer

##### No flips for you!

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.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 therelative velocityof the air and the car, which is the same in any reference frame.

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.

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 directlyagainstthe 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. Thatneveroptimizes the deflection angle.

I may have forgotten to note that I've assumed dp<p (which should be clear from the formula) because otherwise Pierce could just easily turn the asteroid around, and we don't need any dramatic "just missing the earth" bit.

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.You 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.

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.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.

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.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.

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).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.

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.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.

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 asteroidYou'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 isnothingin my calculation that talks about the asteroid hitting the center of the earth.

**hits**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)

*does*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.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.

You can't convince me because you're making an easy error of conception. The 'wind' problem is fraught with this.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.

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.