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.
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.
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.
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.
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.
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 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.
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.
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.
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...
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.
Yes!
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).
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.
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.
A perfectly perpendicular push is x=0. Cos(90 degrees)=0, which means there is no deflection at all.
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.
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.
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.
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.
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.
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.
Nope! Flagpole in city frame. No up/down forces. Lateral forces are wind - being attached to the flagpole = 0.
Car frame: forces don't change. The 'head wind' the car feels isn't present for the flag, which is moving precisely along with and at the same speed as this 'headwind'. So, up/down forces are still zero. Side forces are also the same, ie: wind - being attached to the flagpole = 0. If the wind shifted to the down and right, the flagpole is now pushing up and left, but a change of frame doesn't affect the sum of the forces, so that's impossible.
Sum. The. Forces.
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.
