SKUZZY! I WANT LaTeX!
In the end, we want the final velocity vector to equal the velocity which the bomb would have had assuming no wind at all.
x(T) = x0(T) where x0 = integral[v0 dt] and v0 solves dv/dt = - k v^2 initial condition v = vinitial, x = xinitial and xinitial is fixed. Note this is a no-wind model, but it suffices since we can always shift our frame of reference so that there is no wind, resulting only in a change in the velocity vector v, which is what we have control over in the first place.
To solve this rigorously, we have to consider forces. Knowing the drag equation (force is proportional to the square of velocity) and knowing that force is the first derivative of velocity:
dv / dt = - k (n1 - v)^2 in wind layer 1
dv / dt = - k (n2 - v)^2 in wind layer 2
... etc
where k is the constant appearing in the drag equation and the n1, n2, ... etc are the wind speeds in each layer. To get the final velocity vector, we have to solve the first equation and find the final velocity, then take that velocity as the initial velocity for the second equation... and so forth until we have a velocity function v(t) over the whole range. We then have to integrate to get x(t) = integral[v(t) dt]. We then have to choose vinitial so that
x(T) = x0(T) where x0 = integral[v0 dt] and v0 solves dv/dt = - k v^2 initial condition v = vinitial, x = xinitial
These being nonlinear diffeqs, "all bets are off" mathematically.
However, my intuition says that there should only be one velocity heading vector at which this could work since we only have one variable under our control (vinitial) to solve for one variable we want (x(T)).
Welcome to classical mechanics, where we can't solve the sun-moon-Earth problem without a computer!
Actually 2 Is my hunch, one representing net up wind and the other net downwind...
Layer 1 = 0>5 20kts 360
Layer 2 = 10>15 0kts 180
Layer 3 = 15>20 10kts 270
Layer 4 = 0>5 10kts 090
Layer 5 = 20>25 10kts 360
Lets look at a simplistic example. Assume the bomb is at terminal velocity the moment it is released from 25k, the aircraft is traveling directly into the wind (layer 5), and the wind layers are all of the same magnitude, 5k separations, and their vectors represent the 4 cardinals of the compass, with the last (bottom layer) being the same direction as the initial layer.
Assume: The bomb sight is perfectly calibrated to the ground speed, such that if wind was consistent from drop alt to target, the bombardier need simply place his cross hairs on the target and pickle off bombs. The aircraft is flying due North into the wind with zero cross wind component relative to its trajectory. The bomb sight is a self compensating mechanism for head and tail wind components of any wind vector.
As the bomb falls through the first layer (20>25k...aka layer 5)...the bomb tracks perfectly to target.
When the bomb falls through layers 3 and 4, it drifts left then right the same distance; both vectors cancel themselves out during our bombs journey.
Upon entering layer 2, the bomb picks up what in essence is a 10kt tail wind. As the bomb enters the final layer, it encounters a 10kt relative head wind.
This bomb having transited through 5 different wind layers will still impact the target perfectly, as each of the four lower layers will have canceled themselves out, and the bomb sight was perfectly calibrated for the layer in which it dropped.
If the aircraft were flying directly down wind, the effect and result would be identical, hence the 2 solution theory.
Now consider this scenario:
Wind 0-5k 0 @ 0kts
Wind 5-10k 225@10kts
Wind 10-15k 180@15kts
Wind 15-30k 090@25kts
I propose that their exists 2 unique headings whereby the cross wind of this layering scheme is washed out, leaving only head or tail wind variables on the table to consider. Bomber pilots who are adept at bombing in the wind will recognize the benefit of calculating those two headings...
Of course without some form of drag value to calculate the speed of the bomb as it passes through various layers, its impossible to calculate.
Writing a computer program that does this would be simple if the values were known.
Oneway
