Crashondeck,
Your last paragraph is the correct and complete solution to the mystery - I'd delete the rest if I were you...
In your 30kt wind example, you have missed the fact that momentum is a vector quantity. The change in momentum is from 70,000kt.kg in one direction to 130,000kt.kg in the opposite direction. The change is therefore:
70,000 - (-130,000) = 200,000
not 60,000 as you stated. If you change the example to a 30kt tailwind, you get a change from 130,000 to -70,000:
130,000 - (-70,000) = 200,000
As you stated in your last paragraph, the change in momentum is completely independent of the reference frame - you could be moving with the wind, fixed on the ground, or watching from a passing train - the change in the aircraft's momentum will always be 200,000kt.kg, and is the result of the same forces (tilted lift vector) acting in the same directions over the same periods of time, regardless of what the air is doing relative to the ground.
A bit of algebra may help further: stick with a 1000kg aircraft flying at 100kt IAS in a headwind of W knots:
Starting momentum = 1000 x (100 - W) = 100,000 - 1000W
After the 180 turn = 1000 x (-100 - W) = -100,000 - 1000W
The change is therefore:
(100,000 - 1000W) - (-100,000 - 1000W)
The two 1000Ws cancel out, leaving just
100,000 - (-100,000) or 200,000!
There is no "W" left in the answer, which is the key - the wind has no effect.
cbl.