On Wed, 3 Nov 2004 at 12:33:39 in message
.net, John T Lowry
wrote:
A trip of 100 nm over the ground, in an hour, if into a 10 knot direct
headwind, would be a trip of 110 nm relative to still air.

Are you sure about that?

Aircraft flying at 200k effective speed over the ground 190 knots
agreed?
Time taken = 100/190 = 0.5263157 hours
Effective distance at 200k TAS is 200*0.5263157 = 105 .26 nm

try a very low flight speed of 50k

Effective speed 50 -10 = 40k
Time take = 100/40 = 2.5 hours
effective distance = 50*2.5 = 125 nm

Try 1000k
effective speed over ground 990k
time taken = 100/990 = 0.10101010hours
effective distance = 1000*0.1010101 = 101.1nm

If you imagine that the headwind is the same as the aircraft speed than
it makes no progress at all and covers an infinite distance through the
air to cover that 100nm.
--
David CL Francis