David CL Francis wrote in message ...
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

What you say confirms what John T Lowry said. He said that if you
spend an hour taking that 100nm (measured on the ground) trip into
a 10nm headwind, you add 10nm to your total air distance travelled.
Your calculations show that if you spend approximately one-half
hour going into that headwind, you'll add approximately one half
that distance, or about 5 nm.

Actually, John's original statement could have been generalized to
say that a trip of X nm over the ground in an hour, if into a 10 knot
direct headwind, would be a trip of X+10 nm relative to still air.

It could be further generalized to say that a trip of X nm over the
ground in T hours, into a Y knot headwind, will be a trip of X + T*Y
nm relative to still air. Using your numbers, we get X=100, T=.526,
Y=10, X + T*Y = 105.26nm effective distance, just as you claim.

--Rich