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#3
November 4th 04, 11:13 PM
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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 
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