On Fri, 05 Nov 2004 22:14:57 GMT, David CL Francis
wrote:
On Fri, 5 Nov 2004 at 11:16:17 in message
, Todd Pattist
wrote:
David CL Francis wrote:
That does not answer the fact that the original statement by John
appears to me to be wrong.
It's not wrong.
Nice to meet you here again Todd. I agree I was wrong, in that the
problem he postulated was different from what I assumed. I think my
calculations were right though.
I apologise to those concerned
However John's actual statement now I read it more carefully seems to
imply that given the wind speed you must find the TAS at which you
must fly to get there in an hour!
Is this the calculation that is intended? Even more trivial than my
calculations!
John 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."
Thus we have an unknown cruise TAS cruise speed
Let that be V
We have a 100 nm distance and a head wind of 10k
We have a time of flight of exactly one hour
Therefore 100/(V-10)= 1
and V -10 = 100
it follows that V = 110k
So at a TAS of 110k you travel a ground distance of 100nm against a wind
of 10k and surprise, surprise you than fly 110 'air' nm
More or less self evident so I am unclear what that achieves?
You're wrong, here's why:
you're still wrong and that is why you're not achieving anything.
in the real world your cruise speed remains constant so what happens
is that the 110 nautical miles that the wind makes the 100 miles seem
like, takes longer to fly. engine running for longer equals more fuel
burn from the fixed tankage in the aircraft hence the need to pre calc
the usage and plan for it.
if you are actually a pilot you are an accident waiting to happen.
Stealth Pilot
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