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 wrote :

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

That is still mathematically correct I hope?

John assumed an airspeed that would take 1 hour exactly to
fly 100nm in one hour into a 10 knot headwind. During that
hour, the air moved 10 nm so the aircraft covered 110nm of
air and 100 nm of ground.

So he did, but that means he is implying a cruise speed that will get
you there in one hour.

Perhaps I am naive but it seemed to me you needed to calculate the
effect of head winds on a given cruise speed not a cruise speed to give
constant time? So if your aircraft had to fly against a 50k wind it
simply has to increase its cruise speed to 150k and it covers a 100nm in
one hour just the same. Is that useful thing to know? In that case it
is of course true that the aircraft would have covered 150 'air' miles.

You assumed a speed that would take 1 hour to fly 100nm in
no wind, and then calculated the time. (actually you chose
200 knots, I'm not sure why). The assumptions were
different. For John's case, the aircraft was not flying 100
knots - it was flying at 110 knots and covered 110 air nm
and 100 ground nm.

I chose three different speeds I believe, 200k, 50k and 1000k just to
show the differences.

I apologise again to those concerned for my error in reading the
original statement..

More interesting would be to know the galls per hour consumed at various
speeds and then create a table of cruise speeds against head wind
components (Positive or negative) and then determine the best cruise
speed for each wind speed for minimum Gallons per mile perhaps?

--
David CL Francis