"Edwin Johnson" wrote in message
. ..
On 2006-10-09, Tony Cox wrote:
Here's a problem which seems to have a non-trivial solution.
At least, I've not been able to find a definitive answer to it, but
what do I know??

Rule of thumb responses are interesting, but better would be
a full mathematical treatment. Presumably, a proper treatment

Tony, in "Mountain Flying Bible" the author Sparky Imeson was once
discussing this and a physics professor handed him a formula which is
published in that book. It is the beakeven wind speed for taking off uphill
into wind and downhill with a tailwind. Hopefully it formats correctly here.
If wind is less, takeoff downhill and if more take off uphill.

Vbe = (s * d) / 5 * V

I looked through my copy of the MFB before posting,
but didn't find anything. Good book, but the information
in there is so disorganized its not surprising I drew a
blank. Sparky could do with a better editor!

When I have a spare moment, I'll see if I can verify
it. Of course, its only for take-off -- landing may be
different.


Whe
Vbe = breakeven speed in knots
s = slope up in degrees
d = POH distance to liftoff with 0 slope and 0 wind in feet
V = volocity of liftoff speed in knots TAS

OK, lets do a reality check. My 182 at full weight
supposedly takes 900' to lift off on a summer's day
at 61B. Runway 15-33 has a 2 degree slope. Lift-off
speed is 60 knots.

Vbe = (2*900)/5 * 60 = 21600

Seems a bit high to me. Perhaps Sparky meant

Vbe = (2*900)/(5*60) = 6 knots.

Well, I suppose its possible, but I'd have thought the
figure a little on the low side.


Sparky says that as a rule of thumb, if wind is less than 15 kts take off
downhill. If more than 15 kts take off uphill, provided obstacle clearance
can be maintained.

In this case, Sparky's ROT wouldn't work out too well...