"Tony Cox" wrote in message
ups.com...
No, I think the inverse relationship is the right one, although
I'm not making any claims that Sparky's formula is correct.
I agree that a higher take-off speed means you're less concerned
about what the wind is doing. In the limit with a truly phenomenal
take off speed, who cares at all about wind or even what grade
you're on? You're cranking out lots of power, ignoring wind
and grade and so the break-even speed would be close to zero.

I think you're misinterpreting the break-even point. You are right that,
"In the limit with a truly phenomenal take off speed, who cares at all about
wind", but the formula indicates that as airspeed increases, one must be
concerned with ever-decreasing winds.

That is, the point at which the break-even is near 0 occurs when takeoff
speed is very high. According to the formula, the break-even point is the
wind speed ABOVE which it's important to be taking off into the wind. So
using the inverse relationship, the conclusion is that for airplanes that
can basically ignore wind speed, the wind speed is much more important than
slope even when there's practically no wind.

That doesn't seem right to me.

Still, this is all probably moot since the formula seems to have problems
whether you believe that the V is a denominator or numerator.

I've got the 1998 3rd edition right here, and I still can't find
the formula. In the chapter on "Takeoff", S claims that 1%
downslope is equivalent to 10% more runway, that winds
over 15 knots take off uphill (no grad specified), and that
increased drag on a 1% uphill grade results in 2 to 4 %
increase in takeoff distance and subsequent climb. No idea
what to make of all that. Rather disappointing, I'm afraid.

I agree. Three different rules of thumb, none of which are even close to
being equivalent to each other. They can't all be correct.

It makes me discount his supposed formula, as reported,
even if I could find it.

Yup. Looks like you're back to square one.

Pete