"Tony Cox" wrote in message
oups.com...
[...]
d = v**2 / (2*a)

This part, I don't get. Using the two equations you provide previously,
I
get "d = v*t/2", which is consistent with my distant recollection of
basic
high school physics.

Look at it this way...

d = a*t*t/2 = a*t*a*t/(2*a) = v*v/(2*a)

Okay...not sure why I didn't see that earlier, since I had a suspicion
that's what you had done.

Still, I don't see the value in writing it that way. Probably if I had
bothered to actually do the derivation, I'd understand. If I had to guess,
I'd say it has something to do with keeping acceleration in the equation so
you can include the gravity adjustment for the slope. But with "t" in the
equation, I'd think you could do that as well, and with a slightly simpler
starting equation.

Oh well...doesn't really matter. In algebra, the whole point is that
everything comes out the same, no matter how you "phrase the question".
And since I'm too lazy to bother with redoing the derivation starting with
the alternative equation, I don't have any business questioning which
version is easier.

[...]
Remember, its increasing TO speed *with the TO distance
constant*. It means the energy gained/lost by going downhill/uphill
is the same, but the effect of wind is less significant because
you're accelerating more quickly through the range where
windspeed *would* be significant.

Ahh, right. Thanks. I knew I must be missing something. I guess that does
stand to reason, that an airplane with better thrust spends less time
arguing with the wind.

Okay...well, looks like Sparky's formula is right, and all his other advice
is off-base. Good to know.

Pete