On May 28, 10:10*am, Nine Bravo Ground wrote:
On May 28, 6:02*am, Andy wrote:

On May 27, 10:49*pm, Tim Taylor wrote:

Anyone have a good set of equations or example of how to do this
simply?

The fastest speed through the air mass will give the fastest speed
over the ground. *The wind does not change the speed to fly. *It only
impacts best glide speed to a landing.

So add 0xW for a headwind and subtract 0xW for a tailwind.

Andy

I think Tim means STF for final glide where you are flying in
reference to the ground, not the airmass. John Cochrane's analysis
shows that you have different lift strength targets for upwind/
downwind turnpoints as well, though I don't know if this extends to
STF. John?

I have the final glide formulae in a spreadsheet, including effects of
wind and wing loading, if you are interested.

9B

If you want to derive the formula you need a little bit of 1st year
calculus plus some algebra. Derive a line passing through
the point (-headwind, -MC) that is tangent to your polar. The slope
will be equal to the 1st derivative of the polar at the speed to fly.
I use it often enough when analyzing the performance of gliders
I fly (I've made more than a few prayer wheels in my day).

As far as speed to fly, Andy is correct. Fly through the airmass at
your MC speed. John's paper says you should nudge your MC
a bit up or down when you're flying into our out of an upwind
turnpoint (read the paper for details).

For final glide, THEN you can take the headwind into account.

-- Matt