On May 28, 7:10*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

To clarify, the ground reference STF is reserved for trying to
maximize distance, not speed. This means that your 4-knot final glide
is at the same speed irrespective of wind up until the best glide STF
accounting for wind exceeds the McCready STF - at that point you won't
make it home into the wind unless you speed up. That would only apply
in situations where you make a downwind turnpoint under weak
conditions with enough altitude to get home but without strong enough
lift to make sustained headway - that's only happened to me once - 1.5
knot thermals and a 40 mph headwind (I landed).

I think you could use a version of this logic in making an upwind
turnpoint - though again I think the situation would be rare. In this
case you are calculating your angle over the ground to see if you can
make the turnpoint before you need to take a thermal. I suppose it is
possible that the optimal solution is that you need to fly faster than
McCready speed to make the turnpoint, but I think that means that
you'd be unable to make sustained headway given the thermal strength.
Maybe if you were trying to duck into a turnpoint at the edge of a big
downburst or something.

9B