Andy Durbin wrote:

Now assume a flight condition that resulted from a high g pull up that
approached stall speed. I’ll assume the speed is 40kts, that
the wing tips flexed up 6 feet, and that as the pilot pushes forward
to avoid stall the wings return to normal deflection in 1 second. The
wing tip angle of attack change due to the downward motion can be
calculated from the forward speed of 40kts = 67.5 ft per second, and
the downward speed of 6 ft per second. Inverse tan of 6 / 67.5 is 5
so the tip angle of attack was increased by 5 degrees as the wings
unflexed. The effect reduces to zero at the root where there is no
deflection.

I think the problem here might be picking the right numbers: A high G
pull up would no longer be "high G" at 40 knots (near stall speed), as
the G loading would already be reduced to 1 G. At one G, the wings will
not be flexed upwards. So, I think the wings will return to their normal
position during the speed reduction that occurs after the pull-up is
initiated; that is, more slowly than the 1 second used in the calculation.

To get a 2 G load (a guess - I don't know how much it takes to bend the
wings up 6 feet) on the wing that stalls at 40 knots would require 56
knots. Perhaps there would still be some effect, but it would also be
reduced by the increased speed used (56 knots).


If the numbers are valid then it remains to be decided if wing flex
induced angle of attack changes of this magnitude would have an effect
on stall and stall recovery characteristics. I expect that they
would. Others disagree with me.

It might be a difficult effect to determine experimentally: a pilot
would have detect that the tip stalled when he reduced the G loading.
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Eric Greenwell
Washington State
USA