Kurt,
"Well, my Time Machine design is a standard glider weighing about 2 to 2.5
grams."
That gives an idea of the average loading in the steady glide. If there was
some way to estimate the load distribution over the wing, the force acting
on the flap in the down position could be estimated.
"I have loaded the wings of this design by putting weight at the root of the
wing and supporting the wing 1/2" from the tips on each side. To get up 40
ft a wing needs to hold 75 grams of weight, otherwise they break on launch."
This may be a good estimate of the launch load that breaks the wing. I
think the load on the wing may be more than that. With the load concentrated at
the tips, it would break with less load than if the load were distributed all
over the wing. How can you sandbag a glider wing? Support the fuselage
with the glider upside down. Perhaps load with little blocks of lead.
Distribute the load elliptically along the span and according to some plausible chord
wise distribution, at least so they balance near the CG, even though we
don't know the CP location in the launch. This might even give some idea of the
flexing, bending and twisting of the wing and flap. But let's go with the 75
grams of lift at launch. The weight force of 75 grams is 980 x 75 = 73,500
dynes. Applied to a 2 gram glider, that would produce an acceleration of a
= F/m = 73,500/2 = 36,750 cm/sec^2 = 1,206 ft/sec^2. Yes, 37.5 g's. If we
assume that the launch speed is 90 miles per hour, 4,023 cm/sec, we may
estimate the radius of the flight circle as v^2/a = 440 cm = 14.5 feet. Does this
combination of numbers seem plausible? (v^2 is v squared.)
I said "net lift" because when the glider is going much faster than its
steady glide speed, it will be accelerated upward. That can reduce the attack
angle on the wing and increase the negative attack angle on the tailplane. The
setup of forces then may become, front to back, first the CG pulling outward
around the circle, next the lift on the wing, pulling inward, then the
negative lift on the tailplane, pulling outward. These three forces must add up,
so the full wing lift is the sum of the accelerational force on the CG and
the negative lift on the tailplane. It's hard to know what the forces actually
are. With the wing attack angle reduced, you would expect the CP to move
aft, as required. Your observations of the bunting also suggest the tailplane
was negatively loaded in climb, implying the above force setup was operating.
But most of the load on the wings would be opposed by the centrifugal
force, not the tailplane load. It is too far away and must be much smaller to
make the balance. The tailplane may be snapping from a dynamic load rather than
a static load.
The loads at the instant of release of a catapult launch can be
significantly different than suggested above. A line through the hook and the CG will
show the attack angle during launch, which might be quite large. There are
other forces to consider; the rapid accelerational force and the sudden increase
in drag. The shape of the break may tell you which way the load was
applied. Often a little chip will pop out on the inside of a break.
There is so much going on here that we cannot see. We must work from
inferences.
Gary
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Received on Sun Feb 11 2007 - 21:53:49 CET
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