Hi Gary,
My Cat 1 or 2 models do not climb up in any kind
of half loop. My feelings is the shortest distance
between two points is a straight line, therefore I
design and trim a model so it travels in basically a
straight line, from the release to the top of the
flight path. At the top of the flight path the model
is designed and trimmed to "snap" where it makes a
quick left turn, levels out and is at glide speed. If
you never seen this it hard to imagine that a model
can do this but it can. I've seen many good modelers
achieve this type of launch.
Your analysis is interested and I believe
correct given your basic assumptions.
Kurt
--- dgbj_at_aol.com wrote:
> 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
>
>
> [Non-text portions of this message have been
> removed]
>
>
Kurt Krempetz
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Received on Mon Feb 12 2007 - 05:53:27 CET