


Discussion
Simple Load Test for DLG Wing
I'd like to loadtest DLG wings after joining the panels: Suspend the tips, add a weight to the panel joint  bag of BBs or equivalent  and wait a few seconds.
Does anyone have a simplified method or empirical data on how to calculate the required static load that would represent an inside looping at max lift, i.e. during launch rotation, disregarding loads generated by flutter, torsion, etc.? 



Joined Jan 2008
5,153 Posts

Charles:
Try this: 10 Gs for a 9ounce airplane would represent roughly 90 ounces, or a bit more than 5 and a half pounds. Spread that load over the whole span, and that would probably be near the max load a wing would see. That would be my firstorder estimate of the load, as a semiretired and notverygood aeronautical engineer. For a "better" estimate, I would assume rotating through 90 degrees in one second, at a release velocity of 80 mph. Translate that into the acceleration towards the center of the arc, and that would be (since it is pretty near a symmetrical pullup) pretty close to the Gload being pulled by the wing. My two cents. I'm too tired to reinvent the math after all these years. Yours, Greg 


United States, CA, Tehachapi
Joined Jun 2011
3,781 Posts

I have some tabulated data from tests of various foam cores under bending loads. I never finished the testing because my testing apparatus broke and I had already gotten most of the information I wanted, but when I get home I can see about uploading it. The test was just to see how different core materials affected the bending strength of the wing.
As far as the maximum bending load that a DLG sees, if it comes out of the hand at 80mph (36m/s) and rotates upward with a turning radius of 7m, then the centripetal acceleration is a=v^2/r=36^2/7=182.7m/s^2. Newton's second law is F=ma, so for a 300g glider F=(.3kg)(182.7m/s^2)=54.8N. That load is distributed across the span of the wing, but we can approximate the bending load by assuming the load acts at the center of the wing panel. The maximum bending load at the root of the wing is then M=F*d=(54.8N)(.75m/2)=20.56N*m. Now, we could analyze that much more accurately if we knew the exact lift distribution of the wing and the exact rotation radius and speed on launch, but that should be a reasonable 5minute backoftheenvelope type of calculation. 



Quote:
Thanks for the input  I'll assume a centripetal acceleration of app. 20g, and reduce the fueslage to a pointmass of 140 grams, for a net load of app. 28N, which approximates Greg's test load of 5 lb. A lingering question remains: I'm not an areodynamical engineer, and realize that a wing can't generate more lift than it's capable of doing at a certain maximum angle of attack. However, if you dive the DLG, and then pull up sharply, is the path really an arc? Won't the plane retain significant downward momentum, so that the wings act as a speedbrake, with huge areodynamic loads that exceed lift by a substantial margin? 




Charles,
If you're implying that the AoA of the airplane exceeds say ten degrees, then something has to make it increase to that level. I'm assuming that even if you violently pulled the elevator, the horizontal would likely stall trying to overcome inertia and create an AoA high enough that would damage the wing. I guess what I'm saying is that a control input has to be rapidly applied and the surface has to be able to continue creating force when it too is operating at a very high AoA. Besides the force calculations above... we have to take into account whether or not the control inputs can cause that type of imbalance. I always roughly approximated 5lb as well and assumed it to be overkill. 



The wings have a reasonably high L/D even when past stall AoA, so the lift is always the predominate load. That said, the resultant force is not perpendicular to the fuselage or wing bottom surface. When they load test homebuilts, they usually set the wing at stall AoA (upside down of course), and add load that approximates the lift distribution along the span. You might want the wing at 12 to 15 degrees AoA, so there is actually a load pulling the wings forward.
Stan Hall wrote some good articles for Soaring magazine in the 70's on doing load and flutter tests for homebuilt sailplanes. I might still have them somewhere. http://esoaring.com/stan_hall_collected_works.htm You can assume the lift distribution is elliptical, if you want to distribute the load, and be very close. Wings can generate far higher max Cls when they are pitched up rapidly because the air doesn't separate into a stall immediately. I have some papers with max Cl increased by about 25% by rapid aircraft pitch rates or gust loads. Kevin Edit: Here is a good article on load testing wings: http://lightaircraftassociation.co.u...%20Testing.pdf 


Joined Apr 2009
6,654 Posts

Another point  we don't set up DLGs so that the elevator can pull enough to stall the wing in flight where there is otherwise sufficient airspeed. These are gliders after all... So, although the wing could produce higher lift coefficients than the normal stall lift coefficient, it wouldn't because there isn't enough elevator throw to get there.
Gerald 


United States, AZ, Arizona City
Joined Sep 2001
5,779 Posts

Charles,
Assuming that you are making the wing for yourself: Have you ever broken a wing? If not, then don't worry. If you are worried about creases, that's another issue. If you are worried about whether others might break it, let George, Charlie, Donnie, etc. throw it. If the break it, beef it up. If they don't, it's fine. Some things can be over analyzed. Gary 



Quote:
As my wings get lighter, the chances increase for a failure under load, and some of us have seen a carbon stressedskin structure fail catastrophically  kapow! Carbon don't wrinkle like Kevlar. The last set of fullcarbon panels weigh 95 grams, and are stiff as planks, but are they strong? A failure, i.e. during launch, could put other pilots in jeopardy. See you at the AZ Open! 




Quote:
The energy will burn off a DLG quickly as it tried to generate high G, because the mass is so low. The time at a high G load will be very small, not the 3 seconds usually used. The design Gload for turbulent air for airplanes with more mass is usually: Gmax = (max air speed/stall speed)^2 So for a glider with a 20kph stall, and a throwing speed of 140 kph (38m/sec) (140/20)^2 * 1.0 = 49g Pick your numbers. Fortunately, given the weight of a DLG, this probably isn't much of a design challenge. Kevin 



Joined Apr 2009
6,654 Posts

Not really true, by simple geometry of a DLG.
Let's take a look at that 49g loop. a = rw^2. a is the acceleration. r is the radius. w is the angular velocity (in radians) reference to the center of the loop. 49g > 480m/s^2 which is the acceleration. r and w are related, by velocity. A circle has 2*pi radians. One radian along the circle is equal in distance to r (how a radian was defined IIRC). w = v / (2*pi*r). So, a = r*(v/(2*pi*r))^2 = v^2 / ((2*pi)^2)*r). Solving for r, we get r = v^2 / ((2*pi)^2 * a) r = v^2 / 39.48*a ... approximately. v is 38 m/s. r = 1444m^2/s^2 / 39.48*480m/s^2 = 0.076m Clearly we have a problem. Though the wing could theoretically generate the force at that velocity, it isn't going to be able to make a loop of 0.076m radius. Why? It has a horizontal tail. That tail isn't mounted roughly perpendicular to the wing, as in leading edge nearly straight down. That would be required for the horizontal tail to point its zero lift line into the appropriate direction for this radius loop. Perhaps I made a little math or arithmatic error somewhere, but in any event this should show the point. A DLG can't pull that number of G's, primarily because it isn't fast enough for the resultant loop to be big enough to be geometrically consistent with the physical geometry of a DLG. Gerald 


United States, NY, HastingsOnHudson
Joined Aug 2007
79 Posts

Max launch height
If the Ev (J) at launch is 1/2mv^2(l) and this equals potential energy at launch height Ep (J) = mgh + 1/2mv^2(h), then:
h = ((v(l)^2 v(h)^2)/2g Assuming velocity at height is 2m/s and 38m/s launch you get 73.5 m height, with an average launch height of 60m this represents 82% efficiency, do you think this is possible? 82% Seems kind high to me? Bellisimo 
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