Feb 03, 2013, 04:26 PM Thermal, where art thou? 85045 Joined Oct 2005 2,969 Posts Discussion Simple Load Test for DLG Wing I'd like to load-test 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.?
 Feb 03, 2013, 04:40 PM Registered User Joined Jan 2008 4,808 Posts Charles: Try this: 10 Gs for a 9-ounce 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 first-order estimate of the load, as a semi-retired and not-very-good 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 pull-up) pretty close to the G-load being pulled by the wing. My two cents. I'm too tired to re-invent the math after all these years. Yours, Greg
 Feb 03, 2013, 05:48 PM Registered User Joined Apr 2009 6,522 Posts You need to subtract the wing mass from the plane mass. Gerald
 Feb 03, 2013, 07:00 PM Registered User United States, CA, Tehachapi Joined Jun 2011 3,534 Posts Good point Gerald. That would reduce the bending moment to 12.3N*m. With another 3oz of ballast it would be 18.4N*m.
Feb 03, 2013, 07:15 PM
Thermal, where art thou?
85045
Joined Oct 2005
2,969 Posts
Quote:
 Originally Posted by bwill6 Good point Gerald. That would reduce the bending moment to 12.3N*m. With another 3oz of ballast it would be 18.4N*m.

Thanks for the input - I'll assume a centripetal acceleration of app. 20g, and reduce the fueslage to a point-mass 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?

 Feb 03, 2013, 07:39 PM Father of Fr3aK, DLG Pilot USA, OH, Worthington Joined May 2002 7,363 Posts 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.
 Feb 03, 2013, 09:50 PM Registered User Joined Apr 2009 6,522 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
 Feb 03, 2013, 10:19 PM Registered User United States, AZ, Arizona City Joined Sep 2001 5,427 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
Feb 04, 2013, 07:29 AM
Thermal, where art thou?
85045
Joined Oct 2005
2,969 Posts
Quote:
 Originally Posted by GaryO Charles, Assuming that you are making the wing for yourself: Have you ever broken a wing? If not, then don't worry. Gary
To quote former Intel CEO Andy Grove: "Only the paranoid survive!"

As my wings get lighter, the chances increase for a failure under load, and some of us have seen a carbon stressed-skin structure fail catastrophically - kapow! Carbon don't wrinkle like Kevlar. The last set of full-carbon 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!

Feb 04, 2013, 09:46 AM
Joined Jan 2007
3,852 Posts
Quote:
 Originally Posted by G_T 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
With the low stability margins DLGs are usually flown with, sudden elevator commands could result in a high enough pitch rate to get to any AoA. And we have no control over the air, especially in the turbulent boundary layer next to the ground where DLGs have high speed on launch. Turbulence could increase the AoA to max Cl, especially during the pull-up when the glider is already at a positive Cl.

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 G-load 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