This took a little bit of research and phoning some friends, because this is a load case I’ve generally taken for granted (in my defense some engineers generate loads and some consume them for new designs, I’m a consumer
![Smile :) :)](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png)
).
One of the first things you make when generating structural loads is the V-n diagram (V for velocity, n for vertical acceleration aka Gs). You can actually have accelerations in all three Cartesian x,y,z directions so we call vertical acceleration Nz, but it’s shortened to n here. (Edit: the graph calls it “load factor”, but it’s the vertical acceleration Nz, aka “n”, in units of “G’s” which is actually a unit less scaling factor.) You can actually correlate the diagram below with color ranges on the airspeed indicator:
View attachment 70116
(Source:
Flight Envelope | Support)
Point A is Vs.
Point B is Va - The speed at which you can make full control deflections without negative consequences. For example max elevator deflection will result in an accelerated stall before you can develop sufficient G to over-stress the airframe.
The horizontal lines to the right of point B and point C are your positive and negative G limits respectively.
Vno - the line between the green and yellow arc has to do with gust envelope protection which is on a separate chart, but suffice to say if you hit unexpected turbulence and encounter a large vertical gust in the yellow arc you may not be protected from a structural failure, hence Vno must only be exceeded in smooth air.
And finally Vne is the never exceed speed, the point after which flutter may be encountered, parts may depart the aircraft, etc.
I bring up the V-n diagram because the FARs and Mil-Specs for airworthiness certification apply a maximum symmetric positive and negative acceleration (the lines to the right of points B and C on the V-n diagram) and NOT a rolling G.
The next thing we need to touch on is what spanwise lift distribution actually looks like in symmetric vs rolling applications.
Say this glider is making a wings level pull into a loop and the pilot pulls aft stick right to the positive max G limit in the airworthiness certification. That distribution of lift FORCE (note G is acceleration, now we are talking about the force of lift acting on the wing in lbs) will look something like this:
View attachment 70119
These wings are just fancy looking cantilever beams with an applied elliptical lift force distribution, so the designer/stress analyst is going to size the wing spar(s) to a couple of key load cases described in the airworthiness regulations but generally positive max up bending (at symmetric max positive G) is going to be the sizing load case.
Now telling you guys to not throw full aileron deflection in at max positive G isn’t going to surprise 90% of you because you know not to do that over maneuvering speed anyway, but the fighter pilot community can’t accept that because specific dogfighting maneuvers actually require it. “Flat Scissors” comes to mind:
Scissors (aeronautics) - Wikipedia
So let’s say you’re in a fighter trying to get someone off your tail and you throw in full aileron deflection to reverse your turn, what happens to the spanwise lift distribution shown in the glider picture above?
Recall that an aileron works by changing the camber of the airfoil on the wing. Down aileron = more cambered airfoil = more lift force. Up aileron = less cambered airfoil = less lift force. Look in the opposite direction of your aileron input and that’s the down aileron side, because that’s the side that’s making more lift and “picking that wing up” to roll, like this:
View attachment 70120
(source:
Ailerons)
Note that this is a simplified view of the spanwise lift distribution, it’s actually still elliptically distributed like in the glider picture, except now at the inboard edge of the aileron there’s a big step input on the down aileron side with even more lift than before. Think of these big arrows as resultant vectors showing the sum of the distributed forces in the glider picture. And if you integrate the area under the curve of the lift distribution, your total lift force on the down aileron wing is going to be greater in the rolling case than in the symmetric pull case.
So I think this is the crux of the problem. The engineers are in compliance with the load requirements as long as the wing is designed not to break at the max positive G with symmetrically applied lift force. But throw aileron in and now your lift becomes asymmetric, and the lift on the down aileron wing exceeds the max design value and could result in structural failure.
From a design standpoint there’s a few ways you can tackle this. If your airplane truly needs to do this maneuver then you dream this up as an extra design load case and it becomes your new sizing load case (above and beyond the max requirement in the airworthiness regs). All your structure is going to weigh more than what’s technically required though, and you have to go back to your customer and say “your max payload weight has decreased by X amount, can you live with that?” Inevitably they’ll probably decide they don’t want to live with that, at which point your other option is to:
a. Design your wing for the max positive symmetric G value. The wing attach point must react all the load distribution at the attach point with the fuselage, so now you have a max reaction shear and moment value.
b. Add full aileron deflection and see how much your lift force increases on the down aileron side, and see how much that increases your reaction shear and moment value.
c. Dial back the vertical G force (Nz) applied via aft stick while rolling until the total lift force and reaction shear and moment on the down aileron wing fall back to the original values from part a.
Essentially you end up with two limitations, your max symmetric G value and a reduced max rolling G value and the pilots must know and memorize both. When getting ready to roll in a maneuver like a flat scissors the pilot has to remember to back off the back pressure from the max pull value to the rolling value, reverse the turn with full aileron deflection and then load back to max G once the stick is centered. I think this is what
@MikeD and
@///AMG are talking about, but you guys correct me if I’m wrong. Fundamentally by doing that you guys are ensuring that the force on the wings never exceed the design limit load case, even when a deflected aileron is applying additional lift on one side.
Hope this helps, I had a fun time learning about it myself!