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Va: maneuvering speed ad nauseam
After the reading the Va threads of my past questions, I found a
wonderful way to confuse myself. I notice that I've become quite good at confusing myself and making things complicated. To give myself some credit, I did search old messages but found no resolution. So here we go... I am sure most know the typical textbook definition of Va...goes like..."the minimum speed at which the wing can produce lift equal to the design load limit" or "the speed at which the pilot can use full control deflections without over-stressing the airplane". Essentially, it's the minimum airspeed that, coupled with a control deflection to give you the critical angle of attack and CLmax, will result in + 3.8 g's. Pulling any harder won't help since you'll stall the airplane. Pulling with all your might as speeds below Va will result in the airplane stalling without reaching the limit load factor + 3.8 g's. In essence, Va is the stall speed at the design load factor of + 3.8 g's. Now, the above seems to be what's commonly accepted. Here's my question for this thread: Idealize the wing as a cantilevered Euler beam representative of the wing spar ("the wing"). Assume the lift load to be a distributed elliptical spanwise, transverse load, acting at the centroid of the section. Further assume no other external loading such as drag loads. The predominant stresses are bending (axial) stresses at the outer regions of the spar caps and shear stresses in the spar web. Assume that the failure mode is via the former. Alright, so here, clearly, the failure of the wing is due to excessive loading. The distributed load, expressed in X number of pounds per inch, was too great. In fact, for the sake of simplicity, let's make the distributed load a point load in pounds. Now, we can say that the failure of the wing was due to excessive FORCE which induced excessive stresses in the structure. Consider that a certain airplane weighed at maximum takeoff weight is designed to withstand + 3.8 g's (its design load). Actually, airliners that I am familiar with are tested to ultimate load, or 1.5*design load (+ 5.7 g's before permanent deformation). For now, we'll assume that at + 3.8 g, the plane's wings break off. That would equal to a total force on both wings of 3.8 x 2550 lb or almost 10,000 lbs. The thing that bothers me about Va is that it equates to a number of g's ("design load") AND that Va is being rescaled for weight. By doing so, Va becomes more of an acceleration criterion rather than a structural criterion. It appears as though Va limits positive g acceleration to + 3.8 g, not load itself. In other words, Va adjusted for say, a lower weight, tells the pilot "You will not exceed 3.8 g for your current weight, as you will stall first". If the current weight was 2,000 lbs, the total load on the wings would only equate to 7,600 lbs at + 3.8 g's, lower than the design limit of 10,000 lbs of + 3.8 g's at max. takeoff weight. The acceleration on the airplane would be a "limit load acceleration" but would not produce a limit load condition, per se, structurally. If Va was truly a structural consideration, it would not change (regardless of weight), since no airspeed below Va coupled with any non-stalled AOA, could produce limit loads of + 3.8 g as tested at max takeoff weight of 2,550 lbs. There may be severe flaws in my reasoning...please no flames and be nice. It's a holiday. Alex |
#2
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The wing is not the only thing in the aircraft that might break.
Rework it assuming that the first thing to break is the engine mount at 3.8g |
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#4
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Andrew Rowley wrote in message . ..
The wing is not the only thing in the aircraft that might break. Rework it assuming that the first thing to break is the engine mount at 3.8g Andrew, I think I understand now. Let me see if I get it... To rework the analysis, I'll assume that the weight of the airplane is mainly supported by the wings, and that other components such as the engine mount are essentially invariant in static 1g weight. In other words, by overloading the airplane with other bodies, ballast, and such, the weight of the engine mount does not change. If that is so, I can see how such *individual* components may be "g limited" with respect to their own inertia. Does this follow your reasoning? Thanks, Alex |
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The engine mount on most light aircraft is designed to withstand 9
G's minimum. Where do you get this figure? |
#8
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(Dan Thomas) wrote
Cessna also states in the 172 POH that it's designed to 150% of the G figures given, or 5.7 G's. I think the 3.8 figure would be the yield point, where things begin to bend, and the 150% figure would break them entirely. Or something like that. The C-172 is probably wing limited, since at 200-300 under maximum gross weight, it falls in the Utility Category with +4.4g allowed for the limit load and +6.6g for the ultimate load. Bob Moore |
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#10
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Andrew Rowley wrote in message . ..
(Koopas Ly) wrote: I think I understand now. Let me see if I get it... To rework the analysis, I'll assume that the weight of the airplane is mainly supported by the wings, and that other components such as the engine mount are essentially invariant in static 1g weight. In other words, by overloading the airplane with other bodies, ballast, and such, the weight of the engine mount does not change. If that is so, I can see how such *individual* components may be "g limited" with respect to their own inertia. Does this follow your reasoning? almost... the engine mount needs to support the weight of the engine multiplied by the number of Gs. so if the engine weighs 100kg at 3.8G the mount is supporting 380kg. At 5G it has to support 500kg. So force on the wings at high Gs may go down as your total weight goes down, but the force on other items (engine mount, baggage floor, basically any part of the airframe that has to support something) does not. Thanks for the clarification. What is the critical "item" that will break at 3.8 g's on a light GA airplane like a C172? Is it the engine mount? I see that's it's often mentioned as the "weakest link". It's my opinion that the definition of Va taught to private pilots leads to a false understanding that the wings will break at 3.8 g's. The nuance we've discussed is not stressed. Why? I don't know. Best regards, Alex |
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