1,250 lb is 567 kg. MTOW is 1317 kg according to Wiki. Assume a hopper that's half full, say 300 kg and a total weight around 1200 kg. Dropping that load will give you 1200/(1200-300) = 1,3 G instantaneously. If we're using MTOW and a full hopper, it goes up to 1,76 G. Not a problem on its own, but it all adds up.

The string on a helium balloon doesn't break when you let go of it.

But it isn't the g that kills the aeroplane, its the bending moment of the wings; g is just a convenient means of quantifying the bending moment for a given condition, and for advising limitations. Reduce the mass to a theoretical 1lb, and it could survive 1200g. 1.3 g on a 900lb aeroplane gives the same "weight" as 1.0g on a 1200lb aeroplane.

The increased normal-acceleration may, of course, introduce a change of attitude/alpha; and therein may lie the problem.

Judging by the way the wing folded, it will be the combined shear stress and internal bending moment at the inboard wing fittings that need to be looked at. The PA-25 has a strutted wing which could mean a local bending moment of zero at the bolts, leaving the shear stress as the remaining factor. A change in G-loading will influence this and I was just trying to illustrate that dumping a hopper full of chemicals could have contributed significantly to the total stresses in the structure, even though at first 'lightening the load' appears to be a positive thing. I think that's as far as we need to go with the theories...

I think the detached wing will keep on flying so its initial motion will initially be up won't it?. If you were to model it as two flexible wings suspended by a spring at each cp with a bag of flour at the centre the wings would bend upwards at the tip. Drop the bag of flour and the bending moment at the root/joint unloads rapidly. Yes the cg would accelerate upwards but can't see the harm from that.

I have only flown pax to their destination, thus have sofar strived to arrived with the same payload as I departed with, so I am no expert.
Having said that, I believe the release of payload in flight would not lead to an increased G load on the aircraft. If we assume the aircraft maintains straight and level flight with no change in speed there cannot be any load change on the aircraft. In order to maintain that with a decrease in weight from dropping the load the pilot will have to reduce angle of attack, and that will reduce wing load, with a reduction of the load on the wing attachments, not increase it. If the pilot allows the aircraft to drift up because of the release of weight, he would experience that G load, but the wing attachments would not, as the lift produced by the aircraft would remain constant.
This looks like a case of the pilot pulling up too fast, and loading up the wing beyond what it could handle. That could be because the CG changes when dropping the load. And maybe the wing spar wasn't as strong as it should be.

You are pulling 2G, but at half the weight. Loads would be the same at time =0+ seconds. Would go down from there.

I agree the G load due to pitch input is additive.

@IFMU Just do a simple thought experiment on the load factor formula: n = lift / weight.

Assume 1 for lift as it remains the same throughout the exercise. Further assume weight = 1: n = 1/1= 1. If you shed half the weight, your new weight now is 0.5. so n becomes 1/0.5 = 2. So you just doubled the load factor by sheeding half your weight.

That explains why the load factor momentarily and appreciably increases when you suddenly dump a largish amount of liquid from an airplane. that may well increase it beyond its structural limits.

And for any fellow physicists out there: yes it pains me to talk about weight. It's mass, really, we should be considering, not weight.

The pilot may feel 2g, but the structure doesn't. Dumping from the fuselage, wing bending load is reduced, in fact.

Exactly!

If the mass of the fuselage is n and the force required to "accelerate" it at 1g is X, the force required to accelerate a fuselage with a mass of n/2 at 2g is also X. Of course the wings aren't massless, so reducing the fuselage mass by 1/2 would result in an acceleration of less than 2g, and reduce load on the wing spar even when the acceleration is taken into account.

Ahm no. No. NO.
Imagine a balloon filled with helium with a weight attached to it floating around. Cut the string, and the weight falls down and the balloon shoots up. That is what happens when an aircrafts releases its payload in flight. Either the aircraft goes up, or the load factor is reduced. But absolutely never will the load factor increase from a decrease in payload.

Wings are a source of weight and lift. Fuselage is a source of weight. Payload is a source of weight. Ignore the horizontal stab and the negative lift in normal flight.

Lift = W_wing + W_fuselage+W_payload. If Lift remains constant after release of payload then W_wing + W_fuselage must both go up, redistributing W_payload between them by acceleration in proportion to their weights.

Since the connection between wing and fuselage also took the full W_payload, then after redistribution some of that weight will be in the W_wing and so the connection between the wing and fuselage will be less. The excess lift will not be able to put all of the weight into accelerating just the fuselage.

Imagine at the extreme a flying wing so W_fuselage = 0. The redistribution from dropping the payload is from wing to wing+; the bending in the middle goes down as the weight is more evenly distributed across the span by acceleration of the entire wing due to the now excess lift.

The connection at the wing root held until the outboard segment twisted up. It looks like the plane was pulling up after the dive to release the powder.

It looks like the forward strut failed; the most common failure mode would be buckling. It's supposed to have mid-strut braces to prevent that, but they may have been damaged previously.

Since the strut was under compression at the time maybe the outer strut fitting failed from a fatigue crack. The one on the fuselage is at such an angle I would not look there first.

See https://www.federalregister.gov/docu...ng-lift-struts

Similar to the crash of a Hercules doing firebombing a few years ago where a wing also came off. A sudden major reduction in weight and a sudden increase in wing loading during a pull up.

If a wing is providing a lift force to carry a certain weight, and that weight is removed; the stress and strain in the wing structure will be reduced. Like the string on Hans Brinker's balloon: the tension in the string, that was holding the weight up, suddenly reduces to near zero when the string is cut.

I suspect that wing failure in accidents of this nature is actually caused by a rapid unload, then re-loading of the wing, as the payload is dumped, followed by a max rate pull up. The fin of that A300 was snapped off by a rapid load - unload - load sequence applied by one of the pilots.

If you have an aircraft such as a water bomber or a crop sprayer that is subjected to rapid pull ups fairly often - in a series of alternately maximum and zero payloads, i.e loading / unloading / loading; and this is coupled with ageing and possible corrosion of the internal structure - which might not be inspected or spotted during inspections - you might have a recipe for serious structural failure?

PS; physicus, on planet Earth, the weight of an object is the attraction force produced by the action of the Earth's gravity on the mass of that object.

I stand corrected. Again. :)

And here I was finally hoping to have found a good example of where intuition misleads us... well, it doesn't. If the mass is dropped from the fuselage, then indeed as others have said, the load on the wing-fuselage junction will in fact decrease momentarily. If, however, the mass is dropped from the wings, then the load on the wing-fuselage junction will momentarily increase. In both scenarios the physics follows intuition.

Uplinker because gravity is not a constant but a function of where in XYZ space you are located, it is scientifically inaccurate to refer to the weight of an object without also knowing what the local conditions of the gravitational field are - hence mass being the preferred term. Interestingly, this varies by about 0.7% on the surface of the planet (owing to different rock densities in earth's crust), and of course decreases as you gain altitude and go further away from earth's surface. Other bodies like the moon and the sun play a role as well, and not an insignificant one (c.f. the tides). So, a 1kg mass on a mountain in Peru will measure about 995 grams on the same scale it reads about 1005 grams at the surface of the arctic ocean.

In aviation world, this is often evidenced by the "pragmaticists" in FAA world who for example refer to maximum take of weight (MTOW). Whereas in the "idealist" EASA lands, the same number is referred to as maximum takeoff mass (MTOM).

Just an interesting sidenote...

Do water bombers and crop sprayers contain their water or pesticide payloads in or under the wings then?

I thought they had tanks in the fuselage?

It is convenient to refer to Weight vs Lift even in accelerated frames of reference, neglecting the precise contribution from gravity acting on mass. To do otherwise seriously complicates notation in the local aircraft coordinate system.

That analogy doesn't work. Or if you do want to stick with this: When you cut the string, the lift on the balloon is appreciably more than its weight (I know, it should be mass...) which equates to a L/w of more than 1. If you were to be inside that balloon, you would feel that as a perceived G-force.
Not exactly, it will change, but it's a bit too simplistic to call that a reduction or any kind of beneficial effect without knowing the specifics of that change. Others have commented on the W_wing and other relations. Let's look at what happens inside a wing. I can't provide the shear and moment diagrams for these wings right now but think of the wing as a simple beam, supported at the inboard end by the fixings to the fuselage (I'm ignoring the strut for now). When that wing is being used (prior to the drop), there is a force, trying to bend the beam upward, which is distributed over the entire span and we call that lift. The structure weight of the wing and any other bits (fuel inside it) are pulling the wing in the other direction. These forces do not cancel each other out as the lift force is significantly higher than the others (half the total weight) and therefore there is a distinct bending moment and other forces on the attachement fittings. From this also follows that the beam itself needs a distinct amount of rigidity and strength to cope with these forces on it. If a significant load is dropped from the fuselage, the attachment load will change rapidly. Combine this with a pull-up that will increase the lift force on said wing and the attachment fittings will see a rapid change of stresses/moment.

As pilots, we are all trained to respond to configuration changes and other effects, so bearing in mind that I have never done any flying like this, the normal reaction of the pilot on dropping a load from a hopper would be to push the stick forward, decrease AOA and thereby keeping the aircraft on its level flight path. That avoids the aeroplane moving up, although you can still get a momentary flight path deviation of course. In this case (we're veering away from the theory again, apologies) the drop and pull-up appear to be (difficult to see on this video) timed pretty close together and that's why I started wondering about the drop and the pull-up ganging up on an unsuspecting bit of metal in the airframe.

Typical prune thread.... The aircraft suffered structural failure due to faulty/defective parts. An Ag aircraft is designed to drop it's load either slowly or rapidly. Gender reveal stupidity is American, there have been a few aircraft crash doing this.