G Load during payload drop

Question

Some days ago I saw the accident of the airtractor that, just after a drop, suffered a spar failure in both the wings. Luckily the pilot survived.

If load factor in his simplest formula is:

n=lift/weight

So if an aircraft experiences a sudden loss in weight, is natural to think that the aircraft will experience a temporary increase in G load during the maneuver. My doubt is about where that forces are going to be applied.

Let's now suppose that during the complete maneuver, the aircraft is going to maintain straight and level flight without losing or gaining a feet of altitude. In this case, will the complete aircraft be under that load (like during a turn), or the load is going to be applied only on the wing(s).

Thanks in advance for the answers!

Answer 1

There are a couple of misconceptions in your question that I believe are leading you to a false conclusion that G-loads from releasing payload may have literally ripped the wings off this airplane:

1. Spraying liquid will not cause a "sudden" loss of weight. While fire fighting aircraft are capable of releasing large quantities of liquid all at once, agricultural sprayers like the one depicted are designed to disperse over a much larger area. Hence the gross weight lost on each pass is pretty negligible.

2. Losing gross weight will not cause a sudden and appreciable increase in load factor. Whatever the math might say, I can tell you from the personal experience of standing in the cargo bay of C-130s as thousands of pounds have rolled out the back end, that there is very little perceptible change in felt G forces. (Conversely, I have been walking forward in the cargo bay when the pilot put on a roughly 2G turn and my knees buckled, sending me to the deck and breaking a tooth...)

So, while load factor will change somewhat when weight is changed suddenly, it would be well within structural limits that the aircraft is designed for, and would in no way account for the type of dramatic spar failure as depicted in the photo.

If you are interested in a similar incident, a C-130 tanker tragically and dramatically had its wings snap off in a similar manner. More info in the link here: C-130 crash

I don't recall all the details of this, or the incident you describe, but generally there is corrosion, and/or fatigue cracking, and when the catastrophic failure finally occurs it is as a result of the pilot applying G loads during the pullout, not because of the gross weight change from dropping payload.

Answer 2

Your skepticism is justified, and it's not just a question of how significant the change in G-load created by dropping a payload would be. Dropping weight from the fuselage creates no increase in the stress on the wing-fuselage junctions. In fact, dropping payload from the fuselage decreases the stress on the wing-fuselage junctions. (Dropping weight from the wings, however, would increase the stress on the wing-fuselage junctions.)

Assume an aircraft weighs 10,0000 pounds without payload, and can carry a 10,000 pound payload. Assume the weight of wings is negligible, so the fuselage weighs 10,000 pounds without payload. Assume the payload is carried in the fuselage.

In straight-and-level flight with no payload, 10,000 pounds of force must be exerted on the fuselage by the wings. So 5000 pounds of force must be transferred through each wing-fuselage junction.

In straight and level flight with payload, 20,000 pounds of force must be exerted on the fuselage by the wings. So 10,000 pounds of force must be transferred through each wing-fuselage junction.

Assume the plane is in straight-and-level flight with payload, and then the payload is dropped instantly. The weight of the aircraft is cut in half. The airspeed and angle-of-attack of the wings will initially stay the same, so each wing is still generating 10,000 pounds of force, all of which is transferred through the wing-fuselage junction. There is no increase on the stress on the wing-fuselage junction. The aircraft is now in a +2 G condition and will accelerate upwards. I.e. the flight path will curve upwards. This upward acceleration will create stress on the mounting points of heavy components such as the engine, battery, pilot's seat (including pilot), etc, but will not increase stress on the wing-fuselage junction.¹

The upward acceleration will have some tendency to decrease the angle-of-attack of the wings, decreasing the total lift force, and decreasing the total amount of force transferred through the wing-fuselage junctions, and decreasing the total G-loading, but this will be offset by the plane's inherent pitch stability dynamics (assuming that the payload was carried at the CG), which will tend to pitch the nose up to conform to the upward curve in the flight path and restore the original angle-of-attack.

At any rate, we can see that there is no tendency for the wing-fuselage junctions to fail in a way that would cause the wings to rise upwards relative to the fuselage. The same amount of force is being transferred from the wings to the fuselage as was the case before the drop, but that force is now being used to accelerate the fuselage upwards rather than to support the payload.

Now assume that the weight of the wings is not negligible. Assume the wings weigh 500 pounds each and the fuselage weighs 9,000 pounds without payload. Now in straight-and-level flight with a 10,000 pound payload in the fuselage, 19,000 pounds of force must be exerted on the fuselage by the wings. So 9,500 pounds of force must be transferred through each wing-fuselage junction.

Now when the payload is dropped instantly from straight-and-level flight, and the wings continue to fly at the same airspeed and angle-of-attack and generate 10,000 pounds of lift each, the aircraft will be in a 2-G condition and will accelerate upwards. 1000 pounds of lift force will be "absorbed" by each wing, and 9000 pounds of lift force will be transferred through each of the two wing-fuselage junctions to the fuselage. So dropping the payload from the fuselage has actually decreased the stress on the wing-fuselage junctions, because some of the lift force from the wings is being used to accelerate the wings upwards, rather than to support the payload in the fuselage!

Now let's change the picture and assume that the payload is carried in the wings, not the fuselage. Let's assume the wings weigh 500 pounds each empty, and 5,500 pounds each loaded. Let's assume the fuselage weighs 9,000 pounds. In straight-and-level flight in the loaded condition, the wings are generating 20,000 pounds of lift total, of which 9,000 pounds are being used to lift the fuselage. 4,500 pounds of force must be transferred through each of the two wing-fuselage junctions.

Again, assume the plane is in straight-and-level flight with payload, and then the payload is dropped instantly. The weight of the aircraft is cut in half. The airspeed and angle-of-attack of the wings will initially stay the same, so each wing is still generating 10,000 pounds of force, so the aircraft is accelerating upwards in a 2G condition. Of the 10,000 pounds of force being generated by each wing, 1000 pounds is "absorbed by" the wing and 9000 pounds must be transferred through each of the two wing-fuselage junction to the fuselage. The force on each of the wing-fuselage junctions has doubled.

Take-home lessons--

• Dropping weight from the fuselage does not increase the stress on the wing-fuselage junctions. Rather, if the weight of the wings is not negligible, then dropping weight from the fuselage decreases the stress on the wing-fuselage junctions as the aircraft accelerates upwards.

• Dropping weight from the wings does increase the stress on the wing-fuselage junctions as the aircraft accelerates upwards.

You also asked--

Let's now suppose that during the complete maneuver, the aircraft is going to maintain straight and level flight without losing or gaining a feet of altitude.

This would require the pilot to shove the stick forward to reduce the angle-of-attack, and lift force generated by the wings, at the instant the payload was released. ("Shove" only applies to our thought experiment where the entire payload is released instantly; if the payload is released more gradually then a more gradual forward movement of the stick would be required.) Reducing the angle-of-attack of the wings doesn't create any unusual loads on the aircraft. And the aircraft would remain in a 1-G condition.

Footnotes--

1. Note that for a given total lift force generated by the wings, the higher the total mass and weight of the aircraft, the lower the G-loading, and thus the lower the stress on the mounting points of heavy objects of fixed weight such as the motor, battery, occupants' seats (including occupants), etc. That's why for many aircraft, the published maneuvering speed is lower when the total mass and weight of the aircraft is lowered. Lowering the allowable maneuvering speed also lowers the total amount of force that can be generated when the wing reaches the stall angle-of-attack while flying at the maneuvering speed, which keeps the maximum possible total G-loading constant, which keeps the maximum possible stress exerted on the mounting points of heavy objects of fixed weight constant. (See related ASE question Why does maneuvering speed vary with weight?). However, this approach doesn't make sense if 1) the critical points of concern that will fail first are the wing-fuselage junctions and 2) any added weight is added to the fuselage, not the wings. In this case, as this answer shows, as long as the weight of the wings is non-trivial, then to keep the stress exerted on the wing-fuselage junctions constant, the maximum allowable lift force should be decreased as the fuselage loading is increased, which means that the maneuvering speed should be decreased as the fuselage loading is increased. The opposite is true if the extra weight is being added to the wings rather than the fuselage. See also this related ASE answer -- How does an aircraft's weight affect the V-n diagram?.