Originally Posted by
FlexibleResponse
An interesting analogy for those of us who have dropped bombs (and it would include other loads such as water bombers and helo external loads)…
Imagine a Lancaster bomber of a nominal weight of 45,000 lbs carrying a Tallboy bomb of 12,000 lbs for an AUW of 57,000 lbs .
The bomber is flying and trimmed in straight and level in (unaccelerated) flight at 18,000 feet and at an airspeed of 150 KIAS.
To do so, the power is set at say, 85% power and the wing AOA is generating lift force of 57,000 lbs to equal the total weight in level flight.
The crew then release the bomb over their target without changing any of the control positions or power setting. (Let’s ignore any trim changes due to changes in CG or airflow effects).
On releasing the bomb, the aircraft now weighs only 45,000 lbs…but crew maintains the power, speed and pitch attitude which are still set for a lift force of 57,000 lbs.
Will the aircraft:
A. Continue to fly straight and level at 150 KIAS;
B. Immediately descend with the bomb release;
C. Immediately climb as the lift force of 57,000 lbs far exceeds the new a/c weight of 45,000 lbs (increase in g); or
D. Enter a victory roll?
Now try the same bomb release in accelerated flight pulling up or recovering from a dive with increasing AOA and ask the same question re g force change.
The answer to the accident has been revealed in some of the posts but it is still possibly not clear to all of us.
@blancolirio in his video (above posted by @megan) referred to and explained the phenomena quite well using the terms of change of maneuvering speed with weight change.
Expanding on his line of reasoning, Lift required for level flight is equal to Weight, sometimes expressed as L = Nz x W (where N = g and z = vertical plane and W = weight).
So a Pawnee flying at say 2900 lbs at 124 mph Max weight man speed) in level flight is creating 2900 lbs of lift at the current AOA (L = 1g x 2900).
If the weight of the Pawnee is reduced by 1200 lbs of hopper emergency jettison and the AOA remains constant, the aircraft wings are still generating 2900 lbs Lift, but the Weight has reduced to 1700 lbs.
ie, 2900 = Nz x 1700
That translates to Nz = 2900/1700 which gives Nz = 1.7g
However, the pilot pulls back the stick to almost the stall AOA to achieve the max g of 4.5g at 124mph (the max man speed for 2900 lbs) which is expressed as L = 4.5g x 2900, which gives L = 13.050.
But, he reduces the weight of the Pawnee during the pullup with the emergency jettison to 1700 lbs (13,050 = Nz x 1700).
That translates as Nz = 13,050/1700 which gives Nz = 7.67g
Which clearly is an overstress far in excess of the aircraft ultimate g limits
Now I am not that expert with math so I would kindly ask the Flight Test Engineers on the forum to cross-check my reasoning and figures for accuracy.