Hawker Hunter Crash at Shoreham Airshow

[Trim Stab]

CliveWatson,

It is actually fairly simple to do an approximation to the mathematics of a loop. Acceleration is a vector, so the acceleration due to gravity can be separated from the centripetal acceleration of the loop.

If we assume the loop is a perfect circle of radius R, the distance from the top of the loop to the bottom of the loop is S, and the velocity at the top of the loop is v, and the time from top of loop to bottom of loop is t, then we find that t=pie*R/v, and 0.5gt^2+2R=S. Substitute t, and we get a quadratic equation for R. Solve that, and put in the actual figures you quote, then we can work out out that the centripetal acceleration at the top of the loop would be 1.9g, so (subtracting gravity) he would have had to pull 0.9g at the top of the loop, which is pretty reasonable and would not have risked an inverted stall (though I don't know the stall speed of a Hunter so can't be sure of that). As the aircraft descends, potential energy is converted into kinetic energy so that o.5V^2=gS+0.5v^2, where V is the velocity at the bottom of the loop. Chug away at the maths of that and he would have been doing about 250knots at the bottom of the loop and would have to pull about 9g to maintain a constant radius loop, which is quite high.

Now note that I said "constant radius" loop. This is not necessarily the case. Also, drag would have varied around the loop as different elevator inputs were used, so V might be a bit lower. But nevertheless, i think he might have found himself in a situation where he was having to pull a lot of g at the bottom of the loop, which is more or less what is observed in the videos.

Does anybody know approximate stall speed of the Hunter at the mass and configuration of the day? I can use it to work out how close to the stall he would have been at different bottoming out scenarios (i.e. 500ft, 200ft, 0ft).