Pilot Bails Out At Low Level

[Don Fry]

Go to Piza, take Swmbo. Might get a really nice Christmas present. And you will not be happy till you time that drop.

[Don Fry]

Or a cheapskate might do Alton Towers, and a McDonalds. But the Crimbo present might be inferior.

[PETER BRUCE - Eastchurch Gap]

Hi Onetenor. I can confirm the camera was rotating and helicoptering all the way down which was clearly visible as the camera landed only around 200 foot away in front of me...

PS Don't want to go to Piza...

Regards Peter

[brokenenglish]

Peter, IMHO, the thread will die because you're asking for precise answers, based on pure guesswork, with an impossibility of quantifying the forces involved.

An object fell from a plane. The only available data is the weight of the object. All other data is just guessed or ignored...

In fact (again IMHO), the thread is just an excuse for "waffling on" and throwing in a few terms that sound knowledgeable (sorry, but it's true...).

Most of the "calculations" are based on falling in a vacuum (aerodynamic flight would be impossible!).

You have no ballistic data: the falling orientation of the camera, the frontal area presented to the airflow (probably not constant), and hence the retarding forces countering gravitational acceleration and fall velocity.

As I mentioned earlier, you just don't have the data to obtain the answers you're looking for, other than guesswork and (sometimes irrelevant) opinions.

[PETER BRUCE - Eastchurch Gap]

OK - I accept what you say ans as some of the conclusions were so far out it makes sense that the equations put forward apply to objects in a vacuum- hence the unrealistic height assumptions.

Although a precise figure is not possible I do feel that as the full camera details were given together with the time taken to fall I have today confirmed my own assumption and conclude that with the model following the same line and the angle and size of the model in the sky I feel that the height was indeed around 100 to a max of 150 feet. Perhaps a bit of waffle but theory can't always explain visual happenings but I do feel it has presented some healthy conversation which I have honestly enjoyed. Regards Peter

[Former Member]

[This posting has been removed]

[PETER BRUCE - Eastchurch Gap]

Hi Tom. Can't argue with that at all🤡

[The Wright Stuff]

Hi chaps,

Apologies if I am appearing to resurrect a dying thread. It is indeed interesting and not at all a hopeless ask. However, it is not a one line answer and this is the first chance I’ve had to respond. I’m afraid I don’t live online 24/7 😊.

This is a very common problem for engineers and scientists to solve. The difference is that (unlike O-Level type questions) there isn’t a precise answer because we can only ever get increasingly close approximations based on the accuracy of the information we put into the equations. This doesn’t mean we can’t calculate it: the natural way a scientist would present the answer is to give an interval: i.e. we are confident that the ‘right’ answer falls within that interval. If that interval is too big, we improve our model and estimates of the inputs until the interval is sufficiently small for our purposes. Given the range of guesstimates of the height in the thread so far, I suggest just being able to narrow it down from "somewhere between 100 feet and 245 metres" would be a reasonable ask.

OK, so back to the problem. We solve for the height, speed or acceleration as a function of time by using an equation of motion. For the simple case of a fall in a vacuum, the only force acting is gravity. It acts on the mass, m, so the force is m*g. The equation of motion in this case is just Newton’s Second Law: F = m*a. Substituting for F gives the trivial (but insightful) result of:

m*a = m*g or just a = g

In other words the mass is irrelevant (look up feather and hammer dropped on the moon), and the acceleration is constant and equal to the gravitational acceleration. Of course, then we can use the formula for constant acceleration to find speed (integrate once) or distance (integrate twice), which is where s = u*t + ½*a*t^2 comes from.

As we’ve already said, however, this is not accurate for a non-vacuum. We need to add air resistance. This is a force which opposes the motion and is proportional to the square of the speed. i.e. double the speed, and the resistance increases by fourfold, and so on. We do this by adding a second term for the total force:

F = m*g – k*v^2

Where the k is a constant of proportionality that we don’t yet know. We can solve this equation using Newton’s second law again:

m*g – k*v^2 = m*a

However, this time, due to that pesky second term, the mass does NOT cancel. This equation is much harder to solve than before, because the rate of change of speed at any given time depends upon the speed itself, so it’s a sort of circular argument. This is a differential equation, and there are standard ways to solve it at advanced A level or degree level maths. However, this equation is exactly what the link I posted earlier solves. You could just as effectively plug the numbers into the ln(cosh) equation shown in the graphic, but this saves the effort.

So we know the time (7 seconds) and the mass, but we don’t know the air resistance coefficient. It turns out (standard physics) that this constant of proportionality can be written in terms of real physical parameters like air density, area, etc, and with a new constant: the drag coefficient: Cd, where:

k = ½*Cd*p*A

with p (rho) as the air density: let’s say 1.2 kg per cubic metre at sea level, and A is the cross-sectional area of the camera = width * depth.

In principle, Cd is a constant for any given shape, that we can look up. According to this link, the value for a rectangular box is 2.1. The cross-sectional area is a problem, however, because it depends upon the way the camera is pointing when it falls.

Here, we could increase the complexity of our model, and describe how the object tumbles. However, we could get an estimate of upper and lower bounds by looking at the extreme different orientations. The largest cross-sectional area it could present is 0.115 * 0.048 = 0.00552 m^2, whereas the smallest it could present is 0.048 * 0.04 = 0.00192 m^2.

Using the formula above, then, the largest value of k possible is:

k = ½*2.1*1.2*0.00552 = 0.00696

and the smallest value possible is:

k = ½*2.1*1.2*0.00192 = 0.00242

Using these values in the online calculator gives original heights of between 86.8 metres and 129.9 metres.

In old units, this value would be between 285 and 426 feet.

Of course, if we want a more precise answer, we need to put more accurate numbers in. Donald's post above is quite right: we could calibrate the drag coefficient experimentally, by dropping it from a known height, and then applying the same equations in reverse to find the drag coefficient more precisely.

So I apologise if anyone expecting a precise inch-perfect answer is disappointed by this result. However, I think we can be sure from this analysis that the initial height was more than the OP’s guess at 100 feet.

I hope this helps.

Edited By The Wright Stuff on 22/05/2017 10:39:56

[PETER BRUCE - Eastchurch Gap]

WoW Now that's what I call an answer.Thanks very much indeed for taking the time to present such a detailed examination. As the originator of the video and post I suspect I now need to go to Specsavers as it is clear I have really confused the issue with my initial 100 foot estimate...

Thanks again for answering the thread which I did feel had been left rather "in limbo".

Your a Star -

Regards Peter

[The Wright Stuff]

No problem Peter.

I have to admit, looking at the footage of the ground from the camera at the beginning of your film, it looks higher than 100' to me, so I'm reassured by my answer.

I forgot to mention that the same analysis predicts that the camera hits the ground at between 14.5 m/s and 24.4 m/s (32 mph to 55 mph). These sound fast enough to be believable, but slow enough to be survivable, although I must admit the bottom end of that range (and therefore the lower end of the height range) sounds more likely to me.

I will sign off by saying that although I am a professional physicist, I do not specialise in fluid dynamics so I may know a little more than many, but I am by no means an authority on this. It is possible to calculate the attitude of the camera if you could assume laminar flow, and knew the centre of mass (the air resistance force acts to rotate the camera about it's centre of mass) so the relative positions of the centre of pressure and centre of mass become important. Then there is the effect of the uneven surface of the 'box' producing eddies and turbulence.

Much of this is common to understanding the aerodynamics of a wing in flight, so I'm a bit surprised you didn't get more responses along these lines, but I hope this is sufficient for now! Happy flying!

[Josip Vrandecic -Mes]

It is a pleasant feeling that my smallness is in the company of decent and educated people .... even at the forum .
Peter & The Wright Stuff, Thanks for the right and complete information...Sir !

[Mike Etheridge 1]

I am also impressed Wright Stuff,and as I suggested my basic calculations were inadequate and were only useful to pass your 'O' or 'A' level physics or applied maths. Obviously other factors needed to be considered.

No doubt people involved with ballistics or the design of bombs and missiles would need to know about advanced calculations. However what is also evident is that experiments are also necessary regardless of calculation as was the case with BW's bouncing bomb.

[PETER BRUCE - Eastchurch Gap]

Hi Mike. Funny you should bring up BW and the bouncing bomb and it took many physical and frustrating tries before the actual skip method in practice worked and I am sure BW was a clever bloke even by modern standards. Theoretically and practical are often far apart but as was shown above you have to apply the variables which was the frustrating part if I remember with BW. Saw the film as many others no doubt have. Many lessons...

I still need to go to Specksavers😄

[David Ovenden]

I am glad this thread didn't just fizzle out. Maths never was my strong point but I always liked Physics. Thanks TWS for taking the time to respond. We may not have the definitive answer, but we (well, me anyway) learnt something from the process.

[onetenor]

I still think that the "HELICOPTERING " indicates drag of some sort and also stability of sorts so 100 ' is till plausible IMHO #

The fact it was "HELICOPTERING " indicates drag of some sort and also stability of sorts(. It wasn't tumbling  . means 100' to be quite feasible IMHO This is on reflection from my OP

John

[onetenor]

Sorry thought top message got wiped off

[The Wright Stuff]

Posted by onetenor on 22/05/2017 20:50:05:

I still think that the "HELICOPTERING " indicates drag of some sort and also stability of sorts so 100 ' is till plausible IMHO #

The fact it was "HELICOPTERING " indicates drag of some sort and also stability of sorts(. It wasn't tumbling . means 100' to be quite feasible IMHO This is on reflection from my OP

John

 

I agree that the 'helicoptering' is a worthwhile observation, onetenor, and while it certainly indicates drag and contributes to stability, I'm not convinced it actually produces lift.

I would concur that the helicoptering allows the orientation to present a bigger cross-sectional area to the air, and hence bias my calculation very strongly towards the lower end of the height range I calculated. So I'm going to still maintain that it was over 200 feet!

Edited By The Wright Stuff on 23/05/2017 10:06:12

[PETER BRUCE - Eastchurch Gap]

I have to say that as time has gone by and I have flow the model a lot more I still think my original assessments were about right. The rotation of the camera as it came down I think had a much greater effect than most credit but unfortunately as the act can't be reciplicated again the original video is all we have to go on. Got the grey matter going and I have not forgotten this issue as I can't remove the question from my mind as I hate things that appears to be simple but in actual fact are a right pain in the @£&?✈️

[PETER BRUCE - Eastchurch Gap]

Eight months now - and the question still pops up every now and them...🤔

[David Ovenden]

Peter, this was certainly one of the most interesting threads that I followed last year! It is interesting how perceptions of an event can vary widely and we can come up with very different "explanations" and rationales for them.

[PETER BRUCE - Eastchurch Gap]

Hi David. As I said before this question still comes back to haunt me which I find most annoying as it just “pops up” in my head every now and them and seems to have no trigger except the event itself. I am fully aware that with all the variables there are there is no definitive answer which makes it all the more annoying but the brain teaser is still teasing (annoyingly) me in particular. Glad you and others enjoyed the thread - it would seem (like me ) we find it of interest that so many variables produce such a wide range of answers... Regards Peter