Hello, I am e new forum member, with only one question: a Lancaster bomber flies at abou 16600 feet with a speed of about 240 mph and its nose down about 10 degrees. At that point it looses one of its engines (plus propellor). At what distance from the point down on the ground where the aircraft lost the engine wil the engine hit the ground? I understand that you are experts on ballistics of shooting, but I hope you can help me with this calculation. Thanks, Jan
Jan,
interesting question, could be a question of my father too! I think I can make an educated guess and produce a spreadsheet for you.
However, I have first to pounder about the issue - what is the drag of an motor, and then make some calculations. This costs me about one or two weeks - I have so many other things to do - but I will do it.
It would greatly help if you have some data from similcar cases, even bomb trajectories.
Robert
Dear Robert,
Thank you for your reply. I might be of your fathers age. I tried the use of formula's I found at Wikipedia (try Google with 'ballistiek'): (distance over the ground) Sx(t)=v.cos(a).t and (height) Sy(t)=1/2g.t2 . Because I have the height and 'dropincrease' (g) (= valversnelling = 9,81 m/s2), I can calculate t and apply that in the distance formula. The outcome is about 1000 metres. But this might all be too rough and wrong, so when I found your BfX site through Wikipedia, I gave it a try, realising that gunshooting is not 'engines falling of airplanes'.
I appreciate your input, kind regards, Jan
Jan, after some thoughts, it isn't too difficult.
You made an error in your calculations. The better formula for the height as function of time : z(t) = z0-vz*t-0,5*9,81*t^2
vx=106 m/s
vz=18,6 m/s
In absence of air resistance the engine falls 30,2 s and travels 3198 m
With air resistance ... I made the following calculation ...
the engine falls about 34 s and travels about 2820 m
(It was done by a computer program that is not downloadable)
included a spreadsheet with the calculations
Robert
m m m s m m/s m/s m/s
x y z t v vx vy vz
0 0 0 5059,68 0 107,2896 105,6596299 0 -18,63064352
1 200 0 5006,632916 1,91011851 110,0991639 103,7476291 0 -36,85451598
2 400 0 4917,130465 3,856500547 115,684911 101,7493745 0 -55,04601183
3 600 0 4789,722193 5,842786049 123,5918684 99,61176624 0 -73,16041258
4 800 0 4622,758705 7,874016067 133,3068689 97,28770152 0 -91,13629582
5 1000 0 4414,321708 9,956771129 144,3379402 94,73571952 0 -108,8971277
6 1200 0 4162,139533 12,09939126 156,2491124 91,91865116 0 -126,3516787
7 1400 0 3863,480602 14,31232618 168,6631164 88,80195378 0 -143,3926771
8 1600 0 3515,013049 16,60867491 181,2484026 85,35206006 0 -159,8937437
9 1800 0 3112,612007 19,00500448 193,7007394 81,53484683 0 -175,7044257
10 2000 0 2651,084582 21,5225951 205,7235961 77,31419936 0 -190,6428928
11 2200 0 2123,761651 24,18937213 217,0078103 72,65058728 0 -204,4854075
12 2400 0 1521,865416 27,04300366 227,2089519 67,49953732 0 -216,9509629
13 2600 0 833,4792567 30,13609712 235,919091 61,80985336 0 -227,6781929
14 2800 0 41,76410461 33,54545389 242,6270797 55,52136465 0 -236,1890723
Dear Robert, thank you for all this work, I do understand the outcome, although it seems a bit a long distance between losing engine and impact. I see answers in m/s en metres. Questions: did you recalculate airplane speed from mph (1,6 km/u) in km/u or m/s, and airplane height from feet to metres (0,3048)? And did you incorporate the 10 degrees nose down angle of the airplane? The engine with propellor will have an enormous drag, does this influence the outcome, or is it incorporated in any formula?
Kind regards, Jan
Jan,
working in the metric system has many advantages, so I did convert everything to that system first (see the Excel spreadsheet that was included in the previous post):
convert to metric velocity components
height dropped 16600 ft 5059,68 m vx= 105,6596299 m/s
velocity dropped 240 mph 107,2896 m/s vz= -18,63064352 m/s
angle -10 degrees -0,174532925 rad
Then without air drag one can almost compute the distance where the engine hits the ground without any computers ... if one rounds off the numbers t^2=5000 *2 /10=1000, t=32s. Distance travelled=32*100=3200m
The 10 degrees is taken in account (vx vz), but that does not changes the impact zone much.
More accurately, but still for the case of no airdrag, one has to solve an equation like
z(t)=5059-18,6*t-0,5*9,81*t2
if you do that then
Z(t) z -5,50723E-06 m t 30,27447353 s
x(t) 3198,789668 m
What about air drag?
I have made an estimate of what the ballistic coefficient (measure of drag) could be, put it into the computer program and got the table. I changed the values of the coefficient and looked at their effect. Then, convinced, I fixed the coefficient and presented the results in the previous post.
This is the educated guess, there could be some more drag, but there has to be a lot of drag before the engine hits the ground at 2km. If the motor started spinning then this might be the case. However, I think that it is very likely that you find the motor at 2,5-3 km.
Where was the motor lost?
And if the engine does not deforms hitting the ground, it could travel as much as 16 m in the soil!
(based on a very naive estimation).
Dear Robert, thank you very much for all your input and work. I have one final request: the airspeed I gave you is based on some technical data: a Lancaster with bombload had an indicated (cruise) airspeed of about 180mph, during this specific flight there was a tailwind of about 50mph, so the speed I gave you was the estimated groundspeed (with a 10 degree nose down).
These data are all my estimates, I was not on board, it could be that I am wrong. But the aircraft was already hit by a nightfighter and by flak, it had problems with two engines, there were fires on board and than the engine in question fell off. So the indicated airspeed might have been much lower.
Therefore I would like to ask you to do your last calculation again, with all the same parameters, but with another (ground)speed: 200 mph.
For the moment I think it is necessary to keep you as 'neutral' as possible, but I will tell you the story behind my request after this. Thank you, kind regards,
Jan
200 mph
without drag 2692 m
with drag 2387 m
x y z t v vx vy vz
0 0,00 0,00 5059,680 0,000 89,408 88,050 0,000 -15,526
1 200,00 0,00 4998,807 2,292 94,227 86,438 0,000 -37,513
2 400,00 0,00 4885,409 4,629 103,463 84,688 0,000 -59,436
3 600,00 0,00 4717,281 7,018 115,917 82,721 0,000 -81,203
4 800,00 0,00 4491,775 9,469 130,467 80,473 0,000 -102,692
5 1000,00 0,00 4205,627 11,994 146,228 77,893 0,000 -123,754
6 1200,00 0,00 3854,735 14,610 162,521 74,936 0,000 -144,214
7 1400,00 0,00 3433,860 17,340 178,802 71,558 0,000 -163,859
8 1600,00 0,00 2936,196 20,211 194,594 67,713 0,000 -182,433
9 1800,00 0,00 2352,714 23,262 209,427 63,355 0,000 -199,615
10 2000,00 0,00 1671,090 26,546 222,788 58,425 0,000 -214,990
11 2200,00 0,00 873,873 30,139 234,058 52,859 0,000 -228,011
12 2400,00 0,00 -64,949 34,162 242,440 46,577 0,000 -237,923
Dear Robert, thank you very much, this is of great help. The syory behind my request concerns a Lancaster from 101 Squadron, that was shot down during the Nürnberg Raid 30/31 March 1944. It crashed near Eisenach and with its full bombload still on board it exploded with great force just before impact, killing all crew. They found only pieces. There are, however, stories from locals that mention an engine lost before the explosion. I have been at the crash site a couple of times. A small monument was erected in 2008 by local investigators. The story has intriged me for a couple of years now. Again, thanks for your kind help, can I do anything in retrun? Kind regards :-\, Jan