An aircraft hangar is semi-cylindrical, with diameter 40m and length 50 m. A helicopter places an inelastic rope across the top of the hangar and one end is pinned to a corner, a A. The rope is then pulled tight and pinned at the opposite corner, B. Determine the lenghth of the rope.

So, first I find the diagonal line from A straight to B.


The answer is 64.03124......

Then, I find out that there is a semi-circle shape. So I find the length of the arc.

r=32.05162....... Length=2*Pi*32.05162..... =201.16008m......

But the correct answer is 80.30m

Can anyone tell me where i did wrong?

An aircraft hangar

  • 2
    $\begingroup$ The curve would only be a semicircle if it was wrapped round a sphere. your curve will be part of a helix. $\endgroup$ – Paul Oct 21 '14 at 11:55
  • $\begingroup$ I still don't understand how to solve the problem $\endgroup$ – Chun Jeff Oct 21 '14 at 11:58
  • $\begingroup$ @ChunJeff Still Don't understand? $\endgroup$ – user171358 Oct 21 '14 at 12:04

You could Construct a Right angle triangle with following dimensions

$$h=50$$ (Since It's Length of Cylinder) $$b=\pi\times20$$ (Since $20$ is Radius and If I'm not wrong Circumference of Semicircle would be $\pi r$)

so Length of rope will be $$L=\sqrt{h^2+b^2}$$ $$L=\sqrt{2500+400\pi^2}$$ $$L\approx80.30$$

Approximated Here

  • $\begingroup$ to find the circumference is 2*Pi*r , And how the radius is 20, i am so confused $\endgroup$ – Chun Jeff Oct 21 '14 at 12:05
  • $\begingroup$ @ChunJeff Because If You try to Flatten the Curvature You'll get a sheet of dimensions I mentioned in my answer $\endgroup$ – user171358 Oct 21 '14 at 12:07
  • $\begingroup$ If I'm not mistaken, It's semicircle. And It's diameter is $40$, So radius is $20$ $\endgroup$ – user171358 Oct 21 '14 at 12:11
  • $\begingroup$ Oh, i see what u mean thanks $\endgroup$ – Chun Jeff Oct 21 '14 at 12:12
  • $\begingroup$ If you're happy with my answer, You can express it by clicking on tick mark beside it. $\endgroup$ – user171358 Oct 21 '14 at 12:13

Hint: imagine the semicylindrical surface of the hangar as planar and apply Pytagora's theorem. If you flatten it to get a rectangle, the curve that you search becomes the diagonal of that rectangle.


This question is a simple Pythagoras problem. First, we find the length of the arc whose diameter is 40m. Then we assume it to be a side of the quadrilateral whose side is 50 m. Next, we use Pythagoras theorem to find the diagonal. The answer will be near to 80.34..

  • $\begingroup$ Really?, Yippie, I found a bug in Google Calculator $\endgroup$ – user171358 Oct 21 '14 at 12:06
  • 1
    $\begingroup$ He meant 80.34.. i guess $\endgroup$ – Chun Jeff Oct 21 '14 at 12:13
  • $\begingroup$ @ChunJeff How come Google always get answers right? $\endgroup$ – user171358 Oct 21 '14 at 12:14
  • $\begingroup$ I think You need to correct the mistake $\endgroup$ – user171358 Oct 21 '14 at 12:17

The rope does not form a semicircle. Rather it is an ellipse. So a simple semicircle calculation will be wrong.

The best suggestion above is to flatten out the roof of the hangar. Imagine just lifting the metal off and squashing it flat. You have a rectangle. Length = length of hangar. Width = half the circumference of the circle at the front of the hangar.

Then use Pythagoras to find the length of the rope, as diagonal of that rectangle.


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