Math CentralQuandaries & Queries


Question from Huw:

hello my name is Huw,
My problem is very similar to one on your data base, however when trying to calculate it i could not work out the answer.
the problem is as follows:
two gate posts 48cm apart and a perfect arc of steel is to be made joining the two posts on top. this steel arc is to measure 12cm higher than the top of the gate posts in its centre point.
the question similar to this was asked by "daryl" and answered by "Penny"
If you could reply to this i would be very grateful.
Kind regards Huw


Your question is very much like the question that Daryl asked a while ago.

I drew a diagram and labeled some points. The dimensions are in centimeters.


C is the centre of the circle that forms the arc and r is its radius. The triangle ABC is a right triangle, the length of AB is 24 cm, half the width of the arch, and the length of BC is r minus the the height of the arc above the top of the posts, which you want to be 12 cm. Thus, using Pythagoras' theorem

(r - 12)2 + 242 = r2

Expanding gives

r2 - 24r + 144 + 576 = r2

and hence

24r = 720 so that r = 720/24 = 30 cm.

Hence the length of BC is 30 - 12 = 18 cm and the angle BCA is cos-1(18/30) = 0.9273 radians.

The length of an arc is r times θ where θ is the angle that subtends the arc, measured in radians. Thus the length of half your arc is 30 times 0.9273 = 27.82 cm. Hence the arc at the top of your arch measures
2 times 27.82 = 55.64 cm.

I hope this helps,

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