Date: 07/11/99 at 23:29:29 From: Trotter Terry Subject: Circumferences What is the formula to calculate the length of a 1" OD hose that is wrapped 50 times around a 10" diameter cylinder without riding up the cylinder - in other words it stays flat, with the circumference growing larger with each wrap? Thanks for the help!
Date: 07/12/99 at 08:33:13 From: Doctor Rick Subject: Re: Circumferences Hi, Trotter. Thanks for your question! If I understand you correctly, the hose forms a 2-dimensional spiral like a watch spring, not a 3-dimensional helix like a garage-door spring. (We've gotten more questions about the helix; this is the first I've seen about the spiral.) I can give you a good approximation by considering the area covered by the hose, considering it in 2 dimensions. The thickness of N turns of a hose of diameter d is Nd. The outside diameter of the area covered by the hose is D + 2Nd, where D is the diameter of the cylinder that the hose is wrapped around. The area covered by the hose is the area enclosed by the outside circle minus the area of the inner circle: A = (pi/4)((D + 2Nd)^2 - D^2) = (pi/4)(4NdD + 4N^2d^2) = pi*Nd(D + Nd) where D^2 means D squared. The length of the hose is this area divided by the width (diameter) of the hose, d: L = A/d = pi*N(D + Nd) Putting in the numbers (d = 1", D = 10", N = 50), we find that L = 3.14*50(10" + 50") = 9420" = 785' I hope this answers your question and others like it, since I've given you a formula. If it raises further questions, feel free to ask. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
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