Taping a CylinderDate: 01/29/2001 at 05:28:37 From: Gerald Robinson Subject: Helixes If I want to wrap sticky tape around a cylinder to cover it, what is the relation between the diameter of the cylinder, the thickness of the tape, and the angle between the diameter of the cylinder and the length of the tape? Date: 01/29/2001 at 12:31:33 From: Doctor Rob Subject: Re: Helixes Thanks for writing to Ask Dr. Math, Gerald. I drew the following diagram of the cylinder sliced up one side PQ and unrolled: p' U' Q' o---------o---------o---------o--o |\ \ \ \ | |A\ \ \ \| | \ \ \ _,o T | \ \ \ _,-' | | \ \ S o' | Pi*d | \ \ \ | | \ \ \ | | w _,o R \ \ | | _,-' \ \ \ | o---------o---------o---------o--o P h U Q The relation between the width of the tape w, the diameter of the cylinder d, and the angle A is given by sin(A) = w/(d*Pi), A = arcsin(w/[d*Pi]). The total area of the tape must equal the lateral area of the cylinder, which is Pi*d*h = L*w, if you don't waste any tape (possibly by cutting a triangular piece PP'R off at the beginning and using it at the end to make UQTS and TQ'U'). Here L is the length of the piece of tape, and h is the height of the cylinder. Thus L = Pi*d*h/w. Probably you would waste the pieces adjacent to PP' and QQ' whose total area is PR*P'R = w*sqrt(Pi^2*d^2-w^2), using one continuous piece of tape, so that Pi*d*h = L*w - w*sqrt(Pi^2*d^2-w^2), = w*(L - Pi*d*cos(A)), L = Pi*d*h/w + sqrt(pi^2*d^2-w^2). If you need further assistance, write again. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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