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### Length of Coiled Belt

```Date: 04/11/2003 at 23:36:31
From: Manny Rivas
Subject: Length of coiled belt

Working with conveyor belts we use a formula that gives a very close
approximation of the length of a coil of belt, but I would like to
know the reasoning of it. The outside and inside diameters of the coil
(in inches) are added and then multiplied by the number of wraps, this
is then multiplied by a constant (.1309), and the result is to be read
in feet.

It works, but I would like to know why. I tried to reason it out but I
am stumped.
```

```
Date: 04/12/2003 at 23:38:30
From: Doctor Peterson
Subject: Re: Length of coiled belt

Hi, Manny.

We have a number of answers to questions about the length on a roll
of some material; one of them may give a formula like this based on
the number of turns.

The cross sectional area of the coil is the difference between the
circle formed by the outer diameter, D, and that formed by the inner
diameter, d:

A = pi (D/2)^2 - pi (d/2)^2 = pi/4 (D^2 - d^2)

The cross-sectional area of the belt straightened out is

A = TL

where T is the thickness and L is the length. Setting these equal,

TL = pi/4 (D^2 - d^2)

But T can be found by dividing (D-d)/2 by the number of turns N:

(D-d)/(2N) L = pi/4 (D^2 - d^2)

Solving for L,

L = pi/4 2N (D^2-d^2)/(D-d)

Since D^2 - d^2 = (D-d)(D+d), this is

L = pi/2 N (D+d)

That is your formula; to convert inches to feet we have to divide by
12, and (pi/2)/12 = 0.1309.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry

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