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### Hose Length

```
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/
```
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry

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