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### Circumference of a Tube

```
Date: 10/11/2001 at 14:29:20
From: Ernie
Subject: Circumference

How can I predict the circumference from a tubed length of stock?

Maybe a different way to ask the same question would be, What effect
does the thickness of a tube have on the circumference when it's
formed into a circle?
```

```
Date: 10/11/2001 at 19:32:48
From: Doctor Douglas
Subject: Re: Circumference

Hi Ernie, and thanks for writing.

If we bend a thin wire of negligible thickness with length L into a
circle, the circumference of the formed circle is L and its radius
(the distance between the center of the circle and the wire) is
L/(2*pi).

Say you cut a length L of tube from stock. Let's say that the stock
has diameter D. What I mean by this is that the overall diameter of
the tube (including the interior and the wall thickness) is D.
Or to put it another way, its cross section has diameter D. Now if we
bend this tube around into a circle, we expect that to a first
approximation, the circumference of the formed circle is approximately
equal to L.

Now, as you bend the tube around into a circle, the material has to
deform somewhat in order to make the ends join up. In particular, the
part of the cross-section that is close to the center of the circle
[i.e. at radius r < L/(2*pi)] is compressed, and the remainder of the
material [r > L/(2*pi)] is stretched. What happens to the actual
circumference depends on the shape of the cross section. In the case
of a circular cross section, the compression and the stretching
exactly cancel (to a first approximation), and the result is that
along the axis of the tube, there is no change in length: in other
words, the measured distance along the tube axis is still L, just as
in the case of the wire.

Now we can also see that the distance from the center of the object
to its _outermost_ edge is L/(2*pi) + D/2.  The circumference measured
at its largest point is (2*pi) times this, or L + D*pi.

- Doctor Douglas, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 10/12/2001 at 13:21:03
From: Ernie
Subject: Re: Circumference

I think I understand most of your answer. For example, if tubed stock
(rubber) = 36" in length (L) and 3" in width (D)...

L/(2*pi)+D/2 = 7.2297" center to outside edge (r)
Cir = (2*pi)(7.2297)
Cir = 6.2832*7.2297
Cir = 45.43"
```

```
Date: 10/12/2001 at 14:23:51
From: Doctor Douglas
Subject: Re: Circumference

Yes, your formula is absolutely correct.

C = 2*pi*[L/(2*pi) + D/2]
= L + D*pi

For the values in your problem, L = 36.0, D = 3.0,
so C = 36.0 + 3.0*pi = 36 + 3*3.1416 = 45.4 inches, which
agrees with what you calculated above.

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

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