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### Length of a Copper Helix

```
Date: 11/15/98 at 10:35:38
From: Roland Kim
Subject: Calculating the length of two twisted copper wires

Hi Dr. Math,

I have been searching for a formula I could use to determine the length
of straight wire required if it is to be twisted together with an
identical piece of wire to yield a helix pair of certain length.

In other words, if I have two pieces of wire that were .040 inch in
diameter, and they are twisted together so that the distance in one
complete spiral is 1 inch, and the total linear length of the twisted
pair is 100 feet, how long does each piece of wire have to be to meet
this requirement?

I have searched through some textbooks, but I can only find formulas
that show how to locate points on a helix. If you can point me in a
particular direction on how to find a solution I would be grateful.
Thanks.

Roland
```

```
Date: 11/16/98 at 12:19:55
From: Doctor Peterson
Subject: Re: Calculating the length of two twisted copper wires

Hi, Roland. If we picture one of your wires making a helix of radius R,
with one turn taking H inches, it will look like this:

|<--R-->|
|       |
******|****** |
**             **
| ******o****** |----
|      o|       |   ^
|     o |       |   |
|.  o   |       |   |
ooo.    |       |   |
|    .  |       |   |
|      .|       |   |
|       |.      |   |
|       |  .    |   |
|       |    .  |   H
|       |      .|   |
|       |       |   |
|       |       |   |
|       |       o   |
|       |       o   |
|       |      o|   |
|       |     o |   |
| ******|****o* |   |
**      |  o   **   v
******o******------

The length of the wire can be found by unwrapping it from the cylinder
to form a right triangle:

o
o  |
o     |
o        |
o           |
o              |
o                 |
o                    |
o                       |
L  o                          | H
o                             |
o                                |
o                                   |
o                                      |
o                                         |
o                                            |
o                                               |
o                                                  |
o                                                     |
*********************************************************
2 pi R

We'll get:

L = sqrt(H^2 + (2 pi R)^2)

as the length of one turn of wire, so the ratio of the length of the
twisted pair to the length of the wire will be:

L / H = sqrt(1 + 4 pi^2 (R/H)^2)

The radius of the cylinder about which the center of the wire will
spiral will be about the same as the radius of the wire itself:

********   oooooooo  ********
***        ooo       *ooo       ***
**         oo   **   **    oo        **
*          o       * *        o         *
*          o         *    R     o         *
*          o         *----------o         *
*          o         *          o         *
*          o       * *        o         *
*          o     *   *      o         *
****       oooo*     **oooo      ****
********   oooooooo  ********

So in your example, with R = .020 in and H = 1 in, we get:

L / H = sqrt(1 + 4 pi^2 .020^2) = 1.0079

and for a 100 ft pair, each wire is 1.0079 * 100 ft = 100.79 ft.

That doesn't sound like much difference. It will increase if the wires
don't touch tightly all the way around. You may have to adjust the

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Euclidean/Plane Geometry
High School Geometry

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