# Spiderweb Coil Design

# Spiderweb Coil

Given desired inductance, how to use a KL-1 slide rule to find number of turns needed to construct a spiderweb coil for receiving shortwave radio signals.

Scroll to **Eqn. 6** at the bottom to skip all of the derivation.

## Basic quartic equation

**Eqn. 1**

**Eqn. 1**

where:

x = number of turns we're solving for

w = diameter of wire (in inches)

r = inner radius of coil (in inches)

L = desired inductance of coil (in uH)

*See this page** for a bit of background about where this equation comes from.*

### First, we make some assumptions about the coil form size & wire gauge

r = 1 inch (small coil for say, shortwave radio bands)

w = 26 AWG magnet wire (about .02 inch diameter)

So, w = 0.02 and w^2 = 0.0004

Eqn. 1 becomes:

**Eqn. 2**

**Eqn. 2**

## Further simplification for shortwave radio coils

A neat trick for some shortwave radios is that the coil can act as both the antenna AND the inductance for the "LC tank" to home in on the desired frequency. We use a variable capacitor (either air variable, or varicap diode tuned with a varying reverse bias voltage) and have a little spiderweb coil to form the oscillating tank and the few feet of antenna to capture the signal.

Assumptions:

- for low number of turns (say < 30) for coils with up to 50uH inductance

- and thin wire about 26 AWG (~ .02 inch diameter)

- we can ignore the (4x10^-4)x^4 term

From graphs of the equation (see below), we can see that the 4th power term can be ignored for the realistic root that interests us (the zero-crossing for positive x values).

The blue curve is the full quartic equation with all terms. The red curve is the equation with the 4th power of x dropped. They only differ significantly for large negative x values (negative number of coil turns!?). Note: these particular graphs are for the above choices of coil inner radius (r) and wire diameter (w), as well as a nominal choice of desired inductance of 10uH -- we can see that this calls for about 10 turns of wire in the coil (when we're using google search to graph it for us instead of our trusty slide rule ;)

**Left:*** wide view of graphs, ***Right: ***zoomed in on positive values of interest*

With that assumption, Eqn. 2 becomes:

**Eqn. 3**

**Eqn. 3**

Which is now a cubic equation that will be easier to handle on a slide rule.

## Finding the root(s) of this cubic Spiderweb Coil Equation

nsg.upor.net/slide/sreq.htm shows a method for finding the roots of a generalized cubic equation of the form:

First, we divide both sides by 8x10^-2 to get an x^3 coefficient of 1, yielding:

**Eqn. 4**

**Eqn. 4**

Then, we massage it into the form:

**Eqn. (a)**

**Eqn. (a)**

by substituting in

which will always yield a new equation with no x^2 component (a so-called "depressed cubic"). We'll just have to remember to subtract B/3 from the root we find below in order to get the true root of the original cubic equation for the number of turns in the coil.

As the link above shows, this can be rewritten into the form:

**Eqn. (b)**

**Eqn. (b)**

which is solvable on a slide rule using a "search" method.

Note: I'm not a mathematician, but a bit of research shows this type of curve is known as "Newton's Trident" AKA "Cartesian Parabola", after Descartes, who also called it a "Parabola of the Second Order".

This gives (after collecting terms and naming 'y' back to 'x'):

**Eqn. 5**

**Eqn. 5**

And dividing each side by x:

**Eqn. 6**

**Eqn. 6**

Now, finally, we can choose a desired inductance, do a little addition/subtraction/approximation in our heads and find the number of coil turns required!

Here's the "program" to run on our analog computer (slide rule):

## Find Root of Massaged Cubic Equation

## Example

If we want to construct a small spiderweb coil with an inductance of 4.5uH, we set L=4.5 and **Eqn. 6** becomes:

q = about 8585

p = about 900

On our KL-1 slide rule,

1) bring 8.59 on C under the backside DI scale's index/pointer

2) search with the arrow/cursor for where the value on A and DI add up to 900

3) under the arrow, read the value on the C scale

4) subtract the B/3 (= 16.7) value we used above to alter the equation.

Doing this, we read a value of about 23 on C (A of 520 + DI of 380 = 900).

### 23 - 16.7 = 6.3 turns

Note that there's also another root that can be found, with the cursor over 120 on the A scale, 790 is on the DI scale, also giving the sum of 900. And on the C scale just below 120, we find the square root equal to about 11. But remember that this massaged equation (to remove the x^2 term) means we have to subtract 16.7 from any roots we find, meaning 11 is actually 11-16.7 = -5.6. Yes, -5.6 turns of wire. ;)

You can see the roots in the two graphs below, with the original cubic equation and the massaged equation with x^2 removed and both sides divided by x to get a "Newton's Trident" equation from the cubic. In the latter, we see the actual answer (23-16.7=6.3 turns) along with the "virtual" root trying to join the party.

Original cubic equation

Massaged "Newton's Trident" equation