Inductance of a Layer Wound Coil 
Written by Bryce Ringwood  
This program provides a means of determining the number of turns required, for a given inductance of a multi layer coil. Accurate to about 1% for normal not thin not fat coils. UsesUsed to calculate coils for radio circuits in the LFMF range.
Data Entry SectionEnter L or N as zero. Press "Calculate" Button to get the result. In this program, SI suffixes can be used on input, and are used on output. For example, you can enter 1k for 1000 Ohms  and so on. !Be a bit careful here 10m = 10 millimetres; 500u=0.5m=0.5 millimetres; 2.5u=2.5 microHenries.! Note that you might have to mess about to find the outside diameter. Remember that a ferrite core can double the inductance.
Additional Results
Example  450kHz I.F Coil for a Valve Communication ReceiverA coil has to be rewound for a 450kHz IF transformer for a short wave radio. The required inductance is 62.5µH. The coil former has an outside diameter of 7.6mm. How many turns are needed at a winding length of 12.5mm and an estimated outside diameter of 11mm ?  See the sample data. The inductance required has been divided by 1.5 because the coil has a ferrite core. TheoryThere is no theoretical basis for the following formula, as far as I know.
The formula was named after a Harold A Wheeler who worked it out empirically.
Formulae\displaystyle{{L} ={ {(31.6*{N^2}*{Ra^2})}\over {6*Ra + 9*Len + 10*(Rb  Ra))}}}
The above formula for Inductance in µH is in metric units. $Len$ is length of winding pile.
