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Subject: Re: Coil winding formula ?
From: ChrisAtUpw@.......
Date: Mon, 24 Jul 2000 11:53:23 EDT

Dear Arie,

<< Hi, A long time ago, far, far away, I used to remember a simple formula
 that gave the length of wire for a coil, given the number of turns, internal
 diameter, length of coil and gauge of the wire. But alas, I can't remember
 it, so if anyone knows such a "winding" formula, I would be most 
    The fields on the coil axis are easily calculated. Off-axis fields 
involve solving elliptic integrals. I will give the field H in Oersteds, all 
dimensions in cm, the current i in amps and Pi= 3.142. To change from 
Oersteds to Tesla, multiply by 10 ^ -4

    For infinite solenoids H = 0.4 x Pi x N x i       N is in turns per cm

    For shorter solenoids H = 0.2 x Pi x N x i x (cos(theta1) - cos(theta2))  
  N is in turns / cm. theta1 and theta2 are the angles between the axis and 
the ends of the coil at the point being considered. You may need to be a bit 
careful if using a computer programme to check how theta = arctan(a/d) is 
evaluated as d goes through zero and becomes negative at the end of the coil, 
where a is the coil radius and d is the disance from the point considered to 
one end of the coil. 

    For short large diameter coils you have a choice. H = 0.2 x Pi x Nt x i x 
sin(theta) / a    where Nt is the TOTAL number of turns and theta is the 
angle the coil makes with the axis at the point d from the coil centre. The 
alternative formula involves substituting a and d to give sin(theta). H = 0.2 
x Pi x Nt x a^ 2 x i / (a^2 + d^2)^1.5  The '^' is used for 'raised to the 
power of' and 'x' for 'multiplied by'.

    For Helmholtz coils, if EACH has Nt turns separated by distance a between 
 the coils, the formula is H = 3.2 x Pi x Nt x i / (5^3)^0.5 x a  where 
(5^3)^0.5 = sqrt(125) 

    For the length of wire used, this is 2 x Pi x a x N per cm for a solenoid 
or 2 x Pi x Nt x a for each Helmholtz coil. 

    For a short fat close wound coil of length l, internal radius a1 and 
external radius a2, the wire length is 2 x Pi x N  x l x (a2 - a1) x N x (a1+ 
a2)/2 = Pi x N^2 x l x (a2^2 - a1^2)

    If the dia of the wire is d cm and the coil is close wound, N = 1/d turns 
/ cm, but 'achieved values' may be a few % less.

    Always allow some excess for connections at the ends. The dia of 
enamelled wire is usually quoted as the dia of the copper, but the thickness 
of the enamel or the actual turns / cm, in, etc may be quoted. For fine 
cotton coated enamel wire, I suggest that you see just how many turns you can 
actually wind per cm. I found that it varies a bit. If you 'wave wind' the 
coil for low interwire capacitance, you need to be able to estimate the 
'filling factor', which is determined by the coil winding machine. Check in 
the instruction manual.

    Hope that this helps,

    Chris Chapman

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Larry Cochrane <cochrane@..............>