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 appreciative. 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 __________________________________________________________ Public Seismic Network Mailing List (PSN-L)

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