## Maths formula needed please

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- Maths formula needed please

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Maths formula needed please

14-01-2012 4:54 PM

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coil former size is 150mm length, by 30 mm diameter

I need to have 150 turns of wire on the coil former.

wire size is 22SWG or 0.71mm

is there someone out there who can give me the formula to work out the length of wire needed, please.

Thanks.

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Re: Maths formula needed please

14-01-2012 5:06 PM

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Then 150 turns requires 150 x 30.71 x PI mm or 14.47 metres - say 15 metres to allow for it not being dead tight to the former and the helical effect

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Re: Maths formula needed please

14-01-2012 5:06 PM

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(or pi*d)

If the coil former is 150mm long you'll get 150/0.71 = 211 (roughly) turns along the length so no multiple layers to confuse things.

pi=3.1412 roughly

d=30mm (ignoring the wire thickness)

So 150*30*3.1412 = 14135mm

So less than 15 metres or 47 feet.

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Re: Maths formula needed please

14-01-2012 5:15 PM

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Re: Maths formula needed please

04-06-2012 11:52 AM

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I have been struggling to do a "reverse" of the formula given....

I now need to know the length of the former required to acommodate a specified length of wire....

and despite several/many many tries to change the formula layout -- transferring known quantities across the = sign and dividing etc.... I don`t seem to be able to get a coherent answer....

so here are the known quantities

Length of wire..... 23.35 feet (7.11708 mtr)

Dia of former..... 27mm

(number of turns = irrelevant in this case)

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Re: Maths formula needed please

04-06-2012 12:49 PM

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You get the length of wire per winding (27 + 0.71) x 3.1416 = 87.05 mm where 0.71 is the diameter of the wire

Number of windings 7117 / 87.05 = 81.75 mm - assume 82 mm

Length of former 82 x 0.71 = 58.2 mm

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Re: Maths formula needed please

04-06-2012 12:59 PM

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I don`t need the number of windings... (turns) I need to find the Length of the Former for the specified amount of wire...(23.3 ft or 7.1017mtr) (sorry about the mm error)

so what is the formula, using the given amounts above....?

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Re: Maths formula needed please

04-06-2012 1:25 PM

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To combine the whole lot together you get

Length of former = Diameter of wire x number of windings = Diameter of wire x (length of wire / length per winding) = 0.71 x (7117 / ((27 + 0.71) x 3.1416)) = 58.04 which is a bit less than previously as you need to round up the number of windings which in Excel means that the formula is

+0.71*round((7117 / ((27+0.71)*3.1416)),0) giving the result 58.22

But it is much better to do it in steps

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Re: Maths formula needed please

04-06-2012 2:03 PM

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