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Re: Need Formula for length of spiral



Original poster: "Zoran Tukovic by way of Terry Fritz <twftesla-at-qwest-dot-net>" <ztukovic-at-globalnet.hr>

The formula is:

L = 1/2 * (2pi * N)^2 * A

Where is:
L = length of wire (in meters)
N = number of turns
A = space between turns (in meters)

Regards,
Zoran Tukovic



----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Thursday, February 14, 2002 7:36 PM
Subject: Re: Need Formula for length of spiral


> Original poster: "Jim Lux by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<jimlux-at-earthlink-dot-net>
>
> A further aspect which occurred when driving to work this morning..
>
> Consider the length of each turn (spiral or not, but concentric circles,
in
> any case).  They're going to be some form of 1+2+3+4+5+.... which is a
> N^2+N kind of thing...
>
> Another analogy (due to Leibniz, and the Chinese, much earlier). Imagine
> you are painting in a circle with concentric rings of constant width.  The
> area (or circumference, since it is constant width) of a given ring is
> proportional to its radius.
>
> You know that the sum of all areas of the rings is the area of the circle,
> which is proportional to r^2, so therefore, the sum of the circumferences
> must also be proportional to r^2.
>
> A spiral winding typically doesn't start at the center of the circle, so
> you're really working on an annulus (a circle with a circular hole in the
> middle), but the idea of dependence on r^2 still holds.
>
> And, just to beat this into the ground somewhat more, I considered the
case
> of the spiral where r = k * theta.  There are other spiral forms, which
may
> have different relations (because the assumption of equally spaced rings,
> as above, doesn't hold). (examples: r = r0*k^theta (equiangular/log
spirals))
>
> For those more analytically inclined..
>
> consider a small length of the winding ds.  ds = r*dtheta where r =
> r0+k*theta, so, the total length is
>
> integral[theta start, theta end] of (r0+k*theta)*dtheta
>
>
>
> Tesla list wrote:
> >
> > Original poster: "Jim Lux by way of Terry Fritz <twftesla-at-qwest-dot-net>"
> <jimlux-at-earthlink-dot-net>
> >
> > since the length is essentially integrating a linearly varying function
> > (radius) there will need to be some squared term in the equation.
Granted,
> > for small enough ranges of parameters, a linear approximation will
probably
> > work.
> >
> > ----- Original Message -----
> > From: "Tesla list" <tesla-at-pupman-dot-com>
> > To: <tesla-at-pupman-dot-com>
> > Sent: Wednesday, February 13, 2002 8:37 PM
> > Subject: Re: Need Formula for length of spiral
> >
> > > Original poster: "Steve Stuart by way of Terry Fritz
<twftesla-at-qwest-dot-net>"
> > <sstuart-at-glasscity-dot-net>
> > >
> > > Try:
> > >        L = (Do - Di) / 2 * 1.6 * pi * T
> > >
> > > Where:
> > >        L  = conductor length
> > >        Do = outside diameter
> > >        Di = inside diameter
> > >        T  = number of turns
> > >
> > > It will give you a pretty close approximation
> > >
> > > 73 de Steve
> > > ·¸¸·´¯`·¸¸·´¯`·¸¸·´¯`·¸¸·´¯`·¸¸·´¯`·¸¸·
> > > w8an-at-w8an-dot-net
> > > http://www.w8an-dot-net
> > >
> > > Tesla list wrote:
> > > >
> > > > Original poster: "John Tomacic by way of Terry Fritz
> > > <twftesla-at-qwest-dot-net>" <tesla_ownz_u-at-hotmail-dot-com>
> > > >
> > > > Hi everyone,
> > > >
> > > > Does anyone have a formula that I can use to calculate the length of
> > wire
> > > > required in a flat spiral coil? I have the formula for inductance,
> > however,
> > > > I really need the wire length.
> > > >
> > > > Thanks,
> > > >
> > > > John
> > > > SST coiling in Ottawa.
> > > >
> > > > _________________________________________________________________
> > > > Join the world's largest e-mail service with MSN Hotmail.
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> > >
>
>
>