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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>
Here's another way..
At some point, the turn lengths will be a series 1,2,3,4,5,6,7,8,...,N
(perhaps with an offset, but still a simple arithmetic progression).
The average length of a turn will be (N+1)/2 ( for instance, average of
(1,2,3,4,5) is 3 (or (5+1)/2)
The total length of the spiral will then be N (number of turns) * Lave
(average length of turn), so N* (N+1)/2 which just so happens is the sum of
integers from 1 to N. Interestingly, Gauss' recognition of the pairing you
pointed out enabled him to calculate the sum of 1 to 100 so quickly that it
got him beaten for cheating in school. That is, the sum of 1 to N is
(1+100) + (2 + 100-1) + (3 + 100-2).... + (50+ 100-49), or 50 * (101) =
5050. He didn't actually do 99 additions, which the teacher had expected.
For a close wound coil, the length of the wire would vary as r^2, just
because it is constant width, and the area of the circle covered varies as
r^2. The area covered by a wire is just L*W... So Area=pi * r^2 =
Lwire*Wwire. Rearranging: LWire = pi * r^2/Wwire.
With respect to the division into rings, the unstated assumption was that
all the rings are the same width. If the width is allowed to vary, then all
bets are off.
If anyone is winding equiangular spirals (an interesting construction
exercise, and it might look real cool.. something to think about), then the
length probably doesn't vary as r^2, but as something else. You're
essentially looking at integrating k^(theta) dtheta instead of theta dtheta.
Make use of b^x = exp(x*ln(b)) so now you can integrate using integral(
exp(ax) ) =1/a*exp(ax), where a=ln(b)..
length of wire for a log (equiangular) spiral is therefore proportional to
exp(N)
Tesla list wrote:
>
> Original poster: "Pete Komen by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<pkomen-at-zianet-dot-com>
>
> Sorry Jim, but the r^2 doesn't make sense to me. I suggested that the
> length of the wire is PI * Average diameter * number of turns. Take the
> case of three turns: the center turn length is PI * Diam., the outer turn is
> PI * (diam + delta); the inner turn is PI * (diam - delta). Add them all
> together and the + and - PI * delta terms cancel leaving 3 * Diam * PI.
>
> R^2 gives units of area (a unit of length squared). For a close wound coil
> the area of the windings divided by the width of the wire would give a close
> approximation of the length but not r^2. Note that the width is arbitrary.
>
> You said, "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."
>
> I agree with the first part, but the conclusion is wrong. Consider your
> painting with three wide rings of constant width. The total length of the
> rings is proportional to r^2/width (of the ring). Since the width term is
> not fixed your conclusion is invalid. In any circle of radius R, any
> arbitrary number of rings could fit, thus making the total length (or
> average circumference) of the rings arbitrarily different from R^2.
>
> Regards,
>
> Peter Komen
>
> -----Original Message-----
> From: Tesla list [mailto:tesla-at-pupman-dot-com]
> Sent: Thursday, February 14, 2002 11:37 AM
> To: tesla-at-pupman-dot-com
> 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.
> > > >
> > > > _________________________________________________________________
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> > >