[Prev][Next][Index][Thread]
RE: formulas (fwd)
---------- Forwarded message ----------
Date: Thu, 16 Jul 98 20:01:01 -0400
From: B**2 <bensonbd-at-erols-dot-com>
To: tesla-at-pupman-dot-com
Subject: RE: formulas (fwd)
Antonio Carlos M. de Queiroz, Terry, All,
Ever come across the derivation for the exact solution to the
differential equations for tuned coupled circuits in mathcad?
Barry
-----Original Message-----
From: Tesla List <tesla-at-pupman-dot-com>
Sent: Thursday, July 16, 1998 12:03 AM
To: tesla-at-pupman-dot-com
Subject: Re: formulas (fwd)
---------- Forwarded message ----------
Date: Tue, 14 Jul 1998 22:29:44 -0700
From: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br>
To: Tesla List <tesla-at-pupman-dot-com>
Subject: Re: formulas (fwd)
Jim Monte wrote:
> The formulas you gave are accurate, but not exact. If you compute
> the arc lentgh for the wire, you come up with a log term. The
> program I sent in an earlier post uses the exact formula and works
> for soleniodal coils, pancakes, or anything between these, including a
> straight piece of wire.
I keep that my formulas are exact. For a solenoid it is just a question
of "unrolling" it. The expression gives the diagonal length of a rectangle
with width n*2*pi*r and heigth m. The formula is correct for any n, r, or m. I
verified the formula with a line integral too:
Wire length=sqrt((n*2*pi*r)^2+m^2)
The formula for the spiral coil I derived with an integration (rather long
to post here), and appears to work correctly for any n, r1, or r2:
Wire length=sqrt((n*pi*(r1+r2))^2+(r2-r1)^2)
I didn't try a derivation for a conical coil. Maybe this case includes a
log term, or maybe we are talking about the same expressions written in
different ways. I would guess that the general formula is (not checked,
but reduces to the other cases with r1=r2 or m=0, and works for n=0):
Wire length=sqrt((n*pi*(r1+r2))^2+(r2-r1)^2+m^2)
Can you comment on the derivation of your expression (off list if there is
too much math)? I can't see from where the expressions came.
Antonio Carlos M. de Queiroz