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# Re: Help with this equation!

```Subject:  Re: Help with this equation!
Date:   Fri, 30 May 1997 20:56:27 +0000
From:  "Bert Pool" <bertpool-at-flash-dot-net>
To:  Tesla List <tesla-at-pupman-dot-com>

> Date:          Fri, 30 May 1997 09:25:06 -0500
> To:            tesla-at-poodle.pupman-dot-com
> Subject:       Re: Help with this equation!
> From:          Tesla List <tesla-at-pupman-dot-com>

> Subject:      Re: Help with this equation!
>        Date:  Fri, 30 May 1997 10:45:29 +1000
>        From:  Ian Holland <ian-at-digideas-dot-com.au>
> Organization: Digital Ideas Pty Ltd
>          To:  Tesla List <tesla-at-pupman-dot-com>
>  References:  1
>
>
> Tom Heiber wrote:
>
> > This is an equation I got from Bert Pool's file.
> > =======================================================================
> >     EQUATION 10: CAPACITANCE OF A TOROID
> >                                                     ___________________
> >                                                    /    2
> >          C =(1+ (0.2781 - d2/d1)) x  2.8  x      /  2 pi  (d1-d2)(d2/2)
> >                                                /   -------------------
> >                                             \/      4 pi
> >
> >      C = capacitance in picofarads (+- 5% )
> >     d1 = outside diameter of toroid in inches
> >     d2 = diameter of cross section (cord) of toroid in inches
> >
> >     Equation courtesy of Bert Pool
> >
> > =======================================================================
> >
> > CAN ANYONE help me solve for D1 as in D1=........
> > I have been trying to do this without any results (I suck at math)
>
> Tom,
>
> Hi from downunder!
>
> My initial thought, given the nature of the equation, was that you could
> only solve for d1 &/or d2 by an iterative approach, however I have found
> an approach which may suit your needs:
>
> Taking "Bert's #10" equation and cleaning it up we have,
>
> C = 1.4 (1.2781 - d2/d1) . sqrt( PI (d1-d2) d2 )
>
> If we substitute D for d2 and A for the aspect ration of the toroid,
> (i.e. A = d1/d2), we can put the equation in a form where only one
> diameter term exists and it is then easy to solve for D:
>
> C = 1.4( 1.2781 - D/(AD) ) . sqrt( PI (AD-D)D )
>   = 1.4( 1.2781 - 1/A ) . D . sqrt( PI(A-1) )
>
> Hence D = C / ( 1.4( 1.2781 - 1/A) . sqrt( PI(A-1) ) )
>
> Just select a value for the aspect ratio that appeals to you, say 3, the
> capactiance you want and out pops D.  For mathematical reasons A must be
> greater than 1 (or else solution is not real) and should be greater than
> 2 for practical reasons (or the toroid intersects itself.
>
> Hope this helps.
>
> What is the origin of this equation?  While it is relatively
> straightforward problem to derive the eqn for the capacitance of an
> isolated sphere (one simple integral), my maths is far too rusty to do
> so for a more complicated shape like a toroid.  Is there a derivation of
> the equation from Bert's list anywhere?  (I would be surprised if it
> came from a textbook in its current form given the number of terms that
> can be cancelled directly or eliminated by simple rearrangement).

I based the equation upon formulae for area of toroids and the center
disk, then modified the standard equation for an isolated sphere.  I
then compared results to tables of known capacitances of real
toroids, and added a "fudge" factor to the equation to make it match
real world figures.  It is accurate to within about 5% for toroids up
to about 4 or 5 feet in diameter.  I'm terrible at math, and didn't
even try to reduce the equation to its most simple form.  I'm glad
Ian and Malcom have done the reduction for us!

Bert Pool

>
> Best regards from one of the many generally silent but avid followers of
> the contents of this list,
>
> Safe and sucessful coiling to all.
>
> Ian.
> --
> -----------------------------------------------------------------------
> Ian Holland                               E-mail:   Ian-at-DigIdeas-dot-com.au
> Carnegie, Victoria AUSTRALIA              Callsign: VK3YQN
> -----------------------------------------------------------------------
>

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