Re: Calculating resistance

[Bernard Cleyet <anngeorg@PACBELL.NET>]

"I don't know if these formulae match those given in the reference
mentioned by Bernard Cleyet or not.  But if these don't they might
still be shown to be equivalent by making appropriate mathematical
manipulations involving inverse hyperbolic function identities."


They do, but that leads to another problem (for me)  The soln. assumes
infinite in length cylinders.  Therefore, I think the formulae are only
accurate in three D.  e.g.  long cylinders buried in the earth.

bc

P.s. Harnwell uses complex variables to solve the two dimensional problem
(only applicable when the third is infinite in length).

David Bowman wrote:

> Regarding the dialog between Ludwik K. & John D.:
>
> > > How to solve this problem? Find the resistance R between

cut ---

> I didn't bother trying to look up a previous solution, and instead
> tried my hand at coming up with the exact solution.  (It was a *whole
> lot* easier than finding the exact capacitance of a parallel circular
> plate capacitor, which I have yet to make much headway on, BTW.)  The
> result is essentially identical to the problem of finding the
> capacitance of two infinitely long parallel conducting circular
> cylinders, i.e. the capacitance of 'twin-lead' cable problem.
>
> The exact solution for the resistive sheet is problem given below:
> Let
>
> R = the net effective resistance between the dots
> a = radius of one dot
> b = radius of the other dot
> L = distance between the centers of the two dots
> d = thickness of the sheet/resistive film
> [rho] = resistivity of resistive medium in the sheet
>
> To simplify the solution equation and to allow it all to fit on one
> line I define two dimensionless auxiliary quantities A & B
> defined as:
>
> A == arccosh((L/a + a/L - (b^2)/(L*a))/2) and
> B == arccosh((L/b + b/L - (a^2)/(L*b))/2)  .
>
> Then the exact value for R is given as:
>
> R = [rho]*(A + B)/(2*d*[pi]) .
>
> Also, if we want a formula for the capacitance *per unit length* C'
> of two parallel conducting cylinders where the distance between
> them is greater than the sum of their radii, the exact solution is:
>
> C' = 2*[pi]*[epsilon_0]/(A + B) .
>
> In the special case that both cylinders have the same radius a (= b)
> the result for the capacitance per unit length boils down to:
>
> C' = [pi]*[epsilon_0]/arccosh(L/a)
>
> and the formula for the resistance between two *equal size* conducting
> dots on an infinite (otherwise uniform) resistive sheet boils down to
>
> R = [rho]*arccosh(L/a)/([pi]*d) .
>
> I don't know if these formulae match those given in the reference
> mentioned by Bernard Cleyet or not.  But if these don't they might
> still be shown to be equivalent by making appropriate mathematical
> manipulations involving inverse hyperbolic function identities.
>
> David Bowman
> David_Bowman@georgetowncollege.edu