conductive sheet - resistance between 2 arbitrary points
sunshine
I was wondering if anyone already wrote a program (e.g. matlab) for
the following problem:
- a thin conductive sheet with given conductivity or sheet resistance
- rectangular shape width x height
Calculate the resistance between 2 arbitrary points on the sheet, e.g.
between a corner (0|0) and a point on the sheet (X|Y).
I would need it quite urgently.
Thanks,
David
Bossavit
Alas, your question does not make real sense: It all depends on the
detailed structure of the contacts. Suppose they cover an area of 1
square millimeter (or, in the context of your query, span a length of 1
mm along the perimeter of the sheet): You get a very different result
than if it were, say, 10 square millimeters.
What *does* make sense is, for a conductive volume, the resistance
between two disjoint patches P0 and P1 of its surface, assuming both are
*equipotentials* (which is approximately the case if soldering and
connecting threads conduct much better than the conductive volume
itself). But it's intuitive that the larger the areas of P0 and P1,
the smaller the resistance. Such a problem *can* be solved by finite
elements, and you may perhaps find Matlab code for that. It's not so
common software, I am afraid.
Salmon Egg
<f9ace192-3e26-41a5...@y34g2000yqd.googlegroups.com>,
sunshine <david.le...@gmail.com> wrote:
I have not seen an earlier post show up here. It indicated how to solve
the problem using the Schwartz transformation. I was thinking about it
again and maybe another approach is better and suitable for solutiong
with a spreadsheet.
I repeat that the two dimensional current flow is closely related to two
dimensional potential distributions. Consider a box with the dimensions
of your rectangular sheet. Place cylindrical conductors with the same
shape of your contacts. Connect these conductors to a potential
difference source. The two dimensional distribution is found from these
conductors and their infinite two dimensional array of images in the
walls of zero potential walls. The effects of the remoter images
decreases with distance.
Bill
--
Most people go to college to get their missing high school education.
Salmon Egg
<f9ace192-3e26-41a5...@y34g2000yqd.googlegroups.com>,
sunshine <david.le...@gmail.com> wrote:
Forget the program. I believe the problem has been solved in closed form
using conformal mapping. It does require knowing how to use elliptic
integrals. A good place to start is with Smythe's Static and Dynamic
Electricity.
I can outline how to solve the problem, but will not do it for you.
First realize that solving a two dimensional potential static potential
distribution also solves the conduction problem.
Consider two finite cylinders with a potential difference between them.
Place them over a conducting plane. The potential distribution above the
plane will be the potential from these "wires" and their images in the
plane. It will be a sum of four complex logarithmic terms.
Use a Schwartz transformation to bend the plane into a rectangular box.
Voila, the potential distribution of two wires in an infinitely long box.
Dr J R Stockton
4g2000yqd.googlegroups.com>, Wed, 24 Jun 2009 18:29:40, sunshine
<david.le...@gmail.com> posted:
Consider the conductance between inside and outside of a circular
element centred on your point XY, radius r, thickness dr. ISTM to be
proportional to r. The resistance is thus proportional to 1/r.
Integrate dr/r from the point XY outwards to a convenient small distance
R, and IIRC you get infinity. Therefore the answer to the question as
stated is that the total resistance in each case is approximately
(infinity + something + infinity).
Therefore, if that is right, don't use point contacts; use finite ones,
of known size.
--
(c) John Stockton, near London. *@merlyn.demon.co.uk/?.?.Stockton@physics.org
Web <URL:http://www.merlyn.demon.co.uk/> - FAQish topics, acronyms, & links.
Correct <= 4-line sig. separator as above, a line precisely "-- " (SoRFC1036)
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Roland Franzius
Apprximately you ask for a solution of the 2-d potential equation on a
rectangle with two opposite poles and zero normal derivative at the
boundary.
To get an idea take the real part of the corresponding elliptic function
with poles on that rectangle. Then its easy to calculate the current
across any closed equipotential contour, which tend to become circles
approching the singularities.
For contacts not fitting the elliptic functions you habe to use a
complete set of eigenfunctions for the rectangle with isolated egdes
with two holes and fit the fourier series to the given total current
across any line separating them.
So its easier for non pointlike contact contours to use the matlab tool
for solving 2d-boundary problems.
--
Roland Franzius
Salmon Egg
Salmon Egg <Salm...@sbcglobal.net> wrote:
This post showed up much later than a subsequent post of mine outlining
how to solve the same problem using images. Both ways will work.
I get the impression ghat the OP seems to like numerical computer
solutions in preference to analytic ones. This can be achieved using a
two dimensional Laplace equation solver. Many thermal design programs
have such capability. Use temperature as a proxy for potential and heat
flow for a current proxy. Then it is just a matter of getting the
constance right.
Dirk Bruere at NeoPax
You could try asking in sci.electronics.design
--
Dirk
http://www.transcendence.me.uk/ - Transcendence UK
http://www.theconsensus.org/ - A UK political party
http://www.onetribe.me.uk/wordpress/?cat=5 - Our podcasts on weird stuff
Salmon Egg
Salmon Egg <Salm...@sbcglobal.net> wrote:
I have enjoyed contributing to this thread. It reminds of of earlier
days when I was deeply involved in various calculations. It has reminded
me of the sometimes painful days of going through Smythe's book. Painful
as it was at the time, it provided a sound basis for part of my
professional life. Now, on this newsgroup, I can pontificate a bit
without being responsible for going through the details left to others.
Getting back to the problem, I mentioned the use of conformal Schwartz
transformation that bends a ground plane into a rectangular box. That
involves the elliptic integral form.
Then after thinking about it some more, I suggested the infinite set of
images approach to solve the problem.
I now remember that elliptic functions, and consequential elliptic
integrals, arise out of doubly periodic functions. Thus, the Schwartz
transformation leads to elliptic integrals. The infinite array of images
leads to elliptic functions and back to elliptic integrals. Presumably
each method will give the same answer.
Nick Cramer
Loeb Julie of Julie Research Lab was doing this 50 years ago.
--
Nick, KI6VAV. Support severely wounded and disabled Veterans and their
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Salmon Egg
Nick Cramer <n_cram...@pacbell.net> wrote:
> sunshine <david.le...@gmail.com> wrote:
> > Hi,
> >
> > I was wondering if anyone already wrote a program (e.g. matlab) for
> > the following problem:
> >
> > - a thin conductive sheet with given conductivity or sheet resistance
> > - rectangular shape width x height
> >
> > Calculate the resistance between 2 arbitrary points on the sheet, e.g.
> > between a corner (0|0) and a point on the sheet (X|Y).
> >
> > I would need it quite urgently.
>
> Loeb Julie of Julie Research Lab was doing this 50 years ago.
Analog computers of this nature were in use at that time. Teledeltos
(sp?) paper used for FAX of that day was common. Water tanks with
electrolyte was used. By tilting the pan, cylindrical symmetry solutions
could be found. Additional probes injecting current enabled the
simulation of space charge. These techniques were used for the design of
electron guns.
Salmon Egg
I mistakenly gave it an incorrect date field when posting it the first
time. My apologies for the mixup.
-- jt]]
In article <20090626184242.429$s...@newsreader.com>,
Nick Cramer <n_cram...@pacbell.net> wrote:
Nick Cramer
> [[Mod. note -- This is a repost of this article with the correct date;
> I mistakenly gave it an incorrect date field when posting it the first
> time. My apologies for the mixup.
> Nick Cramer <n_cram...@pacbell.net> wrote:
> > sunshine <david.le...@gmail.com> wrote:
> > > Hi,
> > >
> > > I was wondering if anyone already wrote a program (e.g. matlab) for
> > > the following problem:
> > >
> > > - a thin conductive sheet with given conductivity or sheet resistance
> > > - rectangular shape width x height
> > >
> > > Calculate the resistance between 2 arbitrary points on the sheet,
> > > e.g. between a corner (0|0) and a point on the sheet (X|Y).
> > >
> > > I would need it quite urgently.
> >
> > Loeb Julie of Julie Research Lab was doing this 50 years ago.
>
> Analog computers of this nature were in use at that time. Teledeltos
> (sp?) paper used for FAX of that day was common. Water tanks with
> electrolyte was used. By tilting the pan, cylindrical symmetry solutions
> could be found. Additional probes injecting current enabled the
> simulation of space charge. These techniques were used for the design of
> electron guns.
My recollection is that it was used to maintain the temperature of an
Eppley standard cell.
John Polasek
It will be instructive to imagine how you would find the resistance
between any two points. The simplest thing is to get a sheet of the
material and a Simpson 260 or other ohmmeter and take the resistance
between two points. You could make a grid of points and plot the
result with the hope of developing some function.
But think what is going on. First, in order to read ohms, the meter
injects current into the sheet to be received by the other prod. The
meter scale is calibrated in inverse current.
The prods are quite pointy so that initially there would be an
infinite current density at each of the electrodes. This would appear
to comprise a singularity which the digital computer program might not
be able to handle.
You might recognize that with a Simpson and the sheet, you actually
have an analog computer, which in this case is vastly superior to the
digital model, especially considering that there is no programming
required.
From
he
Ken S. Tucker
On Jul 17, 10:51 am, John Polasek <jpola...@cfl.rr.com> wrote:
> On Wed, 24 Jun 2009 18:29:40 +0200 (CEST), sunshine
>
> <david.leuenber...@gmail.com> wrote:
> >Hi,
>
> >I was wondering if anyone already wrote a program (e.g. matlab) for
> >the following problem:
>
> >- a thin conductive sheet with given conductivity or sheet resistance
> >- rectangular shape width x height
>
> >Calculate the resistance between 2 arbitrary points on the sheet, e.g.
> >between a corner (0|0) and a point on the sheet (X|Y).
>
> >I would need it quite urgently.
>
> >Thanks,
> >David
In Purcell's "Electricity & Magnetism" book, problem 4.30,
pg.424, is an infinite lattice of 1 ohm resistors, set on a
2D "square mesh". He goes on to state the answer of the
resistance across any resistor is 0.5 ohms.
I hesitated to post that as it doesn't directly address the
question, but, I think, the problem can be solved if one
quantizes the grid as Purcell writes.
Regards
Ken S. Tucker
John Polasek
For extra credit calculate the sheet's resistivity in ohms per square
of this 1 ohm manifold.
John Polasek
Ken S. Tucker
I think we can extrapolate from Purcell's problem,
by using a limit process to move from his original
Grid as defined to a continuum with supposition.
As stated in the problem, there are 4, 1 ohm resistors
attached to each "node", such as cartesian graph paper
has, and the resistance across any resistor is 0.5 ohms.
In a line. the resistance between node n0 and n2 should
be (using series calculation), 2r, (O=0.5 ohm),
n0---O---n1---O---n2
which in Purcell's problem is O+O=1 ohm.
I'll orientate the Grid so that the resistance is measured
along a line of resistors.
Let's suppose, for units sake each of Purcell's 1 Ohm
resistors is 1 cm in length between nodes, such that
L=1cm, and further stipulate using - in place of Purcell's
1 ohm resistors - new ones of 0.5 cm that have a
resistance of 0.5 ohms, and that would produce the
same resistance at 1 cm distance by using,
o=1/2 ohm in the infinite grid, so 4 resistors make
o+o+o+o = 1 ohm
Now, it appears we are prepared to enable the length
of our resistors to tend to infinitesmal, though our 1cm
resistance remains constant at 0.5 ohms on the infinite
grid, I think that's obvious but interesting.
What I find interesting is the initial cartesian nodal
system Purcell spec'd is the definer of the resistance,
and the resistance is independant of scale, as far as
I can see.
Regards
Ken S. Tucker