The resistor cube problem

Guy Macon · May 25, 2004

Activ8 <reply2group@ndbbm.net> says...

Is that *really* a dodecahedron flattened out? Where's Bucky when I
need him?

Will Lisa do?

http://homepage.smc.edu/nestler_andrew/dodecahedron.jpg

--
Guy Macon, Electronics Engineer & Project Manager for hire.
Remember Doc Brown from the _Back to the Future_ movies? Do you
have an "impossible" engineering project that only someone like
Doc Brown can solve? My resume is at http://www.guymacon.com/

Richard Henry · May 25, 2004

"Richard Henry" <rphenry@home.com> wrote in message
news:zRwsc.24252$PU5.16197@fed1read06...

"Chris Carlen" <crcarle@BOGUS.sandia.gov> wrote in message
news:c8tsim13152@news3.newsguy.com...
Frank Bemelman wrote:
"Chris Carlen" <crcarle@BOGUS.sandia.gov> schreef in bericht
news:c8ta9v0183o@news4.newsguy.com...

Fun problem. Wasn't what I had originally planned for my first hour
of
the day.

Your next assignment is the dodecahedron

If you can draw it, I'll solve it.

/------------------- 1 ohm ------------------------------\
/ \
K-- 1 ohm --L-------- 1 ohm --------O-- 1 ohm --Q-- 1 ohm --S
| | | | |
| | | | |
| 1 ohm 1 ohm | |
| | | | |
| | | | |
1 ohm C-- 1 ohm --E-- 1 ohm --G 1 ohm |
| | | | | |
| | | | | |
| 1 ohm | 1 ohm | |
| | | | | |
| | | | | |
A-- 1 ohm --B 1 ohm I-- 1 ohm --J 1 ohm
| | | | | |
| | | | | |
| 1 ohm | 1 ohm | |
| | | | | |
| | | | | |
1 ohm D-- 1 ohm --F-- 1 ohm --H 1 ohm |
| | | | |
| | | | |
| 1 ohm 1 ohm | |
| | | | |
| | | | |
M-- 1 ohm --N-------- 1 ohm --------P-- 1 ohm --R-- 1 ohm --T
\ /
\------------------- 1 ohm ------------------------------/

One pair of "opposite corners" is A and I.

I thought somebody would have solved this by now. I just finished the first
half-coat of paint on the wood trim for tomorrow's project, so I'll give it
a stab.

Basic facts for a dodecahedron:

12 sides

20 vertices

30 edges

each vertex is the meeting point of three edges

Using a symmetry argument similar to that of the analysis of the cube, there
are three equivalent paths for current connected to vertices A and I. The
vertices K, B, and M can be connected by virtual wires with no change to the
solution, as can G, J and H. At the next stage, there are 6 equivalent
paths on each side, leading to the equipotental points S, L, C, D, N and T
on one side and Q, O, E, F, P and R on the other. So far we have accounted
for 18 of the 30 edges. But half of the remaining 12 edges connect vertices
of the same potential, so no current will flow in them. The other 6 paths
complete the connection.

S Q

K L O G

A 3 paths B 6 paths C 6 paths E 6 paths J 3 paths I

M D F H

N P

T R

Equipotential paths (no current)

S T Q - O

L C E - F

D N P R

1/3 ohm + 1/6 ohm + 1/6 ohm + 1/6 ohm + 1/3 ohm = 7/6 ohm

Activ8 · May 25, 2004

On 24 May 2004 19:12:09 -0500, The Phantom wrote:

On Mon, 24 May 2004 15:22:14 -0700, Chris Carlen
crcarle@BOGUS.sandia.gov> wrote:

Frank Bemelman wrote:
"Chris Carlen" <crcarle@BOGUS.sandia.gov> schreef in bericht
news:c8ta9v0183o@news4.newsguy.com...

Fun problem. Wasn't what I had originally planned for my first hour of
the day.

Your next assignment is the dodecahedron

If you can draw it, I'll solve it.

Consider the infinite plane grid of 1 ohm resistors with nodes on
the coordinates with integer values, such as (1,1), (2,2), (1,2) etc.
None of the resistors is diagonally connected, as between (1,1) and
(2,2). All the horizontally adjacent and vertically adjacent nodes
have a 1 ohm resistor connected.

What is the resistance between the points (1,1) and (1,2)?

What is the resistance between the points (1,1) and (1,2)?

1,1 1,2 1,3
___ ___
+-|___|--+-|___|-+
| | |
.-. .-. .-.
| | A | | B | |
| | | | | |
'-' '-' '-'
| ___ | ___ |
+-|___|--+-|___|-+
| | |
.-. .-. .-.
| | | | | |
| | C | | D | |
'-' '-' '-'
| ___ | ___ |
+-|___|--+-|___|-+ 3,3

Taking only mesh A, we get 1 ohm

Connecting mesh D, it's 2 ohms

scap that approach...

ok connect two equal potential points in both meshes A and D (1,2 to
2,1 and 2,3 to 3,2) and you get

2
1,2 ___ 2,3
+-------|___|--------+
| |
| |
.5 | .5 .5 | .5
___ | ___ ___ | ___
1,1 -|___|---+---|___|-----|___|--+--|___|- 3,3
| |
| |
| |
| |
| ___ |
+--------|___|-------+
2

that's 1.5 ohm or 3(.5)

and since a single 4 R mesh (like A only) would be 1 ohm,
(2,2)-(1,1) = 1,1 and (3,3)-(1,1)=2,2 so j=k is 1 or 2 i.e., the
dimension, n

n R
_____
1 1
2 1.5 crap! need one more

nah. skip the subtraction and go with n=j=k

n R
_____
2 1
3 1.5

R = .5n -- a wild first guess which would make the next higher
dimension grid

R = 2

1,1 1,2 1,3
___ ___
+-|___|--+-|___|-+ R +
| | |
.-. .-. .-.
| | A/ | | B | | R
| | | | | |
'-' '-' '-'
| ___ | ___ |
+-|___|--+-|___|-+ R +
| | |
.-. .-. .-.
| | | | / | | R
| | C | | D | |
'-' '-' '-'
| ___ | ___ |
+-|___|--+-|___|-+ R +
| | |
.-. .-. .-.
| | | | | | /
| | C | | D | | R
'-' '-' '-'
| ___ | ___ |
+-|___|--+-|___|-+ R + 4,4

er, it's a bit late for me to stare aat this any more to connect 1,4
with 4,1. I'd rather do a mesh or node analysis at this point.
--
Best Regards,
Mike

Activ8 · May 25, 2004

On Tue, 25 May 2004 07:07:26 GMT, Robert Baer wrote:

The Phantom wrote:

On Mon, 24 May 2004 17:15:07 -0700, Chris Carlen
crobc@BOGUS_FIELD.earthlink.net> wrote:

The Phantom wrote:
On Mon, 24 May 2004 15:22:14 -0700, Chris Carlen
crcarle@BOGUS.sandia.gov> wrote:

Frank Bemelman wrote:

"Chris Carlen" <crcarle@BOGUS.sandia.gov> schreef in bericht
news:c8ta9v0183o@news4.newsguy.com...

Fun problem. Wasn't what I had originally planned for my first hour of
the day.

Your next assignment is the dodecahedron

If you can draw it, I'll solve it.

Consider the infinite plane grid of 1 ohm resistors with nodes on
the coordinates with integer values, such as (1,1), (2,2), (1,2) etc.
None of the resistors is diagonally connected, as between (1,1) and
(2,2). All the horizontally adjacent and vertically adjacent nodes
have a 1 ohm resistor connected.

What is the resistance between the points (1,1) and (1,2)?

Yes, this is another category of interesting resistor problems. Don't
give me any hints. Though I am not sure when I will try it. It has
been in the back of my mind since seeing it here a few times.

What is the resistance between the points (1,1) and (1,2)?

Yes, of course; typos happen, eh?

What is the resistance between the points (1,1) and (2,2)?

Don't you mean some other points?

Good day!

The cube of 1 ohm resistors, a resistor on each edge, whit the
question as to the value of resistance across the internal body diagnal
can be made more interesting:
State at least three completely differnt ways of solving this problem
and give the electrical and/or mathematical reasons for each step if the
solutions.
For extra credit, state how many different ways there are to solve
that problem - with a brief description of each method (so we know you
are not cheating).

1. Connect points of equal potentials, combine resistors in parallel
and series. The reason you can do this is because equipotential
points can be shorted without fouling things up.

2. apply a current source, write the node equations and solve for
the voltage, divide by current to get R. THe reason you can do this
is bacause KCL says you can.

3. put a voltage across the stinkin' thing, write the mesh
equations, solve for I, divide it into V to get R. THe reason is
KVL.

Extra credit: There are infinite ways to solve this because there's
no telling how many monkeys can guess the correct answer, use a VOM
meter, etc., and who's to say whether any two individual monkeys
arrived at their correct guess the same way, with the exact same
neurons firing identically. My proof I that statistically, anything
is possible no matter how unlikely.

--
Best Regards,
Mike

Activ8 · May 25, 2004

On Tue, 25 May 2004 00:51:14 -0700, Guy Macon wrote:

Activ8 <reply2group@ndbbm.net> says...

Is that *really* a dodecahedron flattened out? Where's Bucky when I
need him?

Will Lisa do?

http://homepage.smc.edu/nestler_andrew/dodecahedron.jpg

When she's 18 and changes that silly hair-do

I see the pentagons now, and it looks like Richard messed up and
left out one face. Or else I've *really* been staring at the screen
too long.
--
Best Regards,
Mike

Activ8 · May 25, 2004

On Tue, 25 May 2004 04:22:14 -0400, Activ8 wrote:

On Tue, 25 May 2004 00:51:14 -0700, Guy Macon wrote:

Activ8 <reply2group@ndbbm.net> says...

Is that *really* a dodecahedron flattened out? Where's Bucky when I
need him?

Will Lisa do?

http://homepage.smc.edu/nestler_andrew/dodecahedron.jpg

When she's 18 and changes that silly hair-do

I see the pentagons now, and it looks like Richard messed up and
left out one face. Or else I've *really* been staring at the screen
too long.

NEver mind. The area outside the whole shebang makes up one face.
--
Best Regards,
Mike

Frank Bemelman · May 25, 2004

"Chris Carlen" <crcarle@BOGUS.sandia.gov> schreef in bericht
news:c8tsim13152@news3.newsguy.com...

Frank Bemelman wrote:
"Chris Carlen" <crcarle@BOGUS.sandia.gov> schreef in bericht
news:c8ta9v0183o@news4.newsguy.com...

Fun problem. Wasn't what I had originally planned for my first hour of
the day.

Your next assignment is the dodecahedron

If you can draw it, I'll solve it.

I was just teasing, but Richard Henry made a nice drawing. Making
the drawing alone is worth a medal.

--
Thanks, Frank.
(remove 'x' and 'invalid' when replying by email)

Frank Bemelman · May 25, 2004

"Richard Henry" <rphenry@home.com> schreef in bericht
news

ZCsc.28713$PU5.23840@fed1read06...

[snip]

1/3 ohm + 1/6 ohm + 1/6 ohm + 1/6 ohm + 1/3 ohm = 7/6 ohm

Fantastic! Great drawing too. My compliments.

--
Thanks, Frank.
(remove 'x' and 'invalid' when replying by email)

Al · May 25, 2004

In article <oc65b094q78skvn9192itej4odbltf7m3l@4ax.com>,
budgie <me@privacy.net> wrote:

On 24 May 2004 20:11:54 +0100, John Devereux <jd1@devereux.me.uk> wrote:

(snip O/P stuff re 1R cube)

You know that the cube is symmetrical, and the two points a and b are
at opposite corners. So there is no "preferred" route from a to b. The
currents from a to V1, V2 and V3 will be equal by symmetry. Therefore
the voltage drop will be equal too. So the voltages V1,2,3 are
equal. There is therefore no difference between this circuit and one
with V1,2,3 connected together. In which case the 3 resistors
connected to a are in parallel and the resistance is 1/3 ohm.

You can make the identical argument from the "b" end, so another 1/3
ohm appears.

Finally we now have the remaining 6 resistors V3-5, V2-5, V1-4, V1-6,
V2-5, V2-6. But these are all in parallel too since we connected
V1,2,3 and V4,5,6. So we have a final 1/6 of an ohm giving a total of
1/3+1/3+1/6 = 5/6.

John and Mark have put their respective fingers on the elegant solution
approach. Symmetry is the key, and using a current source helps visualise the
volt drop across each thrid of the mesh.

BTW, we had this as a problem in high school physics. Several of us gave the
answer "by eye", without resort to pencil/paper. And the slide rule probably
wouldn't have helped ....

OK, what is the resistance between two points 2.54 centimeters apart in
an infinite conductive sheet with a resistance of 1 ohm/square?

Al

--
There's never enough time to do it right the first time.......

Richard Henry · May 25, 2004

"Activ8" <reply2group@ndbbm.net> wrote in message
news:1gqgrnqego834$.dlg@news.individual.net...

On Tue, 25 May 2004 00:51:14 -0700, Guy Macon wrote:

Activ8 <reply2group@ndbbm.net> says...

Is that *really* a dodecahedron flattened out? Where's Bucky when I
need him?

Will Lisa do?

http://homepage.smc.edu/nestler_andrew/dodecahedron.jpg

When she's 18 and changes that silly hair-do

I see the pentagons now, and it looks like Richard messed up and
left out one face. Or else I've *really* been staring at the screen
too long.

The 12 sides:

1. ABCLK
2. ABDNM
3. BCEFD
4. CEGOL
5. DFHPN
6. EFHIG
7. GIJQO
8. HIJRP
9. JQSTR
10. KLOQS
11. MNPRT
12. AKSTM (is this the one you missed?)

Fred Bloggs · May 25, 2004

Chris Carlen wrote:

Hi:

Guy Macon wrote:

"Here is a classic that really shows the difference between algebraic
and RPN: calculate the resistance between points (a) and (b) using
only a calculator - no drawings or equations on paper.

(a)-- *----------- ONE OHM -------------* V1
|\ |\
| O O O
| N N N
| E E E
| |
| O O O
O H H H
N M M M
E \ V2 | \
*---- ONE OHM --------------------* V6
O | | |
H O | O
M N | N
| W | E
| | | |
| O | |
| H | O
| M | N
| V3 | | E
*----- ONE OHM -------------------* V4
\ | \ O
O | O H
N | N M
E | E |
| |
O | O |
H | H |
M | M |
\| V5 \|
*------------ ONE OHM ------------*--(b)"

I found this intriguing because it seemed silly that I could spend all
weekend studying Fourier transforms and "advanced" stuff like that, and
then be put off by a simple resistor problem. So I gave it a try...

I am thogroughly impressed by any folks who can solve this by eye, using
just the calculator's "arithmetic" capabilities whether it be RPN or
algebraic input.

I had to resort to pencil and paper and came up with the following
matrix solution for the node voltages for attaching a 1V source to
terminal A and ground to terminal B:

[ V1 ] [ 3 0 0 -1 0 -1 ]^-1 [ 1 ] [ 0.6 ]
[ V2 ] [ 0 3 0 0 -1 -1 ] [ 1 ] [ 0.6 ]
[ V3 ] [ 0 0 3 -1 -1 0 ] [ 1 ] [ 0.6 ]
[ V4 ] = [-1 0 -1 3 0 0 ] * [ 0 ] = [ 0.4 ]
[ V5 ] [ 0 -1 -1 0 3 0 ] [ 0 ] [ 0.4 ]
[ V6 ] [-1 -1 0 0 0 3 ] [ 0 ] [ 0.4 ]

Thus, R= V / Ib = 1 / (3*0.4/1) = 1/1.2 = 0.8333 ohms

Confirmed in LTSpice.

Fun problem. Wasn't what I had originally planned for my first hour of
the day.

Clues as to the "by eye" solution would be appreciated. I am no genius,
so usually must resort to methodical techniques such as writing out the
node voltage equations which is what I did after puzzling over it for 30
minutes. I thought a lot about superposition approaches, and re-wrote
the circuit in a planar form with one source applied to one of the three
branches attached to "a" with the other two grounded. There are some
symmetries that become more apparent after planarizing that could lead
to some further superposition approaches from that point. But is all
seems like more of a pain in the butt and more prone to error than
writing the node voltage eqns. So while a keen eye might come up with
an elegant solution, sometimes the brute force method delivers a result
long before that spark of insight hits, usually in the shower at the end
of the day (or week).

Good day!

THE only valid analytical solution uses:
View in a fixed-width font such as Courier.

PI <---> TEE TRANSFORM

1 o-+-------/\/\------+-o 2
|\ RB /|
| \ / |
| |_ _| |
| |_ _| |
| R1 \ / R2 |
/ \ / /
RA | RC
/ | /
\ / \
| \ |
| R3 |
| \ |
| | |
| | |
1'o-+--------+--------+-o 2'

RA x RB R1xR2 + R1xR3 + R2xR3
R1= ------------- RA= ---------------------
RA + RB + RC R2

RB x RC R1xR2 + R1xR3 + R2xR3
R2= ------------- RB= ---------------------
RA + RB + RC R3

RA x RC R1xR2 + R1xR3 + R2xR3
R3= ------------- RC= ---------------------
RA + RB + RC R1

Eliminate V6, then V4, and V5 where vertex is center T- the circuit is
then a simple lattice.

Chris Carlen · May 25, 2004

Richard Henry wrote:

"Chris Carlen" <crcarle@BOGUS.sandia.gov> wrote in message
news:c8tsim13152@news3.newsguy.com...

Frank Bemelman wrote:

"Chris Carlen" <crcarle@BOGUS.sandia.gov> schreef in bericht
news:c8ta9v0183o@news4.newsguy.com...

Fun problem. Wasn't what I had originally planned for my first hour of
the day.

Your next assignment is the dodecahedron

If you can draw it, I'll solve it.

/------------------- 1 ohm ------------------------------\
/ \
K-- 1 ohm --L-------- 1 ohm --------O-- 1 ohm --Q-- 1 ohm --S
| | | | |
| | | | |
| 1 ohm 1 ohm | |
| | | | |
| | | | |
1 ohm C-- 1 ohm --E-- 1 ohm --G 1 ohm |
| | | | | |
| | | | | |
| 1 ohm | 1 ohm | |
| | | | | |
| | | | | |
A-- 1 ohm --B 1 ohm I-- 1 ohm --J 1 ohm
| | | | | |
| | | | | |
| 1 ohm | 1 ohm | |
| | | | | |
| | | | | |
1 ohm D-- 1 ohm --F-- 1 ohm --H 1 ohm |
| | | | |
| | | | |
| 1 ohm 1 ohm | |
| | | | |
| | | | |
M-- 1 ohm --N-------- 1 ohm --------P-- 1 ohm --R-- 1 ohm --T
\ /
\------------------- 1 ohm ------------------------------/

One pair of "opposite corners" is A and I.

Lovely!

But you beat me to the solution in your post below...

(Not that solving these things is as exciting anymore)

--
____________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
crcarle@sandia.gov

Chris Carlen · May 25, 2004

Richard Henry wrote:
[edit]

/------------------- 1 ohm ------------------------------\
/ \
K-- 1 ohm --L-------- 1 ohm --------O-- 1 ohm --Q-- 1 ohm --S
| | | | |
| | | | |
| 1 ohm 1 ohm | |
| | | | |
| | | | |
1 ohm C-- 1 ohm --E-- 1 ohm --G 1 ohm |
| | | | | |
| | | | | |
| 1 ohm | 1 ohm | |
| | | | | |
| | | | | |
A-- 1 ohm --B 1 ohm I-- 1 ohm --J 1 ohm
| | | | | |
| | | | | |
| 1 ohm | 1 ohm | |
| | | | | |
| | | | | |
1 ohm D-- 1 ohm --F-- 1 ohm --H 1 ohm |
| | | | |
| | | | |
| 1 ohm 1 ohm | |
| | | | |
| | | | |
M-- 1 ohm --N-------- 1 ohm --------P-- 1 ohm --R-- 1 ohm --T
\ /
\------------------- 1 ohm ------------------------------/

One pair of "opposite corners" is A and I.

I thought somebody would have solved this by now. I just finished the first
half-coat of paint on the wood trim for tomorrow's project, so I'll give it
a stab.

Folks have to sleep you know. It's just 8:xx AM here now.

1/3 ohm + 1/6 ohm + 1/6 ohm + 1/6 ohm + 1/3 ohm = 7/6 ohm

Ah-ha! Now what's interesting is the possible appearance of a pattern.
Does the sequence of equivalent resistance values converge to 1 as the
n of the polyhedron approaches infinity?

Good day!

--
____________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
crcarle@sandia.gov

Jim Thompson · May 25, 2004

On Mon, 24 May 2004 17:57:01 -0700, "Richard Henry" <rphenry@home.com>
wrote:

"Chris Carlen" <crcarle@BOGUS.sandia.gov> wrote in message
news:c8tsim13152@news3.newsguy.com...
Frank Bemelman wrote:
"Chris Carlen" <crcarle@BOGUS.sandia.gov> schreef in bericht
news:c8ta9v0183o@news4.newsguy.com...

Fun problem. Wasn't what I had originally planned for my first hour of
the day.

Your next assignment is the dodecahedron

If you can draw it, I'll solve it.

/------------------- 1 ohm ------------------------------\
/ \
K-- 1 ohm --L-------- 1 ohm --------O-- 1 ohm --Q-- 1 ohm --S
| | | | |
| | | | |
| 1 ohm 1 ohm | |
| | | | |
| | | | |
1 ohm C-- 1 ohm --E-- 1 ohm --G 1 ohm |
| | | | | |
| | | | | |
| 1 ohm | 1 ohm | |
| | | | | |
| | | | | |
A-- 1 ohm --B 1 ohm I-- 1 ohm --J 1 ohm
| | | | | |
| | | | | |
| 1 ohm | 1 ohm | |
| | | | | |
| | | | | |
1 ohm D-- 1 ohm --F-- 1 ohm --H 1 ohm |
| | | | |
| | | | |
| 1 ohm 1 ohm | |
| | | | |
| | | | |
M-- 1 ohm --N-------- 1 ohm --------P-- 1 ohm --R-- 1 ohm --T
\ /
\------------------- 1 ohm ------------------------------/

One pair of "opposite corners" is A and I.

Unless I'm missing something I count only 11 faces... a dodecahedron
has *12* faces, each of which is a pentagon. I know it quite well
because one of my drafting assignments (when I was in Junior High
School) was to draw a dodecahedron at various rotations.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice

480)460-2350 | |
| E-mail Address at Website Fax

Frank Bemelman · May 25, 2004

"Jim Thompson" <thegreatone@example.com> schreef in bericht
news:q4s6b0l6uvcvn4p8srl7f61i0d4ouu972t@4ax.com...

On Mon, 24 May 2004 17:57:01 -0700, "Richard Henry" <rphenry@home.com
wrote:

"Chris Carlen" <crcarle@BOGUS.sandia.gov> wrote in message
news:c8tsim13152@news3.newsguy.com...
Frank Bemelman wrote:
"Chris Carlen" <crcarle@BOGUS.sandia.gov> schreef in bericht
news:c8ta9v0183o@news4.newsguy.com...

Fun problem. Wasn't what I had originally planned for my first hour
of
the day.

Your next assignment is the dodecahedron

If you can draw it, I'll solve it.

/------------------- 1 ohm ------------------------------\
/ \
K-- 1 ohm --L-------- 1 ohm --------O-- 1 ohm --Q-- 1 ohm --S
| | | | |
| | | | |
| 1 ohm 1 ohm | |
| | | | |
| | | | |
1 ohm C-- 1 ohm --E-- 1 ohm --G 1 ohm |
| | | | | |
| | | | | |
| 1 ohm | 1 ohm | |
| | | | | |
| | | | | |
A-- 1 ohm --B 1 ohm I-- 1 ohm --J 1 ohm
| | | | | |
| | | | | |
| 1 ohm | 1 ohm | |
| | | | | |
| | | | | |
1 ohm D-- 1 ohm --F-- 1 ohm --H 1 ohm |
| | | | |
| | | | |
| 1 ohm 1 ohm | |
| | | | |
| | | | |
M-- 1 ohm --N-------- 1 ohm --------P-- 1 ohm --R-- 1 ohm --T
\ /
\------------------- 1 ohm ------------------------------/

One pair of "opposite corners" is A and I.

Unless I'm missing something I count only 11 faces... a dodecahedron
has *12* faces, each of which is a pentagon. I know it quite well
because one of my drafting assignments (when I was in Junior High
School) was to draw a dodecahedron at various rotations.

The 'outline' of the drawing is a face too. 12 faces total.

--
Thanks, Frank.
(remove 'x' and 'invalid' when replying by email)

analog · May 25, 2004

Frank Bemelman wrote:

Jim Thompson wrote:
Richard Henry wrote:
Chris Carlen wrote:

Your next assignment is the dodecahedron

If you can draw it, I'll solve it.

/------------------- 1 ohm ------------------------------\
/ \
K-- 1 ohm --L-------- 1 ohm --------O-- 1 ohm --Q-- 1 ohm --S
| | | | |
| | | | |
| 1 ohm 1 ohm | |
| | | | |
| | | | |
1 ohm C-- 1 ohm --E-- 1 ohm --G 1 ohm |
| | | | | |
| | | | | |
| 1 ohm | 1 ohm | |
| | | | | |
| | | | | |
A-- 1 ohm --B 1 ohm I-- 1 ohm --J 1 ohm
| | | | | |
| | | | | |
| 1 ohm | 1 ohm | |
| | | | | |
| | | | | |
1 ohm D-- 1 ohm --F-- 1 ohm --H 1 ohm |
| | | | |
| | | | |
| 1 ohm 1 ohm | |
| | | | |
| | | | |
M-- 1 ohm --N-------- 1 ohm --------P-- 1 ohm --R-- 1 ohm --T
\ /
\------------------- 1 ohm ------------------------------/

One pair of "opposite corners" is A and I.

Unless I'm missing something I count only 11 faces... a dodecahedron
has *12* faces, each of which is a pentagon. I know it quite well
because one of my drafting assignments (when I was in Junior High
School) was to draw a dodecahedron at various rotations.

The 'outline' of the drawing is a face too. 12 faces total.

Ha, it's a stretch of the imagination to get there, but of course you
are quite correct. You win today's topological trophy. But now Jim
has lost "face" in a public forum.

Perhaps a consolation prize is
in order. Might I suggest a fine Klein bottle decanter for the merlot
(would lead to better concentration and more wine in the food).

budgie · May 25, 2004

On Tue, 25 May 2004 13:12:52 GMT, Al <no.spam@here.com> wrote:

In article <oc65b094q78skvn9192itej4odbltf7m3l@4ax.com>,
budgie <me@privacy.net> wrote:

On 24 May 2004 20:11:54 +0100, John Devereux <jd1@devereux.me.uk> wrote:

(snip O/P stuff re 1R cube)

You know that the cube is symmetrical, and the two points a and b are
at opposite corners. So there is no "preferred" route from a to b. The
currents from a to V1, V2 and V3 will be equal by symmetry. Therefore
the voltage drop will be equal too. So the voltages V1,2,3 are
equal. There is therefore no difference between this circuit and one
with V1,2,3 connected together. In which case the 3 resistors
connected to a are in parallel and the resistance is 1/3 ohm.

You can make the identical argument from the "b" end, so another 1/3
ohm appears.

Finally we now have the remaining 6 resistors V3-5, V2-5, V1-4, V1-6,
V2-5, V2-6. But these are all in parallel too since we connected
V1,2,3 and V4,5,6. So we have a final 1/6 of an ohm giving a total of
1/3+1/3+1/6 = 5/6.

John and Mark have put their respective fingers on the elegant solution
approach. Symmetry is the key, and using a current source helps visualise the
volt drop across each thrid of the mesh.

BTW, we had this as a problem in high school physics. Several of us gave the
answer "by eye", without resort to pencil/paper. And the slide rule probably
wouldn't have helped ....

OK, what is the resistance between two points 2.54 centimeters apart in
an infinite conductive sheet with a resistance of 1 ohm/square?

Do I care?

Activ8 · May 26, 2004

On Tue, 25 May 2004 21:21:56 GMT, analog wrote:

Frank Bemelman wrote:
Jim Thompson wrote:
Richard Henry wrote:
Chris Carlen wrote:

Your next assignment is the dodecahedron

If you can draw it, I'll solve it.

/------------------- 1 ohm ------------------------------\
/ \
K-- 1 ohm --L-------- 1 ohm --------O-- 1 ohm --Q-- 1 ohm --S
| | | | |
| | | | |
| 1 ohm 1 ohm | |
| | | | |
| | | | |
1 ohm C-- 1 ohm --E-- 1 ohm --G 1 ohm |
| | | | | |
| | | | | |
| 1 ohm | 1 ohm | |
| | | | | |
| | | | | |
A-- 1 ohm --B 1 ohm I-- 1 ohm --J 1 ohm
| | | | | |
| | | | | |
| 1 ohm | 1 ohm | |
| | | | | |
| | | | | |
1 ohm D-- 1 ohm --F-- 1 ohm --H 1 ohm |
| | | | |
| | | | |
| 1 ohm 1 ohm | |
| | | | |
| | | | |
M-- 1 ohm --N-------- 1 ohm --------P-- 1 ohm --R-- 1 ohm --T
\ /
\------------------- 1 ohm ------------------------------/

One pair of "opposite corners" is A and I.

Unless I'm missing something I count only 11 faces... a dodecahedron
has *12* faces, each of which is a pentagon. I know it quite well
because one of my drafting assignments (when I was in Junior High
School) was to draw a dodecahedron at various rotations.

The 'outline' of the drawing is a face too. 12 faces total.

Ha, it's a stretch of the imagination to get there, but of course you
are quite correct. You win today's topological trophy. But now Jim
has lost "face" in a public forum. Perhaps a consolation prize is
in order. Might I suggest a fine Klein bottle decanter for the merlot
(would lead to better concentration and more wine in the food).

Check the time of my post where I corrected myself. 50 min prior.
--
Best Regards,
Mike

Terry Given · May 26, 2004

"budgie" <me@privacy.net> wrote in message
news:qhq7b0hioako892cidqoqojcugpo1u9rvq@4ax.com...

On Tue, 25 May 2004 13:12:52 GMT, Al <no.spam@here.com> wrote:

In article <oc65b094q78skvn9192itej4odbltf7m3l@4ax.com>,
budgie <me@privacy.net> wrote:

OK, what is the resistance between two points 2.54 centimeters apart in
an infinite conductive sheet with a resistance of 1 ohm/square?

Do I care?

Yes

Jim Thompson · May 26, 2004

On Tue, 25 May 2004 21:50:08 -0400, Activ8 <reply2group@ndbbm.net>
wrote:

On Tue, 25 May 2004 21:21:56 GMT, analog wrote:

Frank Bemelman wrote:
Jim Thompson wrote:
Richard Henry wrote:
Chris Carlen wrote:
[snip]

Unless I'm missing something I count only 11 faces... a dodecahedron
has *12* faces, each of which is a pentagon. I know it quite well
because one of my drafting assignments (when I was in Junior High
School) was to draw a dodecahedron at various rotations.

The 'outline' of the drawing is a face too. 12 faces total.

Ha, it's a stretch of the imagination to get there, but of course you
are quite correct. You win today's topological trophy. But now Jim
has lost "face" in a public forum. Perhaps a consolation prize is
in order. Might I suggest a fine Klein bottle decanter for the merlot
(would lead to better concentration and more wine in the food).

Check the time of my post where I corrected myself. 50 min prior.

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