12
$\begingroup$

I dropped in on Ernie a while ago and he told me that there was "good news and bad news". I asked him to tell me about it and he explained as follows: "Remember my Casimir-effect electric motor and how it always used to overheat? The good news is that I have increased its efficiency, so now it won't overheat any more". "And the bad news?", I inquired. "Unfortunately, I got a little carried away and improved things too much. If I ran it at 10 volts and 10 amps, I used to get 99 watts mechanical output and some waste heat, but now I get 101 watts of work out of it". "But that's wonderful!", I exclaimed. "Free energy. How can that be bad news?". "The problem is", Ernie explained, "the extra energy has to come from somewhere. Now as soon as I start the motor it begins to cool down. After a while it gets so cold that it seizes up and stops working. What a disaster!".

"But couldn't you just add a heater?", I asked. Ernie paused for a moment. "Of course!" he cried. "You're a genius. And I can use my new Acme Resist-a-Pak; it just arrived in the mail yesterday". He pulled a brightly coloured box from a drawer and lifted its lid. Inside were a number of components and a sheet of paper labelled "INSTRUCTIONS". I read the following:

Contents: Thirty-one high-precision resistors - each precisely 1.000000 Ohms. One programmable voltage supply - 0.000000 to 9.999999 volts. Sixteen orthogonal connector brackets.

To use: No solder required! Just clip resistors and voltage supply together, using the orthogonal connector brackets at the corners, to produce the resistor network of your choice. Set operating voltage on the power supply and operate as required.

Caution: Orthogonal connector brackets can connect two or more components together, but only at 90 degree angles. Do not try to force them into other orientations or damage will occur!!!

enter image description here

"Hmmm", I mused, "if you take two identical resistors (both oriented the same way) then slide them apart and join their ends with one more resistor and the voltage supply (both oriented orthogonally to the other two resistors), you will end up with a square network with resistance of 3 Ohms. You could use sqrt(3) volts to produce 1 watt of heating - just what your motor needs." Ernie did a few thermal calculations and told me that it wouldn't work. The square network would make the sides of the motor a lot too hot, and the top and bottom would be much too cold.

"Well, I continued, "How about making two identical squares of four resistors each (both oriented the same way) slide them apart and connect their corners with three more resistors and one voltage supply (all oriented orthogonally to the other eight resistors). You will end up with a cubic network and I am sure I can work out its resistance and what voltage you need to get 1 watt of heating." I started to calculate the resistance of the network, and had just got the answer when Ernie told me that it wouldn't work. The cubical network would make the outside of the motor a little too hot and the insides would be a little too cold.

"There doesn't seem to be a solution then", I muttered. "But wait" Ernie replied, "your ideas might just work. If I make two identical cubes of twelve resistors each (both oriented the same way) I can slide them apart and connect their vertices with seven more resistors and the voltage supply (all oriented orthogonally to the other 24 resistors, of course). I calculate that the heat will be even enough to avoid hot or cold spots on the motor - the temperature will be just right everywhere. And there are just enough parts in the Pak to do the job." I was about to point out the impossibility of producing that geometry when I got an emergency phone call from home - the cat was stuck in the tree again - and I had to rush off to help. By the time I had solved that problem, and put away the chain-saw and shot-gun, it was too late to get back to Ernie's shed, but I did sit down in the evening and calculated the resistances of my cubical 11 resistor network and Ernie's weird 31 resistor network (just for fun) - then texted him the results.

To my surprise, I got a message from Ernie the next morning to let me know that 'my' idea had worked perfectly - the motor had run all night without over- or under-heating. This all happened a while ago and I can't remember the two resistances I calculated - maybe you can help.

Improve this question
$\endgroup$
10
  • $\begingroup$ Although I'm reasonably certain that your setup violates the second law of thermodynamics, a more pertinent observation is that the connecting resistors and voltage supply cannot be "oriented orthogonally to the other 24 resistors" unless the 24 resistors all lie in a plane, which they clearly don't if they're a cube. Hence is there some unusual definition of "orthogonal"? The figure seems to show connectors for a hexagonal lattice, which wouldn't allow orthogonal connections in the plane anyway. ?? $\endgroup$
    – COTO
    Nov 10, 2014 at 2:59
  • $\begingroup$ @COTO Ernie's inventions often violate thermodynamics, general relativity, and quantum physics (but are seldom in bad taste). He is just that kind of guy. As always, the words and not the figure describe the problem. $\endgroup$
    – Penguino
    Nov 10, 2014 at 3:38
  • $\begingroup$ That is to say ... as always in Ernie's problems $\endgroup$
    – Penguino
    Nov 10, 2014 at 3:39
  • $\begingroup$ I can't get the visualization of the network @.@ $\endgroup$
    – justhalf
    Nov 10, 2014 at 4:52
  • 1
    $\begingroup$ @justhalf I sketched the circuits as I understand them and added them to my answer below. They are behind the first spoiler tag. $\endgroup$
    – Jason Patterson
    Nov 10, 2014 at 14:54

2 Answers 2

Reset to default
6
$\begingroup$

For those who are interested in the problem but having difficulty visualizing it, I have sketched out the circuits as I understand them. The 31 resistor circuit is not exactly as Ernie built, but I ran out of 8 legged orthogonal connectors, so I had to construct it using the boring 6 legged variety and some jumper wire. The colors are unimportant, I only included them to keep the image from becoming completely uninterpretable visually. I have posted visual solutions farther down, this first spoiler box is only the circuits with no additional information included.

Drawing of the circuits

The following is my solution for the 11 resistor problem. I'm still working on the 31, but I'm probably going to have to go to bed for the night.

I found the points of equipotential and remapped the circuit so that they were shorted. From there it was a matter of finding the equivalent resistance of the remaining paths between those points. $R_{11} = 7/5\ \Omega$

Here is a drawing of the solution to the 11 resistor problem. I have separated it from the 31 resistor problem in case someone wants to use this as a very strong hint in solving the second half without giving away the 31 resistor solution.

Solution to the 11 resistor problem

Gotta love nerd sniping.

I couldn't put this down, so here is my solution to the 31 resistor problem.

The 31 resistor problem is very similar, but it has more points of symmetry and more connections, so it's kind of a mess to draw. I'm not entirely sure I did this one correctly, but the answer is reasonable. I got $R_{31} = 15/17\ \Omega$.

Here is a diagram of the solution to the 31 resistor problem. It's ugly, but due to the number of points of symmetry it actually winds up being a pretty straightforward solution.

31 Resistor Solution

I love Ernie problems.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I hurried too much during my first drawing of the circuit and left out 1 resistor. The drawings have been redone with the resistor in place. $\endgroup$
    – Jason Patterson
    Nov 10, 2014 at 18:50
  • $\begingroup$ Nice answer, thanks for adding the diagrams (I produced something very similar when 'building' the problem). $\endgroup$
    – Penguino
    Nov 10, 2014 at 20:24
4
$\begingroup$

Since one of Penguino's comments mentions offhand that Ernie's inventions violate all sorts of physics, I think it should be safe to assume that the "weird 31 resistor network" is what results from taking two cubes of resistors and connecting them orthogonally vertex-by-vertex via a fourth spatial dimension, then replacing one edge with the voltage supply.

So now we just need to calculate the equivalent resistance between two adjacent nodes of a 3D and 4D cube of resistors, with the connecting edge shorted out. Luckily, I managed to find this paper where there are values (pg. 28) calculated for any pair of nodes for cubes up to 9D. We're interested in two adjacent nodes, so we can just read the first row of the table. Then, to take into account the voltage source replacing the connecting resistor, I believe you should get answers of

7/5 and 15/17

for the resistances.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I thought about something like this, but figured it was too "lateral thinking" for a challenge not tagged "lateral thinking". I'm guessing it's an inside joke that these "Ernie" problems always have some egregious violation of physical laws involved? +1 for finding the paper, at any rate. $\endgroup$
    – COTO
    Nov 10, 2014 at 4:54
  • $\begingroup$ The problem here is that this is not a cube of equivalent resistors, nor is the 31 resistor problem a hypercube of equivalent resistors. If you found the resistance along an edge of a cube or hypercube you could treat it as a parallel circuit between that $1\Omega$ resistor and the rest of the cube/hypercube (the part we're interested in.) The problem is clear in your answer to the 11 resistor question. We know that there are $1/2\Omega$ resistances leading both out of and into our power supply, so the total resistance of the cube must be greater than $1\Omega$. $\endgroup$
    – Jason Patterson
    Nov 10, 2014 at 5:12
  • $\begingroup$ Oops... yes, I appear to have messed up on those values. You have indeed calculated the correct values in your answer. I wasn't thinking too closely and just thought "product to sum" on the actual values and subtracted 1 from the sum, and clearly that didn't work out. Thanks for catching that! I'll +1 your answer when I am able to :) $\endgroup$
    – SpectralFlame
    Nov 10, 2014 at 5:33
  • $\begingroup$ You got the expected answers so full marks for that. I would love to give both you and Jason P the green tick, but his diagrams won me over. Note that, in my opinion, Ernie's violations of physical laws are only egregious in the archaic sense. $\endgroup$
    – Penguino
    Nov 10, 2014 at 20:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.