Resistor Cube Question

[Argus25]

Does anyone remember this old question ? About the resistance from opposite corners of a skeleton cube made of equal value resistors.

The interesting thing about this problem, is that there are so many ways to solve it. Some years ago I came up with an oddball method that I got the idea for after sawing up some metal plates, so I called it the "Bandsaw method" :

http://www.worldphaco.com/uploads/A_...SAW_METHOD.pdf

[kalee20]

Yes, I've seen the problem.

It's fairly straightforward by symmetry... Two sets of three corners are at the same potential, so they can be joined. Then, things simplify to give, as you say, 5R/6.

But what if ONE of the resistors is a different value, say 2R? This destroys the symmetry; you can't do this shortcut; you need the general case, which is much harder. I've not seen a solution to this, and to be honest, I've not tried to solve it myself. I'd attack it by loop or nodal analysis, though I'd expect to get quite a lot of simultaneous equations, and life's just too short!

[Argus25]

Originally Posted by kalee20

This destroys the symmetry; you can't do this shortcut; you need the general case, which is much harder.

Yes, these are the versions that end up in Spice !

My version of it just turned out to be one starting with disconnecting links of the same potential, rather than joining them (the corners) and both methods work just fine, that is, as you say, if the resistors are equal. At the time I did it I had not seen the solution by connecting the corners!

[TonyDuell]

Two other 'cube' puzzles, one to do with resistors, the other not.

8 points at the corners of a cube, each point is jointed to every other point by a resistor R (so edges, face diagonals, long diagonals). What is the resistance between 2 diagonally-opposite corners?

You have a solid cube and suspend it by one corner. Then lower it into water so that half the volume of the cube is under water. What is the shape of the 'hole' in the surface of the water?

[ms660]

I'd just measure it with a meter I'll get m'coat.

Lawrence.

[Ancient Geek]

Originally Posted by TonyDuell

You have a solid cube and suspend it by one corner. Then lower it into water so that half the volume of the cube is under water. What is the shape of the 'hole' in the surface of the water?

Since you did not specify that the cube is of uniform density, it will depend on the location of the centre of mass of the cube

[CambridgeWorks]

I think I agree with Lawrence. I must say I do not recall ever having the need to find a solution to a practical problem anything like this. However, it would exercise the grey matter, if that is what you wish. Personally, sometimes I don't count sheep to get off to sleep, I just sometimes use mental arithmetic counting up to the power of 2. After 32768 I slow down a little. Honest! Try it sometime!
Rob

[Argus25]

Originally Posted by TonyDuell

You have a solid cube and suspend it by one corner. Then lower it into water so that half the volume of the cube is under water. What is the shape of the 'hole' in the surface of the water?

Since the axis of the cube is chosen from corner to opposite corner and the cube is lowered into the water to bisect the cube in half, then it would require that the water edges wraps around all six sides of the cube, so my guess the shape of the hole at the water surface is a perfect Hexagon.

[Radio Wrangler]

But that does assume uniform density, as has already been pointed out, and also that the suspension is freely flexible if it doesn't already happen to be in the right direction if rigid.

The trouble with these sorts of puzzles is in always bunging up the holes someone can pick in it.

Still, a hexagon from a cube is nicely counter-intuitive

David

[Argus25]

Originally Posted by Radio Wrangler

But that does assume uniform density, as has already been pointed out,

I couldn't see that there was a density issue to consider at all, because the cube could not be lowered into the water definitely past half its volume unless its overall density was high enough, the uniformity of the density is not important.

So a polystyrene cube wouldn't do it. So I just assumed it was something like a solid metal cube lowered into the water until half the volume of the cube was in the water and half out.

So perhaps the question could be slightly modified to say something like "a solid metal cube" rather than just "solid cube" so it is obvious to anyone that it would sink in the water if it was not suspended.

TonyDuell; is the hexagon answer correct ?

[TonyDuell]

Yes, the shape of the cut is a perfect regular hexagon.

And yes, I believe the uniformity of density does matter. You require that the cube hangs below the point of suspension so that the diagonally opposite corner is directly below the one you suspend it from. If the centre of mass is not on the long diagonal joining those 2 corners then the cube will not hang in the right way, the cut will not be a regular hexagon.

Oh, and I guess we are neglecting surface tension effects of the water.

[Argus25]

Yes, I guess if the density was such that the cube had a focus of mass that was off axis with the axis of the point it was suspended from, it could "hang at a tilted angle" like some things do in biological systems

As for the surface tension effects, lets assume the cube is 50m on each side, so the surface tension effects and the meniscus at the "water to cube face boundary" are too small to count

Hugo.

[dave cox]

Here is a similar but different one (please excuse the drawing tool I used).
What is the resistance from A to B with a resistor value R ?

This is the 'complete' graph K5 (5 nodes with everything connected to everything else via a resistor R) , but what would it be for the graph K<n> ?

dc

[kalee20]

Originally Posted by Argus25

As for the surface tension effects, lets assume the cube is 50m on each side, so the surface tension effects and the meniscus at the "water to cube face boundary" are too small to count

And that the said 50m cube is made of solid osmium too, so that upthrust of water and any associated currents are negligible.

It's great that the intersection with the water surface is a regular hexagon. I'm thinking now, how to calculate the length of a side of the hexagon.

[Ancient Geek]

Originally Posted by dave cox

Here is a similar but different one (please excuse the drawing tool I used).
What is the resistance from A to B with a resistor value R ?

This is the 'complete' graph K5 (5 nodes with everything connected to everything else via a resistor R) , but what would it be for the graph K<n> ?

dc

a) 0.4R ?
b) 2 * R / n ?

[dave cox]

I should confess to not knowing the answer to either part of my 'K5' question

The Ancient Greek's answer for (b) looks good for N { 2, 3, 4 }
Doing it the hard way for N=5 ...

dc

[Skywave]

Originally Posted by dave cox

This is the 'complete' graph K5 (5 nodes with everything connected to everything else via a resistor R) , but what would it be for the graph K<n> ?

When you say "5 nodes with everything connected to everything else via a resistor R" does that mean that resistors R are only connected to the node points (totalling 10 resistors) or that every line in your drawing (including where the lines join) represents a separate resistor (totalling 20 resistors)?

Al.

[short wave]

Saw this some time ago on y tube on a similar theme using a triangle in "2D" and "3D"

https://www.youtube.com/watch?v=OC09Br5JkIk

[dave cox]

To clarify, yes, 10 resistors.
Al, I didn't think of the lines crossing !

Interestingly, K5 is the simplest graph that CANNOT be drawn on a flat sheet of paper without the lines crossing - Kuratowski's theorem. If you should have toroidal paper (!) you can draw K5 and K6 without crossing.

Well spotted, the video is K3 and K4 !

dc

[ionburn]

I am not getting into the solving of this but I have come across and article when browsing that is relevant.

See 'Radio Constructor' December 1963 (American Radio History site if nowhere else). An article therin simplifies the issue.