@weber @plonkster These are reasoned and rational arguments and I appreciate that crimps and other things have to be considered in the real world.

That is why for a bit of fun, like so many riddles,  I attempted to take the puzzle out of a real world context.

 

20 hours ago, phil.g00 said:

crimp symmetry might make a difference, so solid gold, super -conductor , -273 degree Kelvin crimps everybody.

I wholeheartedly admit to a bit of playful denial of reality.

 

13 hours ago, phil.g00 said:

but I remember Y2K and Santa Claus and sometimes I doubt things.

 

However, what you may gather from my opening post this question did start out as a real world issue.

I  inherited the care of an existing  24 string  x 24V  battery bank ( it's 48V now),  going back in time and changing buying decisions or throwing out good batteries were not options.

Those are real world givens, the type of givens that are common enough,  doing the best with what you've got, that is the real world..

 

7 hours ago, weber said:

Chris Gibson deserves to get credit for those diagrams

I C&P'd the googled diagrams off this webpage:

https://www.wanderlodgeownersgroup.com/forums/showthread.php?t=27488

Which doesn't have a diagram my solution.

It does actually reference the the page you mention now that I check.

And it seems Chris Gibson may have amended kis page to include the solution later, as he  in turn gives credit to:

"Many thanks to "smileypete" from www.canalworld.net/forums for this idea."

And I  couldn't be bothered to the check the provenance any further.

So cheers to Smileypete.

But to say, there is nothing new under the sun I suppose.

 

OK. Now that we've got reality out of the way. Back to the fun puzzles.

 

Here's a corrected version of TTTs 8-way. The green numbers are the first pass I mentioned above, and the purple numbers are the second pass.

 

So now, who can balance 3 batteries (with some doubled-up links), and 5 batteries (without)?

 

@weberThere you go, well done.

@weberI 've had a go at the 5 battery setup,  which is easy enough to match cable distance.

I must admit though, I am battling to match the number of crimps in the loop.

I do like a good puzzle, so without giving away the answer, does the solution have an equal number  of loop connection points?

Or we relying on my lossless connections?

Sorry @phil.g00. I should have explained that my particular denial of reality consists in every jumper having the same resistance. That could be taken to mean that all the jumpers are of the same length (e.g. they might be supplied by the battery manufacturer and are capable of skipping a battery, as in the real-life case that prompted this), or it could be taken to mean that varying the lengths of the jumpers has no effect because the cable has zero resistance and all the resistance is in the crimps, all of which are the same. The resistance of the bolted joints has to be zero, so that lug stacking order doesn't matter. I realise that's different from your original criteria.

Edited January 13, 20197 yr by weber

So like this?

The middle battery has an extra crimp.

Edited January 13, 20197 yr by phil.g00

39 minutes ago, phil.g00 said:

So like this?

No, I see 4s and 3s (see attachment).

 

I'm not sure I understand the question. Are you proposing that as a solution, @phil.g00? Because it certainly isn't one. And it's amazingly asymmetrical.

Edited January 13, 20197 yr by weber

Ah, I get you now.

You can think of these numbers as representing volt drops in millivolts if every battery was carrying 1 amp and every jumper had a resistance of 1 milliohm.

I analysed my solution in the case where all the resistance is in the bolted joints. There is a lug stacking order where the ratio of lowest to highest volt drop is 18:19.

I think I understand the question now,  but how to optimize the connections beyond conventional escapes me.

I'll await the solution to see what I am missing.

I believe that I've worked out the "perfect sharing" for 5 modules, using Weber's constraints. I basically did trial and error, with an eye towards symmetry, and stumbled onto it.

But since Weber and I discussed this years ago, I could have some vestigal memory of how it went, so it doesn't seem fair to claim victory here.

It seems to me that you'd nearly always have buddy pairs, and for symmetry with 5 modules these would have to be on the outside. The connections would be diagonal, of course. Then the main variable is the position of the takeoff points. Again with symmetry, there are only three possibilities, and one (takeoff from the middle module) can be eliminated immediately, since that module will have a second pass total of zero. So that cuts down the search space a lot.

I can't think of a simple proof that symmetry is required; it just seems reasonable that it would be.

I think I am missing something, are we talking about this kind of carry-on?

 

No, there are no unnecessary paths like that. But everything Coulomb says about the solution is true. To more directly answer an earlier question: It does not have an equal number of crimps or connection points in each loop.

Well, if it's not this then I am really not getting the question.

You got it. Well done. 😁

Can you find the optimum lug stacking order? Assume all the resistance is in the bolted joints and that it is the same from lug to lug as it is from lug to terminal.

 

1 hour ago, weber said:

Can you find the optimum lug stacking order?

Yeah, easy enough I think. Just divide and conquer between what current goes on an and what current stays behind at each node.

Minimizing superfluous current travel across junctions  and subsequently the voltage drops --- that's just a exercise.

A worthwhile exercise, I grant you, but too logical to pique my interest as a puzzle.

 

Edited January 14, 20197 yr by phil.g00

        1_    _1__2      5+2+1+1+2+3+5 = 19
       /         \ \
__5__2_\1_    _1_/ /     5+2+1+1+2+3+5 = 19
   /              /
   \_3 _1_    _1_/3_     5+3+1+1+3+5   = 18
      /             \
     / _1_    _1__2_/5__ 5+3+2+1+1+2+5 = 19
    / /          \
    \2\_1_    _1_/       5+3+2+1+1+2+5 = 19