@John Cuthber

Sorry no proof..and it can not work if there's no need.

If we intersect each 3 points with 3 lines forming "Y" @ 120 deg w/ each other,

that will be the minimum connecting length. Unless the 3 points are making > 120 angle

because the lines connecting them is already the minimum.

 

I tried this with the cubic room where a spider web connects the 8 corners .It works

With 13 segments network.All angles of the adjacent web segment are 120 degrees.

e.i. for the minimum length.

 

AutoCAD is the best for this. No math needed. Just click lines to read measurements.

If you copy the sketch I did into a CAD package it will measure the lengths of the lines for you- but it will not tell you if they are a minimum.

If you draft the equations for the total length as a function of the angle (90 degrees in my sketch, 120 in Sensei's) and then differentiate it wrt that angle you will find the minimum length for that layout- So maths clearly helps. but it's not the best layout.

Whilst not the same it strikes me as very similar to the travelling salesman which is np-complete. The fact that even a simplifed version (just the four pockets of the RHS and middle) took me lots of attempts to find my preferred solution which I am by no means sure is the minimal solution makes me think scaling is very very difficult and proof even more so - which again sounds reminiscent to tsp

snooker tabke.JPG

 

 

Why didn't you use the same trick when connecting a1 and c1 connections?

I'm a little surprised that the solution (assuming that's an optimum) isn't symmetrical.

Here is the mirror way done with CAD