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Conduit Section - cables in a tube

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Message 1 of 4
motorhead964
401 Views, 3 Replies

Conduit Section - cables in a tube

Using LT 2009

 

I have to find the minimum inside diameter for a conduit which will hold several cables of similar diameter (0.258").  The number of cables per conduit will be 10, 16 and 20.

 

I have played around with the program but I can't figure out a way to make this work efficiently.  I am trying to account for the fact that different numbers of cables will (or won't) cluster so evenly or uniformly.  That is, with different quantities of cables, there will be some room for the cables to orient unevenly to fill the available spaces.

 

Does anyone know of a way to accurately model this problem?  I have included a .bmp file with 10 and 19 cables in the cluster, just to give an idea.

 

Thanks 

3 REPLIES 3
Message 2 of 4
pendean
in reply to: motorhead964

Wouldn't you just need the exact outside diameter of each cable, and the exact inside diameter of the conduit to draw it just like your image?

Message 3 of 4
motorhead964
in reply to: pendean

I'm trying to find the minimum inside diameter of the conduit to hold the given number of cables.  I guess the real question would be how to properly distribute or arrange the cables, given their number.  Is there any way to do that?

Message 4 of 4
motorhead964
in reply to: motorhead964

Well, I believe I can now answer my own question:  what I am trying to do is referred to as "circle packing," a concept in mathematics.  Here is a website which explains (and calculates, and has images of) drawing circles within circles:

 

http://hydra.nat.uni-magdeburg.de/packing/cci/cci.html#Overview

 

And here is an image of what I'm talking about:

 

http://hydra.nat.uni-magdeburg.de/packing/cci/d1.html

 

Note the glossary on the first link, referring to "rattler" circles, that is, ones which don't fit tightly like all the others.

 

Some of the images for high numbers of circles (i.e., in the hundreds or even 1000) within a circle are kind of cool.

 

That does it for me

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