Geometry question, need outside diameter of 3 grouped objects

[sciguyjim]

If I have 3 tubes, each 1" diameter grouped together at 120° angles, what would be the diameter of a larger tube which could enclose the 3 smaller tubes? There must be a formula, or 2, to calculate this. Thanks.

 

[Mandrake22]

I don't know of a formula but why not fire up "your favorite CAD program", create your four circles and then dimension the "all encompassing" one.

 

[Kiranpatel]

My guess at the answer is that it will be D*(1+1/Cos(30)). I don't have a CAD package to verify it though.

 

[sciguyjim]

I don't have any CAD software, just pencil, paper & a calculator.

 

[sciguyjim]

Kiranpatel, would I solve for "D" in your equation?

 

[prex]

This is quite simple: as the tubes have equal diameters their CLs will be on a regular triangle.
Side of the triangle: d
Height of the triangle: d*cos 30°
Distance of one vertex to center of triangle: 2/3*d*cos 30°
Diameter of the enclosing circle:
D=(2*2/3*cos 30°+1)d=2.155d. prex
motori@xcalcsREMOVE.com

http://www.xcalcs.com

Online tools for structural design

 

[sciguyjim]

So "d" here is the diameter of the small tubes (1&quot. This looks good, at least better than the figures I was coming up with. Thanks a lot.

 

[Kiranpatel]

Sorry, my post was not clear enough. D is the diameter of each of the 3 tubes (1&quot. The term D*(1+1/Cos(30)) represents the diameter of the envelope circle. I note that the number is the same as that derived by prex.

 

[sciguyjim]

I see, thanks Kiranpatel & Prex.

 

[bobc]

http://mathworld.wolfram.com/CirclePacking.html

please try this site, i think it will answer your question.
bob creely

 

[sciguyjim]

Bob, thanks for the great link, that's exactly what I was looking for.

 

Being an X-tube welder I have to agree with Prex.