How Do I Make an Equilateral Triangle from Copper Piping?

[Evastar]

I want to make an icosahedron from copper piping, approx 1.5 metres in diameter.

http://www.rogersconnection.com/triangles/images/icosahedron.jpg

An icosahedron is made from 20 equilateral triangles joined together, but i can't find any 120 degree elbow joints so need to find an alternative way to joint the copper.

Anyone any suggestions for me?

[gazz10]

get a pipe bender, not all that expensive, am sure pick one for the price u going to pay using connections, and then just solder the triangles together

[Evastar]

Thanks for the suggestion, but already thought of that. No way a pipe spring would bend it to that angle, it would get stuck in the joint. Plus the copper is likely to split.

[gazz10]

not a pipe spring a pipe bender, pipe bender can bend to that angle no problem, and a pipe spring could do the same 180 to 120 not all that much.

and why 120...equilateral triangle should be 60

[xant14]

Alright Eva, an interesting piece.
A pipe bender, or spring will bend it no problem, but you will struggle getting the right angle, as it is bent one way and then three dimensionally the other way too. and doing that with a spring, I reckon you'd make a mess.
The way I'd do it, is cut individual tubes, to the exact size, using a jig so none are even a millimetre different. Then you get big ball bearings to make your connection, big enough to take 5 tubes as your pic shows. Tack them on with solder or even glue to start. And work your way untill you have your shape, probably breaking some joints to get it right, then solder the whole thing up.

maybe not what u want, copper and chrome. What's it for? a garden ornament?

Have you ever seen the game magnatron I think it is? I bought my little one it last year, ball bearings and strong magnets, i have made that shape myself. although not 1.5m big.

[father ted]

does it have to be copper?get someone to mig or tig weld some pics of pipe you can notch out the tubing so that it will take the balls ,a good fabricator/welder can do that np

[Mjolinor]

Baking my noodle on this.
Take a ball of radius 1

If you draw an equalateral triangle on a bit of paper then look at it each angle is 60 degrees, if you have one of the points close to you and the opposite side horizontal (triangle upside down) then you rotate the top line away from you what you apparently see is the angle at the bottom getting bigger. When that angle gets to 72 degrees we need to know the apparent length of the other two sides.

Using trig, A/sin(A) = B/ sin(B) so we make the length 1 for easy arithmetic, 1/sin(72) = X/sin(54)

this means that X = 0.851

This is the radius of the circle where 5 of the equilateral triangle sides will fit.

Now we need to convert this to a height above centre on our ball, this is Pythagoras. sqrt(1^2 - 0.851^2) = Height above centre.

It works out at 0.526 above centre.

Where is my mistake as it doesn't seem to work.

[xant14]

yeah I was gonna say that...no I wasn't

[chroma]

Get brake pipe.
Its copper and extremely maleable.
Bend it all by hand

[forntida]

Baking my noodle on this.
Take a ball of radius 1

If you draw an equalateral triangle on a bit of paper then look at it each angle is 60 degrees, if you have one of the points close to you and the opposite side horizontal (triangle upside down) then you rotate the top line away from you what you apparently see is the angle at the bottom getting bigger. When that angle gets to 72 degrees we need to know the apparent length of the other two sides.

Using trig, A/sin(A) = B/ sin(B) so we make the length 1 for easy arithmetic, 1/sin(72) = X/sin(54)

this means that X = 0.851

This is the radius of the circle where 5 of the equilateral triangle sides will fit.

Now we need to convert this to a height above centre on our ball, this is Pythagoras. sqrt(1^2 - 0.851^2) = Height above centre.

It works out at 0.526 above centre.

Where is my mistake as it doesn't seem to work.

You've been looking over Zaps' shoulder. That is plagiarism.

[zaphodbb]

You've been looking over Zaps' shoulder. That is plagiarism.

i agree deeyam cheek hmmm what are we talking about again ?

[Evastar]

Thanks for all the help everyone.

Yes, it has to be copper. I can't go for the lighter copper because it needs to be the size it is and wouldn't be rigid enough in the 10 mm copper.

The reason i say 120 degree angle instead of 60 is because most people making joints count from the outside of the angle, i was looking at 60 deg angles on the net and they were the wrong angle.

We are going to go looking for suitable parts today, so i will keep you all posted.

Any more ideas in the meantime would be great

[Evastar]

We did it!

Or should i say my hubby did it! After hours looking around the plumbers suppliers and B&Q he tried some 90 deg elbow joints, cut the 1/2 inch pipe to size and bent the 90's to 120's/60's by brute force. We had thought they would break or split, but surprisingly enough they didn't! So after making the correct shape he heated them at the joints to melt the solder, and it worked perfectly!

Now to join all the triangles together

[zaphodbb]

We did it!

Or should i say my hubby did it! After hours looking around the plumbers suppliers and B&Q he tried some 90 deg elbow joints, cut the 1/2 inch pipe to size and bent the 90's to 120's/60's by brute force. We had thought they would break or split, but surprisingly enough they didn't! So after making the correct shape he heated them at the joints to melt the solder, and it worked perfectly!

Now to join all the triangles together

well eva we are all waiting for pics of the ic icos hmmm ah the icosahedron ? so when your ready dear ???

[Evastar]

Well it's not finished yet zap, but all the triangles are made and welded, the middle section is put together and bolted with brass bolts, the top and bottom sections are bolted together, the top is just joined on with plastic ties at the moment, needs to be attached properly, and the base needs to be made.

I have a couple of pics of what's done so far, here they are:

Some of the triangles:



The base:



The middle and the top section: