How can I make a pattern to cut a metal pipe at 45 degrees?

How can I make a pattern to cut a heavy metal pipe at 45 degrees? I intend to rotate the pipe and weld it to create a 90 degree elbow. For sake of example, let's say the pipe is 1 foot in diameter.

Reply to
Petrovitch
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What are you going to cut it with? It really isn't that hard. If you draw two lines at 45 degree angles on your welding table and line the pipe up with one you can look down on it from the top and eyeball some dots, then put the pipe in your abrasive chop saw so it looks right and chop it. If there's a little gap, well, you're a welder, right? Fill 'er up regular ..

GWE

Reply to
Grant Erwin

How about a very big joint jigger?

Reply to
Tony

I remember doing this as an exersize in my drafting class in high school, before computers. You end up with a piece of paper to cut out and wrap around the pipe. If you can find an old hand drafting textbook at the library, it would have this section. it only takes a couple minutes with a drafting machine.

I'm sure i could do it in autocad, with great difficulty and much frustration.

Reply to
Karl Townsend

here's the idea. The lower right is the template. The scale is the circumference of the circle. The line on the top of the template is done with a french curve instead of like I did it.

black= construction lines. green= views of part

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P.S. I did this far in autocad for a 12" circle. If you have autocad, email me.

Reply to
Karl Townsend

"Karl Townsend" wrote in message news:BFcHg.13275$ snipped-for-privacy@newsread1.news.pas.earthlink.net...

Isn't there a special name for this?

I used to have an old machineries h/book with sheet metal layouts, templates & that sort of stuff.

Reply to
Borat

Try this site for the procedure.

Reply to
R. Zimmerman

If you are intending to carry fluids inside this pipe you might consider cutting at 22.5 degrees and making two joints. This is also preferable if you are making a barrier. The sharp end of a 45 degree cut joined at 90 is not pleasant to bump into. The pattern development procedure is similar. Randy

Reply to
R. Zimmerman

Three dimensional descriptive geometry is what we called it...

Reply to
Rick

That's really neat Karl. It reminds me of a trick an old guy showed me once. If you need a perfect 90 degree cut in a pipe and you only have a hand hacksaw, wrap a piece of writing paper around the pipe so that the edges line up, tape it in place and use it to guide your cut. I've used it lots of times. Very neat trick.

Chris

Reply to
Christopher Tidy

How did you get to see it? The link is broken and I can't find it by going to the Dropbox.

Bob

Reply to
Bob Engelhardt

Find a copy of the Pipefitters Handbook great for that sort of stuff. You can either make a pattern or do a direct layout with soapstone. Very quick and simple with the tables in the book.

ED

Reply to
ED

For what it's worth, that's a very simple pattern. You make a pattern sheet as long as the circumference of the pipe.

If it had an o.d. of exactly 12", that would be a pattern 37.7" long

Make the pattern about 1-1/2 times the diameter of the pipe; say, 18" wide.

Draw one full sine wave from one end to the other, starting with a zero-crossing at the end of the sheet, and ending with the last zero-crossing at the other end, and the center crossing at exactly the center of the sheet. In other words, you'd have the -say- negative half of the wave on the left hand end of the sheet, and the positive end on the right.

Make the peak-to-peak amplitude of the sine wave to be equal to sqrt(2) * diameter; or in this case, 16.97".

Cut or perforate along the waveform, and wrap the sheet around your pipe. Pattern.

LLoyd

Reply to
Lloyd E. Sponenburgh

Look at this

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--Andy Asberry recommends NewsGuy--

Reply to
Andy Asberry

I should have added that, for any angle cut you desire, make the amplitude of the sine wave equal to the diameter of the pipe divided by the cosine of the angle.

LLoyd

Reply to
Lloyd E. Sponenburgh

Ok... that was really stupid of me... I'm sitting here looking at my own figures and quoting the whole mess wrong.

For 45 degrees, the amplitude of the wave is equal to the diameter -- that is (1).

For any other angle, the amplitude is equal to the tangent of the angle (which for 45 degrees, happens to be 1) divided by the diameter.

Duh..

Sorry.

LLoyd

Reply to
Lloyd E. Sponenburgh

Oh, CHEEZE! I'm having too many finger/brain infarctions today!

The amplitude is equal to the TANGENT of the angle TIMES the damned diameter! ARRRRRRGH!

(yeah... I think I actually typed it correctly that time... DAMN!)

LLoyd

Reply to
Lloyd E. Sponenburgh

and for what it's worth, that would make the amplitude for a 45-degree cut to be equal to the diameter.

Sigh... It SUCKS getting senile!

LLoyd

Reply to
Lloyd E. Sponenburgh

I've got an old Audel's handbook with this graphical solution decribed in it. If you don't find what you're looking for elsewhere, I can scan this and send it to you.

Reply to
Gary Brady

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simple program

Reply to
BIGJIMS

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