I see it now. Applying the 15° after the 60° changes the 60° to 60.8526°. I should have paid better attention to you earlier in the post.
Unfortunely arronbee has not reponded to any of the answers so we don't know any more than the original post.
aaronbee asks if Rhino can unwrap the 3D model or is there some other method to get the tool path.
I believe there is only one interpretation to aaronbee's post. "projected angles".
The two angles combined (60 & 15 deg) equates to 60.293 deg. A drawing using this single angle will give the correct face.
To draw the face when rotating the part, 60 and 7.6307 deg will give the correct angled face.
Last edited by Kiwi; 08-02-2014 at 12:47 AM. Reason: .
I see it now. Applying the 15° after the 60° changes the 60° to 60.8526°. I should have paid better attention to you earlier in the post.
When illustrating the projected angles of the 60°/15° an angle of 29.7074902° is produced.
Thank you. That totally did exactly what I wanted.
And thanks to everyone who responded. I don't think I made it exactly clear that I'm rotating the pipe (x axis) under the torch and the torch only moves across the length of the pipe (y axis). I'm not sawing across it like a chop saw would at some angle.
I took a 1.75" circle in Rhino, extruded a surface, rotated it 15 and 60 degrees, split it, UnrollSrf and ...BAM! ...just what I wanted. Now I have a flat, sine-wave-looking curve that I can use to tell my torch how to cut the pipe.
Also, it is true that I won't get a beveled edge on the cut like you would on a tubing notcher, but this is mating to a flat surface and the wall thickness is like 13 gauge, so whatever gap it has will be minimal and completely acceptable for my application. UnrollSrf was exactly what I was looking for. Thanks.