spiral macro ?

[cyclestart]

Have one to share?

I'm working on a code generator (in python) and finding it difficult to get a smooth loop in from center. A short example using a 1/2" EM with a target bore of 0.9" and a 0.1" stepover.

From cam program

%
G00 X0.0125 Y0 
Z.1
G01 Z-.1 F20
G03 X-.0625 R.0375
X.1125 R.0875
X-.1625 R.1375
x.1094 Y-.1674 R.1875
X.2 Y0 R.2
X-.2 R.2
X.1094 Y-.1674 R.2
G00 z2
%

Using my generator

%
G00 X  0.0000 Y  0.0000 Z.1 
G01 Z -0.1000 F3 
G01 X  0.0250 F50 
G03 X  0.0000 Y  0.0250 R  0.0250 
X -0.0500 Y  0.0000 R  0.0312 
X  0.0000 Y -0.0750 R  0.0563 
X  0.1000 Y  0.0000 R  0.0813 
X  0.0000 Y  0.1250 R  0.1063 
X -0.1500 Y  0.0000 R  0.1313 
X  0.0000 Y -0.1750 R  0.1562 
X  0.2000 Y  0.0000 R  0.1813 
X  0.0000 Y  0.2000 R  0.2000 
X -0.2000 Y  0.0000 R  0.2000 
X  0.0000 Y -0.2000 R  0.2000 
X  0.2000 Y  0.0000 R  0.2000 
X  0.0000 Y  0.2000 R  0.2000 
G00 Z2 
%

The cam generated code blends the arcs, no awkward jogs.

[NateTG]

In order to avoid 'jogs' when you change arcs, you need to make sure that the new radius is initially parallel to the old one. When you 'average' your radius, the machine moves the center of rotation to a place that you're not accounting for.

I'm new to the CNC thing, but I'm trying to figure out how you get 0.9" bore with a 0.2" radius. spiral. Also, I'm assuming you don't really care about the feed rate.

Let's assume, for a moment, that you're trying approximate a logarithmic spiral with final radius $R (at tool center) centered at 0,0 that goes out some $d per revolution. Here's some pseudocode for spiraling outward:

for(n=0; n*d<(R-d); n++) {
   x[0] y[d*n+d/4] R[d*n+/4]
   x[n*d+2/4] y[-d/4] R[d*n+2d/4]
   x[-d/4] y[d*n-3/4] R[d*n+3/4]
   x[-d*n-5*d/4] y[0] R[d*n+4*d/4]
}

You'll need a little bit of massaging to make it start and finish smoothly.

[psychomill]

... but I'm trying to figure out how you get 0.9" bore with a 0.2" radius.

With a 1/2" endmill....just like he said in his post... And his feed is already set at "50". Keep in mind that most machines and controls only resolve to 4 places on the decimal. Within a macro, you can resolve to 8 (thats 8 places total inluding whole numbers) but machine will still round to 4.

[cyclestart]

Originally Posted by NateTG You'll need a little bit of massaging to make it start and finish smoothly.

Finishing smoothly is np. The ugliness is in the first 180 degrees, maybe from a poor choice of starting point. The CAM examples I've found cheat a true spiral by by starting x= "some distance from true center".

Thanks for the reply. Will test that out that math if my brain can decipher it. Formal education was many years ago LOL.

btw: EM is shop shorthand for endmill. Probably what threw you off.

[NateTG]

Originally Posted by cyclestart Finishing smoothly is np. The ugliness is in the first 180 degrees, maybe from a poor choice of starting point. The CAM examples I've found cheat a true spiral by by starting x= "some distance from true center".

Imagine you've got a square peg where one of the corners is on the center of the circle, and you're unwinding a string from the square peg. Every time that you unwind around a corner, your radius is going to change, but the two centers of rotation are going to be in line with each other (along the string).