1. ## O.D. Groove

A few days ago my boss asked me how to calculate a 1" o-ring groove on the outside of a 4" diameter part using a 1" fly cutter on a 4 axis mill using a rotary table and, it has to be the same depth. I've include a crude drawing to help explain.

http://s21.photobucket.com/albums/b2...ringgroove.jpg

2. ## Fly cutter won't work.

A fly cutter will not work. Too hard to explain.
Use a slot mill cutter that gives the correct width of cut or smaller and do a pass for inner and outer edge.

General idea is to cut a circle, like you would on a flat plate, but instead of x and y moving to create the circle you want X and A.

first set zero at the centre of the circle.
move X the 1/2 inch. G1 X-0.5
then for CW G3 I0 J0.
That would cut a circle on the flat.

What you need is for the Y part of the move to be replaced with an A move.
A is in degrees so you have to convert the Y value to degrees. for 4" that is 12.5664 circumference. divided by 1/2" (radius) gives 360/25.1328 which is +/-14.3239 degrees. So for original y going between +0.5 and -0.5, rotary head needs to go Y value multiplied by 28.6478.
You may have to create the circle from small steps of X and A, unless you can slave the Y (scaled ) to the A axis, and disable the Y moves.
I don't know how to do that in G code, but you can calculate your own circle values with sin() and Cos() functions in most controllers.
If you learn how to make your own circles on the flat without G2/G3 then you will only need to substitute the new A value, for what was calculated for Y (scaled)
The formula for a circle is X squared + Y squared = R squared if you can't use sin/cos.
As an algorithm where R is radius
FOR D = 0 to 359 with steps 0.1 to give good resolution better than 1 thou
X=SIN(D) * R
Y=COS(D) * 14.3239
NEXT
I've never tried this, but I think it is close to the mark.
If the rotary part is not digitally controlled, this will be hard work.

3. What make/model of machine control are you using? I don't suppose your machine has Cylindrical Interpolation, does it? If it does, use that. If it doesn't, too bad. You'll need to break the circle up into little-bitty line segments as neilw20 details above.

4. This is what one 90 degree with 1 degree steps would look like. This is for a 1 inch circle on a 4 inch cylinder. The Y axis remains on zero and the X and A axis do the cutting. You do need a full 4th axis to do this.

N10 G00 X.5 Y0. A0. Z.1
N20 G01 Z-.1 F5.
N30 X.4999 A-0.25
N40 X.4997 A-0.5
N50 X.4993 A-0.75
N60 X.4988 A-0.999
N70 X.4981 A-1.248
N80 X.4973 A-1.497
N90 X.4963 A-1.746
N100 X.4951 A-1.994
N110 X.4938 A-2.241
N120 X.4924 A-2.487
N130 X.4908 A-2.733
N140 X.4891 A-2.978
N150 X.4872 A-3.222
N160 X.4851 A-3.465
N170 X.483 A-3.707
N180 X.4806 A-3.948
N190 X.4782 A-4.188
N200 X.4755 A-4.426
N210 X.4728 A-4.663
N220 X.4698 A-4.899
N230 X.4668 A-5.133
N240 X.4636 A-5.366
N250 X.4603 A-5.597
N260 X.4568 A-5.826
N270 X.4532 A-6.054
N280 X.4494 A-6.279
N290 X.4455 A-6.503
N300 X.4415 A-6.725
N310 X.4373 A-6.944
N320 X.433 A-7.162
N330 X.4286 A-7.377
N340 X.424 A-7.591
N350 X.4193 A-7.801
N360 X.4145 A-8.01
N370 X.4096 A-8.216
N380 X.4045 A-8.419
N390 X.3993 A-8.62
N400 X.394 A-8.819
N410 X.3886 A-9.014
N420 X.383 A-9.207
N430 X.3774 A-9.397
N440 X.3716 A-9.585
N450 X.3657 A-9.769
N460 X.3597 A-9.95
N470 X.3536 A-10.129
N480 X.3473 A-10.304
N490 X.341 A-10.476
N500 X.3346 A-10.645
N510 X.328 A-10.81
N520 X.3214 A-10.973
N530 X.3147 A-11.132
N540 X.3078 A-11.287
N550 X.3009 A-11.44
N560 X.2939 A-11.588
N570 X.2868 A-11.733
N580 X.2796 A-11.875
N590 X.2723 A-12.013
N600 X.265 A-12.147
N610 X.2575 A-12.278
N620 X.25 A-12.405
N630 X.2424 A-12.528
N640 X.2347 A-12.647
N650 X.227 A-12.763
N660 X.2192 A-12.874
N670 X.2113 A-12.982
N680 X.2034 A-13.086
N690 X.1954 A-13.185
N700 X.1873 A-13.281
N710 X.1792 A-13.373
N720 X.171 A-13.46
N730 X.1628 A-13.544
N740 X.1545 A-13.623
N750 X.1462 A-13.698
N760 X.1378 A-13.769
N770 X.1294 A-13.836
N780 X.121 A-13.898
N790 X.1125 A-13.957
N800 X.104 A-14.011
N810 X.0954 A-14.061
N820 X.0868 A-14.106
N830 X.0782 A-14.148
N840 X.0696 A-14.185
N850 X.0609 A-14.217
N860 X.0523 A-14.245
N870 X.0436 A-14.269
N880 X.0349 A-14.289
N890 X.0262 A-14.304
N900 X.0175 A-14.315
N910 X.0087 A-14.322
N920 X0. A-14.324

• Originally Posted by Stickmchn
A few days ago my boss asked me how to calculate a 1" o-ring groove on the outside of a 4" diameter part using a 1" fly cutter on a 4 axis mill using a rotary table and, it has to be the same depth. I've include a crude drawing to help explain.

http://s21.photobucket.com/albums/b2...ringgroove.jpg
Is your machine able to do Cylindrical Mapping? With this you simply write the program for the X Y plane and then tell the machine to map it onto the cylinder.

• ## O.D. Groove

We recently did some work milling this type of groove and used a Excel spreadsheet to calculate 1/2 degree moves. If it is of interest let me have your email address and I will send it along with some instructions. We now have a VMC that does helical interpolation in 4 blocks per rev instead of 720.

• Attached is a VB program which will generate code.

• ## Cut and paste is great!

Originally Posted by Kiwi
Attached is a VB program which will generate code.
Well done.

• To determine the angle is the 1 in measured around the arc or square to the radius?

• ## Maybe neither!

Originally Posted by Kiwi
To determine the angle is the 1 in measured around the arc or square to the radius?
They might want it line up with a hole in something.
The it would be a projection of the hole, which might mess the math slightly.
As the diameter of the shaft gets smaller, then the angle becomes larger.
Now we need to know how deep the cut is. It's 3D world. (LOL)

• I noticed other examples used the arc length to determine the angle where I used the angle of a 1in flat sq to the axis.
This is how I see the outcome:
Arc length would need a 1in O ring but groove undersize.
Flat angle method gives a 1in dia groove but needs a larger than 1in O ring.