[uperez]

Hey fellas. I'm trying to offset-turn a part.

This is the code that i made.

M35 (c-axis engage)
T0303
G98G0Z2.C0.M8
X90.Y0.
#100=2.
N100G17G3I-3.J0.W-0.2H-360.F3000.
#100=#100-0.2
IF[#100GE-150.]GOTO100
..
..

Machine is Puma3100LY with Fanuc I-series control.

Now for some reason the machine is popping out an alarm "0020 over tolerance radius".
I tried to run the code omitting the H-360. for c axis and it worked fine.
I don't see any reason why the arc calculation should be affected by this since it's not lying on the same plane where the circle is.
And the manual says 2 more axes can be added as long as it's not on the same plane.

I can mill the jaws to offset the part but that would be an additional set up and there's very small stock to clamp the part, only 12mm so I can't imagine turning this without the tailstock engaged and I won't be able to do that using offset jaws.

Anybody got any idea about this?


Uly

[uperez]

Anyone?..

[fordav11]

over tolerance of radius means your arc does not intersect or touch your end point.
try using R instead of I J and it'll probably work but the arc will still not be correct. but it should work without an alarm at least.
normally on a G2/G3 the X (or U) is also specified on the line. If you omit the X (U) it assumes the start point in X is the X end point as well. if the tool can't reach the end point with the specified I J arc center coordinate you'll get that alarm.
also you're in G17 which is the XY plane and you have a W end point (Z) so technically you are doing helical interpolation.
what exactly are you trying to achieve? your program seems overly complicated and convoluted.

[uperez]

over tolerance of radius means your arc does not intersect or touch your end point.
try using R instead of I J and it'll probably work but the arc will still not be correct. but it should work without an alarm at least.
normally on a G2/G3 the X (or U) is also specified on the line. If you omit the X (U) it assumes the start point in X is the X end point as well. if the tool can't reach the end point with the specified I J arc center coordinate you'll get that alarm.
also you're in G17 which is the XY plane and you have a W end point (Z) so technically you are doing helical interpolation.
what exactly are you trying to achieve? your program seems overly complicated and convoluted.

Hey. Thanks for the reply. What I'm trying to do here is off-center turning (eccentric). Yes this is helical interpolation with the addition of the c-axis rotating 1 rev per pitch of the spiral. I have tried running it without the H-360.(-360 degrees increment in c axis) and the code worked fine. So I was wondering why adding a c axis move affects the calculation of the arc when it's not even on the same plane. I have made this work by using G1 and breaking the spiral along with the c axis movement in to small segments. But it would be damn lot easier if it could be done using G2/G3. Then I wouldn't have to try and see how long segments are to run fast but smooth enough also.

[ProProcess]

Unless I'm missing something here, to do what you want, you'll need to use milling interpolation.
With MI, you would position "X" at the [radius + eccentric], then G2/G3 with an "I-radius".

[ad64075]

I've done this many times. You NEED to use polar interpolation. If you try to generate an arc using G2/G3 in normal contouring mode, you will get an elliptical shape because the machine it seeing only your X-Y plane (G17) with the center point remaining stationary. During "offset turning," your center point is constantly moving, therefore the use of polar interpolation is necessary.

The following program, for example, is for a Fanuc 18i, generated to create a 1" diameter, 1/4" off center, on 1-1/2" material.


G50S3000
G0G28U0.V0.
G53E23.25
G28W0.

N1T0101
G96S650M4P1
G0G54Z.1
X1.6
M8
G1Z0.F.01
X.6F.005
G0X1.6Z.1

(BEGIN POLAR)

M23
G28H0.
G50C0.
G0X1.5Z.1C0.
G1G98Z.01F10.
G112
G3X-1.C0.R.625Z0.F400.
X1.5C0.R.625Z-.002
X-1.C0.R.625Z-.004
X1.5C0.R.625Z-.006
X-1.C0.R.625Z-.008
X1.5C0.R.625Z-.01
X-1.C0.R.625Z-.012
X1.5C0.R.625Z-.014
X-1.C0.R.625Z-.016
X1.5C0.R.625Z-.018
X-1.C0.R.625Z-.02
X1.5C0.R.625Z-.022
X-1.C0.R.625Z-.024
X1.5C0.R.625Z-.026
X-1.C0.R.625Z-.028
X1.5C0.R.625Z-.03
X-1.C0.R.625Z-.032
X1.5C0.R.625Z-.034
X-1.C0.R.625Z-.038
X1.5C0.R.625Z-.04
X-1.C0.R.625Z-.042
X1.5C0.R.625Z-.044
X-1.C0.R.625Z-.046
X1.5C0.R.625Z-.048
X-1.C0.R.625Z-.05
X1.5C0.R.625Z-.052
X-1.C0.R.625Z-.054
X1.5C0.R.625Z-.056
X-1.C0.R.625Z-.058
X1.5C0.R.625Z-.06
X-1.C0.R.625Z-.062
X1.5C0.R.625Z-.064
X-1.C0.R.625Z-.066
X1.5C0.R.625Z-.068
X-1.C0.R.625Z-.07
X1.5C0.R.625
X-1.C0.R.625
X1.5C0.R.625

G113

G0X4.M9
G28U0.V0.M24
G53E23.25
G28W0.
M30

[uperez]

I've done this many times. You NEED to use polar interpolation. If you try to generate an arc using G2/G3 in normal contouring mode, you will get an elliptical shape because the machine it seeing only your X-Y plane (G17) with the center point remaining stationary. During "offset turning," your center point is constantly moving, therefore the use of polar interpolation is necessary.

The following program, for example, is for a Fanuc 18i, generated to create a 1" diameter, 1/4" off center, on 1-1/2" material.


G50S3000
G0G28U0.V0.
G53E23.25
G28W0.

N1T0101
G96S650M4P1
G0G54Z.1
X1.6
M8
G1Z0.F.01
X.6F.005
G0X1.6Z.1

(BEGIN POLAR)

M23
G28H0.
G50C0.
G0X1.5Z.1C0.
G1G98Z.01F10.
G112
G3X-1.C0.R.625Z0.F400.
X1.5C0.R.625Z-.002
X-1.C0.R.625Z-.004
X1.5C0.R.625Z-.006
X-1.C0.R.625Z-.008
X1.5C0.R.625Z-.01
X-1.C0.R.625Z-.012
X1.5C0.R.625Z-.014
X-1.C0.R.625Z-.016
X1.5C0.R.625Z-.018
X-1.C0.R.625Z-.02
X1.5C0.R.625Z-.022
X-1.C0.R.625Z-.024
X1.5C0.R.625Z-.026
X-1.C0.R.625Z-.028
X1.5C0.R.625Z-.03
X-1.C0.R.625Z-.032
X1.5C0.R.625Z-.034
X-1.C0.R.625Z-.038
X1.5C0.R.625Z-.04
X-1.C0.R.625Z-.042
X1.5C0.R.625Z-.044
X-1.C0.R.625Z-.046
X1.5C0.R.625Z-.048
X-1.C0.R.625Z-.05
X1.5C0.R.625Z-.052
X-1.C0.R.625Z-.054
X1.5C0.R.625Z-.056
X-1.C0.R.625Z-.058
X1.5C0.R.625Z-.06
X-1.C0.R.625Z-.062
X1.5C0.R.625Z-.064
X-1.C0.R.625Z-.066
X1.5C0.R.625Z-.068
X-1.C0.R.625Z-.07
X1.5C0.R.625
X-1.C0.R.625
X1.5C0.R.625

G113

G0X4.M9
G28U0.V0.M24
G53E23.25
G28W0.
M30

I'm not very familiar with polar interpolation so please correct me if I'm wrong, in using polar interpolation all y axis movements shall be programmed as radial c and x axis movements as diametral x. And since all supposed y axis movements are translated to c, the tool will stay in center(?) with respect to y. If this is true then using a turning tool, in your program it will cut 1/4" above and below center of the eccentric diameter at 90 and 270 degrees respectively.

The reason to have circular motion with the turning tool is for the tool tip to always follow the center of the eccentric diameter, having the same distance at any given angle, meaning the tooltip will always be on center of the eccentric.

More clarification please XD. I want to know the easiest way of doing this.

Uly

[ad64075]

I may have confused the masses slightly... I stated the program example created a 1" diameter... It actually creates a 1.25" diameter. Anyhow...

.
G1G98Z.01F10. (positions tool tip .010 in front of part face)
G112 (activates polar-probably different code for your machine)
G3X-1.C0.R.625Z0.F400. (X-1. represents where tool tip WOULD end IF part did not rotate)
(C0. represents the Y axis value)
(R.625 represents the arc radius)
(Z is Z, F is in degrees per minute)
(therefore, this first line of code represents only 180' of a circle!)
X1.5C0.R.625Z-.002 (this will complete the second half of the circle)



X-1.C0.R.625Z-.004 (and so on and so forth...)