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How Do You Figure C Axis Feedrate Live Milling Lathe?

munruh

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Jan 3, 2011
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Kansas
When you are milling a shape with live milling on a lathe using polar interpolation (lathe does not have Y axis), how do you figure feedrate? It does not seem to be plain IPM.
 
It's in Degrees per Minute.

Plane angle divided by 60=DPM

But when you have a Million diameters, it's pretty time consuming to try and figure out Machine time. If your simply wondering why IPM takes so fucking long, you need to switch to DPM.

R
 
What control is this? I recently completed a polar interpolation project on a Fanuc w G12.1(polar on) / 13.1 (polar off) and I wrote the X/C(Y) portion of the program with G98 IPM feed rates and it worked splendid...
 
When you are milling a shape with live milling on a lathe using polar interpolation (lathe does not have Y axis), how do you figure feedrate? It does not seem to be plain IPM.

Hello munruh,
The Feed Rate in Polar Interpolation will vary (C or X depending on the mode of feed rate) depending on how close the cutter is to the centre of the part. For the same length of Linear Interpolated move (X and C together), the C axis will move through a greater angle the closer the tool is to the Centre in the X axis. For example, a line of 43.489 length, Linear Interpolated in one axis (combination of X and C to give the appearance of one axis), at Diameter 90.0 the C Axis will move through 51.684deg. At Diameter 20.0, the C axis angular move will be 130.706deg. Because of this, an alarm is raised if the C Axis Feed exceeds a max value set in parameter. Accordingly, when determining the feed rate to use, the relative position of the tool to the spindle centre in X must be taken into consideration. One feed may not fit all.

Regards,

Bill
 
Hi Bill!

Isn’t the coordination of the X/C axes in polar mode handled by the control in such a manner to maintain the programmed G98(IPM) feed rate similarly to how RPMs are handled by the control to maintain the G99(ipr) feed rate in G96(CSS) mode??

Just like you said - the C axis will have a higher change in angle as well as a higher angular velocity the closer the cutter is to the centerline of the part, but that is done automatically in order to achieve/maintain the programmed feed rate, right? (aside from the preset max feed rate limit, which I imagine is always bumped up against if you are, say, slotting straight across centerline).

I mean if we have to slot straight across a 1” bar and I wanted to figure out how to program the feed rate on a 3/8” 4-flute em running 200SFM @ .002ipt wouldn’t it still be 2000RPM x .008ipr = 16ipm and look something like this

M23S2000;
G12.1;
G1G98F16.;
X1.5.C0.;
X-1.5C0:
G13.1;

And thusly take about 1.5/16 minutes to complete??

I have only done this on one machine with a Fanuc 32i control so maybe that’s a different animal but unless I totally overlooked something (which is what I’m worried about now!) I didn’t have to take the X position into account when figuring out what feed rate to program....

OR wait....are you just saying you have to watch out when getting close enough to X0. that you are exceeding the max C-axis feed rate???
 
OR wait....are you just saying you have to watch out when getting close enough to X0. that you are exceeding the max C-axis feed rate???

Hello Nerdlinger,
Yes. I thought perhaps the OP may be seeing a change in the C axis velocity and taking it as the incorrect Feed Rate. Its common when the X axis is going to be close to the centre, to calculate the Feed Rate of the C axis taken into account. I include this in my Post and output a comment to the program to alert that the Feed Rate has been changed to accommodate the max C axis Feed Rate set in parameter.

Regards,

Bill
 
Hello munruh,
The Feed Rate in Polar Interpolation will vary (C or X depending on the mode of feed rate) depending on how close the cutter is to the centre of the part. For the same length of Linear Interpolated move (X and C together), the C axis will move through a greater angle the closer the tool is to the Centre in the X axis. For example, a line of 43.489 length, Linear Interpolated in one axis (combination of X and C to give the appearance of one axis), at Diameter 90.0 the C Axis will move through 51.684deg. At Diameter 20.0, the C axis angular move will be 130.706deg. Because of this, an alarm is raised if the C Axis Feed exceeds a max value set in parameter. Accordingly, when determining the feed rate to use, the relative position of the tool to the spindle centre in X must be taken into consideration. One feed may not fit all.

Regards,

Bill

Thank you, Is there a formula for figuring this?
 
Thank you, Is there a formula for figuring this?

If your referring to an algorithm to calculate a max Feed Rate that will avert an error being raised due to the C axis Max Feed Rate being exceeded, then the following is it:
Metric Example:

F < (L × R ×180/pi)(mm/min)

Where:
L = Distance (in mm) between the tool centre and Workpiece Centre when the tool centre is the nearest to the Workpiece Centre

R = Maximum cutting Feed Rate (deg/min) of the C axis set in parameters.

Regards,

Bill
 








 
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