Deriving radius from I and J addresses

[eliot15]

Two questions:

First, given a circular interpolation command like G3X..Y..I..J.., if XYIJ are all known, is it possible to compute the radius represented by these values?

Second, is this the best area on CNCZone to post questions regarding CNC programming, or is there a more appropriate area??

Thanks

[ger21]

I moved it to a more appropriate forum.


It's easy to find the radius using Pythagorean Theorem. I and J give you the center, and X and Y give you the end.

A^2+B^2=C^2

If IJ mode is incremental, you'll need to do some basic math to get the center.

[eliot15]

Thanks much for the reply, and the move.

Not quite following the Pythagorean Theorem suggestion, however. Maybe if I throw out some numbers, if you have a minute we could work an example.

G41 X-.2706 Y-1.738 F12. D10
G3 X-.2657 Y1.6832 I-19.4299 J1.738

NCPlot tells me that I just cut a 3.4256" long arc with a center located at X-19.7005 Y0.0, and a radius of 19.5075.

Are you saying that I can plug any/some/all of these values into Pythagorean Theorem and arrive at 19.5075??? (In absolute mode).

Thanks

[BlueChip]

If you have a lot of these to do ... or if you're looking for an automated route ... KipwareXC can perform these calculations automatically. INC I and J and K to ABS I and J and K or reverse ... and R to either INC or ABS I and J and K and reverse.

Kentech Inc. - Real World Machine Shop Software for more info and videos.

[eliot15]

Nope. Just wondering if there is a way to calculate the radius without going the automated route. But thanks...

[eliot15]

Sorry...I get it now...

19.4299^2 + 1.738^2 = Radius ^ 2 or I^2 + J^2 = R^2

Thanks again...

[ger21]

Originally Posted by eliot15 Thanks much for the reply, and the move. Not quite following the Pythagorean Theorem suggestion, however. Maybe if I throw out some numbers, if you have a minute we could work an example. G41 X-.2706 Y-1.738 F12. D10 G3 X-.2657 Y1.6832 I-19.4299 J1.738 NCPlot tells me that I just cut a 3.4256" long arc with a center located at X-19.7005 Y0.0, and a radius of 19.5075. Are you saying that I can plug any/some/all of these values into Pythagorean Theorem and arrive at 19.5075??? (In absolute mode). Thanks

Looks like Incremental IJ, so the center of the circle is:
X = -19.4299 + (-.2076) = -19.7005
Y = 1.738 + (-1.738) = 0

Base of the triangle (distance between center and end) is
-.2657 - (-19.7005) = 19.4348

The other Leg is
1.6832 - 0 = 1.6832

Radius = Square Root ((19.4348^2) + (1.6832^2)
Radius = 19.5075

Simple.

Edit: You did it for Absolute IJ, but the code is incremental IJ

[eliot15]

Can you explain that briefly. I'm in G90, so in general my code is in absolute mode. But I've heard that I & J are almost always in incremental mode. Does the control just know to read them as incemental values, unless you do something to tell the control to read them as absolute values???

I'm using a FANUC control. Is it the case that it ONLY reads IJ as incremental values?? And older controls read IJ as absolute values maybe?????

I'm running out the door and don't really have time to think about this at the moment. Maybe I'm overlooking the obvious.

Thanks again...

[sinha_nsit]

With the default parameter setting, IJK are always incremental, being the position of the center with respect to the start point (current position of the tool).

With IJK, R is not specified. If you specify R also, IJK information is ignored.

[ger21]

I'm using a FANUC control. Is it the case that it ONLY reads IJ as incremental values?? And older controls read IJ as absolute values maybe?????

Never used a Fanuc, but some controls allow you to specify which you want to use.

[mmcp42]

the standard (NIST RS274) defines I and J as offsets to the centre of the arc
so even in absolute X and Y, I and J are relative to the current X, Y position

HTH

[eliot15]

Thanks much for the replys.