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For a simple 90 degree outside radius add the the tool nose radius to the radius specified by the print. Then program the I and K values and start and end points with the new value (print radius+ tool nose radius).
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I have a question on I and K values, I am trying to cut an outside radius on a vertical lathe that requires I and K values, what I was wondering because I am not able to use radius comp for the tool nose how does that effect my values does the start point of the radius get smaller in the x and does the z get more minus? and what about the I and K values do they change with the tool nose radius?, and if so how much if I am using a #3 radius tool? Any help in this area would be more than greatly appreciated, I am lost.
For a simple 90 degree outside radius add the the tool nose radius to the radius specified by the print. Then program the I and K values and start and end points with the new value (print radius+ tool nose radius).
The starting X-value will be smaller by 2 times the insert radius. The ending Z-value will be bigger by the size of the insert radius. The I or K value for a 90 degree outside corner will the size of the insert radius plus the part radius.
Example: Turning 1.000 diameter with a .03 corner radius and an insert radius of .031 (like a CNMG-432 insert).
X.878 Z0
G3 X1. Z-.061 K-.061
'I' doesn't need to be used because it is zero, and zero is understood if not programmed.
I and K are incremental values. Any angle other than a 90 degree requires you to do a little trig. You need to know the length of the legs from the center of the part radius to the center of the tool radius. I and K need to be signed. The sign is determined by which direction the part radius is in relation to the insert radius.
Another example. Say you've already figured the front dimensions for an OD 15 degree angle, and now need the I and K values for swinging a radius at the rear of the angle. Lets assume a .031R insert with a .01R on the part.
.01+.031=.041
COS(15) x .041 = .0396
SIN(15) x .041 = .0106
I must admit here that I simply multiply Tangent of the angle times the result of the Cosine value. A bit quicker.
So the block would be X... Z-... I-.0396 K-.0106 because the center of the part radius is below (negative direction) and to the rear (negative direction) of the center of the insert radius. Had you been figuring it for the ID, then it would have been X... Z-... I.0396 K-.0106 because the center of the part radius would be above (plus direction) and to the rear (negative direction) of the center of the insert radius.
Of course there is more trig to be done to find the starting and ending positions of the insert at both locations.
This example is for machines I am familiar with. What I just said may not be true for all lathes.
Hope this helps.
I think this information will really help me out, I am not doing a perfect 90 degrees so I am going to give this a shot and see if I can figure it out with your information, I really hope I can take your guys information that you gave me and figure it out, I will let you know. Thank you very much!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Why don't you post the program section for the angle once you have it figured out. Let us know exactly what you are working on. (Angle, part radius, tool radius, part diameter, how the angle is called out, etc.) I'm sure there would be more than one of us that would be willing to double check your program. I enjoy helping when I can. I am 100% certain that there are plenty of others here who feel the same way.Originally Posted by rod88
the tool nose radius is a #3 (.0468) I am cutting a tapered face that leads into the radius, the angle of the face is 101.6 degrees that leads into the radius of .5118, here is what I have written with the information I have recieved from you guys, hopefully it checks out ok, any more help would be appreciated.
g01x25.1518z6.2424f.025
g03x26.063z5.7036i-.1065k-.5487f.02
g01z------
Does the angle start at o,o?
I am not sure if I follow what your asking, my zero for the z is the parts rest pad and my x zero is the center of the chuck, this is a very large part please let me know if this is what you were asking.!!
0,0 means X0Z0. I assume the 101.6 angle is off the face of the part. I'm also going to assume this angle is measured from the X-axis. Since you said this is a very large part, I am going to assume we are talking inches.
From your last comment I know that the face of the part is not zero. What I don't know and can't assume is what dimension the face of the part is. What is this dimension? What is the X-dimension where this angle starts at? Is it 0, 5.25, 15.625, 25.0, etc.? Can I assume that the diameter of the part is 26.063? This would indicate that you aren't swinging a radius at the start of the angle. Only at the end of the angle.
You are correct we are talking inches.
The taper on the face starts at 21.500 the print calls the face to be 101.6 degrees, actually is 11.6 for programming, but the face tapers down in the z as I feed out in the x. the print call out is the projected point of intersection @ x26.048 z9.92911 the locators changed the z to 6.15911 the Raduis is tangent to the taper and the o.d.. I am doing a outside radius feeding down. Hope this helps
I also forgot to tell you that the outside diameter of the part also tapers down x 26.181, z 2.530.
Let me see if I understand what the part looks like. Back near the chuck we have a dimension of X26.181 Z2.530 on the OD. This tapers to an intersection point of X26.048 Z9.92911. This is where the face angle of 11.6 degrees starts and will have an X-value of 21.5 at the face. Correct?
Man, why couldn't you give me an easy one?
EDIT: To me when you said you are facing down it means that you are making a facing pass from the OD down to X0. In this case the G-code should be G2 not G3. Or does facing down to you mean that you are turning towards the chuck? (Z-dimensions getting smaller.) Your program example suggests the latter.
Last edited by g-codeguy; 06-07-2009 at 07:28 PM.
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