Radius turning questions

[washintonian]

I start a new job on monday that will include more programing than I have done before. I am good enough in programing to do chamfers tapers etc...
but I have never fully understood g02/g03 programing. I have spent the last week reading in these forums a couple hours each night but all i have gained is more confusion...

Could someone walk through programing these two examples step by step using g02/g03 without using R please?

Cut a full .5" radius on the end of a 1" part

and

Put blend radius on a chamfered end of a 1" rod
Ex: to do a .032" by 45 deg chamfer on a 1.5" diameter part with the parts face at 1" using a .016 TNR I would use:
G00 Z1.0 X1.436
G01 Z.9586 X1.5
I get those numbers using this spreadsheet I setup working out tnr comp.

How would I put a .005 blend radius at each end of the chamfer? I cannot find a working formula that I can make sence out of. So any help would be greatly appriciated.
Desmond

[dcoupar]

Just out of curiosity, what kind of control are you going to be programming? Fanuc?

Rather than spend time calculating intersections and compensation amounts, you might want to learn to use the control's tip nose radius comp function (G41/G42) if it's so equipped. Also, some controls have the option of Chamfering and Rounding built in. And some have a feature called Direct Drawing Dimensions Programming, which makes life real easy for manual programmers.

[angelw]

I start a new job on monday that will include more programing than I have done before. I am good enough in programing to do chamfers tapers etc...
but I have never fully understood g02/g03 programing. I have spent the last week reading in these forums a couple hours each night but all i have gained is more confusion...

Could someone walk through programing these two examples step by step using g02/g03 without using R please?

Cut a full .5" radius on the end of a 1" part

and

Put blend radius on a chamfered end of a 1" rod
Ex: to do a .032" by 45 deg chamfer on a 1.5" diameter part with the parts face at 1" using a .016 TNR I would use:
G00 Z1.0 X1.436
G01 Z.9586 X1.5
I get those numbers using this spreadsheet I setup working out tnr comp.

How would I put a .005 blend radius at each end of the chamfer? I cannot find a working formula that I can make sence out of. So any help would be greatly appriciated.
Desmond

The following formulae will get you the tool rad comp values for X and Z. Note that the X value is multiplied by 2 to get a diameter value. Accordingly, the formula shown in the attached picture can be modified as shown in the following formula to get a diameter value.

X = 2 * (tnr - (tnr * Tan(45 - (angle / 2))))
Z = tnr - (tnr * Tan(angle / 2))

Regards,

Bill

[washintonian]

ya will be done using a fanuc controller

Also I setup the cheat sheet for calculating the tnr comp before I wrote the program into my ti-89 I carry with me. It's how to put the blend radius onto a chamfer I am having trouble with since it's a partial arc it includes more trig than I can do on my own. Not all of the controls at the new plant will use g41/42 I want to understand it well enought tor write the program for my calc to solve blend radius on the floor. The problem I have is that I have not been able to find a acutual equasion to solve the problem.
Thanks

[g-codeguy]

Full radius from OD

X1.Z-(.5 + TNR)
G2X-(2 * TNR) Z0 R(.5 + TNR)

Full radius from centerline

X-(2 * TNR) Z0
G3X1.Z-(.5 + TNR) R(.5 + TNR)

In your example of a .032 x 45 deg. it should be

X1.4172 Z1.
X1.5 Z.9586

in order to be a 45 deg. chamfer.

The math is:

1-.9586=.0414
1.5-2*.0414=1.4172

Also I would never rapid to the face of a part.

Putting a .005R on the part using a .016R tool requires more math than I am willing to explain here. What I did was make myself a little chart. I almost always use a .003 or .005 radius on the part using either a .008R, .016R or .031R insert. These 6 values were soon memorized.

TAN[22.5] * .005 = .0021

Starting position for the criteria you specified is:

1.5 - (2 * .032) - (2 * .0021) - (2 * .016) = 1.3998

so X1.3998 Z1.

Moves from the face are X.0148 (radial) Z-.0062

so 1.3998 + 2 * .0148 = 1.4294
1. - .0062 = .9938

so far we have

X1.3998 Z1.
G3 X 1.4294 Z.9938 R.021

The neat thing about 45 deg. angles is the previous values for X & Z moves are reversed for the upper section (at 1.5). From here work backwards. Figure the ending position first.

1- (.032 + .0021 + .016) = .9499

So ending position is

G3 X1.5 Z.9499 R.021

Now work backwards.

1.5 - 2 * .0062 = 1.4876

.9499 + .0148 = .9647

so putting it all together we get:

X1.3998 Z1.
G3 X 1.4294 Z.9938 R.021
G1 X1.4876 Z.9647
G3 X1.5 Z.9499 R.021


I still have the chart for the different values for part radii and insert radii if you'd like me to email you a copy. I think I have even have an example of how to use the chart.

I had to be careful figuring your example because the face of my part is always Z0.

[ro0osta]

I was looking forward to reading the replies to this gentleman's post due to the fact I have the same question. Unfortunately no one solved his (or mine) problem. He specifically wrote "without using R". I'm going to assume that like myself he works with alot of antiquated conrols that need the "I and K".

[g-codeguy]

I was looking forward to reading the replies to this gentleman's post due to the fact I have the same question. Unfortunately no one solved his (or mine) problem. He specifically wrote "without using R". I'm going to assume that like myself he works with alot of antiquated conrols that need the "I and K".

My apologies. Guess I didn't read the original post close enough.

K equates to Z-axis. I equates to X-axis. These values are obtained by figuring the leg lengths of a right triangle with the hypotenuse being the value from the center of the radius on the part to the center of the insert radius....at the start position for the arc. I've already gone through the math to figure the XZ coordinates.

Signs are required when using I & K. The value when the insert is starting at the face of the part will the same for an OD or ID chamfer. A .005R on part using a .016R insert is going to be K-.021. Normally the sign is derived by determining which direction the part radius is from the center of the insert radius. I0 is not needed. I'm sure you can see why there is no I-value at this insert position.

The I & K to swing the radius at the end of the angle requires simple trig. Nice thing about 45 degree angles is the I & K will be equal. However, the signs will not be the same for OD and ID chamfers. (.005+.016) * (SIN)45 = .0148

OD chamfer will be I-.0148 K-.0148
ID chamfer will be I.0148 K-.0148

On an OD chamfer the center of the insert radius will be on the plus side of the part radius thus the sign of I is minus. The opposite is true for an internal chamfer. K value will be the same in both cases. This is the way it is figured on our lathes.

Were this a 30 degree angle, the simplest way to find the leg values is (.005+.016) * (COS)30 = .0182 * (TAN)30 = .0105 so:

OD = I-.0105 K-.0182
ID = I.0105 K-.0182

If you can't picture this in your mind, then lay it out on a piece of paper.

[fordav11]

you can find the maths calcs for TNR comp here....
Tool Nose Radius Compensation Lathe Cam
Or buy the program its very cheap.

For simple external 90 degree arcs the R = tool nose radius + part radius and Z end point = the same number. X start point = the end X value minus 2x the previously calculated R value. For 90 degrees internal arcs R = part radius - tool nose radius and Z endpoint is the same number. For radius to angle tangents or radius to radius tangents it gets very complicated very quickly and beyond the manual calculation capabilities of mere mortals. The types of jobs you can do with manual calculations are limited because of this complexity.
it's far better to spend a few dollars and buy one of the cheaper CAM programs such as BobCAD or TurboCad and have the computer do all the work for you and spit out nice G-code without errors.