# Thread: Help me understand feedrate for degrees.

1. ## Help me understand feedrate for degrees.

Mr. James Weed's last post, "converting ipm to dpm macro question?" reminded me of my ignorance.

We have a part that was programmed by the seller of our lathe. 17 degree helix over .34 length. Our first ever C-axis program, and obviously not as easy to figure as drilling holes.

His 6 passes were based on these figures.

G98X.5Z.25C25.5
G1X.14F100.S5000
Z.14F10.M88
Z-.495C0F600.
G0X.5

This resulted in about an 8 degree helix, but was used by a sister company like this. So I have to assume it was doing its job as we never got any parts back or rejections. However, last time it was set up, I figured why not make it right?

I wound up with this.

G98X.5Z.05C52.
G1X.145Z.015F100.S5000M88
Z-.62C0F1500.
G0X.5

How does one go about figuring the feedrate for C-axis rotation? I used cycle time and eyeball to get my modified program to cut at about the same feedrate. I have no idea where to start looking for this information.

Thanks.

2. If you have a simultaneous move combining a rotary with a linear axis move, then I would assume that the linear feedrate is 'boss' because both motions must begin and end at the same moment. IOW, the feedrate along a ZC vector will be the same as a feedrate along a single axis Z vector movement.

For a single C axis rotation, there you might have to consult the manufacturer's instructions for C axis feedrate programming.

In a machine like a Haas, there is a setting that one can make where the OD of the part is input for the job at hand. So using this value, the controller can determine (automatically) what the feedrate in degrees would be for a rotary axis, based on the last linear feedrate in mode. So a Haas is nice in that regard, that we don't have to be mixing linear ipm commands with dpm commands.

Perhaps that information will help you figure out what to do. Not all controllers handle this in as neat a manner, so you'd have to read the manual to see what to do.

3. g-codeguy,
What is the od of that part? I have a conversion program in my caculator that is very close(???) problem is i have to figure how to put this program back to formala on paper to share with you. need od to convert helix to a pitch.
On my 2ea allen/bradley lath conversions and my fanuc rotory mill, if you have c axis or b axis involved, then it auto interpolates in dpm . drop b or c axis then its back to ipm. maybe machine builder set it up this way???
I looked at formula on a website just the other day, i'll try to find it again. that formula just figures c axis moves, involve a helix(zc) and its different.

here is a link to site;
http://www.cncsnw.com/4thHowTo.htm

4. Mr. HFD: I've seen your posts before so I know you are one smart guy. This time, tho, your assumtion that the feedrate should be the same as the Z axis feedrate is wrong. Witness the F1500. I think my Z axis is 1430 IPM for max rapid, tho I would have to find one of those pesky manuals to be sure.

We have 3 of this model. Meaning we should also have 3 sets of manuals. Just made a quick run thru the shop & our library. Will try harder tomorrow to find one.

Both: The move is a ZC. I don't see what the diameter would have to do with my feedrate. Since it is brass, I am running RPM wide open. There are 6 vanes at .026+/-.001 thick.

Mr. Weed: I looked at your link. Only one formula looked like it was close, but I don't think it is the correct one tho I definitely could be wrong. It is using Diameter as part of the formula. If I changed the X.14 to X1.5 and ran my current program, I don't see why the feedrate should change. It is a function of the distance traveled in Z while the C axis rotates 52 degrees.

I'm glad I have a couple good mill guys helping me out. Thanks.

EDIT: O.D. is .426 diameter.

5. Gcodeguy,
Yes, you're right about the ZCF1500, that would certainly look like a higher than normal linear feedrate, so perhaps that is the inverse time feedrate. I've not used that, but I have heard about it. Check this link, the third post down discusses this briefly.

http://www.cncmagazine.com/archive01/v3i08/v3i08k.htm

6. Obviously something new for me. If I understand it correctly, time for each cut would be 60/1500=.04 seconds per pass. I can guarantee that each pass takes longer than that, so I guess that isn't the solution either. Interesting information to keep in mind, tho. Thanks.

BTW, I get grouchy quite often anymore. Does that mean I have become good and fast?

7. well the od does matter cause your tool is cutting on the circum of the part. bigger the circum the greater the distance for the cutter to travel. 52 degrees on 1.4 cir is not the same as 52 degrees on 1.5 cir.
1.4=4.3982 cir cir/360=.0122 per degree 52*.0122=.6344 travel around cir
1.5=4.7123 cir cir/360=.0130 per degree 52*.0130=.676 " " "

remember that you may program z .340 but cutter travels more than that because its moving around cir on an helix(pitch)
I do understand that a degree feedrate with just c axis is not the same as z and c going together. with both going then your pitch has to be included as well. I think...

8. Still trying to wrap my mind around the inverse feedrate concept

In the original program, the X axis was being commanded to move at 100ipm, so we could presume that approximately the same feedrate would be desired during rotary motion.

So if we know the OD of the part is 0.426" then a 25.5° move represents a distance (along the part circumference) of
.426*Pi *25.5/360 = .0947"
If this movement were executed along a linear axis at 100ipm, it would take
.000947 minutes. To get the inverse feedrate figure, we take the reciprocal:
1/.000947 = F1056

Since the original programmer used F600, the time allotted for the ZC move is 1/600 = .00167min.
The ZC move is a hypotenuse, and I calculated
SqRt[(.638^2) + (.0947)^2] = 0.642 in
.642in/.00167min = 384ipm equivalent

Note that on a Haas and maybe most controls, you'd need a G93 to turn inverse time feedrate on and G94 to turn it off. When G94 is modal, that will give you your standard ipm.

So the inverse time feedrate is not a constant. It needs to be recalculated for every dissimilar movement involving the C axis (combined with another axis), because the time duration for the cut needs to be known exactly, then the reciprocal is calculated.

There may be limits on what the controller will permit for the feedrate command. This means that it is readily possible with inverse time, to exceed the number of placeholders allowed. So, that is something to watch out for.

On your new trial program, you used a 52° rotation and F1500.
So the distance along the circumference if the part is 0.426 = 0.1933 inches.
Your inverse time F1500 indicates that the move should take:
1/1500 = .00067 minutes

The ZC vector in your code is of length 0.663" if I calculated correctly.
.663in/.00067 = 990ipm equivalent.

I'm not sure how that jives with the real world results. Seems a tad fast, but you'd also need to check that inverse feedrate mode has actually been invoked and is active in what you actually are looking at in your current trial run.

We still are not certain at this time, if in fact your controller is using inverse time, or whether it is using degrees/min format for the rotary axis.

9. Best I can figure out from that Haas Apps's post is that when they say divide the inverse time figure into 60, that they are looking for an inch/second conversion rather than ipm. However, since most of us think in terms of ipm, I don't think it helps matters to use 60 as the dividend when 1 works better.

10. Originally Posted by jamesweed
well the od does matter cause your tool is cutting on the circum of the part. bigger the circum the greater the distance for the cutter to travel. 52 degrees on 1.4 cir is not the same as 52 degrees on 1.5 cir.
1.4=4.3982 cir cir/360=.0122 per degree 52*.0122=.6344 travel around cir
1.5=4.7123 cir cir/360=.0130 per degree 52*.0130=.676 " " "

remember that you may program z .340 but cutter travels more than that because its moving around cir on an helix(pitch)
I do understand that a degree feedrate with just c axis is not the same as z and c going together. with both going then your pitch has to be included as well. I think...
I do realize that a 52 degree arc segment of a 2 inch dia. circle will be greater than the same segment of a one inch dia. circle, as would a 1.001 dia. circle segment. I also realize that the tool isn't just traveling .34 in Z, but also in an arc. Figuring that distance is beyond me.

I've always liked math, and been fairly good at it, but if you've never used something....?

I ran 2 test programs this morning. My part program and one with 4 inches added to the X dimensions. Cycle time was 18 seconds for both. Machine is a Daewoo 200MS. Here are the programs if you'd like to try them.

:2 (PLUS 4 INCH)

N820G54M91 (SPECIAL TOOL)
M35
T0808
G98X4.5Z.05C52.
G1X4.145Z.015F100.M88
Z-.62C0F1500.
G0X4.5
Z.05C112.
G1X4.145Z.015F100.
Z-.62C60.F1500.
G0X4.5
Z.05C172.
G1X4.145Z.015F100.
Z-.62C120.F1500.
G0X4.5
Z.05C232.
G1X4.145Z.015F100.
Z-.62C180.F1500.
G0X4.5
Z.05C292.
G1X4.145Z.015F100.
Z-.62C240.F1500.
G0X4.5
Z.05C352.
G1X4.145Z.015F100.
Z-.62C300.F1500.
G0X4.5M90
G0Z1.
M91
M30

:1 (NORMAL)

N800G54M91 (SPECIAL TOOL)
M35
T0808
G98X.5Z.05C52.
G1X.145Z.015F100.M88
Z-.62C0F1500.
G0X.5
Z.05C112.
G1X.145Z.015F100.
Z-.62C60.F1500.
G0X.5
Z.05C172.
G1X.145Z.015F100.
Z-.62C120.F1500.
G0X.5
Z.05C232.
G1X.145Z.015F100.
Z-.62C180.F1500.
G0X.5
Z.05C292.
G1X.145Z.015F100.
Z-.62C240.F1500.
G0X.5
Z.05C352.
G1X.145Z.015F100.
Z-.62C300.F1500.
G0X.5M90
G0Z1.
M91
M30
%

This is the only job I've every run like this so I am having a hard time visualizing. Plus I am an old dog, and you know what they say about us.

11. HFD. 2 things. I couldn't find inverse time mentioned in the manual, and 18 second cycle time is way too long if it were using inverse time feed. So I guess this means the control is using degrees per minute???

Bringing me back to my original question. How do I figure feedrate? Would a call to Daewoo be appropriate? I've always had good luck tapping into the knowledge on forums.

BTW, G94 is a canned facing cycle. No G93 mentioned.

12. Originally Posted by g-codeguy
I've always liked math, and been fairly good at it, but if you've never used something....?

Plus I am an old dog, and you know what they say about us.
me i don't like math, which is why i try to get formulas programmed into my caculator as soon as possible so i can forget about it.
i'm a old dog also...a tired old dog!

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