[dkirby]

I have been attempting to learn how to run my VM16, which has involved a lot of cutting air.
While doing some MDI operations I slowed the spindle down to 50 rpm and noticed the speed fluctuating.
I reached up and grabbed the spindle and was able to completely stop it's rotation.

I was setting up to notch some tubing so I set it up with a 1.5 dia. end mill and tried to machine the tubing. With a little bit of down force it would stall out the spindle.

Could some one please explain to me why this is happening!!!

Thanks, Dave

[sportybob]

Hi dkirby, At that low spindle speed, it is normal to be able to stop the spindle , by hand. There is no torque developed in the VFD at that low speed.

If you need "low speed torque, I do have a VFD replacement that ( when set up properly) it will provide a LOT of torque at even zero speed. (for 7.5Hp it is < 1K)

PS be sure you are in "low range)

Sportybob

[dkirby]

Thanks Sportybob, I guess my next question is. How many ranges does my machine have, and how do I determine what range I am in???
Is it a parameter adjustment or a mechanical one?

If it makes a difference, here are the mach. specs. Partner VM 16 Serial # 4537

Us new guys sure can be a pain eh!!!!

Dave

[sportybob]

Hi dkirby, most likely you have a 7.5 or 10HP motor that uses a Yaskawa / Magnatek VFD drive, ( look in the electrical panel for a beige box mount to the bottom of the panel. These machines use a 2 step pulley ( hi and low range.) the spindle range is changed by moving the motor / spindle belts to either the upper ( high) or lower (low) pulley grooves. Then changing the control parameter to either 1 (low) or 2 (high) F7(prams) F9 (ctrl)

Running in low range allows the spindle motor to run faster and get it into a better torque range.

Sportybob

[dkirby]

Hey Sportybob, I checked the pulleys on the spindle, and there are only 1 step pulleys on this machine. So I guess high and low range is out of the question. I am assuming that the right pulleys are available. I also looked at, what I imagine to be the VFD drive. It is says EMS model pc3-23p7 on it and it is a 7.5 hp motor.

Just like dragracer06 I too happen to own 2 of these machines. They must have been cheaper in pairs, I guess!!! I checked the other machine and it too has only single step pulleys. These machines were bought from a die shop. Do you think they were only concerned about high speed use?
These machines both came from the factory with 6in. risers. For my needs right now I sure would like to have Low Range available!!

Thanks for your help,
Dave

[sportybob]

Hi Dave , the pulleys should have 4 V grooves in them. one pair of smaller and one pair larger. see the attached drawing. belts installed on the top smaller pair ( spindle side) is high range.Attachment 309054

If you have something different, send a pic, I do have a pulley replacement kit available (V belt to cog belt)

Bob

[dkirby]

Hey Bob, unfortunately I have not learned how to do the digital camera deal yet. Maybe after I learn how to use this VMC it will be my next to do project. Probably by then, they will have found another way to take pictures anyway!!! I am just hoping to live long enough to say that I was able to get this machine to do what I tell it to do. LOL!!!
My machine already has the cog belt drive and there are only 1 sheave on the motor and the spindle.
Did this particular VM16 come with 2 ranges of speed even with the cog belt drive?

Right now I am going to have to learn how to do circular interpolation, to get around the high speed that the spindle requires to operate, for what I am attempting to do.
Does anyone have any other suggestions or work arounds to my spindle problem????

Thanks,
Dave

[Brian L]

Hi Dave,

Circular is pretty easy, say you have 1.5" tube clamped parallel to the X axis, set 0,0 centered on the tube at the right end (tube hanging out the left of the table). Set Z0 at the top of the tube. Code would look like this using a 1/2" HSS end mill:


G00 X.25 Y0 (position to use the shank of the end mill to act as the length stop, no material in the machine yet)
Z-1.9 (moves down so tube will hit shank above flutes)
M01 (option stop the machine, now install tube, then hit cycle start)
G00 X.5 (pulls back to beginning position of arc)
Z-1.6 (comes up to align flutes with tube)
M03 S570 (turn spindle on, this is for a HSS EM, if Carbide at least 3x's as fast)
G03 I-.5 F7 (arcs in a circle to cope tube)
G00 Z0 (pulls back to Z0 to allow you to pull out tube, after spindle is off)
M05 (spindle off)
M30 (loops program back to start)

[sportybob]

Hi Dave, I did some checking on your machine. It has a 10K spindle. That means that the spindle pulley is even smaller and less torque at low rpm's. count the cogs on both the motor and spindle side, we'll figure out the ratio you need for a low range. text me a pic of your spindle pulley set up.

On my gog pulley updates, I have to modify the lower spindle pulley taper bushing to make it sit flush, so the upper pulley will stack and align with the motor pulleys.

sportybob

[dkirby]

Thanks Brian for the program to circ interpolation. I will look at it real close to learn how it works. One of the things that is giving me a problem right now is that I have no spindle torque below 1,000 rpm and there is no speed adjustment via the pulleys. I appreciate the help.

Bob I am going to give you a call within a couple of days. I will then be able to give you some more info. I have a couple of things to get out of the way first.

Thanks, Dave.

[dkirby]

Hi Sportybob, I called you on Fri. afternoon. I will try again soon.

Hi Brian L, I hope you are still with me on this, because I am having problems understanding the code for circular interpolation.
It just happens to be 1.5" tubing that I am trying to cope, and I followed you through the XO YO centered on the tube at the right end. This too is exactly what I want to do. I made up a stop on the vise for the length of tube hanging out, so I don't need the Z coming down for that.
I would like to do a .750 radius notch, which to me, looks like a Y move has to be shown somewhere doesn't it?
In your example you show I-.5 which is an X move right?
How do you get a 180 or 360 degree arc move without showing a Y move?
My understanding is X goes with I and Y goes with J, is that right?
In your code the Y is at O and does not move, right?
I am really clueless here Brian and I need every little detail explained, maybe i will have to get you to type slower just for me. LOL!!!!
Here are my specs. I am using 1.5" tubing. The start of the cope will be in from O on X .350 and I am using a 1" inserted cutter to do a .750 radius.
I have tried and tried and tried some more and I still can't seem to get my head wrapped around circular interpolation!!!!!!
Someone help me please!!!!
Thanks, Dave

[Brian L]

Hi Dave,

You have a good grasp, maybe I should have explained a bit better. When you are doing arcs, with g-code, if the ending position is the same as the starting, as in the case of a full circle, you don't need to list the X and Y values, they didn't change. So the code is move positive in the X direction to put the edge of the cutter on the 3 o'clock position of the circle (that was the X.5 move). Then I wanted a full circle, so G03 I-.5, thats the same start and end point, with the center of the circle the I-.5 distance from the start point.

Now, given you want to use a 3/4" cutter, which might not be a good idea as you have speed issues already, but here is how I would do it:

G00 X1 Y.375 (moves far enough away from the part to clear the end of the tube before it moves down)
G01 Z-1.6 F10 (moves down)
X0 (Moves in X to start of arc position)
G03 Y-.375 J-.375 (this time a half circle instead of a full circle, so you need the position of the end of the arc)
G00 X1 (move out away from the tube)
Z1 (pull up)


Hopefully that makes sense for you. Arcs are pretty well explained in the manual, with some practice you should get them under control. Practice cutting air far away from your vises and tables, with large exaggerated moves, like 5" diameter circles or something, so you can see what moves are doing what (run single block).... oh, you can also use "Verify" and watch things on the screen.

[dkirby]

Hi Brian, Yes, I do have torque issues at low speed, but I have a 1" cutter that is 2" long and has carbide inserts along it's whole length so I felt that I should be able to use a speed high enough to get around that problem. I did make a parameter adjustment after Sportybob's advice and hopefully that is going to help some. What do you think?

Now on to the circular interpolation part again!!! In my description of what I am doing I gave you some wrong specs.
The tube is 1.320" in diameter and the cutter is 1" in dia. and the arc radius is .750" Which I guess doesn't change anything except the clearance positions, right?
So I would do X0 Then do a G03 Y - whatever clearance value I use then J - and the same value as before and that should do a 180 deg. arc. right?
Here is where I am lost. Nowhere in this setup do I tell it the arc radius, so how do I end up with .750 rad. notch.
Would I do X .250 to bring 1" cutter back for a .750" rad. and then Y- J- Or am I still out to lunch here???

Thanks Brian for your time and patience,
Dave

[Brian L]

In mild steel tube you can run that 1" cutter at least 760 rpm, depending upon the grade of inserts, so if that is high enough to get you some torque... you will have to see.

Doesn't matter what tube diameter you want to use, just be centered over it in the Y direction. The code I just gave you was wrong, too early in the morning and my math was bogus. OK, you want .75 radius, minus 1/2 the cutter diameter, so .25 is the amount off center to swing the arc. Just change it to:

G00 X1 Y.25
G01 Z-1.6
X0
G03 Y-.25 J-.25
G00 X1
Z1

You are stating the arc radius with the J value, it is the value from arc start to arc center, at the centerline radius of the end mill's path. In this case .25" plus half the cutter, .500" is going to give you the .75" radius you want.

[dkirby]

Brian, many winters ago I spent some time around Phoenix International Raceway and every morning there was nothing but glorious Now if this conversation was taking place a couple of months from now, I am pretty sure I would have more understanding of your plight!! LOL

Now that I am seeing numbers that I can relate to , things are starting to sink in. I tried drawing it out beforehand and I thought that the X value did not change, but I didn't understand how the arc was created with the Y.
You have explained it well enough that I think that I may have a grasp on it FINALLY.
I hope to put it to use today some time.
I will be back with some results.

Thanks again, Dave

[sportybob]

Hi Dave, It appers that someone had exchanged out the "2 step V belt pully with a "single cog belt" they did not care about low speed torque. just "high" speed.
You will need to add another pulley set for "LOW" range to get some torque on the low end.
we will need to know what your current ratio is. ( # for tooth on the spindle to # of teeth on the on the motor.
When using 2 steep "cog pulleys, there is some machining required to get the hi / low range pulleys to align properly.

My kits have been machined.. (5 HP / 7.5 HP or 10 / 15 HP ) kits are available.


PM me for more info.

Sportybob

[dkirby]

Hi Sportybob, a PM has been sent your way.

Hi Brian, I followed your advise on the circular interpolation, and now I am in need of some troubleshooting on your part, if you don't mind.
I tried the 1" cutter with the inserts, but was not happy with the outcome. SO I switched to a solid carbide endmill that I have that is .825 dia. and long enough to do the job.

Here is my setup. The radius of .750 has to be .350" in from the right end. So I centered the tube at the cutter with the offset for a notch at .350 and ran Z down 1.6" to create the start of the cope and remove material and get a visual of the notch.
I then used the following program.
G01 X0 Y .338 Cutter radius of .825 is .4125 The arc radius is .750 .750 - .4125 = .3375
G03 Y-.338 J-.338

The arc looked right but the cutter continued on past the radius that I had already started with the previous Z plunge.
So I reran the program to see what was happening and the readout showed that there was an X move of .338.
How can this be? I programmed the X at 0
I tried various things, thinking it was a cuttercomp problem, but that was not it. My offsets looked to be proper and after many attempts to find out what I messed up on, I guess I need some more advise.

Brian, the way that you explained the math to do circular interpolation was by far the most understandable way to do it that i have been able to find anywhere.
And here I thought I had finally latched on this circular interpolation thing, guess not.
Any idea where I may have gone wrong?

Dave

[sportybob]

Hi Dave, from your PM, you have a 2:1 ratio ( 44/22) so you will need to add / stack another pulley set to the spindle and motor. a 18T on the motor and a 40T on the spindle would get you a 2.2:1 ratio, that would get ya some low end torque.
We will need to check and possible set up the spindle range one parameters in the MISC parameters.

have you tried cutting an arc using the conversational programming?
you can also put the control in block mode and watch the graphics as you "cut or verify" the program. In verify you can select either the programed "part" path, the cutter "path" or both. It is defaulted to the "part" path.
when running graphics verify in block mode the current block move will be displayed on the MDI line. this may help you figure out what is going on.

sportybob

[Brian L]

Hi Dave, sorry to be MIA, trying to set up a laser engraver and having some issues and was tied up yesterday. OK, so you were at 0,0 and plunged straight down then moved Y plus, did a half circle... did you do a G01 back to center i.e Y0 before you pulled up? Post all the code you have exactly as you have it in the machine and I'll see what's up. The machines rarely move odd amounts on their own.... it's always tied to what we put in.... when something weird is happening I always revert back to GIGO, garbage in, garbage out.... it bites me every time.

[dkirby]

First, thanks to everyone trying to help me along.

Sportybob, at this point in time, I can only concentrate on one thing at a time unfortunately. I have not got to the point of doing Conversational yet. I wished that I understood how to do the graphics thing, it may help me to see what is happening. I have just started to learn how to run this machine.

I had already made up a program to do a notching job and it ran fine, and did exactly what I needed done. Here is a brief description of what I am trying to accomplish.
These black pipe tubes are part of a spray boom. they are 1.320" in dia. and 42" in length and there are 3 notches with a .750 rad. at various places along the length.
I have 2 vises set up on the mill
The first notch is moved in to the left .350" with a .750" rad.
The second notch is set at 14" to the left from the first notch and it is .750" in from the inside edge
I then have to move the pipe in the vise to the right by 17" because there is not enough X travel to do the third notch
I made a stop out of a tube 1.5" dia. that now fits in the second notch that orientates the pipe for position at the proper length along the table.
I then move X to the position for the third notch. It's Y position is the same as the second notch.

I wrote the program with all the Go1 , Goo and Z moves and it ran perfectly but I was using a 1.5" endmill, that because of no low speed torque I could not slow the spindle speed down enough for that endmill. I did machine 1 pipe and it fit perfectly in a weld fixture that was in use previously. It was a very simple program and it worked.
So I had to try and learn to use circular interpolation with a small cutter, and this is where my fun began!!!
With the help of Brian L, I thought that I had maybe had a handle on it. Now I know better!!! LOL!!!!

Hi Brian, From memory I will try to give you the code that I was originally using, I have tried so many different things trying to get it right that now I am almost totally confused. Anyways here it is as close as I can remember.
the cutter is .825" solid carbide .4125" rad. .750 - .4125 = .3375 ( Math for circ. interp).
G0 G17 G20 G40 G90
G00 x-17.0 y 7.150 M00 load position from floating zero's
G00 X 0 Y0
G43 H10 M03 S700
G00 Z .100
Go1 Z-1.500 F2
Goo Z. .100
Go1 X0 Y -.338
G03 X0 Y-.338 J-.338 F3
Because I had already plunged for the notch I could see where the radius of the notch was going to be. It looked like the Y position was exactly what I wanted but the cutter continued on in - X past where I needed it to end.

I hope this is clear enough to understand,
Dave