[Cartierusm]

Below are some pics of my home made 4th axis for my Bridgeport. I'm using a 570 oz.in Stepper with 72VDC. I used a lathe head for it so it's interchangeable with my lathe. It's just using a belt reduction for power transmission. The problem I'm having is that with the leverage the chuck provides it moves when I'm cutting something on it. I can move it by hand while it's not moving, it takes a little force to move it, but I can move it no problem. I have it set at full power on my driver when it's at a standstill. Now I understand that there is no real transmission on it and no means of keeping it from antirotation as a worm gear drive would provide, but would this setup in your opinion be that weak? If it works and I can engrave lightly then it's not a wiring problem I'd guess? Any suggestions on how to easily modify this setup without having to purchase and build a worm gear setup?

[neilw20]

Your holding torque is the stepper torque multiplied by your pulley ration.
Make sure the driver you are using does not have the current reduction feature enable, other wise the holding torque will be reduced considerably.
Set the current limit to the highest setting, consistent with not exceeding the specs of the motor, driver or power supply.
You won't do any better than that without more reduction, or a holding brake.

Professional units use a drag brake that is strong enough to take all the cutting loads, and an encoder and a servo to move it with brute force.
That gets rid of the backlash. There is none.

[Cartierusm]

I do have the holding torque set to full and the correct amps. So what you're saying is that I should add some sort of transmission that will allow antirotation?

[neilw20]

Not exactly. What I am saying is you need some minimum value of holding torque, which is dictated by the cutter loads in your case.

As you may have noticed while doing milling, you must clamp work pieces very firmly to resist the force the cutter/drill etc exerts on the job.

It is no different holding something in the chuck. Same rules apply.

If torque needed to be resisted is 10 times what the stepper motor can supply, you need at least a 10:1 reduction to just break (no pun intended) even.

For similar reasons, your machine is big stiff and strong, to resist cutter loads.

Once the job/cutter deflects more than the chip load, then you get chatter, but that's another story.

[Cartierusm]

OK LOL, I understand what you're saying so can this design be altered easily?

[neilw20]

No. It will only take light cutting loads.
What current is the driver set to?
What driver you using? Post some specs.

[Cartierusm]

I'm using a Keling 8078 Driver, http://www.kelinginc.net/KL-8078.pdf
and the current is set to 72VDC and 2.1A...I think there's my problem I thought I had set it to the correct current but I just went and check while writing this post and it's not right. According to the stepper documents the current is 5 amps. Now on my driver it says Peak Current and RMS Current, the stepper doc only says 5A, does that mean peak or RMS? Here is the doc for the stepper http://www.kelinginc.net/KL23H2100-50-4B.pdf. Thanks.

[ger21]

I'm going to guess that even with the correct current, you'll still need more belt reduction, at least a 2 stage to get you close to 10:1.

[neilw20]

Set the driver to 4.9A. That will be within the ratings of the motor.
If you wire up the enable signal, you will be able to 'disengage' the drive to rotate the chuck by hand.
It should work OK for light engraving.
There does not seem to be a way of defeating the idle current reduction, so as a workaround, you could command it to step backwards and forwards 1 microstep while stationary. This will increase the holding torque considerably while only moving an almost imperceptible amount.
Contact keiling to see if the feature can be disabled selectively.

[Cartierusm]

I changed the amps to 4.9 and I can still move it but it's definately stronger. It might work this way.

Neil, why would I want to rotate it by hand?

"There does not seem to be a way of defeating the idle current reduction" Do you mean the holding torque? Because there is a DIP switch on the driver to set it to full or half? If this is not what you mean can you expain further including the part of moving back and forth a bit?

[neilw20]

Set that switch to FULL and try it.
Whichever position gives the higher holding torque, use that one.
Just turn against it with your hand. You should feel the difference.
That was not in the document file I read.

Rotating by hand during setting up a job, getting the chuck to a convenient position to use the chuck key. In your application there may be no need for this, but the driver has the feature.

Unplugging a stepper while the power is on can kill the driver. That's one reason the enable signal is useful.
If you need to disconnect the drive, you can disable it, then unplug it without frying the driver.

[Cartierusm]

OK gottcha, always wondered what that feature was for. Thanks.

So I do feel a difference by changing the holding torque, but does your way of moving the axis a microstep back and forth have any advantages now that I have set the holding torque to full?

[neilw20]

Only if you can measure a drop in supply current once the axis has been stopped for a second or two after moving, do you need to resort to the workaround and that would require lots of fudge lines in your G-Code.

[Cartierusm]

I'll stick with this and go from there. Thanks for the help.

If this won't work in the future, if I plan doing real cutting, how would I go about what Ger mentioned.

My large timing pulley on the axis itself is Pitch Diameter 2.865" and the small one on the stepper is Pitch Diameter 1.194". I had to use a large one on the axis as it had to fit it's shaft. I could go bigger on the stepper but probably not bigger than 3.82". So Ger you mentioned using 3 pulleys. One on the axis, one on the stepper and one I'd guess in between. The one in between obviously would need two pulleys. So if it's possible using the two pulleys I already have could I get the 10:1 ratio?

[neilw20]

I think Ger means is use a double reduction.
2 belts 4 pulleys, 1 idler shaft.
Stepper to idler shaft - belt 1.
Idler to main shaft - belt 2.
To make it a but simpler, you can have a fixed location for the stepper and the bracket and make an adjustable arm for the idler shaft/pulleys.
Really the idler could be just a small and a large pulley mounted on one overhanging shaft.
If each reduction was 3.3:1 you would get 3.3 x 3.3 = 10.89:1
I would go for as much reduction as is practical to fit.
Using tensioned belts you will get almost no backlash.
Think even 5:1 + 5:1 = 25:1. GO for a BIG reduction. Same amount of work. Better result.

[Cartierusm]

Ok I can figure ratios probably by myself but not the placement of the different size timing pulleys.

SO if I can get away with having a 2.865" Pitch pulley on the 4th Axis then can someone help with the rough sizes on the different parts.

So 4th Axis 2.865" Pitch Pulley -- What size connnected to this on the idler bigger or smaller? -- Then on the idler going to the Stepper big or small -- then on the stepper big or small? Thanks.

Also is there some sort of calculation on how to figure out multiple gear ratios?

[neilw20]

Whole numbers (I chose at random) are teeth.
If the first ratio is 57:24 that is 57/24 = 2.375.
If the second ratio is 51:19 that is 51/19 = 2.684.
If you use one driving the other just multiply 2.375 x 2.684 = 6.375.
Or another way (57 x 51) / (24 x 19) = 6.375.

If the first ratio is 57:24 that is 57/24 = 2.375.
If the second ratio is 19:51 that is 19/51 = 0.373
If you use one driving the other just multiply 2.375 x 0.373 = 0.8858.
With the big pulley being the driver it would 1/0.8858 = 1.129 : 1.

Simply multiply all the product (result of multiplication) of the driven teeth(s) and divide the result by the product of the driving teeth to give the resultant ratio.

At the end of the day you just need the biggest reduction possible for the most torque.
That's really easy. Use the biggest pulleys you can fit with the smallest pulleys available.

Don't worry yourself with pitches and all that stuff.
Figuring out what belt lengths to use is best done from manufacturers charts, but really the length of the belt (inches) is simply the pitch (teeth per inch) multiplied by the number of teeth in the belt.
Yep. It is easier in inches.

[garagefela]

Have a look here to easily work out center distances and belt lengths. It is easy to work out yourself when both the pulleys are the same size but gets a bit harder when one is bigger than the other.

http://sdp-si.com/web/email/beltspulleys/index.htm

Cheers M

[Cartierusm]

Thanks. I think I'll have to come up with something else becasue if the pulley on the 4th axis is 2.865" that would mean the pulley on the idler going to the 4th axis would have to be over 8" for the 3:1 ratio. As far as I know they don't make them standard that size in timing pulleys as least not from grainger or SPD, the problem also is a pulley over 7" is like $150, so there's not way I'm putting $300 into this system.

Question, if the 4th axis is holding tight enough for light milling how hard is it to mill my own timing pulley? What bit would I need, right now I'm using a .375 (L) Pitch system? I remember Hass Machine did a video on it, I'll see if I can find it. I could then also make the pulley on the spindle smaller allowing me to make the pulley on the idler smaller as well to get the ratio I need.

[neilw20]

You need to be able to lock the spindle every time you move it.