calculating travel calibration on cnc machine?

I'm trying to make sense of some calibration numbers on a CNC router.

X and Y calibration are listed as 1832.811


The steppers are 1.8°, which I gives 200 full steps per revolution.
There is an 15/54 chain reduction driving a 20DP, 20 tooth pinion on a rack.

Can anyone help me navigate how one arrives at this calibration number? I'm getting all sorts of unhelpful results while attempting these calculations. I realize that whatever number I get from theoretical calculations would have to be tweaked in order to get real world results, but I'm off by stupid amounts.

anchorman said:

I'm trying to make sense of some calibration numbers on a CNC router.

X and Y calibration are listed as 1832.811


The steppers are 1.8°, which I gives 200 full steps per revolution.
There is an 15/54 chain reduction driving a 20DP, 20 tooth pinion on a rack.

Can anyone help me navigate how one arrives at this calibration number? I'm getting all sorts of unhelpful results while attempting these calculations. I realize that whatever number I get from theoretical calculations would have to be tweaked in order to get real world results, but I'm off by stupid amounts.

Click to expand...

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only thing i have done is use 4.0000 gage block against a stop and zero a dial indicator read or zero DRO and remove gage block and move til indicator reads zero and see what DRO says and then adjust til when you move 4.0000 the DRO reads 4.0000
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sure you can do math or keep changing and measuring til you are closer and closer til close enough, eventually you get close enough where you are <.001" and then adjusting small amounts. some DRO needs its calibration check yearly. i just use a gage block and indicator

So, let's see if I can get there (and apologies if in the process I am repeating things you already know).

A DP gear has a circular pitch (distance between teeth along the PD) of π/DP = π / 20 = .1571". A complete revolution of the 20-tooth gear should produce a theoretical movement of 20 * .1571" = 3.1416 = π (no surprise that we got back to π when you consider the formula for CP and the fact the the DP and the number of teeth are the same). Keep in mind that this theoretical value depends on both the gear and the rack being EXACTLY the right size.

Meanwhile, 1 complete revolution of the stepper motor will only give 15/54 of a revolution of the gear, based on the chain reduction, so 1 revolution of the stepper motor = 15/54 * π = .8723" of movement. IF the stepper driver is single-stepping, it takes 200 steps to do a complete revolution of the stepper motor, so each step of the motor will produce a movement of .8723/200 = .0044". Multiplied by 1832.811, this gives a movement of 7.9971" - about 8 times too much.

BUT that number gives a clue to a crucial missing bit of information: almost certainly the stepper driver is NOT set to single-stepping, but to micro-stepping; in this case, it would appear it is using a micro-stepping factor of 8 (each of the 200 steps is divided into 8 micro-steps). Thus, instead of 200 steps per revolution, it takes 1600 micro-steps for one revolution of the stepper motor shaft. (The computer or microcontroller won't know the difference between a micro-step and a step; it just sends out a pulse - so it takes 1600 pulses to achieve a full revolution.) Redoing the math, 1 micro-step (= 1 pulse) provides ( π * 15 / 54) / 1600 = .0005454" of movement. Multiply that times 1832.811, and voila! you get .9996" of movement - very, very close to 1"; almost certainly within the range of rounding errors and/or the difference between theory and actuality.

Of course, the number 1832.811 should have been calculated the other way around. You want to achieve 1" of movement, and you know that each pulse to the 1/8 micro-stepping motor produces a theoretical movement of (π * 15 / 54)/1600", so to determine the calibration, divide 1 in. by ( ( π * 15 / 54 ) / 1600 ) in./pulse and you get 1833.465 pulses per inch. Oops -- just a bit different from the factor they provided.

Again, though, theory and actuality are not quite the same. Presumably they started with 1833.465 and tested the results just the way DMF described. Then they computed the error and made the appropriate adjustment. So let's say they commanded 4" of movement, and came up instead with 4.0014" of actual movement. The calibration is off by 4 / 4.0014, so multiply that times the calibration factor: 1833.465 * ( 4 / 4.0014 ) = 1832.811. Voila!

Thank you! Everything I had been reading about gave micro steps in units of ten, but i guess 8 and 16 are also common? This machine is actually set a little under what it should be, I'm currently aware that it is about 0.0001 off for every inch of travel, which means over a 97" piece of MDF, the machine cuts it at about 96-15/16". It's hard to measure super exactly over those kind of distances with just a tape measure, but some people I work with have laser measuring devices that are more accurate, and I might try to tweak things by mounting measuring against something attached to the gantry at each end of the table and see where that gets me.

I'm also looking at options for timing belts instead of chain drive, so having the problem worked out in front of me, I can easily switch up the numbers for whatever ratios I end up with. I figure that if I have to switch to 3.625:1 instead of 3.6:1 on the reduction drive it isn't going to effect the motor's ability to move and push the machine into a cut, but it will mess up the distance traveled significantly enough. It would be nice if these machines had a DRO and closed loop sensors on the movement, but most folks don't care to have tolerances that tight, and the cost of an 8' DRO scale (plus for the other axes), would probably cut too deep into the profit margin for the machine manufacturer.

Glad it was helpful. No, I wouldn't think a small change in ratio would be a problem for the motors, but yes, you would need to recalibrate. And yes, multiples of 2 are common for microstepping.

To dial this in more accurately than a tape measure, you might consider getting a $15 digital caliper from your local off-shore discount tool house. With care, this might let you tweak it a bit more precisely. Better yet, get a cheap dial indicator, though a cheap one will be limited to 1" of travel. Which brings up a good point: with 96" of travel, you have an advantage in that any error can be greatly multiplied and thus more easily detected.

I pretend to be a machinist in my spare time, which is also helpful when things break at work, so I already have a bunch of nice measuring tools, indicators, calipers, gauge blocks, etc. the problem with these machines is that for what they are used for, you only need a few thousandths tolerance on parts, often times sloppier than that, but the way they’re built, the tolerance problems add up. I can easily live with being off .0001 over an inch, but not so much when it’s cumulative over 96 inches! Thus, even a tape measure could be helpful in calibrating a machine like this. I have a piece of 3/8” rod with a 60 point ground on the end that turned up in my life at some point. It sits nicely in the spindle collet and lets me jog the machine from one end of a board to another, and check the travel distance on the machine pendant/controller vs a standard sheet of MDF, which I can measure close enough with a tape measure to verify that it is indeed 97” long. My short distance tolerances ought to be well within spec if I can get it to do that.

I was looking at more radical reduction ratios in order to try to up the basic tolerance of the machine when I re-do the drives, but for some reason, it’s very hard to find off the shelf timing gears that will allow you to do more than 3.6:1. Most of what’s available has a hub and bore that is too small for the 1/2” shaft on the motors, and normally stocked sizes over 72 teeth don’t seem to exist. Smaller spur gear for the rack seemed like a possibility, but it seems you run into problems with gear life the smaller they get.

You can try this EXcel sheet.

stepper calculator.xls - Google Search