[CAM_MAN]

I really could use some help. I am build ing a router from purchased plans which has already been a nightmare. I have never really got into gearing. The router plans parts list called for certain sized gears and belts for the X and Y axis. I recieved the gears and machined the bores to fit my ballscrews and motors. When doing the assembly of the gears and belts for the Y and Z axis nothing fit right. After reviewing photos that were sent with the plans I discovered that none of the belts or pulleys on the parts list were actually the ones on his machine. I finally got the "correct" part numbers he used on his machine. Anyhow I really thought about the way he had the machine geared and didn't think something was right.

Z axis - 20 tooth on motor 30 tooth on ballscrew
Y axis - 42 tooth on motor 30 tooth on ballscrew
X axis rack - 16 tooth pulley on motor 17 tooth pulley on driveshaft. 12 pitch 12 tooth gear on drive shaft to rack and the rack is a 12 pitch.

My question is that shouldn't the gearing be the same for all axis? I really don't know how to figure out the final ratio for the X axis. I know that the Y and Z ratio are different. Isn't Y reduced one way and the Z reduced the other way? I am thinking he just used whatever pulleys and belts would fit his design.

I am using Tormach 1200 oz. in. motors on all three axis and 5/8" x 5 TPI ballscrews. My goal is to have some descent torque but mainly a high feed rate. I would like to have all three axis be ale to achive the same feedrate. I am pretty much stuck with the rack gearing on the X axis but could still change the gearing on the X axis driveshaft and motor. The only pulley-belt configuration I could get to work with the way the the mounts are machined is 30 tooth on the motor and 20 tooth on the ballscrew or vice-versa. And would that be a 1.5 to 1 ratio? This is the only thing keeping this machine from running. Could somebody please give me some help or just steer me in the right direction?

[HuFlungDung]

The trick is to understand how much movement you get per turn of the motor.
With modern controls, they are usually designed to be able to compensate for different ratios on different axis. Old controls used to have difficulty performing circular interpolation if the ratios were not exactly the same. So the ratios you have may not matter that much if your controllers can properly handle the number of pulses to the axis requiring the most pulses to make a given move.

Your rack gear will move pi linear inches in one revolution of the 12 tooth gear. So it appears that your X axis is plenty fast: (16/17) * pi = 2.956" per motor rev

Your Y axis will move (40/30) * .2 = .28 inch per motor rev

That is a fair discrepancy. I'd say you need to use a speed up ratio on the Y axis, and a speed reducing ratio on the X axis, to bring them closer to one another.

The Z axis is not a major concern, because it is not typically involved in anything except simple linear 3 axis interpolation.

[2muchstuff]

Your Y axis is actually increased or stepped up in ratio to 1:.71428 With a 200 step motor driving a .2" lead screw, you come up with 714.2857 steps / inch. A 1200 in/oz motor on a .2" lead will push 1513.305 lbs.

Your Z axis is reduced in ratio, 1.5:1 With a 200 step motor and a .2" lead you get 1500 steps/ inch. A 1200 in/oz motor on a .2" lead will push 3177.966 lbs.

Your X axis is reduced in ratio, 1.0625 With a 200 step motor driving a 12 pitch gear, you get 212.5 steps per 2.956" or 71.8876 steps per inch. A 1200 in/oz motor will give you 79.6875 lbs. of thrust.

[CAM_MAN]

Thank you very much for the help. I did make one mistake though in my earlier post. It is a 72 tooth pulley on the driveshaft and not a 17 tooth pulley. So what would that ratio be. I am sure this changes a lot. Sorry about the mistake. 2muchstuff what formulas do you use to calculate the ratio and come up with the steps per inch and thrust. Thanks again.

[2muchstuff]

Well that changes things a bunch on the X-axis. The X-axis is reduced by a ratio of 4.5:1 (72/16=4.5). With a 200 step motor driving a .2" lead screw you come up with 4500 steps/inch ( 200 steps x 4.5 ratio x 5 turns/inch of screw= 4500). A 1200 in/oz motor will give you 9533.898 lbs. of thrust.

(1200in/oz)/ 16 oz/lb=75in/lb)
Thrust= .177(load)(lead)
75=.177(load)(.2)
75=.0354(load)
75/.0354=load
2118.644=load in lbs

2118.644(4.5 ratio)=9533.898 lbs of thrust

Now these final figures are theoretical but will give you an idea where things are heading. There still is friction and effiency of the ballscrew to be figured in. 95% effiency is a good place to start for ballscrews.

Ooops messed up, forgot the x was on a rack not a screw.
These figures will work for the axis' that are using ballscrews, just change the figures.

[2muchstuff]

Let's try this again.

X axis is reduced by 4.5:1 (72/16=4.5)

16/72(pi)=.6981" travel/motor revolution
200 steps(4.5)=900 steps/.6981" or 1289.2135 steps/ inch

A 1200 in/oz motor(4.5)=5400in/oz
5400/16 oz/lb= 337.5 lbs of thrust.

As you can see there is a significant difference when using a rack vs. a ballscrew. A screw can multiply the force quite a bit.

[CAM_MAN]

Thank you very much for the formulas and answers. It is greatly appreciated. Now I have a clue of what I am dealing with in the gearing department.