Stepper motor rez

[steeltoes]

I'm trying to design a system around the components I already own. This is my first CNC and need a bit of help.

For the gantry I am using a 200 oz in #23 stepper dual shafted, rigid shaft coupled to each side. On each end I have 1/5 10 tooth timing pulleys. So am I correct in saying for every rev, I'll have 2" of linear travel? Do I need use a reduction of sorts?

thanks in advance

[doorknob]

I don't think that you have provided enough information.

How is the timing belt driving the gantry?

[steeltoes]

Open ended, fixed on each end drive pulley and an idler on either side lower than the drive pulley

[Bear5k]

I'm trying to design a system around the components I already own. This is my first CNC and need a bit of help.

For the gantry I am using a 200 oz in #23 stepper dual shafted, rigid shaft coupled to each side. On each end I have 1/5 10 tooth timing pulleys. So am I correct in saying for every rev, I'll have 2" of linear travel? Do I need use a reduction of sorts?

thanks in advance

This is really, really unclear. First, I assume you are driving a leadscrew of some sort. The number of starts and pitch will be the primary determinant of how much linear travel you get. After that, then you multiply that by your gearing ratio. If you have the same size pulleys on the motor and the screw, then you are getting a 1:1 ratio. If the circumference of the pulley on the motor is 1/5th the circumference of the pulley on the screw, then your travel per revolution of the motor will be 1/5th what it would be if you directly drove the screw with the motor. The good news about gear reduction is that you increase torque and increase resolution when you do this. The downside is that your top-speed is lowered and you will introduce a moderate amount of backlash into the system.

[steeltoes]

Sorry...yeah I wasn't explicit enough

Going to be an XL belt drive system...each side running entire length of 60"...Single stepper/double shafted, mounted on the gantry...rigid 1/2" shafts to the left and right approx 18" in length....on the each end of those there will be 10 tooth timimg pulleys... the belt will be horzizontal, 90 deg around an idler, 180 deg around the drive pulley, then 90 deg around another idler, so the belt would now be on the same plane as it started.


The belt is a .2 pitch, so I would I be accurate in saying that for every rev, the distance travels would be 2" (.2"x10 tooth)

[ger21]

Find the pitch diameter of the pulley and multiply it by pi (3.14). It should be very close to 2", depending on how many decimal places you have for the pitch diameter.

[steeltoes]

Correct, but my initial question was. For every rev, i'll have approx 2" of travel. Will it be accurate enough? Seems like to much distance per rev IMO

[ger21]

Only you can answer that. For me, I'd say no.

[steeltoes]

So if I use a 10 tooth and a 40 tooth I'd have a 4:1 reduction. With that I'd end up with about .5" per rev. I like the idea of a belt type system, but with all the add on stuff acme rods are looking like a better choice. Any recomendations for size and thread on a 60" table?

[Bear5k]

So if I use a 10 tooth and a 40 tooth I'd have a 4:1 reduction. With that I'd end up with about .5" per rev. I like the idea of a belt type system, but with all the add on stuff acme rods are looking like a better choice. Any recomendations for size and thread on a 60" table?

60" table and 200 oz-in motors? You may want to look into rack-and-pinion drive systems with a decent gear reduction.

[steeltoes]

60" table and 200 oz-in motors? You may want to look into rack-and-pinion drive systems with a decent gear reduction.

Only 60" travel length...My usable area is 2'x4'...It is table top model style, designed to cut 1/4 or less ply and foam

[Bear5k]

Only 60" travel length...My usable area is 2'x4'

For a 4' X-axis, you are still most likely better off going R&P. See the CNCRouterParts builds using R&P for some ideas.