can someone look this over i think i may have a error here,
Im planing to use dual drive motor on my x axis with rack and pinion set up
5x9 table.

how does that change this calculation...

rack
S1811Y-RB-1P
24 diametrial pitch 20° pressure angele

Pinion gear = 1” in diameter
One rotation = 1 * pi (3.141) = 3.141 in travel per rotation
Target rapid transverse = 1300 ipm

NEMA 34 HIGH TORQUE STEPPER MOTOR 1810 oz-in,
http://www.kelinginc.net/SMotorstock.html
RPM of pinion to achieve rapid transverse = 1300 / 3.141 = 413.88 ~ 414 rpm
Estimate stepper RPM = 3000 rpm
Required reduction ratio = 3000 / 414= 7.24:1 ratio
Available peak servo torque = 1810 oz-in
Torque output of reduction gearbox = 1810* 7.2 = 13032 oz-in
Torque at radius of pinion gear = 13032/ (7.2 / 2)
.........................................= 13032/ 3.6 = 3620 N

Acceleration = F = m * a

Available peak force at pinion radius = 3620 N

Estimated mass of gantry = ~150 lbs

3620 = 150 * ?
= 3620 / 150
= 24.133 ipm

Target rapid transverse = 1300ipm
Time to achieve rapid transverse = 1300 / 24.133 = 53.86 sec

how do i calculate my resolution?

You'll never get anywhere near 3000 rpm with that stepper. At 300 rpm, it's already lost half it's torque, and by 900 rpm, the torque will have dropped to about 220oz-in.

You'll never get anywhere near 3000 rpm with that stepper. At 300 rpm, it's already lost half it's torque, and by 900 rpm, the torque will have dropped to about 220oz-in.


ok any suggestion for better motor. Not sure if my math/ formulas are right,

Would a servo motor be a better choice?

does any one see any issues with my calculation...see frist post

Can't say, but I'd take a look at :
http://www.mechmate.com/forums/index.php
Maybe sign up and ask there too?

Gerald there is into Rank and Pinion and maybe could help the maths - He has been a sometime poster here too - but been a while I think. He's got some freebie downloads for his big R&P system.

Hope that is useful - Cheers - Jim

Eloid,

To get the true picture of your setup, you need to look at the speed vs torque curves for a given stepper. Sometimes a smaller motor has better performance than a larger one when operating at higher speeds. I used a smaller one below.

I chose to use (2) of the 960 oz-in steppers from your source. I arbitrarily chose a gear reduction ratio of 3:1 operating the motor at 5000 half steps/sec. This gives 750 RPM at the stepper shaft, 250 RPM on the output resulting in a traverse speed of 785 IPM. The choice is kind of arbitrary; you might try different scenarios.

The stepper torque has declined to 354 ounce inches at this speed. With (2) steppers on your heavier axis, you have 758 max ounce inches of torque, and 3x this at the drive shafts
or 2124 ounce inches. At a 1/2 inch radius pinion this results in about 4248 ounces or 265# max of force.

Your large carriage weighs 150#; this is 150/32.2 = 4.66 slugs (mass).
a = f/m so your acceleration will be 265/4.66 = or 57 ft/sec squared.
The time to accelerate to full speed will be t = Vfinal/a, 785 IPM = 65.4 ft/min.
Now t = 65.4/57 = 1.1 seconds to accelerate 150# to 785 IPM. (That's flyin' right along, you know!)

At 758 IPM the out shaft goes 758/3.14159 = 250 RPM, motor at 750 RPM, 750/60 RPS.
Assuming you will use 10x microstepping, the max step pulse rate will be (750/60) x 2000
which is 25000 pulses per second.

The resolution will be running a pinion of 3.14159" circumference from 6000 pulses per pinion revolution, giving .0005236"/pulse, about a half thousandth per pulse.

In my calculations above I neglected the inertia of the steppers (and friction and load force
which are unknown to me).

I hope this helps your choosing.

Regards,
Jack C.

sorry still new to this... i hoping your can break up some the math bit more
understand.... what u say ..but need to know how you get some of your rpm cal and ipm.

thanks


Eloid,

To get the true picture of your setup, you need to look at the speed vs torque curves for a given stepper. Sometimes a smaller motor has better performance than a larger one when operating at higher speeds. I used a smaller one below.

I chose to use (2) of the 960 oz-in steppers from your source. I arbitrarily chose a gear reduction ratio of 3:1 operating the motor at 5000 half steps/sec. This gives 750 RPM at the stepper shaft, 250 RPM on the output resulting in a traverse speed of 785 IPM. The choice is kind of arbitrary; you might try different scenarios.

The stepper torque has declined to 354 ounce inches at this speed. With (2) steppers on your heavier axis, you have 758 max ounce inches of torque,

354x2=708???????

and 3x this at the drive shafts
or 2124 ounce inches. At a 1/2 inch radius pinion this results in about 4248 ounces or 265# max of force.

if was 1"? PINION????

Your large carriage weighs 150#; this is 150/32.2 = 4.66 slugs (mass).

32.2? where did this come from?

a = f/m so your acceleration will be 265/4.66 = or 57 ft/sec squared.
The time to accelerate to full speed will be t = Vfinal/a, 785 IPM = 65.4 ft/min.
Now t = 65.4/57 = 1.1 seconds to accelerate 150# to 785 IPM. (That's flyin' right along, you know!)

At 758 IPM the out shaft goes 758/3.14159 = 250 RPM, motor at 750 RPM, 750/60 RPS.
Assuming you will use 10x microstepping, the max step pulse rate will be (750/60) x 2000
which is 25000 pulses per second.

The resolution will be running a pinion of 3.14159" circumference from 6000 pulses per pinion revolution, giving .0005236"/pulse, about a half thousandth per pulse.

In my calculations above I neglected the inertia of the steppers (and friction and load force
which are unknown to me).

I hope this helps your choosing.

Regards,
Jack C.