Stepper Math questions...

[Aegeon]

If anyone can validate my logic here, it would be appreciated (newbie alert)...

I'm considering a timing belt/pulley setup for the X on a large table. I'm trying to work out general motor requirements so I can keep an eye out. Here's how I understand things...

For example, a certain NEMA 34 stepper is listed with 793 oz-in holding torque. The motor is bipolar, 4.68V, 3A, 1.8deg/step. Looking at its torque chart, it looks like it will do 550oz-in at 2000 half-steps/sec.

My assumptions:
1.8deg/step = 0.9deg/half-step, so 400 half-steps/revolution.
2000 half-steps/sec = 5 revolutions/sec = 300RPM @ 550oz-in.

If you have this setup mounted in a gantry:
Pulley A: 1.00" dia, 3.14" circ, attached directly to motor spindle.
Pulley B: 3.00" dia, 9.42" circ, attached to pulley A via belt.
Pulley C: 1.00" dia, 3.14" circ, attached to same axis as Pulley B.
Pulley C is coupled to the timing belt, to move the gantry.
So, the motor drives pulley A, A turns B at 3:1 reduction, B turns C at 1:1, C rides along the belt to move the gantry.

If the motor turns pulley A at 300RPM @ 550oz-in, pulley A turns Pulley B at 100RPM @ 1650oz-in (3:1 reduction). Pulley B turns pulley C at 100RPM @ 1650 oz-in also (1:1). So, if the circumference of pulley C is 3.14", at 100RPM, the gantry would travel 314" per minute with a force of 1650oz-in.

Since the motor does 400 half-steps/rev, each half-step moves pulley A 1/400th of it circumference. Therefore, pulley B moves 1/1200th of its circumference. Pulley C (and the gantry itself) also moves 1/1200th of its circumference, or 0.0026" with each half-step.

I know all this ignores friction, efficiency, belt stretching, etc., but I'm looking at the basic logic for now. Tell me if I'm way off here...

Thanks,
Aegeon

[musicmkr]

If the gantry had no mass, and your pulley system was 100% efficient then yes that math works out.

Unfortunately your gantry will have mass, and a lot of it. Ignoring all other losses, the 314 ipm speed of your gantry would only apply if the gantry were put in motion by something other than the stepper motor. Moving the gantry requires very little energy (if built properly), accelerating it from a stand still takes a ton of energy.

Keep in mind that the most efficient "Timing Belt" system is only 98% efficient. Now a "do it yourselfer" might get somewhere in the area of 95% efficiency. You have stacked two belt systems end to end, giving you a transmission efficiency of 90%.

High School Physics tells us that Force = Mass x Acceleration. To calculate the acceleration of your gantry you must first know it's mass. Luckily for us on earth an object's mass is equal to it's weight. For a large table type gantry with all of the hardware attached including the X linear drive system, the Z linear drive system plus the weight of the spindle you're looking at a minimum of 25 lbs. More likely your gantry bridge will weight somewhere around 40 lbs.

Since your drive pulley has a diameter of 1" it would therefore have a radius of 1/2". 1/2" is the length of the "moment arm" transmitting the torque to the drive belt. With your above example you can assume a 10% pulley system loss so your output torque would be about 1490 oz-in. 1490 oz-in means that at the end of a 1" long moment arm, the system would exert 1490 ounces of force. Since the moment arm of this system is 1/2" you would get double that or 2980 ounces of force pushing the gantry down the table. That's roughly 185 lbs of force (nice motor you've got there).

If we do some algebra, F=M*A becomes A=F/M (185/40). At this point in the torque curve, your motor would be accelerating your gantry at a rate of 4.6 Feet per Second per Second (ft/s^2). That is assuming that you are not performing a cutting operation, and that there is no friction on the linear rails (there will be).

I would do this type of calculation for several points throughout your torque curve to see if this motor is adequate for your application. Remember that your gantry will loose lots of power to friction, and that any cutting operation will require lots of force to overcome. Depending on the low end of the torque curve you may find that your motor isn't strong enough to get your gantry up to a desired speed before it reaches the other end of the table.

I know you were just trying to "ballpark" your system with the calculations that you gave but I didn't want you to be disappointed if it didn't work out as you had hoped. Even the calculations that I show here are a very simple version of what any engineer would do to design this system, but it will give you useable results to help design your system.

[Aegeon]

musicmkr, thanks for the reply...

here's a little expansion on the above, if anyone wants to comment. Again, I have NO physics experience, so I'm trying to stumble my way through the logic.

Based on Musicmkr's calculations in the previous post, the pulley setup would be acting to provide 2980 oz of force to move the gantry, assuming a 90% efficiency between the two belt systems. My question now comes to friction. Here's what I have so far (sorry for all the metric/imperial changes):

I'm considering using extruded aluminum rail (www.8020.net) with their linear bearings. Basically, a plate mounted on a plastic (UHMW-PE) pad that rides in the top groove of the aluminum extrusion. The literature lists the coefficient of friction of the pad versus the aluminum rail as 0.12 (static) and 0.016 (kinetic).

So, if we assume the gantry weighs the stated 40lbs (18.14kg), and assume all the mass of the gantry is directly above the linear bearing (it wont be, but for simplicity's sake):

The Normal Force of the gantry = Gantry Mass * Acceleration of Gravity = 18.14kg * 9.81m/s^2 = 177.95kg m/s^2 = 178 Newtons
Resistive force of friction = Normal Force * Coef of Friction = 178 Newtons * 0.12 = 21.4 Newtons = 4.8 pounds of force

So, does this mean that 4.8 pounds of force are required just to overcome the resistive force of friction between the gantry and the extruded rail during acceleration? In other words, is it a simple subtraction from the force provided by the motor, prior to the F=M*A equation?

For acceleration with no friction, we had:

A = F/M = 185/40 = 4.6 ft/sec^2

With friction, does this become A = (185-4.8)/40 = 4.5 ft/sec^2 ?

Thanks again,

Matt