Motor math...am I doing this right?

[WoodSnarfer]

I read Arvid's nice paper on motors...and want to try to figure an actual example based on a motor I was looking at.

Okay, if I have a DC servo motor that has these specs:

24 VDC
RPM: 2500
Torque (in-oz): 50
HP: 1/8
Amps: 5.5

This means that at 2500 RPM, at the voltage/amps stated, the motor will generate 50 in/oz of torque, right? So, if I were to gear the motor down (let's say 3:1), I'd be at 833 RPM and 150 in/oz of torque, correct?

I'm more comfortable thinking about in/lbs...so 150 in/oz = 9.375 in/lbs, correct?

Now, let's take this motor and gear and connect it to an 8 TPI ballscrew, with 80% efficiency. To determine the force at the nut, I would use this equation:

screw lead = 0.125 in
efficiency = 0.8
torque at screw = 9.375 in/lbs
(edit: T = Thrust, in pounds)

9.375 = T * [0.125 / (2*pi)] * (1/0.8)

9.375 = 0.024868*T

(edit) T = 377 pounds of thrust

Jog speed at 833 RPM would be 104 IPM...a bit on the slow side. If I ran the motor at 36VDC, maybe I could get around 3600 RPM and end up with a jog of 150 IPM. Torque and power ought to be about the same.

Double-checking the power of the motor, let me convert my torque at the screw (9.375 in/lbs) into SI units (newton-meter):

9.375 in/lbs = 150 in/oz * 0.007062 Nm = 1.0593 Nm

Power = M * rotational speed * 2pi
P = 1.0593 * (833/60) * 2pi
P = 92 watts, or about 1/8HP

Did I get all of this right?

Thanks!
-Chris

[trubleshtr]

Chris, Don't quote me but, Ithink that if the motor is rated for 50 in-oz of torque, that is the peak out of the motor, meaning that is all it could give ever, if it was pulling the full load amps constantly, this would most likely wear your motor out if operated at 100 % of it's capacity continuosly.

[WoodSnarfer]

Good point...

In a CNC application, I'm assuming the motor won't be called upon for full power on a continuous basis. I'm thinking that certain activities (e.g. getting the x-axis to move from a dead stop, or lifting the z-axis, etc.) will be demanding on the motor. Milling or routing any significant depth, at any significant speed will also be demanding.

So, (I think) the idea is to hit your target spec, but have around 2x headroom (someone else mentioned 3x...but that gets really expensive). In my example, I'd say that I wouldn't want the nominal linear motion power demand at the cutting bit to be any more than around 180-185 N of force.

I'm just getting on board with this stuff, so please (anyone) correct me if I'm wrong.

Thanks,
Chris

[arvidb]

Hi Chris, nice to hear you liked my paper

I think you are mostly correct, except: the force at the nut would be 377 pounds - since you use inch-pounds in, you get pounds out.

Be careful not to load the motor too much when running it overspeed (36 V). Remember that power = angular velocity * torque, and the motor is not made to produce more than 1/8 HP or about 92 W. So while running at >2500 RPM, I think you need to derate its torque/current capabilities. But I guess it should be able to take some overload, though... Here we could really use Mariss' (or some other expert's) comments; I'm a bit uncertain.

Arvid

[WoodSnarfer]

Thanks, Arvid...yes I see that I mixed units, so I will note this in my post with an edit (in case someone else reads just that post, in the future). It almost seems impossible that a little 1/8 HP motor could produce 377 pounds of thrust (after gearing)...but I guess the math is okay.

I found a nice website that explains how various screws work, and it gives some neat formulas to play with interactively:

http://www.roton.com/web/application.jsp

Oh boy...the more you learn, the more you realize you have yet to learn!

Next, I'm off to do some acceleration calculations...I wonder how quickly we should expect a 150 pound gantry to accelerate from zero to 2.5 inches per second...

Thanks,
Chris

[trubleshtr]

Good formulas on ball screws, thanks woodsnarfer, these will come in handy for my long X axis.

[arvidb]

Originally Posted by WoodSnarfer *snip* Next, I'm off to do some acceleration calculations...I wonder how quickly we should expect a 150 pound gantry to accelerate from zero to 2.5 inches per second... *snip*

Use Newton's formula for acceleration: F = ma (or, in this case, a = F/m).

F force in newtons,
a acceleration in m/s^2,
m mass in kilograms.

I'm not sure what has to be done to make this formula work with imperial units; perhaps the easiest is to translate your units to SI ones:

m = 150 pounds = ca 68 kg
F = 377 lbf = ca 1677 N
v (target velocity) = 2.5 in/s = 0.0635 m/s

a = F/m = 1677/68 = ca 24.5 m/s^2. So taking only the mass of the gantry in consideration you would reach 2.5 in/s in just 0.0635/24.5 or just under 3 ms! This won't be the case of course, because of screw and motor inertia, friction (which makes the available force less than the ideal, calculated one), motor inductance (it takes some time for the motor to develop full torque), and probably a lot of other things too.

Arvid