# Motor Sizing Help (Basic Rotational Inertia Problem)

1. Oct 1, 2007

### dswarrenga

Hello everyone,

I'm having trouble with an application involving rotating a 10.8lb device through 60 degrees of rotation using a stepper motor (Hurst 4006-004: step angle .1 degrees, Output speed 3.33 rpm, rated torque at 175 pulses/sec of 162 oz-in). It appears that the motor is underpowered and I may need to increase it however when I try applying some basic physics, I don't see where the motor can possibly be underpowered. Maybe someone can help me. Here's what I have:

Moment of Inertia about the rotating axis: 90.8 lb*in^2
Angular velocity .348 rad/sec
Time to reach full velocity, .25 seconds

I of course am neglecting friction in this simplified analysis (mainly because I don't know how to apply it here. It's a aluminum to aluminum sliding contact joint)

Any insight here on how to properly size a situation like this would be very helpful.

Warmest Regards,

David warren

2. Oct 1, 2007

### FredGarvin

Can you describe the set up just a bit, i.e. geometry and how things are connected/restrained?

Friction can be a tough thing to estimate. According to Mark's Handbook, aluminum-aluminum contact without lubrication results in a coefficient of static friction of 1.35. So right there, for two flat pieces, you need to add 35% on to the weight to account for breaking static friction.

3. Oct 1, 2007

### dswarrenga

The motor mounts to the very bottom shaft. When activated, it rotates the entire assembly. I have modelled the coordinate systems as shown at the bottom and have calculated the mass at 10.8 lbs and the Izz at 90.7 lb*in^2. The square plate at the bottom is rigidly affixed to a large box and doesn't rotate (it serves as the bearing). I don't have the true angular accelerartion of the motor as the spec sheet doesn't say so I'm having to guess at it. I know the whole aasembly slews pretty fast through the require 60 degree arc so it has to be pretty fast. Hope this helps.

David Warren

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4. Oct 1, 2007

### FredGarvin

I did some quick numbers, and the torque required to rotate the mass through that angular acceleration is pretty low. I am getting around .027 Ft-Lbf. This doesn't look at the overall power available from the motor however. Does the spec sheet call out a max power?

This, combined with your model, makes me wonder about a couple of things:

1) Like you already mentioned, friction. Is it possible to put a thrust bearing at the bottom where the mount plate is?

2) The load distribution and the transfered load to the shaft of the stepper motor. Are you certain that the mass that you are trying to rotate is not applying a side load on the stepper motor shaft?

5. Oct 1, 2007

### dswarrenga

The motor specs indicate a max torque of 162 oz*in @ 175 pulses/sec, input power 8 watts, voltage 12vdc. Unfortunately that's it.

The printout from the cad program is as follows:

VOLUME = 4.9690188e+01 INCH^3
SURFACE AREA = 6.3583122e+02 INCH^2
AVERAGE DENSITY = 2.1778886e-01 POUND / INCH^3
MASS = 1.0821969e+01 POUND

CENTER OF GRAVITY with respect to CSYS_PART_DEF coordinate frame:
X Y Z 4.0248265e-01 -4.8382887e-01 -5.5962498e+00 INCH

INERTIA with respect to CSYS_PART_DEF coordinate frame: (POUND * INCH^2)

INERTIA TENSOR:
Ixx Ixy Ixz 4.5954778e+02 -1.1090908e+00 2.5133556e+01
Iyx Iyy Iyz -1.1090908e+00 5.0086922e+02 -1.6748325e+01
Izx Izy Izz 2.5133556e+01 -1.6748325e+01 9.0778344e+01

INERTIA at CENTER OF GRAVITY with respect to CSYS_PART_DEF coordinate frame: (POUND * INCH^2)

INERTIA TENSOR:
Ixx Ixy Ixz 1.1809190e+02 -3.2164824e+00 7.5822281e-01
Iyx Iyy Iyz -3.2164824e+00 1.6019358e+02 1.2553534e+01
Izx Izy Izz 7.5822281e-01 1.2553534e+01 8.6491949e+01

PRINCIPAL MOMENTS OF INERTIA: (POUND * INCH^2)
I1 I2 I3 8.4364123e+01 1.1793060e+02 1.6248270e+02

ROTATION MATRIX from CSYS_PART_DEF orientation to PRINCIPAL AXES:
-0.03787 0.99693 -0.06856
-0.16477 0.06144 0.98442
0.98560 0.04858 0.16194

ROTATION ANGLES from CSYS_PART_DEF orientation to PRINCIPAL AXES (degrees):
angles about x y z -80.658 -3.931 -92.175

RADII OF GYRATION with respect to PRINCIPAL AXES:
R1 R2 R3 2.7920664e+00 3.3011108e+00 3.8748099e+00 INCH

I established the new coordinate system along the shaft at the bottom edge with the (+z) axis co-axial whith the shaft and pointing down. I believe we can fit in a new bearing design for improvements. I'm also pretty sure there is some possibility of side shaft loading as the CG of the system is offset from the axis of revolution, however its only off by about .4". I'm not sure how big a contribution that would be. If I'm reading the motor spec correctly, 162 oz*in would relate to about a 10lb force on the @1" shaft. (If I'm following the equations correctly).

6. Oct 5, 2007

### TVP45

I never get anything right using mass pounds, so bear with me. Are you saying the radius of gyration is about 3 inches?

7. Oct 7, 2007

### TVP45

Later on, I realized you probably nailed the problem, Fred. I don't see any mention of a flexible coupler; is there one? If not, you may have enough radial eccentricity to overload the motor.

8. Oct 8, 2007

### dswarrenga

There is no flexible coupler. The shaft of the motor simple is inserted into the bottom of the shaft and pinned in place with a key pin. There is some flexibility permitted, but not a great deal. Unfortunately the person who designed this is no longer availible for consult. I clocked the deflection time for 60 degrees (the range of motion on the motor) from rest at 5 seconds. We definitely aren't talking about a fast moving motor here.

9. Oct 8, 2007

### FredGarvin

Ideally there should be a flexible coupler that is rigid torsionally but allows for misalignment. That would alleviate the side loading issues to a large extent as well.