Recap of the last posting

In our last posting, we were comparing the inertia of a 12” long, 0.25” diameter steel rod (0.0013 lb-in^2) to a 1” long, 1” diameter aluminum coupling (0.00957 lb-in^2).

We asked why the lighter weight aluminum coupling would have a larger inertia than a heavier steel rod and the person in the back said “it’s because the coupling’s radius is larger and inertia is a function of the radius squared” and they were correct.

Let’s continue with our calculations by adding another inertial element.

Adding a roller (rubber sleeve)

Figure 1: Calculating inertia for a primary roller (rubber sleeve)
Figure 1: Calculating inertia for a primary roller (rubber sleeve)

I’m going to add a rubber sleeve over the steel shaft. I’m going to make it 10” long. That’s 2” shorter than the steel shaft; because the shaft needs to be supported by bearings on each end and we have to have some shaft length for the coupling to clamp on.

I Googled “density of rubber” and found that hard rubber is 74 pounds per cu-ft. and soft rubber is 69 pounds per cu-ft. This gives us a range of 0.0399 lbs/in^3 to 0.0428 lbs/in^3.

So someone picks a number for us to use.

Do you like 0.04 lbs/in^3? Great, let’s go with that.

Calculate the volume of the sleeve

I’m going to pick a circumference that is 2” with an outer diameter of 2/\pi or  0.63662”. Why that value you ask?

Well, one revolution would produce two inches of linear motion if a piece of material was sitting on the surface of the roller. We’ll get into that inertial calculation latter.

Using our trusty volume equation again \pi(r^2) \times length we get:

    \[ Volume = \pi\left(\frac{0.63662}{2}\right)^2 \times 10 = 3.1831 in^3\]

Multiply the volume by the rubber’s density and we get a weight (w_1) of 0.1273 lbs.

Using our inertia equation

    \[J = \frac {(w_1r^2)}{2} \]

we get:

    \[J_1 = \frac{(0.1273\times0.31831^2)}{2} = 0.00645 lb-in^2\]

But wait. Our rubber sleeve has a hole in it that allows it slip over the steel shaft. So let’s calculate that inertia and subtract it from our total.

    \[Volume = \pi\left(\frac{0.250}{2}\right)^2 \times 10 = 0.491 in^3\]

    \[Weight (w_2) = 0.491 \times 0.04 = 0.0196 lbs\]

    \[Inertia (J_2) = \frac{(0.0196 \times 0.125^2)}{2} = 0.000153 lb-in^2\]

Subtract  J_2 from  J_1 and we get:

    \[J_T = 0.00645 - 0.000153 = 0.006297 lb-in^2\]

There’s an another way

Now having done all that we could have looked up the inertial equation for a thick-walled hollow tube which is:

    \[J_T = \left(\frac{(w_1 - w_2)}{2}\right) \times (r_2^2 + r_1^2)\]

Where r_1 is the inner radius and r_2 is the outer radius.

Plug and chug our values:

    \[J_T = \left(\frac{(0.1273 - 0.0196)}{2}\right) \times (0.125^2 + 0.31831^2) = 0.006297 lb-in^2\]

The same as we calculated before.

Note that the weight is its actual weight, by that I mean the weight of the missing center is subtracted from the total (as if it was a solid) weight

A more efficient way

Now, I don’t know about you, but the first time I looked at that equation I would have said that it should “r_2^2 - r_1^2” not “r_2^2 + r_1^2”, but you do add the two squared radii together. I’ll leave the derivation of that up to you.

The equation that makes more sense to me is the following:

    \[J =  \left(\frac{\pi\rho h}{2}\right) \times (r_2^4 - r_1^4)\]

Where \rho is the density and h is the height or length of the hollow tube.

This is exactly what we did by using two separate equations. One for the whole rubber sleeve and one for the 0.25” section we removed.

Now that we have calculated the inertia of the rubber sleeve, let’s calculate the total inertia that the motor shaft sees.

Our 1”x 1” coupling has an inertia of 0.00957 lb-in^2.

Our 0.25”x 12” steel rod has an inertia of 0.0013 lb-in^2

And our 0.6366” x 10” rubber sleeve has an inertia of 0.006297 lb-in^2

Since everything is directly attached to the motor’s shaft, all we need to do is add the three inertias together to get our total inertia of 0.017167 lb-in^2

More next time.

In our last posting, we calculated the inertia of a helical shaft coupling. The coupling is used to connect the motor shaft to the load shaft.

We calculated the inertia of:

  • a coupling that was 0.5” in diameter and 0.5” long at 0.0003 lb-in2
  • a coupling that was 1” in diameter and 1” long at 0.00957 lb-in2

The notable difference between the two couplings is that the large one, even though it’s only twice as large in both its length and diameter, has an inertia that’s 32 times greater.

So what, you might ask? Well if we are striving to match the motor’s rotor inertia to the load’s inertia, then a larger coupling might force us to choose a larger motor and larger motors cost more. A large coupling may very well be needed to meet the torque requirements of the application, so don’t just go by the inertia of the coupling when choosing one.

Steel load shaft inertia

Okay, so what’s next, now that we have a coupling attached to the motor’s shaft? How about we connect a steel load shaft to the coupling and see what impact that has on the total inertia.

Figure 1: Calculating the inertia of a steel rod used as a load shaft
Figure 1: Calculating the inertia of a steel rod used as a load shaft

Let’s pick some numbers out of the air and make our steel shaft 0.25” in diameter and 12” long. If you Google “density of steel lbs/in3” you get at the very top of the page:
“Approximately 7.85 g/cm3”. Not quite the units that I wanted, but if you look down the page a bit you’ll see steel has a density of 0.283 lbs/in3. Yea, yea there are different carbon steels and other alloys to consider, but ours is the one that’s 0.283 pounds per cu-in.

The inertia (J) equation for this shaft is the same as the one we used for the coupling in the previous posting and that is:

    \[ J = (mr^2)/2 \]

where w is the material density and r is the shaft radius.

First we need to calculate the weight of the shaft first with:

    \[ Mass_l_b_s = \pi \times  r^2 \times length \times density \]

    \[  3.14159 \times .125^2 \times 12 \times 0.283 =\boxed {0.167} \]

Then we will calculate the inertia of the shaft:

    \[ (0.167 \times 0.125^2)/2 = \boxed{0.0013} \]

Load shaft inertia equals 0.0013 lb-in2

Why does the steel shaft have a lower inertia than the aluminum coupling?

An interesting thing to note, ok I think it’s interesting, is that the one-inch diameter aluminum coupling weighing only 0.0766 lbs (2.18 times lighter than the steel shaft) and yet it has an inertia (.00957 lb-in2) that is 7.36 times larger than the steel shaft.

How can a lightweight aluminum coupling have seven times the inertia of a steel rod that’s 12 inches long and weighs more?

Yes, you got it! Inertia is a function of the radius squared.

  • The coupling has a one-inch diameter, a 0.5” radius. Square it and you get 0.25
  • The steel shaft has a 0.25” in diameter, a 0.125” radius. Square it and you get 0.015625.

This produces a 16:1 inertial factor just by having a larger diameter. So even though the aluminum coupling is lighter than the steel shaft, its diameter is what makes its inertia greater.

I think we’re getting up to speed with this. Or, We’re overcoming the inertia of these calculations.
More next time.

In our last posting, we talked about having a good quality coupling that connected the motor’s shaft to the load’s shaft.

Finding the right coupling

A good quality coupling would be able to accommodate a reasonable angular misalignment between the shafts as well as parallel misalignment. In addition, it should have the minimum torsional windup.

This coupling has an inertia that needs to be taken into account when trying to size a motor. The coupling that I’m thinking about is made of aluminum, has a hole for the motor and load shafts that go all the way through the coupling and it has helical cuts on its circumference that allows it bend and accommodate the misalignment of the shafts.

This cylindrical shape is a good one, to begin with.

Let’s start off by considering it a solid piece of aluminum. We’ll take into consideration the material that was removed from the coupling (the internal diameter) that accommodates the shafts later.

Figure 1: Aluminum helical couplings
Figure 1: Aluminum helical couplings

Google “list of moment of inertia” and you’ll get all sorts of neat places to find the equations that we’re going to use and some of them derive the equation, so I’ll let you decide to do that if you’re so inclined.

Our solid aluminum coupling shape’s equation is:


J = inertia
m = mass
r = radius

I always used a J for inertia because the “I” was used for current and the current’s “C” was used for capacitance and I bet we could go on for quite awhile with this alphabet soup.

I’m going to pick the actual dimensions from an unnamed coupling manufacturer, shown in Figure 1,  shown in Figure 1, Example 1,  and we’ll crunch some numbers and see what we come up with.

The coupling “cylinder” is 0.5” in diameter and 0.5” long.

The first question we need to answer is: how much does this thing weigh? What is its mass?

Calculate the mass of the coupling

All we need to do is calculate the volume and multiply it by its density to get the mass.

I Googled “density of aluminum English units” and found aluminum to be 0.0975 lbs/cu-in. The volume is V = \pi r^2L or:

    \[3.14...\ \times .25^2 \times 0.5 = 0.09817477\ in^3\]

Multiply the volume by the density m=V\rho and we have the mass of the coupling:

    \[0.0982 \times 0.0975\]

    \[m = 0.00957\ lbs  \]

The inertia is J = (mr^2) \div 2 or (.00957 \times 0.25^2) \div 2 = 0.000299\ lb-in^2

The manufacturer defines the inertia as: 0.000310 lb-in^2.

We’re reasonably close

A larger coupling

Let’s pick a coupling that twice the size of the one we just did.

The coupling “cylinder” is now 1” in diameter and 1” long, as shown in Figure 1, Example 2.

How much do you think the mass and inertia are going to change because of this increase in size? Go ahead and pick some values and write them down so you can’t forget them and don’t look ahead for a hint.

Go through the volume/mass calculation again and we get 0.0766 lbs. Hmmm, that’s 8 times more mass than the smaller coupling. Did you think it would be eight times more?

And the inertia calculation produces (drum roll please) 0.00957 lb-in2 or 32 times larger than the smaller coupling. How was your estimate on this one?

Why did it increase by so much? Well, the weight is proportional to the square of the radius. If you double the radius you increase the weight by four times. If you double the length, you double the mass. Thus, the mass is 4 * 2 = 8 times heavier.

I love high powered math.

Now the inertia is proportional to the square of the radius too, so if you double the radius the inertia increases by four times. Thus, 4 * 8 = 32 times more inertia than the smaller coupling.

Oh, by the way, the manufacturer’s inertia value is: 0.00902 lb-in2

More number fun next time.

In our last posting, we talked about how a very torsionally flexible coupling could affect the ringing at the load.

A rubber tube

We proposed a one-inch long plastic tube that fit snugly over the motor’s shaft and the load’s shaft and that the torsional windup of this coupling was zero. The moves were snappy and settled out with very little ringing.

Then we eventually made our plastic tube coupling 12” long. The longer the tube the more torsional windup we had. The motor would take a step, but the tube would just twist or windup and it would eventually turn the load. The load ringing would take a long time to settle out.

Now this discussion was taking place in a frictionless environment.

Figure 1: Torsional windup of a rubber tube and load inertia
Figure 1: Torsional windup of a rubber tube and load inertia

Add load friction

Add some load friction and our 12” hose coupling would have to windup until the torque it was transmitting exceeded the sticky-friction or stiction that holds the load still. The load would eventually move if the motor stepped far enough, but when the motor stopped the load would get stuck at some in-between location. This in-between location would be located at a torque value that is less than the stiction value and with some of the motor’s torque “stored” in the torsional windup.

So do you think a stiffer coupling is best? How about we replace the plastic tube with an aluminum tube that again fits over both the motor’s shaft and the load’s shaft?

Aha, those were kind of trick questions because the answer is yes for the first question and no for the second, but why? Weren’t we trying to get the load to be tightly connected to the motor’s shaft?

If a solid coupling were used

If the load is coupled to the motor’s shaft via a solid coupling then we’d have no windup at all. However, both the motor and the load’s shaft need to be perfectly aligned with each other. The shafts need to be perfectly lined up and parallel to each other and the mounting faces need to be perfectly parallel too. Remember the motor shaft and the load shaft are in fixed positions and anything other than perfect alignment will cause the shafts to bend as they rotate.

So what happens if the shafts are slightly off? Well, something has to “give.” That something is the shafts. In order for them to stay aligned as the shafts turn they will bend. (Wasn’t “As the Shafts Turns” a soap opera?) What happens is the motor shaft or the load shaft or both shafts bend at the point they exit their bearings. This back-and-forth bending motion accommodates the misalignment of the shafts. This back-and-forth bending motion takes place every time the shaft makes a move, be it a complete revolution or a partial one. The shafts are being stressed and will eventually break because of material fatigue. It’s what happens when you repeatedly bend a piece of metal back-and-forth. I’ve actually seen a broken motor shaft stay attached to the motor body via a split ring on the inside of the front motor housing bearing, but the shaft was no longer attached to the rotor.

The bottom line: flexible coupling

Bottom line is that you need a good quality flexible coupling to go between the load’s shaft and the motor’s shaft. This desirable flexible coupling needs to be able to accommodate the angular misalignment’s between the motor and load shafts and have very little torsional windup.

More next time.

We’re continuing our discussion about stepper motor ringing and inertia matching. We had established the following load-to-rotor inertia ratios:

Inertia Mathing
Figure 1: Inertia Matching in stepper Motors


I’m sorry if you Googled “ringing” because of a hearing issue and end up on this blog. You should Google “tinnitus” instead or search the WebMD site.

Now where were we? Oh yea, in our last posting we were talking about load-to-rotor inertia matching. We had established the following load-to-rotor inertia ratios:


We started our stepper motor sizing discussion with the last posting and introduced the NEMA rating for motors. The stepper sizes that we talked about ranged from the smallest NEMA 14 up to the largest the NEMA 34. Other manufacturers might have smaller (NEMA 8 &12) or larger sizes (NEMA 42), but we’ll let those folks write their own blogs.


Our last posting talked about applications that had marginal torque and that caused our “Hybrid” motor to correct its position.

These applications were essential “path critical” and the correcting moves the “Hybrid” control did cause the path or the move time to vary. Proper motor sizing avoids this issue.