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Sizing: Determining the mass of the coupling

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.

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LMD Linear

Liberty MDrive (LMD) Linear Actuator products integrate a 1.8° 2-phase stepper motor, external shaft linear mechanicals and drive electronics to deliver long life, high accuracy, and repeatability in compact, low cost packages.

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