With this posting we’ll begin a discussion about leadscrew/ball screw/linear slide systems.
Picture a screw with a nut on it. No, no not a Walnut. Sighhhh. OK then, picture a right-hand threaded 1/4-20 screw with a 1/4-20 nut threaded on to the screw. A 1/4-20 screw has an outside diameter or major diameter of 0.25″ and the thread of the screw has 20 turns per inch. Thus, the name is a 1/4-20 screw. The valleys of the threads are cut into the screw’s shank and are referred to as the root diameter or the minor diameter.

Orient it so that you are looking at the head of the screw and the thread portion is pointing away from you. Hold the nut with one hand so it won’t spin and turn the head of the screw clockwise (CW) with your other hand. No, it doesn’t matter if you hold the nut with your right or left hand. What happens? The screw moves toward the nut or the nut moves toward the head of the screw. (Do you remember righty-tighty?)

Now rotate the screw counter clockwise (CCW). What happens? The nut moves away from the head of the screw. (How about lefty-loosey?) We just created another rotary-to-linear motion converter.
Let’s get even more creative by attaching a flat plate to the nut. Now, when the screw is rotated, the flat plate/nut moves back and forth. Let’s try to make it even more usable.

Let’s mount the screw in an assembly that has a bearing/restraining support for the screw that prevents it from moving in a linear fashion. We’ll also have the screw’s head line up with the shaft of a stepper motor. We’ll add a mounting bracket for the motor so that the motor’s shaft and the screw’s shaft stay lined up. Remove the screw’s head and add an appropriately sized coupling to connect the two shafts together. Install ball bearing rails on both sides of our flat plat. These bearings will now prevent the nut from turning (Yes, you can let go of the nut) and provide a low friction mechanism for the flat plate to slid on. The flat plate moves toward the motor when the stepper motor’s shaft rotates CW (as seen from the back of the motor).When the motor’s shaft rotates CCW the flat plate moves away from the motor. Google “motorized linear slides” and you’ll find out we didn’t just invent a new rotary-to-linear converter, so don’t file for your patent.

We can still embellish our design by replacing our 1/4-20 nut with a ball nut. (For more info Google “ball nut”.) We can increase the diameter of the screw so it can transmit higher forces. We can have different screw lengths which will provide for different stroke lengths. We can have different threads per inch which can produce fast or slow motion or fine or coarse motion. For example a two pitch ball screw (two revolutions per inch) will move faster linearly for a fixed motor speed than a five pitch ball screw (five revolutions per inch) spinning at the same speed. However, a five pitch ball screw would have finer positioning capabilities than the two pitch one. Using a stepper motor in the full step mode (200 steps per revolution) with a five pitch screw produces 1,000 steps per inch while the two pitch screw would produce only 400 steps per inch. Thus, there’s a trade-off between speed and positional resolution. In addition to “English” screws we can have metric threads of various turns per centimeter too.

More about ball screws and linear slides next time.

One of the challenges that face motion-control engineers is how to properly select a motor. If the motor is undersized or too small, it will not handle load. If the motor is oversized or too large, the motor will be too expensive in terms of purchase point and operation. Accurately sizing the motor can help prevent either of these scenarios from occurring.


One of the primary challenges in motion control is overcoming inertia. Inertia is an object’s tendency to resist changes in acceleration. Overcoming inertia is a constant struggle for engineers, especially on assembly machines that have short motion cycles but require very high speed. A motor needs to supply sufficient force in a linear system, or torque in a rotational system, to change the acceleration of the load in a controlled manner.

The first step in properly sizing a motor is to calculate the load inertia. For example, a rotating mass’ moment of inertia (J) can be determined by:

This describes the moment of inertia for a point mass (m) at specific distance (r) from the rotation axis. The formula can be used to build up moment of inertia for complex shapes like cylinders, disks, spheres, and blocks. Load inertia is reflected back to the motor shaft from the load and the components in between. The value should include any extra mechanical elements that motor is responsible for moving. This could mean parts such as screws, pulleys, belts, or couplings. These additional moving parts will demand higher performance from the system.

Finding the appropriate analytical expression for complex systems depends on the weight, factoring in acceleration due to gravity, rather than just the mass of the parts. The motor-sizing process requires the total system inertia, including both the load inertia and the motor inertia. The tendency of many engineers is to only calculate the actual load and gearbox into the inertia while leaving out the weight and inertia of the belts, pulleys, and other mechanical components.

This method is why many adapt the 10% oversize approach and bump up their motor requirements to the next major size, or use the same frame size with a large torque value. Consequently, it leads to oversizing the motor.

The ratio of load inertia to motor inertia indicates how effectively the motor can control the load. When the inertia ratio is high, it indicates that the system may have a hard time controlling the load. A low inertia ratio obviously indicates the opposite; the motor will have an easier time and be more effective controlling the load. However, a low inertia ratio could mean the motor is oversized for the system, costing more and being larger than necessary.

Ultimately, the rule of thumb is based on performance. Inertia mismatches in servo and motion control can occur as high as 60:1, but the performance can be excellent depending on how well you can tune the system. Auto-tuning drives can compensate for machine resources and vibration to support accurate performance at even high speeds.

Torsional flex or compliance is an important factor in motor sizing. A motor sized at a 1:1 inertia ratio may have problems controlling a load if the system is too loose and has a high compliance. The system could require tightening or the operation parameters may need to be relaxed.

Gearboxes help manage inertia by reducing it via the square of the gear ratio. Gearboxes also cut motor speed, which can be problematic with stepper motors as they typically only run at several hundred revolutions per minute (RPM). Servo motors benefit from the use of gearboxes because they operate at higher speeds—between 2,000 and 6,000 RPM.

Application Requirements

The trapezoidal motion profile shows the acceleration and deceleration areas along with the constant velocity range. (Courtesy of Motion Control & Motor Association)

After determining the inertia for the system, the next step is to determine the operating parameters and torque required. One should start by defining the motion profile for the load. The most basic form is a trapezoidal shape; a burst of sudden acceleration that’s followed by a period of constant velocity and ending in a deceleration. Acceleration and deceleration can be determined by:

Looking at the simple motion profile above, the acceleration and deceleration mirror each other and can be determined by:

Torque Requirements

Once you have the load inertia, acceleration, and deceleration, you can calculate the amount of torque required to position the load. The total torque (Tt) is the sum of the acceleration torque (Tacc) and the load torque (TL). The load torque is the sum of mechanical losses in the system, which is typically loss due to friction and gravity.

To calculate the friction, one multiplies the friction coefficient for the sliding surface and the normal force to said surface. Friction is a design factor that can be easily forgotten in the motor-sizing process. It is helpful in determining the frictional force for a measuring instrument such as a torque wrench.

The root-mean-square (RMS) torque is the torque required from the motor for the application. It considers not only the amount of torque, but also the duration of the torque. To determine TRMS, the torque acceleration and torque deceleration must be calculated:

The last torque value for the equation is the run torque. This is the torque value that maintained constant velocity throughout the run phase and the small span of idle time at the end of the move.

Once you have the TRMS, you can determine the required speeds by referencing speed-torque curves. Manufacturers provide these curves for their motors that describe their performance during operating speed ranges. These data plots are an essential reference to determine if the motor is appropriate for the application conditions.

The speed-torque curve shows the rated torque point of the motor. The low end of the curve is where the motor is safe to operate, but may be oversized (green area). On the high end of the curve, the motor would be undersized and can possibly fail overtime (orange area). (Courtesy of Motion Control & Motor Association)

Infrastructure and Environment

During the sizing process, external conditions must be accounted for when choosing the right motor. One must consider the voltage and frequency characteristics of the power source. The motor you select may vary depending on the power available.

The operating environment also is crucial. Will the motor be exposed to harsh environments such as low temperatures or increased moisture environments? Will it be placed in a remote location making regular maintenance visits difficult? Is the motor needed for wash-down environments? The answers to these questions will have an impact on what motor you choose.

Physical Motor Aspects

An easy way to control the load is to increase motor power. However, physical limitations may prevent you from bumping up the motor size. Space restrictions restrict the frame size and length of the motor. Some manufacturers increase motor power by stacking magnet laminations while maintaining the same frame size. This makes the motor longer but does not increase width. However, this could suffer from space restriction as well.

When it comes to stepper motors, oversizing is common practice. Stepper motors are designed with high pole counts—on the order of 50 or more. They can be commanded to advance in discrete steps rather than in continuous motion and are able to operate in open-loop, making them effective and inexpensive.

Many will oversize stepper motors to prevent them from stalling when placed in over-torque mode. There’s typically no feedback-monitoring device on the motor shaft, so the motor stalls without being noticed. By properly sizing the motor, one can run stepper motors in closed-loop allowance. This helps saves money, which can be invested into purchasing an encoder.

Beyond the Basics

Engineers can apply several techniques to shrink the size and cost of a motor. The first is to add a gear reducer. The gearbox may reduce motor requirements. A worm gearbox is about 30% efficient, while a planetary gearbox is 85% efficient. Using a high-efficiency gearbox can reduce the load on the motor.

One can also consider replacing servo motors with stepper motors. After calculating the required torque, a stepper motor can be used in low-speed applications. With the right frame size, a stepper motor can work on its own without the need for a gearbox, providing significant cost savings.

When choosing a motor, a common mistake is to opt for a motor with continuous duty torque, which is equal to the max torque requirement. Motors have two modes: continuous duty mode/peak or overload mode. The motor can operate at peak torque/current for brief intervals without harming the motor windings. Many applications consist of brief and rapid moves.

Choosing a motor at max torque requirement means overpaying for the motor. When properly sized, one can select a motor that’s able to operate in the overload torque range for brief periods of time. This allows users to select a small motor and save money.

When selecting a motor using this method, pay close attention to duty cycle. The intervals of peak current should be short so as stay under specifications, and occur at a low-enough frequency to enable the motor windings and electronics to cool off. Duty cycle from the standpoint of the gearing is important to prevent premature wear and failure.

View the original article and related content on Machine Design

Recently I came across a simple infographic from Power Jack Motion that did a nice job of visualizing the difference between alternating current (ac) and direct current (dc) motors. This article will follow and present the information contained therein.

What is Motion Control, Anyway?

Motion Control is a sub-field of automation encompassing the systems or sub-systems involved in the moving parts of machines in a controlled manner. Using a computer to control an actuator offers benefits, but today getting data and feedback is increasingly important.

Some companies are already seeing the advantage of adding sensors and feedback to a production line. In addition, this extra information can let machines communicate to each and connect to the internet to take advantage of the Industrial Internet of Things (IIoT).

Today motion control often includes a blend of electronics and mechanical components divided into three parts, which have been seen as analagous to how the human body operates.

Motion Control Humans
Processor or Motion Controller Brain
Actuator Muscles and joints
Sensors Senses

To elaborate on this further, the infographic added more detail.

Processor or Motion Controller

There are three types of motion controllers: standalone, PC-based, and individual microcontrollers. This is becoming more and more of an important consideration when setting up production lines. To prevent incumbent inertia and future-proof a production line, the manufacturing brain is evolving.

Previously programmable logic controls (PLCs) were the go-to controller—a robust, centralized control that could be put into a protective enclosure and have everything connect to it. However, as production line advance, the amount of cables, complexity, and cost has increased. Today a PLC that focuses more on software might be better to handle the automation advancements. The programmable automation controller (PAC) does just this. While it expands the capabilities of a PLC, the difference in terms is debatable.

There are individual microcontrollers that allow a production line to add features while not affecting legacy equipment, or letting companies start taking advantage of the IIoT with less commitment. To combat legacy equipment, some older PLCs may continue to run while a new automation process is added.

For example, a new inspection system could accept or reject parts on a line. The legacy equipment might not see this. The legacy equipment only sees a part going in one side and out the other. It can be blind to the new equipment—so long as it knows what to do with the output from the new equipment, it can continue to run as it previously has.

Amplifiers and Drives

There are many types of amplifiers and drives, and there is a lot to dissect here. But when looking at advanced automation, servos look like they are able to add more value and feature to a line when compared to stepper motors. However, this can add more cost and complexity. Servo drives and servo amplifiers transform a low power current/voltage applied to the servo motor windings to produce torque. Different amplifiers include analog servo drives, sinusoidal, trapezoidal, digital servo drives, single phase (brushed) servo drives, three phase (brushless) servo drives, and electric vehicle motor control.

Key features of ac and dc motors
Low power demand on start Easy installation
Controlled acceleration Speed control over a wide range
Adjustable operational speed Quick starting, stopping, reversing, and acceleration
Controlled starting current High starting torque
Adjustable torque limit Linear speed-torque curve

AC synchronous motors

Feedback Sensors and Mechanical Components

Feedback sensors provide motor location. There are several types, including:

  • Quadrature encoder: Gives position relative to the starting point
  • Potentiometers: Gives analog position feedback
  • Tachometers: provides velocity feedback
  • Absolute encoders: for absolute position measurements
  • Resolvers: convert mechanical motion into an electric analog signal to find an absolute position

Overall, motion control is complex, and gaining an understanding of everything is difficult. Simple inforgraphics such as the one described in this article help to simplify things, making it easier to determine which expert or company you should be talking to.

View the original article and related content on Machine Design

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.