## Physics and Mechanics: How to Account for Belt and Pulley Inertia in System Design

Newton’s first law of motion states that if an object is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by an unbalanced force. In the engineering and manufacturing industries, we are intimately familiar with this concept.

In order for a motor to be able to adequately accelerate or decelerate a load, it has to first overcome the load’s inertia. Inertia, the tendency to do nothing or to remain unchanged, is the resistance to a change in motion we experience in mechanics. It’s what we’re always working against to power vehicles and create new things.

### Inertia in Belt-Driven Linear Motion Systems:

For engineers and manufacturers who deal with belt-driven linear motion systems, this lesson translates to a simple fact: motors we use must be able to overcome both the inertia of the applied load *and* the inertias of the belt, pulleys, and motor coupling.

### Estimating Inertia: A Step by Step How to From the Experts at Illinois Pulley & Gear

It’s possible to estimate the amount of each component’s respective inertia with accuracy sufficient for most purposes. This can be accomplished through the use of a standard inertia equation. First, you’ll need to know that the amount of inertia associated with an object depends on the axis around which it rotates.

Because the applied load and belt rotate around the driven pulley’s axis together, we can begin our estimation by grouping the applied load and the belt together. Gather data you’ll need from manufacturer information: belt manufacturers will generally provide weight information in mass units. The mass of the belt can be calculated by taking the mass per unit length by the belt’s total length. Model the applied load and the belt as a point mass, and calculate their inertia using the following:

### J_{L} = mr^{2}

- Let “J
_{L}” represent the inertia of belt and applied load (kgm^{2}) - Let “m” represent the mass of belt and applied load (kg)
- Let “r” represent the radius of driven pulley (m)

Next, treat the pulleys and coupling as solid cylinders rotating about their own axes. Calculate their inertia using the following solid cylinder inertia equation:

### J_{p} = ½mr^{2}

- Let “J
_{p}” represent the inertia of the solid cylinder, or the pulley and coupling (kgm^{2}) - Let “m” represent the mass of cylinder (kg)
- Let “r” represent the radius of cylinder (m)

Of course, in many driven actuators, pulleys will have different masses, and in turn, different inertias. This can occur if one pulley is driven while the other pulley is idler. This solid cylinder inertia equation will be sufficiently accurate for most uses.

However, if you need the highest accuracy for pulley and coupling inertia measurements, calculate with the consideration that these components have a center bore. This equation is great for higher accuracy needs:

### J_{ph} – ½ m(r^{2}_{0} +r^{2}_{i})

- Let “J
_{ph}” represent the inertia of the hollow cylinder, or the pulleys and coupling (kgm^{2}) - Let “m” represent the mass of cylinder (kg)
- Let “r
_{o}” represent the outer radius (m) - Let “r
_{i}” represent the inner radius (m)

Often, belt driven systems will use a gearbox. Gearboxes are widely loved for their abilities to increase torque and reduce speed. Interestingly, they also reduce the inertia of the load directed to the machine’s motor. When this happens, make sure to divide the total inertia of the entire moved mass by the square of the gear reduction. (The total mass includes the applied load, belt, pulleys, and coupling.) Then, add the inertia of the gearbox back in.

This equation produces the total inertia directed back into the machine’s motor. The total can be used for motor selection and sizing choices. Need an equation? Mathematically, it looks like this:

### J_{L} + J_{p1} + J_{p2} + J_{c}

### J_{total} = _______________ + J_{g}

### i^{2}

- Let “J
_{total}” represent the total inertia directed to the motor (kgm^{2}) - Let “J
_{L}” represent the inertia of the belt and applied load (kgm^{2}) - Let “J
_{p1}” represent the inertia of the first pulley (kgm^{2}) - Let “J
_{p2}” represent the inertia of the second pulley (kgm^{2}) - Let “J
_{c}” represent the inertia of the coupling (kgm^{2}) - Let “i” represent the gear reduction
- Let “J
_{g}” represent the inertia of the gearbox (kgm^{2})

## Still Need Some Expert Help with Your Belt Driven Linear Motion System?

As you have already learned, determining which gears and pulleys are the right ones to use is critical. Failure to use the correct sized pulleys, for example, can cause you to drive your pumps incorrectly. Finding that you could use some help from a specialist? Contact the experts at Illinois Puelly & Gear.

Illinois Pulley & Gear (IPG) manufactures a large variety of high-quality timing pulley stock and timing belt pulleys. Made on-demand, IPG is able to produce virtually an unlimited variety of products.

All IPG pulley stock is precision machined from bar stock at their Schaumburg facility. They do not use extrusions or imported pulley stock. Pick your material, tooth profile, and any number of teeth, and IPG will produce it.

## Illinois Pulley & Gear: Timing Belt Pulley Manufacturers

At Illinois Pulley & Gear, we are passionate about producing high-quality products that are built to last. Every product is USA-made to order. You will always receive the highest-quality product, made according to standard or custom specifications.

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Our clients find that Illinois Pulley & Gear is the option that makes the most sense economically, without compromising whatsoever on effectiveness, function, or efficiency. We are client-oriented and ready to listen.

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