Pulley Speed Formula: Pitch Diameters, RPM and Belt Speed

Define the driver and driven pulley

For an ideal belt drive without slip, the belt travels at the same linear speed around both pulleys. Let D1 be the driver's pitch diameter and n1 its speed in RPM; let D2 and n2 describe the driven pulley. Equal belt speed gives D1n1=D2n2, so n2=n1D1/D2. Both diameters must use the same units. Identify which pulley receives power from the motor before applying the ratio: reversing driver and driven labels reverses the calculated speed change. The equation describes an ideal kinematic relationship, not the full mechanical design of a belt drive.

Calculate a simple speed reduction

Suppose the driver turns at 1,800 RPM and has a 100 mm pitch diameter, while the driven pulley has a 300 mm pitch diameter. The driven speed is 1,800×100/300=600 RPM. The larger driven pulley turns more slowly because it requires more belt travel per revolution. The reduction ratio n1/n2 is 3, matching D2/D1. As a reverse check, 600×300/100 returns the original 1,800 RPM. A result of 5,400 RPM for this arrangement would indicate that the diameter ratio had been inverted.

Find a required driven pitch diameter

Rearrange the same equation as D2=n1D1/n2. If a 1,500 RPM driver with an 80 mm pitch diameter should ideally produce 500 RPM at the driven shaft, the required driven pitch diameter is 1,500×80/500=240 mm. This is a target from the speed ratio, not an instruction to select a component by its outside diameter. Actual belt systems distinguish pitch, datum and outside dimensions. Consult the manufacturer's terminology and available sizes, then recalculate the speed using the actual chosen effective dimensions rather than retaining an unobtainable ideal diameter.

Calculate the belt's linear speed

With pitch diameter D in metres and shaft speed n in RPM, ideal belt speed is v=πDn/60 metres per second. For the 100 mm driver at 1,800 RPM, first convert 100 mm to 0.1 m. Then v=π×0.1×1,800/60=3π, approximately 9.4248 m/s. The 300 mm pulley at 600 RPM gives the same belt speed. The RPM values differ while the linear belt speed agrees. A speed-unit converter can change 9.4248 m/s into another linear speed unit, but it cannot convert RPM to m/s without a relevant radius or diameter.

Combine stages using intermediate shaft speeds

Consider two reductions. The first stage uses a 100 mm driver and 200 mm driven pulley, reducing 1,800 RPM to 900 RPM. A second driver on that intermediate shaft has 80 mm pitch diameter and drives a 240 mm pulley, reducing 900 RPM to 300 RPM. The total reduction is 2×3=6, and 1,800/6 confirms 300 RPM. Track which two pulleys share a shaft and therefore share its rotational speed. Multiplying every diameter in a drawing without identifying the paired stages and shaft connections can produce a ratio that does not correspond to the actual transmission.

Keep ideal speed separate from capacity

Slip, belt condition and the effective operating geometry can make a measured speed differ from the ideal result. A toothed synchronous drive uses its compatible pulley tooth counts to establish the nominal ratio; a friction drive requires attention to the relevant effective diameters. Even a correct nominal speed ratio does not establish allowable power, belt tension, minimum pulley size, shaft loading or guarding. Manufacturer design guidance addresses these additional choices. Use the simple equation to check the kinematics, then use the applicable component data and design procedure for the actual installation.

Use the local calculator for arithmetic

The linked Basic Calculator can evaluate the multiplication and division in the worked examples. It does not model slip, select a belt or recognize a drive layout. Write the units and the driver/driven definitions beside the expression before entering it. For the first example, 1,800×100/300 gives 600, which you then label RPM. Independently compare the linear belt speeds at both pulleys using πDn/60. Matching belt speeds and a consistent reverse ratio provide useful checks that the numbers were assigned to the correct pulleys.

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