Stepper motor gear ratio

I am currently looking at the motor above for a project where I am reeling in a 20lbs on a pulley, I was told that it could work at a certain gear ratio. The radius of the pulley will be 2 inches/5.08 cm. Is it possible to reel in this weight? If so, what are the critical parts and things I should know in order to be able to reel this weight as effectively as possible?

I have forwarded your request to the Product Specialist and will post the reply when I receive it.

Hello @zenkebede,

Welcome to the DigiKey community.

Our first consideration is torque. Given your design, the motor must provide:

torque = 20 * (2/12) = 3.3 lbf = 630 oz-in

DigiKey does sell stepper motor with this capability, but they are massive devices. Most require a matching motor drive which further adds to the cost.

You may be money ahead if you reduce the length of the torque arm. Instead of using a 2-inch pully, a smaller motor with a 1/4-inch shaft could perform the same task with less cost.

Please tell us more about your project so that we can assist in narrowing down the possibilities.

Best wishes,


P.S. Some motors are equipped with an optional electric brake. This would hold the shaft in a fixed position upon loss of power. An alternative is to look for a motor with a worm gear head. Recall that a worm gear acts as a one-way gear that moves only when driven by the motor.

Hi, the torque you listed is 630 oz-in. On the specifications for the motor listed the holding torque is rated 679.82 oz-in. Is there a certain value of holding torque near the maximum holding torque where the motor would be ineffective?

My apologizes, the project is about reeling a point mass segmentally while it is in oscillation as a pendulum. I would like to reel in the cable at a maximum instantaneous speed of 0.5 m/s. Just to confirm would I be able to reel in the cable with a tangential speed of 0.5 m/s given a shaft of 1/4 inch?

Hello @zenkebede,

A few thoughts:

  1. What is the nature of the windlass (winding drum) upon which you are gathering the excess cable. I assume the device will not oscillate directly from the windlass as the oscillation would be disturbed by the small changes in the fulcrum point - each oscillation would partially wrap and then unwrap.

Perhaps it is better to provide a bushing through which the cable passes. The resulting fulcrum would be stationary. The motor could then be moved to any convenient location.

  1. How will you measure the length or changing length of the cable? If it is measured by the steps of the stepper motor, then how will you account for variations in the windlass such as when a cable wraps on top of another cable? Stated another way, the small shaft diameter lowers torque but greatly complicates the calculation especially as cable accumulates on the windlass.

Perhaps it would be better to use an independent sensor such as a quadrature encoder. This separation of motor from measurement would greatly simplify the design allowing you to use something like an electric boat lift.

  1. As for holding torque of a stepper motor, visualize the stepper motor as a very slow machine. It will advance and then hold each step for a given holding torque as long as the winding is powered. As the stepper speed increases the available torque is generally reduced. The actual torque and available power are a function of both motor and drive.

Please forgive me if I have misunderstood your project and level of expertise. these things are hard to know beforehand.

I’m interested to know more about your project.