Being a non-engineering type, I'm trying to calculate what power and torque I would need from a DC stepper motor to execute a particular task. I'm certain that my question will open me up to potential observations regarding my lack of knowledge in this field. That said, if your critical observations use humor to make the point, I will appreciate them for that. However, I do expect to gain some good knowledge from this group. Here goes nothing.

I am building a device that will be used to lift, and hold 200 pounds approximately 2 feet in elevation. The mechanical elements of this device are 100% the same as a standard worm-shaft driven, scissor jack for a motor vehicle.

Envision a letter "A" as the two arms on the top of the scissor, and the left arm is anchored to the substrate, and the right arm is attached to the threaded block that travels along the worm-shaft. As the worm-shaft is rotated, the threaded block will travel along its axis which will in-turn cause the "A" to get taller and skinnier. This achieves the lift needed for the device.

Question 1:
In order to achieve the required power and torque to lift and hold 200 lbs without having to manually turn the worm-shaft (as one would when jacking up a car), should I couple the motor to the worm shaft concentrically, or use one gear on the motor shaft, and one on the worm-shaft?

Question 2:
Depending on the direct-couple method, or the gear ratio method, how do I determine the power and torque of the motor?

I really appreciate any help I can get with this challenge.

Scissor Jack

  • $\begingroup$ Any particular you chose a Stepper Motor. Did you consider a DC motor? I am assuming you will use wall power. $\endgroup$ – user8055 Dec 23 '19 at 18:48
  • $\begingroup$ I figured stepper motor because if I recall, they generate more torque. But frankly, the application calls for the most economical combination of size, weight and technical specifications (torque and power), so if DC is better, then that'll be how I'd go. $\endgroup$ – Michael Chippendale Dec 23 '19 at 21:54

The length of each arm of your jack should be comfortably bigger $12"(inch)> 2ft/2$ because if they are close to 12" they lock at horizontal position.

We chose 18" and set the pegs at the lowest level so the arm is 20 degrees when not deployed meaning the jack can expand from 6" to 36" with a travel course of 30" which is more than you need but is more practical.

The horizontal force at the worm-screw is $$F=200/2*cot(20)= 100lbs*2.74= 274.7 lbs$$

From here on F becomes smaller if we lift the load.

Let's say the diameter of your worm-screw is 1/2" and its pitch is 1/32", meaning the slope of the thread is approximately $$ 1/2*\pi*32= 1/50$$ and the force on it is $$f_{screw}=274.7/50= 5.49 lbs \\ \text{and the torque needed is}\\ T=5.49*1/4"= 1.373 lbs.ft $$

This is the case with no friction and 100% efficiency. I would use a load factor of 1.5 and check the manufacturer's recommendation.


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