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Torque and Linear Motion Formula

Periodically, I get questions from users of the site regarding torque and how much is required to

move certain masses and a recent user emailed me specifically and it gave me the happy

motivation to add this page. :) These masses are things like the gantry, or the z-axis assembly

with the router mounted moving on the y-axis, or the just the router mount and router within

the z-axis assembly. The question is usually in the form of, can my ??? oz-in motors be used on

your CNC machine.

Before the formula can be used to determine torque, there will be certain aspects of the machine

that needs to be considered. FIrst, are the lead screw alignments pretty good? Is there anti

backlash mechanisms on the machine which will add friction to the linear motion? The next

thing to consider is the properties of the machine, because you will need this information in the

formula. Ok, so let's give it a try:

To determine torque, lets gather some information. By the way, we are talking about torque

during a continual turning motion, not at a holding position. Let's say we are trying to figure out

the necessary torque to move the gantry. This motion is on a horizontal plan, so gravity is less

important than with a vertical situation, like the z-axis lead screw. We will need to know the

number of inches pre revolution. This is pretty easy since it's the screw # of threads per inch.

You will also need to know the pi. Pi is half of a circle in radians, or 3.1415927 rounded to the

ten millionth. The formula is as follows: Torque = ((weight)(inches/revolution))/2(pi)(efficiency).

You are wondering where do I add in things like friction. There is also a formula to determine

the weight that considers friction. Should I expose this formula?!? Hmmm... Well, it doesn't hurt...

much. Weight = (load being slid)(Coefficient of sliding friction)+force against sliding

mechanism. You can use this formula to your hearts content, but I will not use it here for

simplicity.

Ok, so, let's say we have a gantry that weighs 50 lbs and the number of inches per revolution is

one tenth (1/10 or .1). This is a standard thread pitch for 1/2 ACME screws (sometimes 1/8,

depending on your preference). Therefore, the torque (lbs-in or pounds at one inch)

requirements will be: torque = (50 (1/10))/2(pi) or 5 / 6.2831854. We will leave out the

efficiency of the screw/nut combination for simplicity. We get a value of about .7958 lbs-in. Now

your thinking, that number seems WAY low. That's because it's in pound inches and not ounce

inches, which is the typical unit in the US for these ratings. There are conversion charts across

the internet to convert this number to oz-in, or Nm (Newton Meter). This conversion happens to

be 12.7328 oz-in or torque necessary to move the gantry in constant motion. Now remember,

this may seem like a low torque requirement, and there is no factor of safety built into this

number. There are those characteristics like friction, misalignment, and torque reduction at high

speeds, so work in a bit of wiggle room here. The number also shows how powerful a screw

mechanism is, and how easy it is to move heavy loads with this mechanism. When looking

around for motors, look at their datasheets which should contain a torque curve related to RPM

(revolutions per minute). This is where this formula and applying it to the properties of the CNC

machine being considered should be used. I would love to receive some feedback on this topic

especially if I need correction by the math engineers out there, or if more clarification and

simplicity is desired.

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