Acme leadscrew vs timing belt getting really high numbers

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The discussion revolves around comparing the axial force produced by a leadscrew versus a timing belt system in a CNC project. The leadscrew calculation yields an unexpectedly high axial force of 967.7 lbs, while the timing belt system produces a more reasonable 62.67 lbs. The discrepancy raises questions about the mechanical advantage of the leadscrew, particularly in relation to its efficiency and the number of rotations required for linear movement. A key insight is that while the leadscrew offers significant mechanical advantage, it also requires many rotations to achieve linear displacement, leading to energy losses. Ultimately, the conversation highlights the importance of considering energy output and efficiency when evaluating different linear positioning methods.
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Hey all,

I'm on summer break so my engineering brain is almost completely turned off. So I would not be surprised if I am making some kind of silly error. I'm trying to compare different methods of linear positioning for a CNC project.

------------------------- Part 1

OK, so here's the scenario.

We have a stepper motor that is able to produce 30 in-lbs of torque.
We have a leadscrew whos manufacturers states that 0.031 in-lbs of torque will "lift 1 lb"

(Not used in calculations, but it is a 3/8-12 acme two start with 43% efficiency)

If I do:

30 in-lbs * ( 1 lb / 0.031 in-lb) = 967.7 lbs axial force

This seems astronomically high.------------------------ Part 2

We have a timing belt arrangement producing linear motion by being constrained (from translation) by a drive pulley on the same stepper motor, a free spinning pulley bringing the belt into proper tension, and a cart attached to one "side" of the belt.

The motor produces 30 in-lbs of torque.
The drive pulley has a diameter of 0.900".
The system is 94% efficient.

30 in-lbs * (1/0.450 inch) * 94% = 62.67 lbs axial force

This seems reasonable to me. The answer from Part 1 absolutely does not.------------------------ Part 3

Leadscrew has 0.166 inches of linear movement per rotation
Timing belt assembly has 2.827 inches of linear movement per rotation

The ratio between the two (TB:LS) is 16.87:1 (ignoring efficiency so far)

The ratio between Part 1 and Part 2 (TB:LS) is 15.44:1 (not ignoring efficiency)How is it possible that the linear displacement to angular displacement ratio is vastly different than the axial force to torque ratio? The leadscrew is magically exceeding it's mechanical advantage?
 
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I was ignoring that frictional forces at the screw will increase as the axial load increases.

Anyone know how I can take this into consideration when I go to calculate the axial force produced at the nut with 30 in-lbs of torque at the screw?

Here's the data sheet: nookindustries dot com/pdf/NookInchAcmeScrew.pdf

Page 4 is what you want. On my pdf viewer it's listed as page 21, not sure why.

3/8-12 two start rod with a plastic nut is what we're going for

Thanks all
Luke

PS: sorry for the 10 post link workaround but I figure that's for spam and I need to link to the datasheet for this question
 
Taiden said:
How is it possible that the linear displacement to angular displacement ratio is vastly different than the axial force to torque ratio? The leadscrew is magically exceeding it's mechanical advantage?
Putting torque on a threaded screw provides a huge mechanical advantage. Needless to say, there's a conservation of energy, so although there is a very large mechanical advantage for the lead screw, you have to turn the screw a very large number of times. The screw has to be rotated 12 times to move it just 1 inch.

Take your torque of 30 inch pounds and determine the amount of energy it puts out after rotating 12 times. In other words, your motor is producing 30 inch pounds of torque which is equivalent to a force of 30 pounds, 1 inch from the center line of the shaft. So the total distance the 30 pound force rotates through in one revolution is 30 pounds times the circumference (2" * pi). That gives you 188.5 inch pounds of energy. Multiply by 12 and you have 2262 inch pounds of energy produced. If we use that energy to rotate the screw 12 times, we will have produced some force over a distance of 1 inch. Per your calculation, the screw has produced 967.7 inch pounds of energy, the rest we can assume has been given up as heat due to inefficiency. Now if we divide 967.7 inch pounds by 2262 inch pounds, we get an efficiency of only 43% which is exactly the number given by the power screw manufacturer. Seems to work out! :smile:
 
THANK you for that post. Our engineering professors don't teach us how to solve problems, they teach us how to follow book steps. I never once considered using energy to find the solution but it's obvious now how easy and effective it is. I'm going to look more into this tomorrow afternoon when my brain is awake so I can do it out and follow it 100%

Thank you again
Luke
 

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