- #1
Taiden
- 9
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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?
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?