There are two main questions that we can answer with respect to motor torque and the mechanical advantage of lead screws, 1) What torque motor do you need to lift a particular weight, or 2) What maximum weight will my motor torque be able to lift.
This formula uses Newtons (N) as it's final unit. Use this with the included radius (R) to determine the torque. Newtons can easily be converted to lbs or ounces using online conversions.
Effort = Sf + (Load/(2 x pi x (R/p) x Se))
where:
p = pitch of the screw
Se = screw efficiency = Standard lead screw will be between 20% (.2) and 40% (.4)
Sf = static force. This is the force that is needed to start the movement. The number may be eliminated, but it is good to use a number in the 5 N to 20 N range.
Load = the expected load that the effort will need to carry (i.e., the router and the included axis assembly that the motor will need to lift)
R = radius of the lead screw
This formula is based on the "law of the machine"
The final effort amount with its unit of newtons and R will be the torque. For example, if the effort comes to 100 N (newtons) and the R is .5 inches, then you can assume that the effort is 50 N-in since it would take twice the effort to turn form the one inch mark from the center of the shaft.
Example:
Load = 90 N (20.2 lbs)
R = 1 inch since that is the length from the center of the shaft that the motor is rated
p = 1 inch / 13 = .08 inches
Effort = 5 N + (90 N / (2 x 3.14 x (1 / .08) x .2))
Effort = 5 N + (90 N / (6.28 x 12.5 x .2))
Effort = 5 N + (90 N / (15.7))
Effort = 5 N + (5.73 N)
Effort = 10.7 N = 2.4 lbs = 38.4 oz-in
I am putting the oz-in on the end because the formula considers the distance from the center of the shaft to be one inch.
Therefore, a 425 oz-in motor would be able to lift a 20.2 lb Router with its accompanying assembly. If the assembly and router is heavier, plug in the numbers and determine the effort required.
With a bit of algebra, the formula can be rewritten to find the load:
Load = (Effort - Sf) x (2 x pi x (R/p) x Se)
Another formula that does not consider friction at all:
Effort = (Load x p) / (2 x pi x R)
Lets see if we get similar results:
Effort = (20 lb x .08 inches) / (2 x 3.14 x 1)
Effort = 1.6 / 6.28 = .255 lbs = 4.08 oz-in
The results from both formulas appear to be very small because a 13 TPI screw will have enormous mechanical advantage.
It is evident that the first formula that does consider friction that we are loosely estimating is far more conservative than the second formula. Either way, even the most conservative formula shows that the 425 oz-in motor will handle very large weights. If you are using a lead screw with only two turns per inch, .5 inch pitch, you can determine the requirements with the first formula.
Example for a 10 TPI 5 start (2 turns per inch) lead screw:
Load = 90 N (20.2 lbs)
R = 1 inch since that is the length from the center of the shaft that the motor is rated
p = 1 inch / 2 = .5 inches
Effort = 5 N + (90 N / (2 x 3.14 x (1 / .5) x .2))
Effort = 5 N + (90 N / (6.28 x 2 x .2))
Effort = 5 N + (90 N / (2.512))
Effort = 5 N + (35.83 N)
Effort = 40.828 N = 9.18 lbs = 146.88 oz-in
Customer Response:
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I need the calculation to determine the stepper motor torque to find the load that it can withstand in horizontal position using a lead screw at 1/2" diameter with 13 TPI.