In our last posting we found that a gearbox reduces the load inertia that is attached to the gearbox’s output shaft by the inverse square of the gear ratio (gr) or:
J reflected = J load/gr2
The inertia reduction can come in very handy when trying to size a large inertial load so that it works well with a smaller motor. In addition to reducing the load inertia by the inverse square of the gr, the motor’s torque is increased by the gr minus the efficiency of the gearbox. This all sounds like magic. We can get good load-to-rotor inertial matches and increased torque all at the same time. The down side is that we need to purchase another component, but that may have added benefits related to the overall combined size of the motor and gearbox when compared to a larger stepper that can handle the load directly.
One of the things to remember is that the motor is going to have to spin faster in order to have the output shaft of the gearbox spin at the desired speed. Why is this important? Do you recall what the speed-torque curve of a stepper looks like? Of course you do. A stepper has lots of torque at low speeds and the torque falls off at higher speeds. I’ve chased my tail a few times by incrementally increasing the gr only to find out that the motor’s torque has fallen off at the higher speed and that we didn’t gain anything. Also remember that we’re losing a little bit of the motor’s output torque because of the gearbox’s efficiency.
A less expensive gear reduction system is pulley system. Picture a pulley attached to the stepper motor’s shaft that’s 1.0” in diameter connected via a belt to a second pulley that is 3” in diameter. We have a gr of 3:1. Typically that’s as high a ratio that I would go with pulleys.
The larger pulley gets larger really quick with higher ratios. For instance, a 5:1 ratio would have the larger pulley grow to 5”. Why is this an issue? The inertia of a disk is proportional to the square of radius. We just increased the radius from 1.5” (for the 3” pulley) to 2.5” (for the 5” pulley)
1.52 = 2.25 versus 2.52 =6.25 that’s almost three times more inertia.
Let’s play with some numbers.
We’ll assume that we have a 1” diameter pulley attached to the motor shaft that is 0.375” thick or length. Calculating its weight:
W pulley 1 = π*r2*length*density = weight in pounds.
3.14159*0.52*0.375*0.098 = 0.029 pounds
J pulley 1 = Wr2/2 = 0.029*0.25/2 = 0.00363 lb-in2
Repeat the calculation for the larger pulley:
3.14159*1.52*0.375*0.098 = 0.26 pounds
J pulley 2 = Wr2/2 = 0.26*2.25/2 = 0.293 lb-in2
And you remember that a belt is like a thin ring and its inertia is:
J belt = Wr2 where r is the radius of the smaller pulley.
Assuming the belt weighs 0.25 pounds we get:
J belt = 0.25*0.52 = 0.0625 lb-in2
J reflected total = J Pulley 1 + J belt + J pulley 2/gr2 + J conveyor belt/gr2 (used in our last posting)
Plug in our values:
J reflected total = 0.00363+ 0.0625 + 0.293/9 + 0.529/9
J reflected total = 0.1575 lb-in2
Where J reflected total is the inertia reflected to the motor’s shaft.
An interesting item to note is that the pulley system’s J reflected total is 0.1575 lb-in2 while the gearbox with a 5:1 reduction from our last posting produced a J reflected total of 0.0362 lb-in2
Both the pulleys and the gearbox did their job reducing the load inertia (0.529 lb-in2) that is seen by the motor.
Read the next blog post: Designing with Motion Control in Mind
1 thought on “Reducing Load Inertia”
The inertia reduction can come in very handy when trying to size a large inertial load so that it works well with a smaller motor. In addition to reducing the load inertia by the inverse square of the gr, the motor’s torque is increased by the gr minus the efficiency of the gearbox. This all sounds like magic. We can get good load-to-rotor inertial matches and increased torque all at the same time.