Now that I have the CIM motor data, I was wondering if somebody could help me interpret it. For example let’s say that I want to see how much RPMs it will have if X torque is applied. So I look at the torque first to see where x is found, then go up to find RPMs (since RPM is dependant upon torque). Ok that seems like it is right.
But now lets say I wanna find what the max torque of a gearbox will be. For simplicity lets say the ration is 10:1. So that means the torque is multiplied by 10 on the output. So if I have a (once again for simplicity) a 1000 ounce load, I would work backwards and say that 1000 / 10 = 100. This means I look to see where 100 oz/in lays on the x axis then go up to find the RPMs at that load(before the gearbox)? then I divide RPMs by 10 to find the RPMs coming out of the gearbox?
I have never really been formally taught this skill, so any help is appreciated.
The RPM the motor will turn depends on how much voltage you supply to it. However, it also depends on how much load is applied to the motor…and the current the motor draws (amps) is also related to how much load is applied.
Also, the motor will work best if you keep the load to about 20% of the maximum (stall) torque, and the voltage to about 80% of the maximum voltage. So if the motor can produce 350 oz-in torque at stall, you would try to limit the load to 70 oz-in or less. If the RPM at maximum voltage is 5300, you would want to run the motor at about 4300 rpm (or less).
To calculate the maximum torque you can get from the motor at it’s design conditions, multiply the torque times the transmission ratio. So if you have a 10:1 transmission, you would get 70 x 10 = 700 oz-in torque. Also if you are running the motor at 4300 rpm, the transmission would reduce that by a factor of 10, so the output shaft would turn at 430 rpm.
If you would give us some hints about what you are trying to do with the motor, we might be able to provide some more useful information.
Thanks for that Ken, I looked it over and did learn some stuff. But Let me aska more specific question about all of this now based on what I am trying to do.
Right now in theory what my robot will have is 6 inch wheels and will be driven by 2 CIM motors running through banebots 12:1 transmissions (I know I have seen all the horror stories about them). So I figured I would start out by calculating the max weight the robot would be able to move. So from looking at the CIM motor specs it looks like stall torque is at around 340 oz-in.
I started out by figuring out the toque after the gear ratio, which is 12:1 so 340 * 12 = 4080 oz-in. then I wanted to convert the units to what is more useful to me. 4080 oz-in * (1 lb / 16 oz) * (1 wheel / 3 in radius) = 4080/48 = 85lbs per motor.
This would turn out to be 170 pounds with a 1:1 outside of the gearbox.
As for sprockets I will porbally be using a 28:12 ration (unreduced representation, 12 tooth drive, 28 tooth wheel) which will result in 170 * (28/12) = 397 pounds
Does this all sound right and make sense how I did it? (in a perfect system)
Also instead of going through and figuring out efficiency stage by stage is there a number people like to use for a overall drive system efficency?
That sounds about right. For the planetary transmissions, figure about 90% or so effieciency per planetary reduction stage…so you’d have somewhere around 80% total efficiency with the transmissions and gear reduction.
Also keep in mind that you won’t get stall torque out of the motor for very long! it will soon pop the circuit breaker/fuse if you have the electrical system properly built.
And “the max weight the robot would be able to move” depends on how that weight is supported…if it is on wheels, then it takes a lot less force than you have available to move it, but if you are trying to slide it across a very tractive surface, it will take even more force.