Hello,
I would like urgent help regarding the procedure to calculate/determine the torque of a CIM Gearmotor (specifications mentioned below) for the Gear ratios of 27:1, 36:1, 48:1, and 64:1.
Link for the CIM Gearmotor Specsheet: http://content.vexrobotics.com/docs/217-2000-CIM-motor-specs.pdf
Regards,
Lunabot
It is difficult to answer your question outright, without more information.
I can try to describe some of the principles involved…
The motor will exert as much torque as it can to overcome whatever load you put on it, until that load exceeds its stall torque. The more load you put on it the slower it goes. Once you hit the stall torque… it stalls, and stops moving.
The stall torque of the gearbox is equal to the motor stall torque multiplied by the gear ratio.
The motor motor load, the higher the motor’s current draw.
Typically in FRC… max loading for a CIM motor is calculated based on an approximately 40-amp load, since we run these motors through 40 amp circuit breakers (which can actually run at 45 or 50 amps for limited periods of time).
Torque is proportional to current draw. At stall, the motor draws the stall current. The motor draws 40 amps when it has a torque applied on it equivalent to (Stall Torque) x (40 amps / Stall Current).
Does this make sense? If you provide more details on your application I’m sure someone can help you more.
-John
Thank you very much for your quick reply John. Main application is building a robot for the NASA Lunabotics Competition, as a junior team member I was given the task to determine the maximum torque we can get out of that gearmotor based on the Gear Ratios and cost. Your explanation is very clear and helpful, I really appreciate it! 
Can please explain how increasing the torque would be beneficial for a robot, in terms of movement. Thanks again!
Short Answer: More torque = faster acceleration = less motor load = less current draw = robot draws less current for less time = less tripped breakers = less battery drain.
Longer answer…
When the motor starts moving, it instantaneously jumps to its stall torque. This torque then accelerates whatever mass the motor is spinning, the motor speeds up, and the torque decreases to “zero.”
If the motor isn’t connected anything, the motor only needs to accelerate its internal rotating mass, so it jumps to free-speed very quickly.
If the motor is accelerating a robot, things get more complicated…
Think of your robot as a 1-wheel’d system for a minute. All the weight is evenly balanced over 1 wheel.
At the moment the motor starts spinning, the stall torque of the motor is applied to the wheel (multiplied by the gear ratio) as a torque on the wheel. The wheel applies this torque on the ground as a linear force at the distance from the center of rotation equivalent to the radius of the wheel (T = F x D, F = T / D).
Now… if this force is greater than the force of friction on the ground, the wheel breaks traction and accelerates the rotational mass of the wheel until it spins. How fast does it spin? Well… good question. In this case the load on the motor is equal to the force due to friction of the wheel on the ground (acting as a brake on the wheel) multiplied by the radius of the wheel, divided by the gear ratio.
The robot is accelerated forwards by this frictional force, based on F=MA where M is the mass of the robot, F is the force of friction at the wheel, and A is the rate the robot accelerates.
At some point… the robot will accelerate to the point where the speed of the robot matches the slipping wheel. At this point the wheel stops slipping, and the robot accelerates based on the load of the motor. Which we’ll talk about next…
Think about the robot with infinite traction.
The force that the wheel applies on the ground is less than the frictional force between the wheel and the ground. This means, the force caused by the torque of the motor is the force which accelerates the robot.
Now, as the robot accelerates, the wheel spins faster and faster. As this happens, the torque load on the motor decreases linearly with the speed of the motor until the speed of the motor is equivalent to its free-speed (well, close to it – accounting for frictional loads of the robot gearbox and wheel which will cause it to stop accelerating at some percentage of its free speed – probably about 80% for FRC robots).
Also remember… load and current are proportional. So when the motor is at stall, it is drawing stall current. As the load decreases and the motor spins faster, the current draw decreases. The faster the robot accelerates, the quicker the current draw decreases, the less likely you are to burn up a motor, pop a circuit breaker, or drain your battery. (With a 40-amp circuit breaker like we use in FRC – CIM motors are more than tough enough to survive. The breaker trips before the motor is damaged).
-John
Also…
Think about your robot pushing against a stationary object (like a wall).
In this case, if your robot has more traction than torque at the wheel, the wheel will stall, the motor will stall, and it will draw a lot of current. Probably bad.
In this case, if your robot has more torque than traction, the wheel will be spinning, and the floor will be acting as a brake on the wheel. You know exactly how much this load is, because you know the force of friction between your wheel and the ground. The motor will be loaded at some amount, depending on your gear ratio, and will draw some current depending on this load.
In FRC, if we’re designing robots to get into pushing matches with other robots, we design them such that when they’re pushing against a wall – the motor draws about 40-amps continuously.
This way, we know that even when we’re pushing – we won’t pop circuit breakers.
This is another reason to design for increased torque.
-John
A few other notes:
If you have multiple motors driving the same wheel… the load will be divided evenly among those motors.
If you have multiple wheels, the weight of the robot is spread over those wheels. The weight resting on each wheel will obviously determine the friction force between that wheel and the ground.
Multiple wheels connected to the same motor/gearbox will contribute their loads together.
Non driven wheels, contribute no load, but represent traction which isn’t being used to help accelerate the robot, or push obstacles.
-John
As a point of comparison, a couple years back my Lunabotics team used BaneBots RS-775-18 motors (running at 12V, and 1 for each of our 4 wheels), all running through a BaneBots P60 gearbox (I want to say 64:1 or 256:1)–and had one of the faster Lunabots if we kept our speed controllers alive.
Which brings up John’s 2nd “extra” post. We had something like 20A (or 10A, in some cases!) fuses on the robot, one per wheel, and never blew one. OTOH, we rarely moved on the regolith because our speed controllers couldn’t take the current, and shut down to prevent frying–and didn’t restart until the robot did. A change in speed controllers–to something we could change the safety settings on (:rolleyes:) resulted in a lot of motion until we put ourselves in a sticky situation and couldn’t get out… without frying speed controllers. ::ouch::
Plan for pushing, as John points out–you’ll need it.
A “torque-ier” drivetrain will have greater acceleration. Your robot will reach its maximum speed faster with more torque. however, as Torque and RPM are inversely related, the more torque you have, generally the slower your top speed. That is, unless you add more motors, increasing the resource pool of your drivetrain in a different way.
Think of it like a car transmission.
You wouldn’t want to accelerate out of a stoplight while in 4th gear, because 4th gear is designed to give you a high speed, at the expense of low torque. It would take all day to get up to 100MPH starting in 4th.
1st gear has a large ratio, supplying the torque you need to break static friction and get the car rolling from a standstill. However, due to the ratio, 1st gear has a very low top speed and could only carry your vehicle to maybe 15MPH before reaching dangerous engine RPMs. In 1st gear you would never reach 100MPH within the limits of the system. You shift from 1st into 2nd, then 3rd, then finally 4th, to reach your top speed in efficient time*. A Continuously Variable Transmission (CVT) is most ideal, because you can have an infinite number of ratios to maximize efficiency*.
Torque is also why a semi truck is powered by a 12 cylinder engine instead of a 4 cylinder. While both operate at similar RPMs, a more powerful motor will output more torque. Torque allows the massive, heavy truck to haul a loaded trailer at the same speed as a Prius. Force=mass*acceleration. A truck uses its force (torque=force at along a lever) to accelerate a great mass.
Alternatively, a Veyron uses the torque from its 16 cylinders to accelerate the relatively light vehicle quickly and achieve blisteringly high speeds.
Basically, more torque=more acceleration. Too much output torque, you compromise top speed. Too little output torque, you compromise acceleration. You could direct drive your robot with CIMs, but it would take you months to reach 5000 RPM (if you could move ever at all). You could gear your robot 5000:1, and you would have an ungodly amount of torque, but you could only rotate your wheel once every minute! Of course these are both exaggerations, but there is a careful balance between getting enough top speed, and being able to practically reach that top speed on a 54’ field
*When FRC teams use shifting transmissions, they are not using them to improve acceleration, but to allow a selection of high torque or high speed, to deal with match conditions. Depending on the gearbox, acceleration to top speed is only improved by a small fraction of a second when shifting like a car. Their high speed gear may not have the torque to push another robot, so they shift low to assist with dealing with opponents.
I am bumping this thread since much of the motor basics are covered well and we hope to extend the basic concepts. Please excuse the thread hijacking.
For our off season project this year, we are considering experimenting with a poor man’s Continuously Variable Transmission (CVT) gearbox. The basic concept is to use two motors but to intentionally mismatch free speed by about 2x. The 1st motor would be geared for high torque and would reach its free speed when the wheel was at 1/2 its top speed (lets call this the shift RPM). The 2nd motor would be geared for top speed. The PWM commands to the ESC for the 1st motor would be twice the value of the 2nd motor up to the shift RPM. The 2nd motor will continue increasing PWM values up to full voltage at full speed past the shift RPM. We will have one of these gearboxes on each of four wheels.
We understand that past the shift RPM the 1st motor will act as a generator and load the 2nd motor. The question is how severe is this load? What if we set the ESC to coast and signal the PWM to zero above the shift RPM? Will the flyback diodes in the 1st ESC effectively still load the 2nd motor? Will this damage the 1st ESC (we have blown HBridge with flywheel shooters)? Is it better to continue to drive the 1st ESC to max voltage?
More specifically, what we are considering is:
1st motor: RS775 12:60 (32dp) 14:72 (24dp) direct drive 6" wheel ESC=coast
2nd motor: MiniCIM 12:72 (24dp) same 72T gear ESC=brake
So the combined available stall torque (before gearing) is about the same as a full size CIM. During a pushing match we would drive all motors at full voltage.
We have looked at JVNs motor combiner xls - it predicts favorable results, but does not solve this exact configuration. It appears to assume matched free speed and therefor does not appear to comprehend the 1st motor loading the 2nd motor.
These guys pretty much have everything covered. I’ll just drop this handy reference sheet here which has a couple of equations for calculating maximum torque and other specs:
Motor reference sheet
Maybe you team can get some use out of it.
Edit: (Also, didn’t realize this post was 5 months old . . .)
Thanks for the quick response; it is important to understand the raw motor performance in designing/choosing a gearbox. We have been using a similar table published by JVN.
The task at hand is to combine two dissimilar motors from the table to create a new, wider range super motor.
Has anyone ever run a test:
How long can you stall a 775 at full voltage before the factory-installed smoke escapes?
I cant find it now, but I believe I saw 30Amps for over a minute. Keep in mind that the RS775 stall current is rated at 18V as 130A. If we design for enough torque to slip wheels when pushing against a wall, the spreadsheet predicts 45A and moderate motor RPM (internal fan running). I dont think we will violate any non-smoking laws, but thats why we are trying this off season.
Any thoughts on back driving the “generator” if the ESC is in coast mode?
Are you implying you’re planning on using a 775-12 for this application???
The 775-12 draws 30 amps stalled at 12 volts.
The 775-18 draws 87 amps stalled at 12 volts.
I should have specified. We are planning RS775-18 with a 20:1 ratio (plus the added torque from the 6:1 MiniCIM).
Since there is enough torque to overcome the CoF (1.2) of our wheels, the motors never stall and the current per motor is 45A when pushing a wall (with the wheels slipping on the carpet). It is hard to find dynamic CoF for FRC wheels. The only one I did find was 85% of the static CoF so that percentage is what I have been using as an estimate.
BTW: you have a similar issue with full size CIMs - if you ever let them stall you draw too much current and pop breakers. You simply have to design to slip the wheels.
At 12V, a 775-18 motor draws 45 amps at 6385 RPM
At 12V, a MiniCIM motor draws 45 amps at 3008 RPM.
But if you have them geared 20:1 and 6:1, respectively, they can’t simultaneously be at those speeds.
6385/20 = 319 RPM
3008/6 = 150 RPM
A bit off topic, but while I’m not sure about full voltage, I found this year that they’re actually quite tolerant of stalls at partial voltage; they can go a surprisingly long time at 50% - I believe we went the better part of a match with one stalled in such a manner when we had a limit switch fail during the DC regional, with no obvious damage to the motor.
That’s an interesting way to look at it. Since I don’t yet have a spreadsheet to dynamically calculate operating points, I used JVN table - which does include voltage droop (which looked like about 7.5V).
We did intend for the RS775 to reach max RPM when the MiniCIM was at 1/2 max RPM, so your estimates make sense. Clearly the wheel RPM will be the same for both and currents will split accordingly. The biggest concern I have is what happens after the RS775 exceeds it free RPM. I know it will behave like a generator and load the MiniCIM.
I do have some serious unanswered questions with this configuration. If we put the ESC in coast does that sufficiently break the generator current path such that it does not significantly load the driving motor? I don’t think anyone has tried coasting the motor after it exceeds free RPM but instead continue to drive at full voltage. If we back-drive the RS775 to twice it’s free RPM (coast or driven) will the generator blow the ESC?
775-18 free RPM is 13000. With 20:1 reduction that’s 650 RPM.
MiniCIM free RPM is 6200. Half of that is 3100 RPM. With 6:1 reduction that’s 517 RPM. So you’re a bit off with your gearing.
That’s an interesting way to look at it. Since I don’t yet have a spreadsheet to dynamically calculate operating points, I used JVN table - which does include voltage droop (which looked like about 7.5V).
I was just taking the numbers you posted to see where they would lead. The motor calculator I posted here makes this easy to do.
Here are the results using 7.5V instead of 12V:
At 7.5V, a 775-18 motor draws 45 amps at 1407 RPM
At 1407/20*6 = 422 RPM, a MiniCIM draws 48 amps at 7.5V
See example calculation attached.
I do have some serious unanswered questions with this configuration. If we put the ESC in coast does that sufficiently break the generator current path such that it does not significantly load the driving motor? I don’t think anyone has tried coasting the motor after it exceeds free RPM but instead continue to drive at full voltage. If we back-drive the RS775 to twice it’s free RPM (coast or driven) will the generator blow the ESC?
Good questions. Maybe the engineers from VAX and CTRE can answer for their respective motor controllers.
There’s also the question of the 775-18 spinning at 13000*2 = 26000 RPM. Can its rotor, fan, and the bearings handle this? And even if the the motor leads are open-circuit (zero generated current), how much load does the fan put on the MiniCIM at that speed, through a 20:1 gearbox?
*
MCALC example.pdf (30.4 KB)
MCALC example.pdf (30.4 KB)
Good questions. Maybe the engineers from VAX and CTRE can answer for their respective motor controllers.
There’s also the question of the 775-18 spinning at 13000*2 = 26000 RPM. Can its rotor, fan, and the bearings handle this? And even if the the motor leads are open-circuit (zero generated current), how much load does the fan put on the MiniCIM at that speed, through a 20:1 gearbox?
Good point on the fan - I had not thought of that. The 26000 RPM is only 33% higher than the 19,500 free RPM at the rated 18V. The robot will not be at this velocity very often or for long so I think there is sufficient margin.
I assume that the ESC have flyback diodes, so it will never truly be open-circuit. I am not sure what load that will impart on the MiniCIM. If the ESC blows, it expect it will be the flyback diodes that smoke.