As the title says, we are currently thinking of using a 100-1 gear ration on a CIM to lift our robot using winch style system. I feel like this is over the top and will make our robot climb too slowly.
How large is your winch drum?
As mentioned in previous threads, there’s so much more than your gear reduction. Your spool diameter, your rope diameter all factor in as well. Don’t go on a hunch, find the equations and calculate out your speed. Shouldn’t take more than 10 minutes.
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It’s about 9 inches in long and about 2.5 inches wide
Then yes lol it’s overkill. I’d go around a 63:1 to be safe
Assuming the 2.5" dimension is the diameter of your drum, you’ll be within the realm of a reasonable climb (albeit perhaps a bit on the slower side). This can depend on how your rope will wrap around the drum (if it’s a thicker rope wrapping concentrically on top of itself, your effective winch diameter will be increasing, meaning you’ll reel more rope in per revolution as you increase). If you have significant rope pile up, you may be glad you had that additional reduction. If your rope doesn’t build up, then you may want a slightly more aggressive gearing.
That being said, be careful to consider how you’re getting that reduction. A 100:1 reduction with a VersaPlanetary mounted to a CIM will push it beyond the suggested output shaft loading.
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There are numerous calculations tools available here to help you make a factual decision rather than an opinionated one. Try JVN’s Calculator located in the media section.
Does that calculator not include the VersaPlanetary gearbox or am I just missing it?
In an application like a climber, I like to try to optimize for running the motors near peak power. Since it’s only running for a few seconds per match, in optimizing like this, we get the best possible performance given the motors we select, which in this case is the fastest possible 1 CIM climb.
Peak power on a typical FIRST motor is usually around 1/2 the max RPM, and 1/2 the stall torque, which makes sense when we consider the motor curves, and the formula for power through an output shaft:
Power = rotational velocity * torque
According to VEX, at peak power, the CIM motor has an output torque of 172oz-in (10.75lb-in) at 2655rpm. Given that the diameter of the drum is 2.5in (radius of 1.25in), and the maximum lifting weight of the robot fully loaded with bumpers and a battery should be less that 150lbs, we can do some math:
Final Ratio = Final output torque / motor torque = 1.25in * 150lbs / 10.75(lb-in) = 17.4 : 1
Without knowing the specifics of your climber, like efficiency or how the rope spools up, I can’t offer much more than saying that a perfectly efficient system with a 1.25in radius drum will lift a 150lb robot as quickly as possible when geared to a CIM motor at 17.4:1.
Some important things to consider are the efficiency of the gear drive (I might add a 10-20% safety factor for this), the time that the motor can sustain the given loading situation (available on the same VEX Pro motor page), and how other current drawing systems on the robot might affect this.
Using the aforementioned JVN calc, a CIM driving a 100:1 reduction and a 2.5" drum would:
lift 154# 36" in 5.82 seconds. It would use a tiny fraction of the motors available power (14.28 amps of the motor’s 131 amps at stall) and would stall at 1365# of pull. I hope your rope could stand that.
If 5" drum was substituted, it would:
lift 154# 36" in 3.3 seconds. It would use a more reasonable fraction of the motors available power (25.86 amps of the motor’s 131 amps at stall - putting it in the more efficient range) and would stall at 682# of pull. Still plenty of power left, but less chance of pulling things apart.
If your gearbox and drum diameter are already firm, then you could substitute a Mini-CIM for the CIM and still have about the same lift time as with the CIM (it’s a faster but less powerful motor) but have less chance of breaking something
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For a better understanding of motors and gearing selection, see excellent videos by Ken Stafford, including this one:
*A CIM running at 14.28 amps and 12 volts is running at 63.7% efficiency, and outputs 109 mechanical watts
(which is 32.4% of max power).
A CIM running at 25.86 amps and 12 volts is running at 63.5% efficiency, and outputs 197 mechanical watts
(which is 58.5% of max power).
*
As I read the CIM motor’s data sheet (under the Special Features heading), 12 V, 27 A, 64 oz-in, 4320 RPM is the Nominal loading condition; the motor is tested by running at this voltage and torque for 6 minutes, resting for 30 minutes, and repeating that cycle 1000 times.
During such a test, speed will decrease somewhat as the motor’s innards (i.e., armature coils, brushes, and magnets) get warmer. My testing suggests that 30 minutes is not enough time for a CIM’s innards to return to room temperature, unless cooling is accelerated.
Operating a CIM motor beyond its Nominal loading condition is ok for shorter durations; however, it will require more time to cool down after overloading.
Cool down time can be reduced by forcing the innards to release their heat. Finned heat sinks are one method. Some teams have used internal forced cooling, running air through the 10-32 threaded mounting holes via vented screws.
