# Using 2 CIMs to lift 250lbs

My team is considering using 2 CIMs to lift our robot during the climb, with an approximate 250lbs needing to move. I am of the belief that 2 CIMs should be able to lift the robot at 475 RPMs, geared at an 11:1 reduction without stalling. I’ve provided a CAD of the current gearbox we are considering using. Its ratio is roughly 1:1.5 going into a 20:1 Worm, and it outputs about 380RPM. According to my estimates as long as the RPM is under 480 it shouldn’t stall the motors.

Is this feasible? Will the motors be able to handle it or did I mess up my math?

EDIT: Taking the input everyone gave I proposed a different gearing further down the thread at 1750 RPM moving a 1 inch drum with a torque of .9 Nm/s

Other items needed: Winch size.

How big your winch is will play into the gearing.

Once it comes out of the gearbox at ~380 RPM it spins a 4in stealth wheel that spools up 6 feet of rope

250lbs on a stealth wheel wont end well. Id recommend using some sort of lathed down drum, a plastic wheel wont cut it.

I’d recommend searching JVN Calculator on the internet on google, its a spreadsheet that has everything in it for calculating speeds, torques, stall, etc.

You say being under 480 RPM

at this ratio and weight wont stall the motors, but are going 380 RPM
, that means you are at 80% stall, which is not good. You want to be below 30% (ishhhh) usually to be safe.

Another question is how you plan on stopping backdrive, I see you have a wormgear, are you afraid of that stipping its teeth under extreme pressure like this?

So when I was calculating the torque on the motor it came out too 2.4 Nm/s at 480rpm. According to andymark the motor stall at 2.42 or 2.44 Nm/s so it’s just barely safe.

Concerning the work gear stripping I just have to trust it to handle all the backdrive.

I’ll look into a different drum if I end up using it, but I’m still not sure if it can even lift the weight yet. Is the torque I calculated correct?

2 CIMS =21.5 ln lb of torque at stall.

Total GR of 1.5*20 (I’m assuming the first ratio is 1.5:1 not the 1:1.5 as you stated.)

Total torque after GB of 645 in lb.

4” Wheel has a 2” radius. Torque =rF.

Output force of 322 lbs.

HOWEVER. This is at stall. And assuming 100% efficiencies. A worm gear can vary quite a bit but I’d wouldn’t want to estimate higher than 60%. By the time you add in other inefficiencies call it 50% at best.

This gives you maybe 160 lbs. but that is still at stall. You really want to be at or below half that. To me the most practical way of doing this is switching your spool from a 4” wheel to a 1” metal spool.

So it’s mathematically possible, but unlikely due to imperfections and inefficiencies?

That’s pretty much the gist.

Even if there’s no inefficiency and everything is perfect, it’s really risky to run 2 CIMs at stall for the electrical system. A CIM

at stall draws well over the current rating of the breaker, and sooner or later it will pop them. Hopefully the breakers don’t pop before you finish the climb, but isn’t it better to not take that risk?

The problem is that at stall, you’re not moving, by definition. And at nearly stall, you’re moving veeeery slowly. What you want to do is look at the power on the motor curve. Lots of stuff in (Newtonian) physics comes down to energy and power. By lifting the robots, you’re adding potential energy with every inch you lift. Power is energy per second, so it can give you a good estimate for how fast you can lift.

If you look at the power curve at ~400 RPM

you can see you’re putting out 100W. (And drawing a LOT of current to do it. Which means it’s actually not going to work.) But you could be putting out 300W if you just geared it so the motor was turning faster.

Practically speaking, you want to gear so the motor is turning rather faster than at peak power. If you’re ideally geared for peak power and something adds more load, you go slower, put out less power, and draw more current. (Which kills your battery and makes you go even slower.) If you’re ideally geared for faster than peak power, then extra load means you put out more power, which means you don’t slow down nearly as much.

A decent place to be is geared for 25% to 35% of stall torque. That puts you close to peak power, but on the fast side. In the world of frictionless gears, that means you’d be lifting that load at around 11 in/s.

^What Kevin said.

I would add that you can lift any weight with any motor through selection of gearing.

For the same weight as 2x CIMs you could use 4x 775Pros, which can deliver about twice the power of the CIMs. Thus the 775s would climb twice as fast, or give you twice the margin on force, roughly speaking. (I’m ignoring some effects in this simple thought experiment.)

The output of your gearbox is 380 rpm. You have 11:1, 1.5:1 and then 20:1 reduction. That’s a 330:1 reduction. To get 380 rpm output, you have to have 125400 rpm input. CIMS don’t go that fast.

Also, a 380 rpm output with a 4" diameter wheel is 4775 in/m, which means that you’re doing about 6 ft/sec.

It’s also worth noting that worm gear drives aren’t very efficient. Your gearbox will be losing a lot of your 2 CIMs worth of power to friction losses.

Does the 250 lb come from lifting your robot plus one other? If so, is your robot going to weigh 90~100 lb or less with bumpers and battery?

Here is a link to the calculator for anybody looking for it.

I can see 250 lb being accurate with three 150 lb robots. Our team considered doing this but didn’t have the time to commit to a lift this year.

3 * 150 = 450 I get it, but that doesn’t change my answer.

Gregg probably your best bet will be to use the andymark sport series planetary gearboxes with the highest ratios you can grab. Most anything else will run the risk of stripping gears.
I don’t know if the VEX rachet will fit on those (assuming so but didn’t look at the diagrams) but is so make use of them. Yes, you can drop after T0 but you don’t want to finish the lift with 5 seconds left and risk a brownout.

Use the calculator others have recommended and keep the current draw as low as possible.

To clear up a a misunderstanding the CIMs do not go 11:1 before entering the gear box, that’s just it’s top speed for torque.

Thanks for all the help guys. After taking the input and revisiting the math I’ve learned it’s impossible due to the current required being too high for the CIM too handle.

Currently I’m considering running the motors through a 2.85 reduction at 1750 RPM on a 1 inch drum. This should give it a torque of .9 Nm/s on each motor, which should be it’s peak efficiency.

*The place to start is to calculate the required power.

Say you want to lift a 250lb load at 1 foot/sec.

The power required is 250 lb_ft/sec = 339 Watts.

The max power of 2 CIMs is 337*2 = 674 Watts (at 12 volts, which may not be available at the end of a match under load).

The transmission power efficiency is the big question. If it’s any less than 339/684 = 50.3%, you will not be able to lift 250 lb load at 1 ft/sec with a battery voltage of 12 volts.

The worm gear alone, if it’s not a quality design, could be less than 50%.

You need:

• more power (3 CIMs?), and/or

• better efficiency, and/or

• slower speed

See my previous post about calculating required power first.

Did you mean 1750 RPM

at the drum?

1750 RPM

on a 1/2 inch radius drum gives a linear speed of 7.6 ft/sec

2 CIMs cannot lift 250 lbs at 7.6 ft/sec.

Or did you mean 1750 at the motor, which would be 1750/2.85=614 RPM

at the drum?

If so, that’s a linear speed of 2.7 ft/sec. 2 CIMs can’t lift 250 pounds at that speed either.