CIM Motors to pull up Robot

I was considering using the CIM as a winch motor for the pull-up bar.

If our 130 lb. robot was directly under the bar and was pulling up from the center of mass, say 12 inches off the ground it would take a maximum torque of 130 lb*12 in = 1560 in-lb.

To keep the motor happy we will keep the load reflected back to the motor between 20-50% of Stall Torque.

CIM Stall Torque = 2.2 Nm which is 20 in-lb.
Running the motor ar 40% Stall Torque gives us Working Torque =20 in-lb
.4 = 8 in-lb

so Gear ratio for CIM woul be 1560 in-lb/ 8 in-lb = 195/1

If I am right, I am looking at too big of a gear reduction to make the motors work.

[quote=Joe P]I was considering using the CIM as a winch motor for the pull-up bar.

If our 130 lb. robot was directly under the bar and was pulling up from the center of mass, say 12 inches off the ground it would take a maximum torque of 130 lb*12 in = 1560 in-lb.
QUOTE]

This is only true if you have a 12" lever arm at the shaft of the CIM motor. 12" off the ground isn’t the right number, because no matter of far you are off the ground, the amount of torque you need to lift the robot remains the same if the speed of going upward remains the same.

It takes 130 lbs to lift up a robot. The stall torque is 20 in-lbs. If you have a winch, and the diameter of the winch is 5 inch, your motor can excert 4 lbs upward force at stall torque. 50% of that is 2 lbs upward force. 130 lbs / 2 lbs is 65:1 a reduction ratio. You should note that this is moving at 50% free speed, so you should check and see if this is too fast or not.[/quote]

the 12" radius he quoted is at the winch

or do you intend to have a 12" lever arm that does the lifting?

I also recommend you look at pnumatics - with the regulators at 60psi you can get a cylinder that can produce much more than 130 lbs of force

all you need then is a latch to hold it extended when the power is cut.

Could you expand on this a bit? Are you saying that a pneumatic cylinder will not hold it’s position at the end of a match? I don’t know since my team has never used them before.

It will as long as you have pressure left in your accumulators.

If you have a retracted cylinder, what exactly will happen (like to the solenoid) after the end of the match?

Pneumatics are interesting.

Quick basic rule:
Pounds per Square inch * Area in sq. inches = Pound Force
PSI * Area[sq in] = F

Attached is a decent cutaway of a cylinder.

On a 2" Bore cylinder, the area (pir²) of the Bore-end is pi(1²), or roughly 3.14 sq. inches.
Your maximum PSI at the cylinders is 60 PSI.

60 * 3.14 = 188.4 lbf

Meaning the cylinder can push OUT 188.4 lbs.

When retracting, the area of the Rod must be subtracted from the total area.
From the Pneumatics Manual, the rod diamter is .625", therfore radius is .3125"

So now you go:

Area Bore - Area Rod
pi*(1²) - pi*(.3125²)
pi - pi * .09765625
pi - .306… = 2.834 sq. inches

now again we do PSI * Area
60 * 2.834 = 170.04

(Note, the values for the other cylinders, as well as this one, are listed on pg. 12 of the Pneumatics Manual This is so you know WHY and HOW MUCH, not just HOW MUCH the cylinders can hold).

As put in the thread about pneumatics holding their air when power is cut…

A double solenoid valve will continue stay in the state which it was at before power was cut. If the tubing was set for Extended, it will stay extended. and vice versa.

A single solenoid valve will always return to it’s default state. The valve uses an electromagnet to switch from ON to “Not ON” (Different from OFF). Note: regardless of how you are positioned when power is cut, the solenoid will go to the “Not ON” state.

cylinder.JPG


cylinder.JPG

Alrighty, I think that this is an important enough issue that I’d like to make a number of quick corrections. One of the most difficult problems I’ve had in the past was properly picking gear ratios… so I’ll make a quick lesson here.

First and foremost… let’s take a look at the CIM motor.

It has a stall current of 114 Amps, and a stall torque of 346.9 oz-in, which we’ll switch over to 21.68 in-lbs since it’s probably more applicable in this situation.

There’s been talk of designing around the max power of motors, or 40% of the motor stall to optimize power, but this entire idea can lead to big trouble!

Designing the gear ratio around the max power is not the ideal situation in FIRST, since we’re limited to the amount of current that the motors can pull without tripping the circuit breakers. Designing a gear ratio around the max power in this example is 114 * 50% = 57 Amps. Since I believe we’re still on 40 amp breakers for the CIM motor, this would be a very poor choice, resulting in your robot being unable to lift itself!

Instead, design your components around a safe current draw, such as 35 Amps.

Solving:
114 * X = 35 A
X = 30.7% of max (stall) torque, which is 6.66 in-lbs.

Now, to look back at the problem, if the winch has a diameter of 5 inches, the radius will be 2.5 inches, which is the moment arm. However, I have a hunch that Ken was just typing a bit too quick, so I’ll assume he meant a 5 inch radius / moment arm.

With this said, we need a torque equal to:
5 inches * 130 lbs = 650 in-lbs.

Since we know that we want to run at 6.66 in-lbs, we’ll need to create an arm gearbox that has 650 / 6.66 ratio, or 97.65 : 1.

Including a moderate factor of safty of 1.3 or so… I’d say you’d like at least a reduction of 125 : 1.

To reduce this massive reduction, simply reduce the size of the moment arm. Half size moment arm = half gear ratio.

I hope this helps! Good luck! :slight_smile:

Matt