Help: Torque to Linear force conversion

We have a cim motor (343.4 oz-in of torque), a mini cim motor (12.4 in-lbs of torque) and a toughbox mini with a gear ratio of 10:71:1. The pulley we have connected to the toughbox mini has a 1" radius. Using these pieces of information I have calculated our linear force to be about 240lbs. Can anybody verify this? Do I need any other information?

The torque numbers you have are at stall (i.e., the maximum possible). Under normal operating conditions, you might get half that. If you’re gearing the CIM and Mini-CIM together, you’ll take a significant hit to efficiency, perhaps about 20% less than perfect. The gearbox itself will take another ~10% efficiency. By those numbers, you’ll get around 130lbf with your pulley.

Thanks!!! If we did 2 mini cims instead would that help the efficiency or result in a net loss due to the weaker motor?

I think that 2 mini Cims would put you around 115lbf at peak power. Do you know what resistive force will oppose your motors? If so you could find the motor speeds. Remember though, even though this is peak power, it’s not peak force, which could be about twice these numbers.

Also keep in mind the current draw. If you are running a CIM at peak power it pulls ~65A, and at stall it is like 130A. Depending how long you are running under these conditions you will have to consider the tripping he fuse.

We are using the motors to pull down our catapult which is connected to surgical tubing at the other end of a fulcrum. The calculated value for the tubing pull force is 516 lbs when the tubing is completely stretched (I realize that is a ton of force). The mechanical advantage that our winch has is 4 due to the length of our catapult arms (the winch will be connected to the end of the catapult arms).

It’s a lot of info to explain let me know if I missed anything :slight_smile:

You need a much greater reduction than 11:1 if you’re pulling 500lbs. In addition, I would be worried about ratcheting on your belt teeth (assuming you’re using timing belts and pulleys).

Not to hijack the thread but is there a calculator out there that we can punch numbers into to figure this information out.

We are looking at using either 2 mini-cims or 2 full size cims in the 2 cim ball shifter. High gear reduction of 3.667:1 and Low gear reduction of 8.333:1

Are hope is to remove the parts for the high gear and use the low gear to winch back a shooter. also is there any advantages to using a larger or smaller pully on the winch.

We are actually using a ratcheting strap similar to one you would use on a trailer. We are connecting the strap to the catapult and then fastening the other end to a wheel and axel which we plan on driving it with the motor gearbox combo as earlier described. I don’t have experience with this kind of thing so any additional guidance would be appreciated. Thanks.

You might be able to use JVN’s motor calculator for basic calculations. The best way, though, is to manually apply torque, speed, and efficiency equations to motor curves. You’ll understand the problem much better.

How are you going to release the catapult, if this is your plan? Will you shift the gearbox into neutral, or just backdrive the motor? If it’s the latter, I can promise you that motor and gearbox resistance will not let your catapult launch with any appreciable speed.

We were going to use a pneumatic cylinder to pull the drive gear off the driven gear (allowing the wheel and axel to spin freely in a neutral position)

You’ll likely want a ratcheting system to take the load before you do that…

I’m afraid that this season we’re going to see a lot of shifting winches that work well at first, and then suffer a bad case of broken teeth and sadness.

You can use JVN’s mechanical design calculator (posted earlier) to solve this.

Use the Linear Mechanism sheet. Set motor specs, drum diameter, etc. as appropriate and set load to 1/4 of the bungee force (since you have 4:1 mechanical advantage). Set Travel Distance to the length of pull in, which you can figure out from your design.

We used a gearbox design similar to that which you speak of (pulling the gears apart) in 2010 on our kicker. We used a window motor which did not backdrive (which will not be nearly enough power this year), and opened the gearbox to kick. We only engaged the gearbox again after the kicker hit the hard stop and was not moving. We needed quite a large piston to keep it together. We ha no reliability issues with the gearbox at all, but we did kill 1 window motor. If I were to do it again I would probably put a ratchet somewhere in the high speed end of the gearing, on the motor side of the opening gearbox, and use a larger motor (probably a miniCIM or CIM for this kind of load and pull in time).

We are ahead of the game on that one! We have planned for the ratcheting mechanism from the beginning so the implementation of one shouldn’t be a problem.

Thanks!! I never thought to do that (In spite of the fact that it was very clearly pointed out to me earlier).

While it will be helpful to use the JVN Mechanical Design Calculator, there are a couple important things to understand about designing a motor setup (# motors, motor selection, gearing, etc.)…

First, a few things to know about motors:

  • @ Stall: no speed, max torque, no power, max current draw
  • @ Max Power: half speed, half torque, max power, ~half current draw
  • @ Free Speed: max speed, no torque, no power, min current draw

You never want to design a system to operate at motors’ max torque, because this is only provided at stall (torque then decreases linearly until 0 @ free speed)… For a “heavy-lifting” operation, I’d recommend designing for a point a little faster than max power… about 55-60% of free speed (which is 45-40% max torque). Designing for this point means that unexpected losses will push the motor closer to max power rather than further.

Also, always take in to consideration some losses… probably about 10-20% loss. Remember to account for this in both speed and torque calculations.

You can also shrink the pulley radius to increase your mechanical advantage… I’d personally recommend a smaller pulley than 2" diam for this reason.

Also, are you sure you really need ~500 pounds of force in your surgical tubing?! That is a phenomenal force! How are you pulling that back currently on your prototype? Perhaps try using more travel in your surgical tubing to enable as much energy storage but with a lower max force… (U = 1/2 k * x^2).

I’m low on time right now, but perhaps I’ll post again later with a worked out example…

Back in 2010, ratchet failure due to broken/gouged teeth was the biggest problem our robot faced (well, second-biggest; our ratchet-tooth-click-counter approach to setting the kicker power was probably the biggest). Our kicker was supremely powerful (could score from the rear zone), but I’ve always avoided similar mechanisms since then for just that reason. Disengaging teeth under that much force without damaging them is not a trivial matter.

We’re using a 3/8" stainless ratchet and pawl. What were you using?

Something similar. I can snap a picture of the assembly and the damage and upload it (either later today or tomorrow) if you’d like.

We were disengaging our pawl to release our kicker, and the primary failure point was the pawl gouging out the ratchet teeth over time.

Yeah, I’d love to see it.

Relieve the load on the teeth before releasing the pawl.

You designed to have enough power to pull it back? Pull it back a half tooth more while disengaging the pawl.

Similarly, if using a dog gear method… Drive the motor forward and backdrive while pulling the dog.