I am working on designing a 4-bar linkage arm and was attempting to do some calculations on which motors and gear ratios I should be using. I calculated it at 60in with 20lbs of force. Below are my calculations with a dual-globe motor setup. I came away with an absurd amount of supplied stall torque. If I have 20000 oz-in of required torque, how much actual torque should I be shooting for? I understand that one needs to supply more than just the stall torque, but I don’t understand how much of a difference there needs to be between the required and supplied torque. Maybe you guys can help me.

Arm Length: 60 in
Arm Weight: 20lbs
Required Torque = Length * Weight = 19200 oz-in

Disclaimer Not all of this may be correct. So please don’t hold it against me if it is

Okay. Now’s a good time to ask this question:

Does having two motors supply more torque? or does it seem that way because there is less load on each motor?

One of our engineers said that our stall torque should be around 25% higher than our required torque. This is to make up for friction. However, that only works if you know exactly where the center of gravity will be and exactly how much it will weigh.

We assumed what the weight will be and where the center of gravity will be so the required torque calculations were definately off.

We aimed to have half of 25% above the required torque (required: 100 lb-ft, stall: 187.5lb-ft)

Also, the center of gravity is roughly the middle of the arm. So if your arm is 60 inches, you should plan on the center of gravity being between 20 and 40 inches from the axel of the arm. (obviously…)

Make sure you are putting enough current into those globes. We made a big mistake this year and calculated what gear ratios we needed based off of the stall current of the FP motors. When we hooked it all up, we were only powering those motors at about 1/3rd the current we had expected and the arm didn’t run. That only set us back about a week

Also, with that ratio. That arm is going to move very slowly. If you are planning on using it for a FIRST application, you might want to lower that ratio, cause as it is right now, you’ve got almost 17 times as much torque as you need. (34 if you lower that length down to 30 inches).

Also. I like working with lb-ft better than oz-in simply because the numbers are smaller and easier to work with

I tend to design for 25% stall torque, which keeps the motors from the kit nice and happy for a long time. With some motors this is overkill, the globe happens to be one of them. We actually stalled one for an entire match (programming mistake) and while it was quite hot after the match, it still worked fine fine. (Almost gave one of our team members an uppercut not ten minutes later, actually) Your current design is very conservative, and will be very slow. If you use a 9:1 reduction I think you’ll be fine. (This loads the motors to 50% of stall, which would eventually melt a FP or BaneBots, but the globe should handle it fine)

Thanks for the help guys. My other question is, if I am given a free speed, how do I much is the actual speed affected? How do I calculate it with different amounts of force?

An easy way to do this is to draw out a linear torque-speed curve:

Plot free speed as a point on the speed axis. Plot stall torque as a point on the torque axis. Connect these two points with a straight line. This line defines the torque at a given speed, or the speed at a given torque. (Two ways of saying the same thing, although the dynamic relationship between the two is somewhat more subtle.)

The equation for that line can give you the speed at any torque:

Ok so I did so recalculating. The speed shown is the free speed at that torque. But here is what I get, can someone make sure my math is right.
**
Stall Torque (oz-in) Free Speed (RPM) Free Current (Amps) Stall Current (Amps)**
Minibike 570.00 5310.00 2.70 133.00
Globe 3194.53 79.87 0.58 21.58
Nippon-Denso 1316.99 92.00 2.50 25.00
Keyang 1699.34 85.00 2.00 20.00
Banebots 125:1 4935 134.00 1.00 42.00

Enter Weight (lbs): 20 lbs
Enter Length (in): 40 in
Torque Required: 12800 oz-in

of Motors: 2.00

Enter Stall Torque: 3194.53 oz-in
Enter Free-Speed: 79.87 RPM
Enter Gear Ratio: 8 :1

Since a number of posts have come up since I started the spreadsheet I’ll let it speak for the details itself.

Summary:
Your torque required seems high… should be force * arm length unless there are other factors… arm angle also plays into this… looks like there is some ‘extra’ math creeping in here, possibly applying gear ratio’s multiple times