Ok so I have 2 CIM motors with a 5:1 gear ratio each with a 8 inch wheel

So what I want to know is how much torque will this have and how fast (in mph) it can go while moving 180 pounds

Each motor will have a Max 30 amp fuse

The free speed of a CIM is 5310 rpm. Reduced 5:1, about 1060 rpm free, 800 rpm adjusted or 13 rev/s.

An 8 inch wheel has a circumference of 8*pi/12 = 2.1 ft, for an adjusted free speed of about 27 fps. (This is way too fast.)

If you assume you run the CIMs at 40A (acceptable with a 30A breaker for short periods), the torque is 0.7 N-m * 2 motors * 5 (reduction) = 7N-m, or 5.1 ft-lb. Through a wheel 1/3 of a foot in radius, the force is about 15.3 lb, producing a theoretical best acceleration of .085 gees, or 2.72 ft/s/s. (5.44 ft/s/s if you have four CIMs, two on each side). Pretty slow.

**From here down added about 3 hours later:** My assumptions above were for an FRC application, where the field is about 50 feet long. At 5.4 ft/s/s initial acceleration, it would take you about 6 seconds to get across the field if uninterrupted (e.g. a trench run) assuming you don’t want to crash into the wall at the other end (if you don’t care about that, make it about 5 seconds). For FRC, to go faster (meaning getting more things done in 150 seconds), it is usually helpful to gear slower. For 8" wheels with 4 CIMs and a full weight robot, you probably want to gear a bit below 10:1 if running full court, or around 15:1 for shorter sprints, or even lower if doing really short sprints (e.g. stuffing the vault in 2018).

If you’re planning to put this thing on a larger course, the time it takes to get up to speed is less important than the speed you can eventually reach. As noted above, this will be about 27 fps (or 18-19 mph).

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