Can anyone validate my motor calculations?

[GeeTwo]

Originally Posted by LoneWo1f1998

I'm assuming that a 3" sprocket translates to a 1.5" Load Torque Lever arm.

Provided that 3" is the pitch diameter of the sprocket, this is correct. If 3" is the outer diameter of the sprocket, the lever arm will be a bit shorter.

Assuming that's right, let's sanity check the numbers...

Stall current is 22A, so you shouldn't be breaker-limited unless you decide to be. That means you get an initial torque (I'm assuming a spike, or that you move quickly from zero to full on a speed controller) of 16.6 ft-lb. Applied through a 1.5" (.125 ft) lever arm, this is just shy of 133 lb at stall or startup, a bit more than the spreadsheet gave. This is probably due to assumed inefficiencies in the spreadsheet. Since you're already past the gearbox for your measurements, your losses are probably far less as long as your chains are straight and nothing is binding.

I'm not sure what you mean by "the whole stack", so I'll assume six totes. A tote weighs 7.8#, so six of them are 46.8#. Assuming 125# of lift, you'd get an initial acceleration of (125 - 46.8)/46.8 gravities, or 53 fps/sec. If you could keep up that acceleration, you'd reach 6 feet of elevation in under half a second, and be going up at 25fps, which corresponds to 200radians per second. So it sounds like you would very quickly approach equilibrium speed, which is about 75rpm * (125-46.3) / 125 or about 45 rpm. At this speed, the chain would move 45 * 2 * pi * 1.5 / 60 or about 7" per second. That roughly agrees with your 2.32 seconds if you were contemplating a 17" lift.

BTW, If you counterbalance your load with about 20-25 lb, you'd nearly double your top speed with a full load. At a price - unless you had a large chain loop or deep cable run so that your counterweight was at the back of the robot, this would move your center of gravity even farther "forward".