I Acceleration on an electric unicycle, how much does the rider have to lean?

[rcgldr]

Can someone here check my math?

On an euc (electric unicycle), the motor has to exact the same amount of torque onto the wheel+tire that the rider exerts onto the euc, otherwise the frame would rotate (due to the motor exerting an equal in magnitude but opposite direction onto the frame than it does onto the wheel, a Newton third law like pair of torques).

Choosing some arbitrary numbers:

rider = 200 lb
euc = 100 lb
rider + euc = 300 lb
tire radius = .83333 foot (10 inches)
rider center of mass 3 feet from riders feet

If rider is standing on a plaform accelerating at 1/2 g (32.174 feet / second^2) rider center of mass is 3 x sin(arctan(0.5)) ~= 1.342 feet ahead of feet and 3 x cos(arctan(0.5)) = 2.683 above feet. The external torques onto the rider cancel: torque from gravity = 200 x 1.342 ~= 268.328 lb ft, opposing torque due to acceleration = 100 x 2.683 = 268.328 lb ft. Rider exerts a linear force and zero torque onto the platform.

For the 300 lb of rider + euc, 1/2 g acceleration requires 150 lb force. With a tire radius of .83333 foot, this translates into 125 lb ft of torque. The rider needs to lean forwards enough so gravity generates a 268.328 lb ft + 125 lb ft = 393.328 lb ft of torque. Rider center of mass is ~1.96664 ahead of contact patch.

For 1 g of platform acceleration, rider center of mass is 1.5 feet ahead and above contact patch. For euc, force = 300 lb, torque on wheel = 300 x .83333 = 250 lb ft. Total torque = 300 + 250 = 550 lb ft. Rider center of mass = 550 / 200 = 2.75 feet ahead of contact patch.

 

[jedishrfu]

It's interesting that you worked your problem in English units and not SI metric units.

Doing that makes it harder for many folks to check due to the conversion factors, constants like g, and due to our science training in the MKS system that makes us familiar with the numbers.