[Richard Wallace]

Originally Posted by Jared

The rotational inertia of the wheels does little to hinder robot acceleration.

For anybody interested, I compared the kinetic energy of the robot moving at 10 feet per second to the rotational energy of the wheels on a robot moving at 10 feet per second. If we compare these energies, we'll see how much of our power goes to spinning the wheels, and how much goes toward moving the robot.

A fully loaded robot weighs 150 lbs, which is 68 kg. 10 feet per second is 3.05 meters per second.

KE = 1/2*mv^2 = 0.5(68)(3.05)^2 = 316.285 Joules
That's the amount of energy it takes to bring your robot up to speed.

For the rotational energy, we've got to find the moment of inertia for the wheels. The radius of the wheel is 4", which is equal to 0.1016 meters, and I'll guess that the mass of the wheel is 2 pounds (probably heavier than the actual wheel), which is equal to 0.91 kg. The wheels are disc shaped, so we can use I = 1/2 * m *r^2 = 0.5(0.91)(.1016)(.1016) = 0.0046968 kg * m^2

One rotation of the wheel causes the robot to travel 8*pi inches = 25.1327 inches = 0.63872 meters/revolution.

3.05 meters/second (divided by) 0.63872 meters/revolution = 4.77517 rev/second = 30.00 radians/second

For the rotational energy, E = 1/2 I*omega^2 = 0.5(0.0046968)(900) = 2.114 Joules per wheel.

For six wheels, that's 12.684 Joules for a robot.

tl;dr, it requires 328.969 Joules to bring your robot to 10 feet per second, and 12.684 of these Joules (3.8%) are used to get your wheels up to speed. This assumes that you have 6 colsons with a diameter of 8" and a mass of 2 pounds.

The fact that the wheels spin when the robot moves makes the robot feel 5.7 pounds heavier to the drive system.

Nicely presented! You might also want to consider the moment of inertia of the CIM motor rotors.

For example (using the motor rotor inertia figure given in another recent thread) if the 8" wheel 6WD robot in your example is driven using CIM motors with 14:1 speed reduction ratio between motors and wheels, each motor rotor will have a "reflected inertia" of J_rotor * (ratio^2) = 0.015 kg-m^2, or about three times the figure you used for one wheel.

For a different example, consider a drivetrain with 4" wheels and half the speed reduction ratio. Will a robot with the same mass get up to speed quicker if it is on smaller wheels?