[dellagd]

Originally Posted by Ether

Yes, trackwidth is the distance between the 2 front wheels. (Or the 2 rear wheels). Here's an easy way to remember: It's the width of the "track" left by the vehicle when traveling forward in a straight line.

Ah, well even if I did find out the dimensions correctly I must have set up the proportion between the carpets velocity and the robot's orientation inversely. The solution below works out, so unless a mathematical equality that makes this give the right answer for the wrong reason is working here and I haven't realized what it is, I must have accidentally flipped that value somehow. It was late, I probably used the wrong angle for being tangential to the robots rotation which lead me to the wrong proportion.

Updated solution:

Spoiler for Solution:

2 + Vcy = Fy
Vcx = Fx
Fy/Fx = -2 / (28/12) #Known angle of vector sum/friction
Vcx/Vcy = 2 / (28/12) #Known angle of carpet velocity

Substitute in for Vcx and Fx:

2 + Vcy = Fy
(2 / (28/12)) * Vcy = (-(28/12) / 2) * Fy

Substitute in for Fy:

2 + Vcy = ((-2^2)/((28/12)^2)) * Vcy
Vcy = -1.153

Into another equation:

Vcx/(-1.153) = 2 / (28/12)
Vcx = -.988

Robot Velocity components are opposite of these.

Vry = 1.153
Vrx = .988

Vr = sqrt(1.153^2 + .988^2) = 1.519 ft/s

Vtangential = r * omega

r = d/2 = sqrt(2^2 + (28/12)^2)/2 = 1.537 ft

omega = Vt / r = 1.519 / 1.537 = 84/85 = .988 rad/s


I confirmed this using the other method and my answer was the same down to 6 decimal places.