[Ether]

Originally Posted by Ether

A more likely cause is that the rectangular bot had poorer weight distribution or the frame wasn't flexible enough to keep traction on all 4 wheels or the rollers were binding. There's no kinematic reason why mec needs to be square in order to rotate properly.

Originally Posted by efoote868

Quickly sketching the vectors on a napkin, it looks like to me that wheels in a square will optimize spinning (maximize total torque on robot) for a particular wheel distance d from center of robot; perhaps you could comment?

Let's consider the ideal case first, in order to understand the physics: mec wheels with no roller friction and no roller axial or radial free play; floor surface is non-compliant (ignore carpet compression and stretching).

Refer to attached sketch


First, the kinematics (ideal):

Let P be the perimeter of the rectangle formed by the centers of the 4 wheels:

P = 2*(Lwb+Ltw);

Let r be the wheel radius

For a given vehicle rotation speed omega_v (radians/sec) of a vehicle rotating in-place about the center of the aforementioned rectangle (i.e. Vx=0 and Vy=0), the wheel rotational speed omega_w (rad/sec) will be given by

omega_w = (1/r)*K*omega_v .... (see page 7 of my mec kinematics paper)

... where K = (Lwb+Ltw)/2 = P/4;

solving for omega_v and substituting for K:

omega_v = (1/K)*r*omega_w = (4/P)*r*omega_w;

So you can see that the vehicle rotation speed for a given wheel rotational speed, is the same for all rectangles with the same perimeter.




Now, the forces and torques (ideal):

wheel torque: tau;

carpet force component in plane of wheel and floor: Ff = tau/r;

total carpet force in direction of mec roller axis: Fr = Ff*sqrt(2) = (tau/r)*sqrt(2);

Let the ratio f be defined as: f = Ltw/Lwb;

theta = atan(f); .... (see sketch)

alpha = pi/4 - theta;

carpet force component in direction of vehicle rotation: Fv = Fr*cos(alpha) = (tau/r)*sqrt(2)*cos(alpha);

distance from center of rectangle to center of wheel: D = (1/2)*sqrt(Lwb^2+Ltw^2);

P = 2*(Ltw+Lwb) = 2*Lwb*(f+1) => Lwb = (1/2)*(P/(1+f)) & Ltw = f*(1/2)*(P/(1+f));

so D becomes:

D = (1/2)*sqrt(((1/2)*(P/(1+f)))^2+(f*(1/2)*(P/(1+f)))^2);

D = P*sqrt(1+f^2)/(4*(1+f));

torque about center of rectangle: Tv = Fv*D;

Tv = ((tau*sqrt(2))/r)*cos(alpha)*(P*sqrt(1+f^2)/(4*(1+f)));

Tv = ((tau*sqrt(2))/r)*cos(pi/4 - theta)*(P*sqrt(1+f^2)/(4*(1+f)));

Tv = ((tau*sqrt(2))/r)*cos(pi/4 - (atan(f)))*(P*sqrt(1+f^2)/(4*(1+f)));

using:

cos(pi/4-atan(f)) = (1+f)/(sqrt(2)*sqrt(1+f^2));

... Tv simplifies to:

Tv = ((tau*sqrt(2))/r)*((1+f)/(sqrt(2)*sqrt(1+f^2)))*(P*sqrt(1+f^2)/(4*(1+f)));

Tv = (P*tau)/(4*r);

Now that I've done all that trig and algebra, here's a much quicker way:

carpet force component in plane of wheel and floor: Ff = tau/r;

total carpet force in direction of mec roller axis: Fr = Ff*sqrt(2) = (tau/r)*sqrt(2);

carpet force component in direction of wheel axis: Fa = tau/r;

torque around center of rectangle:

Tv = (tau/r)*(Ltw/2) + (tau/r)*(Lwb/w) = (tau/r)*(Ltw+Lwb)/2 = (tau/r)*(P/4);

So the torque on the vehicle is the same for all rectangles with the same perimeter.




Now consider the non-ideal case with roller free play, roller friction, and carpet compression and stretching.

This will be an intuitive explanation.

In the case where Ltw >> Lwb (very wide configuration), the fore/aft component of the wheel translational motion is much larger than the strafing component.

In the case where Lwb >> Ltw (very long narrow configuration), the strafing component of the wheel translational motion is much larger than the fore/aft component.

Since a non-ideal mec wheel is considerably less efficient in the strafing direction, you'll see more losses when trying to rotate when Lwb >> Ltw.


Bottom line:

For the small deviations from square typically seen with FRC mec bots, and for properly functioning mec wheels on a frame that's sufficiently flexible to maintain traction on all 4 wheels, the bot should turn fine - and certainly better than a skid-steer with the same dimensions.