Depending on gear reduction, frame and power (# of motors) you may be able to turn.

Our LogoMotion octanum, with a cim & 775 on each transmission, actually handled quite well in traction mode, even though it was in ‘long’ orientation with traction wheels on the ends.

Cantilevered octacanum is far from impossible, but you’ve got to think seriously about how the module is supported and what moments you are putting on the pivot shaft. How is this attaching to the frame?

I like your single piston switching method. Do you have a hard stop for the modules so that the piston doesn’t take the path of least resistance and kick one module out, leaving the other un-switched?

Its hard to see but if you look at the side view, all the way on the outside there are two 1/2 inch holes and right above those there are two small holes next to the vertical line of three holes.
A bolt/standoff through that hole will act as a hard stop for the module

Have you done the math to make sure your modules will actually shift? We are moving to 4 1.5" diameter cylinders because our 4 1 1/16" were having problems. Our setup is different than yours but it seems like you will have a hard time lifting your whole robot.

The issue is that they are not using a 1.5 inch bore. Two 1.0625 inch bore will probably not be able to lift up the robot (and remember to take into account the added weight from battery and bumpers).

2826 did a cantilevered octocanum in 2011, and their design is on frc-designs. One difference I guess is they had the traction wheel geared for speed and had the Mecanum wheel geared to a lower speed for control purposes.

I went through the calculations just now and it seems like it may not be able to lift the robot so we’re prepared to switch them out to 1.5’’ bore.

I can’t figure out how all the forces are acting but looking at how the weight is acting through the mecanum wheel straight up and how the module is pushing sideways it seems like it may roll the wheels outward to switch wheels instead of just pushing straight down to raise the wheel. Would this help at all?

If someone could help me draw a FBD and figure that out, i’ve highlighted where the piston pushes and the rotation points

It’s a torque thing. Here’s a rough verbal sketch of what to do off the top of my head. I"m sure if I sat down and drew it out, I would find an error or nuance. But, here goes…

Draw a vector from the rotation axis of the module to the point where the cylinder applies force. Then, draw a vector from the point where the cylinder applies the force in the direction of the force with the magnitude of the force from the cylinder. Take the cross product of these two vectors. Or, if we make an assumption of planarity and only need the magnitude of the cross product, then Torque = Force * Radius * sin(theta), where theta is the angle between the vectors. This is the “lifting” torque you get from your cylinder.

To determine the torque you need to lift, do the same math, except draw the vector from the rotation axis of the module to the contact point on the ground of the traction wheel. The force then becomes the amount of weight the wheel supports. This is probably pretty close to 1/4 of total robot weight, assuming you have a center of mass close to the center of the robot. The angle between the vectors changes as well, as the weight vector is straight down.

If the torque necessary to lift is less than the torque provided by the cylinder, you aren’t getting the mecanum wheels off the ground.