2583's Octocanum Modules

This year 2583 is using Octocanum for the first time ever

4x1.5" inch Colsons geared to 7 fps and 4" Vex mecanum wheels geared to 20.5 fps (around 15 fps forward after loss of top speed from mecanum)

Gears inside the module get the reduction without any belts or tensioning

One CIM per module
4.1 lbs per module without CIM (6.9 with CIM)

Each side of the octocanum is activated by one 1.0625 bore piston pushing in between them, total of two pistons

Traction wheels deploy on the outside to make it impossible to turn for straight driving in autonomous and to prevent being pushed while shooting

All comments and critiques are appreciated

http://i.imgur.com/oWtz0TIl.jpg
http://i.imgur.com/85rjEakl.jpg
http://i.imgur.com/HfPRTNAl.jpg

Holy image size, Batman!

Is that a mounting hole for a CIM I see there in the side?

yes it is,
a CIM with a 14 T gear mounts to the side there
(fixed the image size)

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.

These are pretty cool. They are also the first cantilevered meccanum wheels I’ve ever seen (not that there’s anything wrong with that).

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?

It pivots around the hole at the top of the module and the standoff on the mecanum side is where the piston will push/pull

I have attached a few screenshots of the side rail including two modules to help visualize it




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

Here are some screenshots of the two positions that the hard stops will create when the piston extends/retracts


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.

I don’t see why. We lift our robot onto traction wheels with two 1.5" cylinders with no problem.

It’s hard to see how much mechanical advantage they have–looks like close to none–but 2 x pi x (1.5 in/2)^2 x 60 psi = 212 lbs; well more than enough.

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).

We’ve cantilevered mecanum wheels a few times.

Whoops, must have missed where they gave the bore, and was going by eye. (And not very well, apparently.)

2 x pi x (1.0625 in/2)^2 x 60 psi = 106 lbs, which is well below comfort without a 2:1 or better mechanical advantage.

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


Try modeling it as a torques around the pivot point.

For extension:

How much torque would the weight of the robot on the mecanum wheels impart?

vs.

How much torque can the force from the piston impart?

For retraction:

How much torque would the weight of the robot on the traction wheels impart?

vs.

How much torque can the force from the piston impart?

Post these same pics with the horizontal and vertical distances between the cylinder point and pivot points noted.

Also x-y distance to wheel shaft from pivot point

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.

Hope that helps.