2583's Octocanum Modules

[Electronica1]

Originally Posted by Tom Ore

We've cantilevered mecanum wheels a few times.

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.

[ferrari77]

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

[Ty Tremblay]

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?

[AdamHeard]

Originally Posted by ferrari77

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

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

[jee7s]

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.

[lcoreyl]

make sure to look at the geometry when the colson is just coming off the ground. The pic you posted is looking at after the module has been shifted, and the angles are different.

Also, unless I'm missing something here, the traction position is the more problematic one. the cylinders are also fighting gravity with the colsons down, but now they are "pulling" so they have less force. Also, it appears the geometry here is worse for the cylinders.

Lastly, how did you calculate your top speed reduction?

[ferrari77]

Here are the screenshots of the dimensions in both positions







I used 150 lbs as the robot weight so that gave me 75 lbs on each side and using 1.0625 bore i got (pi*(1.0625/2)^2*60) i got 53 lbs of pushing force for each side, (106 lbs total and 150 lbs of robot weight)

In the first orientation the calculation is 1.3in*75lbs=97.5lbft of torque from the wheel and 3.21in*53lbs=170lbft of torque from the cylinder which provides almost double the force through the cylinder

In the second orientation the calculation is 1.54in*75=115.5lbft of torque from the wheel and 4.1in*53lbs=217.3lbft of torque from the cylinder which provides almost double the force

The mistake i might have made is, because we are using the same cylinder to push two of the modules is the force cut in half so there would actually be 85 and 108lbft of torque, respectively? This would not give us enough torque and would be problematic

The top speed was calculated using a CIM speed of 5300 and a 90% efficiency. It is around 7 fps for the colsons and 20 for the mecanums

[Ty Tremblay]

In a perfect world, the modules would move at the same time and 50% of the force would go to each module. However, if you don't have even weight distribution front-to-back, you may see one module move before the other. In this case, you'd have >50% of the force going into each module.

Having not seen the entire CAD of your robot, you may not have room for this. You could use one piston per module, and have one end of each piston on a pivot on your frame (Show in yellow). This way the pistons would have a much larger lever arm to work with.

[AdamHeard]

Originally Posted by ferrari77

Here are the screenshots of the dimensions in both positions







I used 150 lbs as the robot weight so that gave me 75 lbs on each side and using 1.0625 bore i got (pi*(1.0625/2)^2*60) i got 53 lbs of pushing force for each side, (106 lbs total and 150 lbs of robot weight)

In the first orientation the calculation is 1.3in*75lbs=97.5lbft of torque from the wheel and 3.21in*53lbs=170lbft of torque from the cylinder which provides almost double the force through the cylinder

In the second orientation the calculation is 1.54in*75=115.5lbft of torque from the wheel and 4.1in*53lbs=217.3lbft of torque from the cylinder which provides almost double the force

The mistake i might have made is, because we are using the same cylinder to push two of the modules is the force cut in half so there would actually be 85 and 108lbft of torque, respectively? This would not give us enough torque and would be problematic

The top speed was calculated using a CIM speed of 5300 and a 90% efficiency. It is around 7 fps for the colsons and 20 for the mecanums

I can't actually read the numbers in the pictures.

[Madison]

The images you've attached are at too low a resolution for me to be able to make out the dimensions.

First, some assumptions:

1) Assume a robot weight of 150 lbs. to account for bumpers and battery.

2) Assume that your weight is equally distributed over all four wheel pods as a simple first pass. That's 37.5 lbs. of weight acting through each wheel pod given the previous assumption.

To calculate whether your cylinders have enough force to keep the mecanum wheels off the ground --

1) Calculate the torque the robot weight acting through the wheel exerts around the module pivot point.

This is weight per wheel (37.5 lbs) x distance from pivot (1.3", I think) -- 48.75 in-lbsf.

2) Calculate the total force exerted by the cylinder.

3.14 * (1.063 / 2) ^2 = .89 in^2

Subtract the area of the cylinder rod --
3.14 * (.3125 / 2) ^2 = .077 in^2

Final area - .813 in^2

Build in a safety buffer for your air pressure.
.813 in^2 x 45 psi = 36.56 lbsf.

Divide in half because the cylinder acts on two modules.
18.29 lbsf.

3) Calculate the torque the cylinder exerts about the module pivot.
18.29 lbsf x 3.21" = 58.72 in*lbsf.

58.72 > 48.75, so you're fine, but cutting things a bit close. You've built in a buffer for your air pressure here, but a wildly uneven weight distribution could make it so this won't work.

Now, when switching to mecanum wheels, gravity will do all the work of getting the mecanum wheel to the ground and the cylinder takes over the instant the traction wheels comes off the floor.

The images you've provided don't show that state and it is less favorable than what you have shown. You can repeat the same calculations for that state, though, to determine if you'll have enough force to lift the robot onto the mecanum wheels. If you have enough to lift (when both wheels are touching the ground), you'll have enough force available to hold it in the position you've shown.

Also -- check your units. You're multiplying inches and pounds-force and labeling the results as foot-pounds. You'll also want to revisit your speed calculations. 5300 RPM is the CIM motor's free speed and it operates at a much lower speed under load.

Edited to add:
Of course, the less favorable geometry exists again when switching from mecanum wheels back to the traction wheels. You should calculate the applied torque in that condition to make sure the retracting cylinder can exert an appropriate amount of force. Given how close things were in the fully switched state, as demonstrated above, you may find that you're sitting on the razor's edge.

[Team 2016]

Interesting concept.

[amreuland]

Here is a video of the Working System

http://www.youtube.com/watch?v=fL7d_D83i48

[ferrari77]

Some additional views of the working system

http://youtu.be/TvRj_4lFalo