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Re: 2583's Octocanum Modules
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.
__________________
--Madison--
...down at the Ozdust!
Like a grand and miraculous spaceship, our planet has sailed through the universe of time. And for a brief moment, we have been among its many passengers.
Last edited by Madison : 04-02-2014 at 19:44.
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