Actuated Drive (Slide / H)

[GeeTwo]

Originally Posted by Owen Busler

Yes thank you for catching that. When we had a stack the omni would slip on the carpet, which is why we switched it out for a traction wheel.

These are my basic calculations:

1.25 inch cylinder
.4 inch in diameter rod inside cylinder
60 psi

(1.25)(pi)= 3.92 square inches
3.92-(.4*pi)=2.67 net square inches
2.67*60=160 lbs

The pivot point was equidistant from where the piston was mounted and the center of the wheel.

The area of a circle is πr², not 2πr. This makes the area π(1.25² - 0.4²)/4 = 1.1 in² net, or 66 pounds. If you'd pushed down with 160 pounds, the robot would have been lifted off of half of the edge wheels, and you should have had enough traction even with a tall stack of totes.

As I understand this both from your last sentence and the subtraction of the shaft, you had the cylinder pulling in order to push the wheel into the carpet; is this correct? Even though the difference isn't usually that large, we've been intentionally trying to load our pneumatics so that the greatest needed force is delivered when the shaft is moving outward; if we'd done that with our arms in 2014, we probably would have saved ourselves some headaches at competition.

[Owen Busler]

Geez i must be out of it today. Thanks for correcting that. It did seem way too high.

But yes what you have stated is correct.