pic: Octocanum Module

[wmarshall11]

Originally Posted by GeeTwo

A 1" diameter cylinder has an area of 0.78 square inches. At 60 psi, this would generate over 45 pounds of lift, which sounds about right to authoritatively lift a 140 pound robot.

I'd be a touch hesitant to use the force generated by the piston to say that this module will actuate. The rotation of the module about its pivot axis makes the problem a rotational one, rather than a purely linear F=mg situation.

From my understanding, actuation depends on the torque exerted by the piston at least cancelling the torque exerted by the ground on the dropped wheel at full actuation, as well as all points before that. Ideally, we'd like the torque exerted by the the piston to be far greater than that of the normal force, since the module should actuate with some vigor and keep the robot suspended on the dropped wheels, even if another robot climbs its way on top.

I haven't taken a look at your CAD, but based on some inspection of the image you've posted:

Assuming that your dropped wheel center is 6 inches from the pivot axis, it drops 10 degrees from horizontally aligned with the pivot and your robot weighs 150lb, each dropped wheel exerts a torque of -39lbf. The maximum torque I can coax out of your piston, at about 4 inches from the pivot axis and assuming it exerts all 47.12lb of force perpendicular to the radius drawn between its standoff and the pivot axis is 15.71lbf.

Take it for what you will, but I'd suggest moving the piston further from the pivot or moving the dropped wheel closer. Increasing your piston bore would also be effective, and will provide a more dramatic (quadratic) effect than changing your radii.