Cam- gear motor - force applied to follower?

[Chris D]

Hi Guys,

I have to actuate a lever and I could use pnuematics but not hydraulics. I would prefer not to use pnuematics though - which leaves me with a CAM and gear motor.

Here is what I want to accomplish:

The device being moved requires 600 pounds of force. There is going to be a lever attached to this which will give me a 3:1 advantage. So, I need to have a way of pushing against that lever with 200 pounds of force. As this is all in the design stage, I have some wiggle room on design at this time.

What I don't have a clue about, is determining the amount of torque required from a gear motor to drive a cam, which in turn pushes against that lever.

The diameter of the cam will be about 1" and, being round, I would only have 1 rotation to go from full lift - to full retract. I need to travel about .120" at the CAM side of the lever (resulting in .040" of travel at the final object).

I was thinking if I mounted the cam on it's own axel, then drove the cam with small gear motor, I could get the lift I need. But I have no way of estimating or calculating the torque required to move such a load.

I realize the trade off between speed of acutation and gearing as well. But if I can determine the required torque, I can select a gear motor with enough speed to be acceptable. There isn't much concern at all for duty cycle, this isn't something that get acutated every 10 seconds - rather, more like every couple of minutes or so.

Can anyone help with this?

As I mentioned, I am flexible within limits - this has to fit in a pretty small area which usually makes pnuematics best.

Thanks guys!

Chris

[fkaCarel]

A cam is a rotating lever. With the required speed a small motor/reductor will do. Distance * mass = required energy. The real problems are in the support of the cam-axle and the handling of the push/pull in the drive.

Carel

[Geof]

I don't want to rain on your parade but I think a cam mechanism might prove more difficult to implement than other possible solutions. Carel's comments are cryptic but correct; you can get an estimate of the torque required by considering the distance through which your final force acts, the amount of rotation and the throw of the cam. If you use a 1" diameter cylinder offset by 0.06" to get the travel you need you can just consider the static situation at half travel and treat the cam as a simple lever. However, it is Carel's other comment that is most significant; your will have to get a little gearbox that can take a 200 lb side thrust on the output shaft and that might prove to be a challenge.

There is also another problem and that is control; a little gearmotor unit will coast a considerable distance when the power is turned off and because your cam works by half a rotation giving the travel then the next half the retract you have to stop the motor exactly on the peak of the cam. You can solve this by using a gearmotor with a built in brake but you cannot avoid needing to mount two limit switches and have a bit of logic in your control.

Air cylinders are much simpler; you can just drive them against a stop and leave the air on. I have never heard of an air cylinder overheating when it is stalled. One suggestion I have is that you should consider having a small bore (3/4"?) cylinder with a stroke of 2 or 3 inches pushing a wedge to operate your lever. A wedge with a rise of 0.120" over a travel of 2.4" gives a 20 : 1 advantage so your cylinder force is only 20 lb assuming no friction. You can keep the friction very low if your wedge runs between two fixed ball races and the lever has another ball race rolling on the slope of the wedge.

[fkaCarel]

I always liked these "sledgehammers". Ready with sensor rails. Bolt-on & go.

Carel