cylinder sizing 1

[grunt58]

Need help sizing a cylinder for an application, a little confused on calculating the force needed. We need to rotate a roller conveyor out of the way so a pallet car can pass by, drawing attached. We believe 2 pneumatic 4" bore cylinders with 24" of stroke and around 80psi will work. If any more info is needed let me know.

Thanks

 

[hydtools]

The pull force of the cylinder will be the area of the piston minus the area of the rod times the air pressure.

Ted

 

[grunt58]

Yes, we figure 2 4" bore cylinders at 80 psi will produce 1625lbf. Need help with the necessary torque to rotate.

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[hydtools]

The torque created by the cylinder will be the force times the perpendicular distance from the hinge to the cylinder force line of action.

Ted

 

[dwallace1971]

You might have a very hard time rotating that to 90 degress in the configuration you have shown. As the frame rotates, the moment arm decreased until, near the end of the stroke, it's very small. It would work better if you moved your bracket attached to the frame out and ended the cycle with the cylinder at some angle to the roller assembly.

"On the human scale, the laws of Newtonian Physics are non-negotiable"

 

[dwallace1971]

Additionally, you might add some carefully sized torsion springs at the hinge to assist your piston.

"On the human scale, the laws of Newtonian Physics are non-negotiable"

 

[Dougt115]

Torque is just the weight or force times the perpendicular distance.

Your design is simple in building but complex in analysis.

Replace the bracket that you were attaching the piston to with a Pulley at the pivot point. Fix a cable to the pulley and run it around the pulley a couple of times and then have the actuator pull the cable. (like a garage door system).

This will create a constant moment arm for the actuator and your calculations will be simplified. You will want a damper on this so the unit never slams down.

Note
If you are feeling adventuresome you can change the pulley to a cam so that the torque exerted by the unit 17,700*cos(raised angle) = Actuator force * cam radius. Rearranging this cam radius = 17,700 * cos(raised angle)/(1625*2) where the raised angle goes from 0 to 90.

 

[grunt58]

Thanks all, I was told to check it and detail it. I will bring it up that maybe this is not the best approach.

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Win 7 Pro X64

 

[MintJulep]

You really need to think about how you will control the closing too.

Right not it will go BANG! Very loudly, and probably only once.

 

[hydtools]

Grunt58, best is in the eye of the beholder.

Ted

 

[chicopee]

It would be to your best interest to have a counterbalancing weight nearly equaling the moment exerted by the weight of the bridge. If you were to study lift and bascule bridges, all of these have counterweights which minimize the amount of external forces to lift them. Additionally, think about whether or not you need to rotate this bridge 90degrees, 80degrees, in my mind seems more than sufficient.

 

[dvd]

The counterweight is helpful for part of the motion, after the mass / inertia has been accelerated. With a counterweight there is more mass and inertia to accelerate and decelerate. So if it is a slow cycle with small accelerations, it might be good. You will need to investigate the accel/decel forces with and without counterweight.

Also, 80 psi is getting kind of high. Most plants I have worked around want 60 psi as the maximum design pressure.

Agree with Mint Julep on the lowering. Think about this one.

 

[desertfox]

Hi grant58

To Orkney the torque out properly you need to show the drawbridge at different angles with the cylindr connected.
At each position the vertical weight of the drawbridge multiplied by its horizontal distance to the hinge which gives the torque the cylinders must exert about the same hinge, so in order to get that you need to find the line distance perpendicular from the hinge to the cylinder line of action and multiply this by the force available in the cylinder.
Again you need to layout the drawbridge and cylinder in different positions because as the drawbridge rises the torque around the hinge decreases but your cylinder mechanical advantage also decreases as the drawbridge comes up and as mentioned by others there might not be enough advantage for the cylinder to do its job.

 

[LittleInch]

I agree with all the others - this unit will simply not lift to 90 degrees as shown. If someone pushed it the last few degrees to fully open , how do you close this thing in a controlled manner?

A counter balance with perhaps a pneumatic worm gear driving a quarter circle gear would seem to be far mor controllable - provide smooth constant speed rotation and be able to inch the last few degrees to stop it clattering into things at the end of the action.

There's a reason that all lift bridges like this use counter weights.

A complete re-design me thinks....

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[hydtools]

Meter-out control for the cylinder in the pneumatic circuit will control the descent of the drawbridge to as fast or slow as desired.
If the drawbridge is raised to less than 90 degrees or the cg is kept to the left of the hinge, gravity will return the drawbridge. The rate of descent can be controlled with a meter-out control on the cylinder exhaust cycle.

Ted