Torque Required for an hydraulic motor

[SkyD]

Hi guys,

One of the designers in my office has asked me to calculate the torque required to turn a tool around. I must admit I'm a bit rusty when it comes to general mechanics like that...

Basically the tool has a prism shape with a horizontal axis of rotation passing through its CG when empty. The tool is loaded by adding the Carbon fibre laminate at the top, creating an imbalance... The whole lot is then rotated so that the laminate is at the bottom...

Are you lost yet?!!!

What I need to calculate is the torque required to turn that thing... Any idea?

Thanks for your help!

Cheers,
SkyD.

 

[MintJulep]

F = ma, or its rotational equivalent, T = I omega

 

[Barry1961]

The amount of torque to spin the tool unloaded will depend on the bearings the tool is mounted on. There will be a certain amount of seal drag and torque required to stir the grease. This can be significant.

If the imbalanced load has to overcome gravity you multiply the distance from the pivot by the weight of the imbalance and add it to the drag above.

The amount of torque for acceleration in ft-lbs is equal to the moment of inertia in ft-lb-sec^2 multiplied by the acceleration in radian per sec^2.

Barry1961

 

[SkyD]

Thanks Guys!

I thought about F=ma obviously or the equivalent... but my reasonning was this: if the speed is constant there is no torque applied... That's obviously in theory with no friction... But obviously if you've ever tried to turn a 4tons tool, you'll know that you can't do it by hand like that...

I guess the seal and grease drag can explain why on it's own. I just thought I'd check cause I was not too sure it was reasonable. Then there is the acceleration.

Alright, I know where I'm goin now!

Cheers guys!

SkyD.

 

[MintJulep]

Speed isn't constant.

You start in one position. Move to another.

At 4 tons it doesn't take much acceleration to make a bunch of force (or torque).

 

[Barry1961]

A rule of thumb for static friction that seems common for wheels with ball or cylindrical roller bearings is a .02 cf. If you are going to be using tapered roller, angular contact or some other type of pre-loaded bearing the coefficient may be significantly higher.

Often when dealing with a large load the drive ends up being sized on how quickly you want to stop when E-Stop is pressed. Stopping a four ton flywheel quickly can get expensive even with most of the mass close to the center.

Barry1961

 

[SkyD]

Thanks guys, all this is very useful! I think I should be able to sort it out now...

your help is much appreciated!

Cheers,
SkyD.