Torque Calculation

Hi,

Ok, so I have a cylinder which rotates about its centre axis. The amount of torque required to turn the cylinder through its axis is 26.5 KN/m. In order to reduce this torque I'm fitting a slew ring and using a pinion to rotate. The distance from the cylinder axis to the centre of the pinion gear is 400mm (which is now where the torque will be applied).

Forgetting the slew ring for a minute. If I was to apply a torque 400mm from the centre of the cylinder what would be the required amount to turn the cylinder? Would this be correct:

26,500 x 0.4 = 10,600 N/m of torque required to rotate cylinder about its axis (applied torque 400mm from axis/pivot)?

Incorporating the slew ring and pinion with a gear ratio of 9.86:1....

10,600 / 9.86 = 1,075 N/m torque requirement to turn pinion gear?

I know we could go a step further and include tooth pitch, tooth length etc, but at this stage I just want to get rough calculations.

Please see attached for clarity.

Thanks

[COLOR=#333333]Torque units should be given as N-m or kN - mm (Force Applied - length) not N/m or Force per / length.[/COLOR]

[COLOR=#333333]So Force applied ( N ) x 400 mm = N - 400[/COLOR]

Thanks Kelly,

I understand. So - not per (/). Newton Meters is newton meters as far as I'm concerned. As long as the right units go into the right equations as they should then we're all good.

But hopefully you get the gist of what I'm asking?
Forgive me, but I don't understand your answer, can you explain? Does it relate to the question I asked?

Thanks again

Torque is a product of a force times a distance. Not a quotient. In your case, kN x m.

Your drawing shows a gear ratio of 148:15 or the 9.86 reduction you refer to in the problem. The torque on the pinion to rotate the internal gear would be 26.5/9.86 kN x m, or roughly 2.7 kN x m. (1991 ft lbs)

This answer does not account for friction in the gear train. It does not account for start-up torque – you need to supply torque to accelerate the cylinder from rest. Also neglected is any start up friction. Sometimes it takes a bit of extra force to overcome friction at rest compared to friction in motion.

So now you have the pinion torque you need a speed to calculate the size of the motor. You will "oversize" the motor a bit to handle the extras mentioned above.

[QUOTE=james_cliff;14369]Thanks Kelly,

I understand. So - not per (/). Newton Meters is newton meters as far as I'm concerned. As long as the right units go into the right equations as they should then we're all good.

But hopefully you get the gist of what I'm asking?
Forgive me, but I don't understand your answer, can you explain? Does it relate to the question I asked?

Thanks again[/QUOTE]


/ = divided by

- = multiply by

so

N / m = newtons divided by meters

N - m = Newtons times meters.

If one asks a math question they should express the math units correctly to avoid confusion..

Thanks for the replies.

So I would be wrong in the calculation I had done.
Knowing I need 26,500 N-m to rotate the cylinder, it was a simple case of multiplying this by the gear ratio to tell me roughly I need 2,700 N-m of torque through the pinion? This pinion will sit directly on the output shaft of a hydraulic motor. So I'll probably factor in 1.5 for start-up and frictional losses and look at actuators with around 4,000 N-m.

Out of curiosity..... imagine I welded a pin onto the cylinder 0.4m from the central axis / pivot. I then applied a 1m long lever to this point (1.4m total distance from fulcrum). Would I be right in saying I need 19,643 Newtons of force on the end of this 1m lever to rotate the cylinder through its axis (27,500 / 1.4)?

Thanks,

Don't let the welded pin and arm fool you. The length of a handle doesn't matter. Torque is the product of the distance from the center of rotation to the point of application of force, regardless of how many "fixed" or welded components are involved.

I will be using the slew ring and pinion, and accept Hudson's answer. I also understand what your saying Jboggs.

But just to satisfy my curiosity....

26.5 KN-m to rotate at axis required. If, theoretically if I was to weld a nut 0.4m outwards from the axis of rotation, and then put a torque wrench on that nut and apply a force. Surely I would be applying a torque to the nut? Even though the nut is 0.4m from the axis of rotation, if it was subjected to a certain torque, the cylinder would rotate around its axis. This torque would be considerably less than the amount needed to rotate directly on the axis? Forgetting about the gearing for a minute, what would be the correct way to calculate the required torque on the nut in this scenario, knowing we require 26.5KN-m directly on the axis and have a lever of 0.4m?

Reading JBoggs post again, I'm beginning to think I might be way off the mark here. Would I be mistakenly thinking there would be an application of torque on the nut, when actually only a force can be applied to it?

FYI, there are a bunch of gear ratio equations and calculators on this website.

See [U][B][URL="https://www.engineersedge.com/gears/planetary_epicyclic_gear_13736.htm"]Gear Ratio Equations and Calculator[/URL][/B][/U]

Thanks for the vote of confidence.

If you weld your nut .4m from the center, you can have a .4m increase in the lever arm so the force applied to the lever will be less. The product of that lower force and the longer lever length will still be 26.5 kN. Switch the lever around so that it passes over the center and you have effectively shortened the lever arm. In that case a higher force will be required but again the higher force times the shorter lever arm will be 26.5 kN x m.

So far we have been discussing a lever arm that is on the radial line from the center of rotation to the welded nut. If you apply the same length lever to the nut at right angles to that line, the required force parallel to the radial line from the center to the nut is the same as if it were on the center of rotation. It is always the distance of the path of that force from the point of rotation times the force equals 26.5 Kn x m.

Thanks for the help all.
Much appreciated.
Jim