Is the relationship between power, torque and RPM a linear relationship?

[Stephen123]

Hi I have a problem dealing with rpm power and torque.

I have a motor powering a gearbox which has a 20:1 ratio.

The power and rpm are not an issue so i have not looked at those figures but the motor produces 240Nm of torque. Since it is a 20:1 gearbox, i have assumed the torque is 20 times greater with the rpm being 20 times slower? Is this correct?

From there the power is transferred through a number of gears. Can I follow the same method above and assume the torque multiplies each time buy the gear ratio?

I.e. the gearbox output shaft is 110mm and connected to a 600mm gear, therefore is the torque at the outer surface of the gear is equal to (240nm*20)*(600/110) = 26181.8Nm of torque?

Thank you.

 

[A.T.]

I.e. the gearbox shaft is connected to a 110mm shaft connected to a 600mm gear,

Did you mean "a 110mm gear is on the gearbox shaft"? It's best to post a diagram, or be more clear what gears are on the same shaft vs. meshed with their teeth.

therefore is the torque at the outer surface of the gear

A torque not "at the outer surface of the gear" but around its axis.

 

[Stephen123]

Did you mean "a 110mm gear is on the gearbox shaft"? It's best to post a diagram, or be more clear what gears are on the same shaft vs. meshed with their teeth.A torque not "at the outer surface of the gear" but around its axis.

Sorry typo, the 600mm gear is on the 110mm shaft. My thought is that if the gear was essentially a wheel, would the torque applied by the 600 diameter be a ratio of 600mm/110mm?

Therefore the torque at the 110mm shaft multiplied by the size ratio would give the torque at the outside of the 600mm?

 

[jbriggs444]

Therefore the torque at the 110mm shaft multiplied by the size ratio would give the torque at the outside of the 600mm?

For a given torque on a shaft, the required force to exactly counterbalance that torque would be different if applied on the outside of the shaft or the outside of the gear. The ratio of the required forces would indeed be the ratio of the radii. 600mm/110mm as in your example. If 600 N of force on the outside of a shaft could keep it from turning then 110 N of force on the teeth of the gear welded to the shaft could do the same thing.

 

[CWatters]

Hi I have a problem dealing with rpm power and torque.

I have a motor powering a gearbox which has a 20:1 ratio.

The power and rpm are not an issue so i have not looked at those figures but the motor produces 240Nm of torque. Since it is a 20:1 gearbox, i have assumed the torque is 20 times greater with the rpm being 20 times slower? Is this correct?

Yes.

From there the power is transferred through a number of gears. Can I follow the same method above and assume the torque multiplies each time buy the gear ratio?

Yes.

I.e. the gearbox output shaft is 110mm and connected to a 600mm gear, therefore is the torque at the outer surface of the gear is equal to (240nm*20)*(600/110) = 26181.8Nm of torque?

No. Both the 110mm and 600mm gear turn at the same rpm and have the same torque. As AT said, torques acts "around the axis" not at a distance from the axis.

 

[sophiecentaur]

By definition:
Power = Torque X rotational speed (rotational speed is specifically Angular Velocity)
So the relationship is Linear. But very few motors will produce constant torque over all speeds so you need to be careful using that equation.
But you can say that
Gear Radius₁ X RPM₁ = Gear Radius₂ X RPM₂
and
Torque₁ / Radius₁ = Torque₂ / radius₂
and many other combinations.

 

[A.T.]

Sorry typo, the 600mm gear is on the 110mm shaft.

If I picture this correctly, then the shaft radius is irrelevant here. You know the torque applied to it (240Nm*20 ignoring losses), so the force at the 600mm gear's circumference is 240Nm*20/0.6m.