I'm having a hard time visualising how the units of torque are

[Jones1987]

I'm having a hard time visualising how the units of torque are measured. I can picture how quick m/s or RPM is. But when it comes to torque, I cannot picture what 'x' amount of N/m does.

Example, for a motor with 3.3 Lb-Inch or 7N/cm. Does it move 3.3lb of weight per inch? Does that mean anything higher and the motor begins to struggle? slip?

Or is it purely the amount of force it exerts per distance? 7N of force per cm travelled?

 

[xxChrisxx]

You may be having trouble becuase it's not N/m it's Nm. Force x distance. It's how much 'twist' you have abount a certain point.

The best way to imagine it is one of those weight balance things where you hang a weight at a distance. This imparts a torque about the fulcrum.

An let's say you attached a motor that produced 10Nm torque to this balance and hung a weight equalling 10N force at 1m. The motor could rotate the balance. If you hung 20N at this distance the motor wouldn't be able to rotate the balance. Or if you hung 10N at 2m you wouldn't be able to rotate it. As in both cases.

20N x 1m = 20Nm
10N x 2m = 20Nm

 

[Jones1987]

Thanks Chris, I follow what your saying.

So for example if I had a motor which could exert a torque of 30Nm (I think its roughly 3kg m), the motor would be able to move a mass of 3kg at a steady pace every 1 meter, any higher amount of mass the motor will become less responsive? Am I thinking along the correct lines?

 

[brewnog]

No, it's not a rate thing. It's how hard your motor can twist its shaft.

Fasten a 20m diameter wheel (10m radius) to your motor, and dangle a weight applying a 3N force off the wheel rim. When the weight is moving upwards vertically, your motor requires a little over 30Nm.

 

[paisiello]

Thanks Chris, I follow what your saying.

So for example if I had a motor which could exert a torque of 30Nm (I think its roughly 3kg m), the motor would be able to move a mass of 3kg at a steady pace every 1 meter, any higher amount of mass the motor will become less responsive? Am I thinking along the correct lines?

Part of the important distinction is that a torque is a force applied perpendicular to the direction of movement you are referring to. Your example has the force applied parrallell to the direction of movement. The perpendicular distance is the distance to an axis of rotation which can be any arbitrary axis you choose (of course the math often simplifies if you pick the right axis).

 

[middlj]

See attatched

 

[Herr-Kuhn]

Very simply torque is nothing more than a twisting force. You can have a torque or moment with no shaft displacement (RPM). Torque is calculated by taking the force*distance that force is acting from the center of rotation.

In terms of rotational power on a shaft, the constituents of power are torque and speed. Long ago it was determined that 550 ft-lbf/sec was one HP and it was measured by the amount of time it took a horse to lift a weight a certain distance. Thus the term "horsepower"

To get power you simply take the product of speed and torque. For HP you can take ft-lbs and RPM and get power through the equation. P=(TorqueXSpeed)/5252. The 5252 takes into consideration that one HP is 550ft-lbf/sec and as well the fact that there are 2pi radians per revolution. It's always best to go back to the basics and do the unit conversion yourself instead of using cook book formulas like the one I gave above.

Think of it this way...if you exert a force on a breaker bar and it is attached to a bolt, you have induced a torque...but there is no power (unless you are spinning the bolt at some speed). Let's say that torque is 300 ft-lbs (you'd have to be very strong to exert that much or on the end of a very long bar). If you had that same torque on the crankshaft of an engine and let's say that it's a really good engine and can turn 12,000 RPMs and the torque at 12,000 RPM is 300 ft-lbs, then that engine's output is: (12,000*300)/5252=685.5 HP. Step back down to 6,000 RPM and 250 ft-lbs and you see the power is much less at 285 HP, even though the torque there is just 50 ft-lbs less than it is at 12,000 rpm. This is how Formula 1 engines which are so small can make so much HP...it's all on the top end and done through moderate torque and high RPM. ON the flip side, when Audi went into diesel for the R10 and the like, the diesels made the same or more HP, but at a MUCH lower RPM...therefore they had to make the components heavier to handle the torques on them.

In the automotive after market there is a lot of arguing over which is better, torque or power. These statements bother me because the people who make them fail to realize that you can't make power without shaft speed AND torque on the shaft. The two are completely inter-related to the question. It's about power band width in the real world and the setup with earlier and broader torque will have a better overall power curve as compared to one with the lower and less broad torque curve. Also, it's typically the torque that breaks parts on a driveline, not the RPM.

Hope this helps to visualize the relationship between torque, speed and power.