Why does torque change from flywheel to drive shaft?

[thecombover]

Some buddies and I were discussing cars and their torque. We were wondering why the amount of toruqe that is exerted by the flywheel is not the same amount of torque at the end of the drive shaft. Does it have something to do with the longer the drive shaft the less torque? or does it deal with the extra mass added by the extra length? If anybody knows it would be great to find out thanks!

 

[daveyman]

I don't know much about cars, but I do know something about torque.

The equation for torque is $$\tau = rF\sin{\theta}$$.

The variable r represents the distance from the center of rotation to the point that is rotating. The variable F is the force vector of the point (mass times acceleration in a particular direction). The sine term represents the angle between the axis and the direction that the point is rotating. In this case, this angle is 90 degrees. Since the sine of 90 degrees is 1, all we have to worry about is r and F.

Now if you imagine a point on the outer edge of the flywheel, you could determine r (the distance from the center) and F (the mass of the "point" multiplied by its acceleration).

If you imagine a point on the outer edge of the drive shaft, you would get different values for r and F. So, according to the definition of torque, you would get a different value.

Although the torque is not the same, power must be (nearly) the same or we would have a violation of conservation of energy. So now we turn to the definition of power (in terms of torque).

$$Power=Torque\times 2\pi \times Rotational Speed$$

So whichever object (flywheel or drive shaft) has less torque, it must also have more rotational speed. This means both objects are capable of delivering the same amount of power and no conservation laws are violated (whew!).

 

[Phrak]

Some buddies and I were discussing cars and their torque. We were wondering why the amount of toruqe that is exerted by the flywheel is not the same amount of torque at the end of the drive shaft. Does it have something to do with the longer the drive shaft the less torque? or does it deal with the extra mass added by the extra length? If anybody knows it would be great to find out thanks!

The flywheel is mounted to the engine crank. Between the crank and the drive shaft is the transmission. The transmission changes the torque developed on the crank and applies it to the drive shaft.

 

[Naty1]

As Phrak post implied, transmissions change gear ratios and those transform torque relationships. Its effectively the same idea as using a long breaking bar on a socket when trying to break loose a tight nut: a longer bar permits more torque but you have to apply more motion to the end to get it. An engine has little torque at low RPM and so high gear ratios enable the engine to rev up more easily to an RPM range where more torque is produced.

HP = (torque x RPM)/5252

Derivation and good discussion at:
http://www.vettenet.org/torquehp.html