# Car Differential

1. Jun 14, 2010

### amushrow

I've been having trouble trying to figure out how forces going through an open differential would be resolved. In this case for example:

[PLAIN]http://img708.imageshack.us/img708/9586/exampleone.png [Broken]

If 200 N.m of torque was applied to the prop shaft then each wheel would get 100 N.m, if wheel A resisted this with a maximum of -90 N.m of torque and wheel B with a maximum of -30 N.m then I would expect wheel A to 'lose' 70 N.m of its torque to the other wheel, so that both side of the differential would have the same amount of resistance.

However in this case:

[PLAIN]http://img8.imageshack.us/img8/3879/exampletwox.png [Broken]

50 N.m at the prop shaft gives each wheel 25 N.m, wheel A resists with -25 N.m but wheel B gains some extra force from somewhere and has -40 N.m acting on it. In that situation what forces would we see on wheel A and the prop shaft? And what about situations where there is no force from the prop shaft in the first place and forces are only acting on the wheels?

I've been thinking on this for a while now and have come up with a few solutions but nothing that fits for every case, I feel I'm just missing something.

Last edited by a moderator: May 4, 2017
2. Jun 14, 2010

### rcgldr

A differential normally increases the torque because it has a gear ratio that decreases angular velocity. With an open diffrential, the torque applied to both tires is always the same (igoriring internal firction and inertia within the differential itself). If one of the tires is slippping, then the maximum torque opposing the engine's torque is limited by the forces applied by the tires to the pavement, and the rest of the torque goes into the rate of angular acceleration x angular inertia of all the rotating parts in the drive train (from the engine to the tires).

Any form of limited slip differential will keep both tires angular velocity similar, and limit the amount of torque sent to the tire with faster angular velocity (either from spinning, or being the outside tire in a turn). Unlike an open differential, a limited slip differential can apply a different torque to each axle.

Last edited: Jun 14, 2010
3. Jun 14, 2010

### jack action

The torque on wheel A and B are always equal and it is always equal to half the torque from the ring gear S.

The power must be conserved, so we can find the relationship between the wheels' rpm this way:

$$P_{S}=P_{A}+P_{B}$$

$$T_{S}rpm_{S}=T_{A}rpm_{A}+T_{B}rpm_{B}$$

$$T_{S}rpm_{S}=\frac{T_{S}}{2}rpm_{A}+\frac{T_{S}}{2}rpm_{B}$$

$$2 rpm_{S}=rpm_{A}+rpm_{B}$$

So if one wheel has its rpm = 0, then the other one goes at a rpm that is twice the ring gear's rpm.

So a differential gives a constant torque to an axle and splits the power between the wheels by adjusting the rpm of each wheel.

In opposition, with no differential, the axle gets a constant rpm ($$= rpm_{S}$$) and splits the power between the wheels by adjusting the torque of each wheel ($$T_{S}=T_{A}+T_{B}$$).