# Automotive Differentials

1. Dec 31, 2011

### bugatti79

Folks,

Is it true to say for standard "open" differentials that when a car is turning a corner that the torque is equal on both wheels but the power will be different based on the expression P=TW, assuming that there is sufficient traction for all wheels, ie no slip?

thanks

2. Dec 31, 2011

### mender

Yes; because of the way open diffs are constructed, the torque to each wheel will always be the same, and because of the difference in distanced traveled by each wheel, the power delivered will be different.

This also applies to slip conditions.

3. Jan 3, 2012

### bugatti79

Thanks. For limited slip differentials, the torque is the same on each wheel in tractive conditions but in slippery conditions torque can vary on each wheel due to torque biasing right?

4. Jan 3, 2012

### mender

Torque biasing is kind of misleading when used to describe how a limited slip or locked diff works; that implies an active control of the torque according to the traction available when in actuality it is a wheel speed control.

All the LSD does is keep the wheel with no traction from moving faster than the one with traction by locking them together either with a ratcheting or a friction-based (worm gear or clutch) system. That has the benefit of not reducing the torque being applied to the wheel with traction to equal that of the wheel with less traction, as happens with an open diff as we already agreed.

Although some of the means of operation of the LSD can involve torque to actuate the locking mechanism, the main design objective is to keep both wheels rotating at the same speed. The different amount of torque being applied at the road surface is because of the differing traction available, not a torque biasing system per se.

Electronic traction control systems apply the appropriate brake to slow a spinning wheel and again synchronize wheel speeds but the result is essentially the same.

Last edited: Jan 3, 2012
5. Jan 3, 2012

### bugatti79

Ok just to clarify I have rephrased the above above based on your reply.

What I find a bit misleading and I believe some other people do too is when they refer to 'power' instead of 'torque' when describing differentials. Part of me is telling me that we can describe differential behaviour using either 'power' or torque because they are related by P=TW..but I dont think that is right...?
You could argue that if a wheel is spinning that it has some torque and also some power and so its suffice to use either for descriptive purposes etc...? Whats your view?

Thanks

6. Jan 3, 2012

### mender

Proper usage of terms is important when discussing things exactly; unfortunately that doesn't happen much outside of forums like this!

Without having an example of the verbiage being used and the context it's hard to say what the best terms to use are but I'm sure you've noticed that sales brochures are more about selling than telling!

7. Jan 3, 2012

### Dotini

One seldom mentioned effect of LSD (in rear wheel drive) is that understeer is generally promoted. This is good if the vehicle in question oversteers, but bad if it already understeers.

Respectfully submitted,
Steve

8. Jan 3, 2012

### bugatti79

I like to think of torque which provides the movement of a component while the definition of power I use to describe the time element in keeping a component moving at a certain speed.

Just my 2 cents :-)

Interesting

9. Jan 3, 2012

### Dotini

The extent to which the LSD locks the wheels together is a major factor in the wheels wanting to turn a corner. If the LSD is 100% locked up and the axle is effectively solid, the wheels will want to run straight ahead and resist turning.

Respectfully,
Steve

10. Jan 4, 2012

### bugatti79

Can you explain why is it not technically correct to say that control of the 'torque' is the same as control of the wheel speed? If one controls the torque of a wheel then this automatically controls the wheel speed..?

Under what conditions? The wheel speeds will be the same on a straight road and hence the torque assuming traction is the same on all wheels and this is equivalent to an open differential right?

Under what conditions do you want wheel speeds to be the same in a LSD..?

11. Jan 4, 2012

### mender

Other way around; if one controls the wheel speed it will indirectly control the torque output to each wheel. Controlling the wheel speed is much simpler and cheaper than controlling the torque bias so that's the usual method.

Ideally the wheel speeds should only be the same when traveling in a straight line, and would differ as the vehicle goes around a corner. However, controlling the wheel speed to allow for the proper difference in distance traveled during the corner while also compensating for differing traction is complicated, so the simpler control scheme of locking the two wheels together is the usual solution.

Here's an actual torque splitting/biasing differential:
http://en.wikipedia.org/wiki/Torsen

12. Jan 4, 2012

### bugatti79

Assuming there is tendency for either wheel to slip on soft ground on a corner. Your comment above wouldnt apply in the case of sufficient traction on both wheel while negotiating a corner..right?

thanks

13. Jan 4, 2012

### mender

Right; for that, an open diff is better. When cornering (once the corner is initiated), having the wheels locked can increase the likelihood of oversteer by forcing one wheel to slip, reducing the amount of lateral traction available as a result.

That's why the brake/traction control system has caught on, it emulates an LSD on demand but doesn't interfere with normal differential action when not needed.

14. Jan 4, 2012

### Tea Jay

I see most of the open diffs really just pushing the gears on the faster side of the axle.

If both are turning at the same speed, both sides are being propelled.

If one side starts to go faster, say because you are going around a corner, or, one tire is on ice and the other on dry pavement, the faster tire is simply still the one being pushed, and the other side is coasting.

If its a LSD, the two sides are held together, but the faster side is still being pushed and the slower side is still coasting if the break away torque setting is exceeded, allowing the two sides to go different speeds. (The two sides are both pushing until the difference between them exceeds the ability to hold them together, then it differentiates as normal.)

If its a full locker, then the two sides can't differentiate even when the stresses build up....and the tires have to chirp or skitter around the turn on pavement, etc.

There are other types, like a Torsen, etc...which have different means of proportioning the differentiation.

I have a really old chevy video illustrating how an open diff works. I like the way it really simplified how the gears worked.

:D

Last edited by a moderator: Sep 25, 2014
15. Jan 4, 2012

### mender

Not quite; in fact there is a small tendency to drive the slower tire more than the faster one because of the friction of the differential gears themselves, but I'm being pedantic.

1. An open diff drives both wheels all the time with the same torque whether in a straight line or cornering. Trying to accelerate with the inside tire on ice and the outside tire on pavement should prove to you that an open diff doesn't allow either tire to "coast".
2. An LSD typically uses clutch packs to resist different tire speeds with no preference to which tire is going slower or faster. Usually there is a break-away torque that is preset with spring pressure. Drive torque (stepping on the throttle) increases the break-away torque value by forcing the spider gears apart, putting more pressure on the clutch packs.
3. A Detroit style locker is the only diff that coasts or overruns on the inside tire. It has a ratcheting mechanism to allow that.
4. A spool is locked 100% of the time and is the one that causes the chirping.

Last edited: Jan 4, 2012
16. Jan 4, 2012

### Tea Jay

1. The tire on ice is faster, and spins, the other one on dry pavement just sits there - I called that coasting.

What do you mean by them getting the same torque...if that means they are both spinning in the above scenario, its the opposite of what I see.

2. - Yup

3. - Yup

4. - Well, there are selectable lockers like air lockers, elockers, etc, that are like a spool when locked, and either open or like an LSD when not locked, depending on model.

17. Jan 4, 2012

### mender

The torque that each tire exerts on the surface whether ice or pavement is the same; the slippery surface limits the torque to both tires, which is why a vehicle gets stuck with only one tire on ice. They are both being driven (no "coasting").

18. Jan 5, 2012

### Tea Jay

If both are being driven, why is the one with traction (On dry pavement for example) just sitting there/not pushing the car, and the one on ice spinning?

The gears are only pushing the faster side as far as I know (As shown in the video I attached). Are you explaining something other than what the gears are doing?

When off road for example, this scenario is pretty common with open differentials. The faster side is the one being pushed by the differential, and, the side that is faster is moving....and the other one is no longer moving because the gears themselves are designed to only push the faster side. I know the differential is getting the rotation from the drive shaft, and proportioning it to the axles. If both axles are being rotated by this, shouldn't both tires be turning?

It sounds like you are saying that the differential is instead taking the rotation of the drive shaft, and applying it to both sides, but that it can't push any harder than the weakest side's torque. Physically, I believe that effect, but, don't see that as a direct gear engagement result, but as the collateral result of what happens if the gears are only pushing the faster axle shaft. I'm more of a suspension guy though, so if you can explain what I'm missing, I'd really appreciate it.

Last edited: Jan 5, 2012
19. Jan 5, 2012

### bugatti79

Its sitting there /not pushing the car because the other wheel thats slipping on ice is dictating what torque is going to be applied to both wheels ( Rememeber in an open diff the torque is always the same on each wheel).

Why is the slipping wheel dictating the torque? Because an open differential will find the path of 'least resistance' which is through the slipping wheel. The least resistance being the torque required to overcome the friction of the ice which is going to be less than the torque required to overcome the friction of the non slippy surface.

Hope this helps.

20. Jan 5, 2012

### Tea Jay

To clarify what I am trying to understand - The differential is pushing on both sides, but only one of them moves (The one on ice for example). The one that moves can spin as fast as the rotation is being transferred, and the other one doesn't have to move at all.

I describe this as the differential propelling the faster side, and letting the other side "coast". (Coasting here is used by me to just mean the side that is not being rotated)

I believe your are pointing out that the "coasting" side is actually getting power too, just not enough to rotate it because it has more traction/resistance to overcome to be able to rotate.

Mechanically, I'm not sure I understand where this energy is going on the "coasting" side that is getting torque as the diff rotates. With a locker, the system winds up if there's a difference between the two sides, which is, as you mentioned, released by the tire's slippage, etc.

So, I'll get it if you explain the difference to me in terms of this example:

The tire that is not spinning is getting rotational input, but not enough to rotate it. The stationary tire is contributing more to getting the rig unstuck than if the rig had one wheel drive on the spinning tire's side.

If the above is true, then I understand what you mean. I just think of it as the diff pushing the faster side, mechanically....as that's the "Path of Least Resistance".