Theory: Using brakes to assist traction with an open differential

[Steel]

Originally Posted by KiwiBacon

Why would they bother to do that? Humvee's have electric diff locks front and back.

Well see, that's news to me. maybe it was some other vehicle, i coulda sworn it was hummers. Ah well, it was a few years ago.

[Rider on the Storm]

Hey...

KiwiBacon: The actual resisting force will not be any higher than required to resist the torque put on it.

Agreed, I should have said the maximum possible resisting torque.

For the rest of the part, lets say that the torque does get divided equally between the two wheels. So, in this situation, the torque provided to the differential is 2Xr, and the torque provided to each wheel is Xr, ‘irrespective’ of the traction available at each wheel. So if Xr torque is provided to the right wheel(which is the one on tarmac), it should be able to move the car ahead, because the same traction is still available, and hence it should offer the same resistance to turn, irrespective of what the other wheel is on, tarmac or slush. This means that the car should move ahead.

But that does not happen in an open diff. So then, the way I see it, there seems something not right. If the above were true, then we would not require limited slip diffs. So I think the torque will not be split equally.

Thanks for bringing it up, food for thought.

Cheers

[GreyGoose006]

wait.
the ammount of torque is dependant on the ammount of braking force too
if you just tap the brakes, the brakes will not put enough torque on the spinning wheel to overcome the traction of the gripping wheel and move the vehicle foreward.
however, with an engine that is capable of putting out enough power, and the brakes applied adequately, the rear wheels will act as if they are locked together.

i guess its all a moot point though because limited slip diffs and 4wd (or even 2wd and decent tires) can get you thru just about any scenario that you are likely to find in normal everyday driving.

[KiwiBacon]

Originally Posted by Rider on the Storm

Hey...



Agreed, I should have said the maximum possible resisting torque.

For the rest of the part, lets say that the torque does get divided equally between the two wheels. So, in this situation, the torque provided to the differential is 2Xr, and the torque provided to each wheel is Xr, ‘irrespective’ of the traction available at each wheel. So if Xr torque is provided to the right wheel(which is the one on tarmac), it should be able to move the car ahead, because the same traction is still available, and hence it should offer the same resistance to turn, irrespective of what the other wheel is on, tarmac or slush. This means that the car should move ahead.

But that does not happen in an open diff. So then, the way I see it, there seems something not right. If the above were true, then we would not require limited slip diffs. So I think the torque will not be split equally.

Thanks for bringing it up, food for thought.

Cheers

The car will only move forward if the traction you get from the slush/mud/whatever is enough to overcome the set in your tyres, friction in the other wheel bearings etc.

While it doesn't take much to move a car on a flat and smooth surface, if you've got a wheel slipping then you're probably not on a flat and smooth surface. Front wheels sinking into snow or mud can take a lot of force to dislodge.

[replicant_008]

The Australian Ford Territory and GM-Holden use a AWD system with open differentials and use the disc brakes to control traction (controlled by a Bosch 5 traction and dynamic stability control system). Differences in individual wheel rotation rates trigger intermittent braking intervention to control traction... both systems also have optional driver selectable hill descent control...

A wheel slipping will have a rotation speed significantly greater than the other wheels so once braking occurs on that wheel, torque is redirected to the other wheels. Mechanical LSD operate locking mechanisms that fix the rotation speed of the two output shafts so that wheels on the same axle rotate at the same speed and receive similar levels of torque....

[corning_d3]

Originally Posted by massiv

I need help with a 4x4 theory. I know that with an open differential, if you install seperate hydraulic brake levers to each rear wheel, and you're stuck, if you apply the brake to the wheel that's spinning, you will transfer the torque to the other wheel. But if you simply apply the regular brakes, this will apply identical force to both wheels, it should theoretically transfer the torque evenly to both wheels right?

Any help is appreciated.

You've just described Jeep's new Freedom Drive 2. It operates on the same principle, although I'm not very fond about it. The new '08 Dodge avenger will have this option. Maybe on a pure off-road vehicle this would be nice, but I don't want my car dragging the brakes and losing horsepower to gain a little traction. That's what my right foot is for...

[KiwiBacon]

Originally Posted by replicant_008

The Australian Ford Territory and GM-Holden use a AWD system with open differentials and use the disc brakes to control traction (controlled by a Bosch 5 traction and dynamic stability control system). Differences in individual wheel rotation rates trigger intermittent braking intervention to control traction... both systems also have optional driver selectable hill descent control...

A wheel slipping will have a rotation speed significantly greater than the other wheels so once braking occurs on that wheel, torque is redirected to the other wheels. Mechanical LSD operate locking mechanisms that fix the rotation speed of the two output shafts so that wheels on the same axle rotate at the same speed and receive similar levels of torque....

There's a small difference between the traction control systems you're describing and the original topic of this thread.

That small difference is traction control applies brakes to individual wheels as needed, not to all four (brake pedal) or one pair (handbrake).
Making it a completely different situation.