Theory: Using brakes to assist traction with an open differential

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.

Welcome to AF.

This may not be the right forum to get an answer so I have moved your post to Engineering/Technical where you have a better chance of getting an answer to your question.

I've never had much success with that method. While it can stop the wheel that's wilding spinning, it doesn't do very well at evening out wildly different torque requirements.

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?


Anyone who has driven a farm tractor in the mud knows about this.

Your theory is correct, and most farm tractors have had separate brake pedals for individual back wheels for about the past 80 years to exploit this principle.

If you get stuck with one spinning wheel, a bit of brake on that wheel will often help you get unstuck.

Of course, many bigger tractors also have a differential lock, too.

When I have gotten a RWD car stuck in the snow, sometimes, I can get unstuck by gently applying the parking brake when one wheel is spinning. This also has the same effect you describe, even though the brakes are applied to both wheels.

...
with the advantage of no front brakes
:)

When I have gotten a RWD car stuck in the snow, sometimes, I can get unstuck by gently applying the parking brake when one wheel is spinning. This also has the same effect you describe, even though the brakes are applied to both wheels.

Was that car fittedwith drums or discs?

Was that car fittedwith drums or discs?
It's actually worked on many cars and trucks over the years..... all drum brakes though. I suspect it should work on disc systems, too.

I should emphasize this SOMETIMES work. Usually it doesn't, then I have to start digging :frown:

It's actually worked on many cars and trucks over the years..... all drum brakes though. I suspect it should work on disc systems, too.

I should emphasize this SOMETIMES work. Usually it doesn't, then I have to start digging :frown:

I guess my problem is, by the time I'm down to trying tricks like this, I'm very stuck.

That and my truck has discs on all corners, whether the self-energising aspect of drums helps them in such situations I'm not sure. But my truck does nothing but wheel-hop the times I've tried it.

The wheel hopping is from the squat/antisquat built into the suspension geometry. When you feed wheel torque straight into the brakes you're creating a vertical component with no weight change to resist it.

I think they teach this method to HMMWV drivers in the military as well, if they get stuck with one wheels spinning, put on a littel brake pressure and hit the gas. IIRC that is.

With the weight distribution as it is in most vehicles,the front brakes will almost always render this unworkable.I have tried it in a stuck semi,and it is even more unworkable.

parking brake even.

I think they teach this method to HMMWV drivers in the military as well, if they get stuck with one wheels spinning, put on a littel brake pressure and hit the gas. IIRC that is.

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

lol

Hi, well in practice we might have different experiences, but this is what I think is theoretically right. I don't think the torque will get split equally.

In the considered situation, lets say the left wheel is in slush and the right wheel is on proper tarmac. The resisting force given by the left wheel will be the coefficient of friction(COF for convenience) between slush and the wheel(which will be very less) multiplied by the load of the car on that wheel. The resisting force given by the right wheel will be the COF between tarmac and the wheel(which will be much higher than in the previous case) multiplied by the load on it. If the forces are represented by X & Y respectively:

a) The resisting torques will be X*r and Y*r respectively, ‘r’ being the radius of the wheel.
b) The ratio of resisting torques of left wheel to the right wheel will be X:Y.
c) The ratio in which the driven torque is split between them will be Y:X, because more the resisting torque, the more the differential will tend to send the torque to the other wheel, which is why your right wheel tends to remain stationary and your left wheel spins in the slush.

In the second case when u apply normal brakes:
a) the resisting forces will be (X+B) & (Y+B), where ‘B’ is the braking force
b) The resisting torques will be (X+B)*r and (Y+B)*r respectively
c) The ratio of resisting torques of left wheel to the right wheel will be (X+B) : (Y+B).
d) The ratio in which the torque will be split between them will be (Y+B) : (X+B).


In this case, the torque supplied to the left wheel, is greater than the resisting torque provided by that wheel.

A bit too mathematical I agree but I hope it explains it.

Cheers

Hi, well in practice we might have different experiences, but this is what I think is theoretically right. I don't think the torque will get split equally.

In the considered situation, lets say the left wheel is in slush and the right wheel is on proper tarmac. The resisting force given by the left wheel will be the coefficient of friction(COF for convenience) between slush and the wheel(which will be very less) multiplied by the load of the car on that wheel. The resisting force given by the right wheel will be the COF between tarmac and the wheel(which will be much higher than in the previous case) multiplied by the load on it. If the forces are represented by X & Y respectively:
Problem right there.

The <b>limit</b> if the resisting force of the tyre on tarmac is it's COF*weight.
The actual resisting force will not be any higher than required to resist the torque put on it.
Since the differential shares the torque evenly, the resisting forces never get higher than the lower of the two wheels grip. In this case the slush.


This error follows through your proof, in your first part (c) you've said the torque is split by the ratio x:y. But it's not, it's split x:x.

With the brakes on you've got (x+b): (x+b) so unless there's a friction bias in your brakes, it's not helping.

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.

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

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.

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.

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....

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...

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.:icon16: