Theory: Using brakes to assist traction with an open differential

[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.

[BNaylor]

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.

[KiwiBacon]

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.

[MagicRat]

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?

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.

[GreyGoose006]

...
with the advantage of no front brakes

[KiwiBacon]

Originally Posted by MagicRat

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?

[MagicRat]

Originally Posted by KiwiBacon

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

[KiwiBacon]

Originally Posted by MagicRat

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

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.

[Steel]

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.

[butch h]

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.

[Steel]

parking brake even.

[KiwiBacon]

Originally Posted by Steel

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.

[GreyGoose006]

lol

[Rider on the Storm]

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

[KiwiBacon]

Originally Posted by Rider on the Storm

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.