Increase Traction w/o a rear locker? How?

[spencyg]

It would be very interesting to attempt the same test outlined above by Michael using the e-brake instead of the brake pedal for application of drag on the diff. It does seem as though it should work, but as you've stated, I'm not quite sure I'd be comfortable taking a stab at scientifically explaining it. From this discussion I could see a significant benefit in seperating the two rear e-brake cables into seperate levers or pedals for the purpose of stopping or slowing the spinning wheel. I think it is undeniable that this would work for increasing traction to the rear in a low-traction situation.

Spence

 

[michaelgroves]

More info on how my 4x4 system works. The computer senses what wheel is slipping and applies brake pressure to the slipping wheel and sends torque to the wheel with traction.

Yup

 

[michaelgroves]

It would be very interesting to attempt the same test outlined above by Michael using the e-brake instead of the brake pedal for application of drag on the diff. It does seem as though it should work, but as you've stated, I'm not quite sure I'd be comfortable taking a stab at scientifically explaining it. From this discussion I could see a significant benefit in seperating the two rear e-brake cables into seperate levers or pedals for the purpose of stopping or slowing the spinning wheel. I think it is undeniable that this would work for increasing traction to the rear in a low-traction situation.

I can't see any significant difference in applying the e-brake versus the footbrake, other than of course that it will only apply to the rear axle. This would make it the correct method to use if one had a torque-biased LSD in the rear axle.

Land Rovers have the parking brake on the transfer box rear output shaft, so that wouldn't be useful in my case.

Regarding splitting the e-brake lines, yes, that would work very well, and indeed, it's quite a common thing to do. They call them "fiddle brakes" here.

 

[Pskhaat]

...senses what wheel is slipping and applies brake pressure to the slipping wheel and sends torque to the wheel with traction.

Without traction, there is no torque...the torque on any shaft will be dependent entirely on the traction at that wheel

An open differential maintains torque to both wheels. If one wheel is stuck and the other is spinning they are BOTH receiving the same amount of torque. The computer applies braking to a slipping wheel but that only affects wheel speed not applied torque.

A locking diff on the other hand in this situation would send nearly all torque to the stuck wheel and thus the dangers of locking diffs especially on steering axles.

Torque is independent of shaft speed.

A truly open center differential will maintain equal torque between the front and rear.

Applying brakes (equal amounts) to each side of the axle with one wheel stuck will decrease the ratio of resistance. The torque to each wheel (or set thereof) remains the same. Anecdotally, as you've said michaelgroves, I think it is too small to be useful or noticed in most situations.

 

[Photog]

Torque = length X force (ft-lb).

If there is no resistance to the power going to the axle, there is no force and no torque. With an open differential, all of the twisting power is being allowed to move through the tire with no traction. This tire has no resistance, and therefor no torque can be applied.

So you are correct, both tires have equal torque - ZERO.

By applying brakes (resistance) to the tire that is slipping, that measurable torque will rise. For the vehicle to move forward, the tire with the traction will need a certain amount of torque applied. The slipping tire will need to have the braking power increased until the necessary torque is reached, for the traction side to move the vehicle.

Now both sides are receiving enough torque to move the vehicle with the traction under the one tire.

The "fiddle brakes" are also called "cutting" or "steering" brakes. They can be operated by cable or by hydraulics.

 

[Pskhaat]

With an open differential, all of the twisting power is being allowed to move through the tire with no traction...So you are correct, both tires have equal torque - ZERO.

Sorry Brian I'm not following you, one can apply an equal amount of non-zero torque both to a moving (flying wheel) and non moving (stuck wheel) object. In this situation torque is indeed being applied equally to both wheels, that is a fundamental design of a differential, yes?

 

[michaelgroves]

With an open differential, all of the twisting power is being allowed to move through the tire with no traction. This tire has no resistance, and therefor no torque can be applied.

So you are correct, both tires have equal torque - ZERO.

Of course, that's only the case when one tyre is off the ground completely. When both wheels are on the ground, and one is slipping, it's equally important to know that both sides - the side that is spinning and the side that is stationary - are still both under equal torque, and supplying equal force attempting to move the vehicle. It's counter intuitive - most people assume that the stationary wheel is doing nothing because it's not moving. As Scott says, neither torque nor force necessarily imply movement.

 

[Photog]

So the goal for traveling is to move forward/backward and not get stuck. So if we make sure there is enough torque to any wheel with traction to keep the vehicle moving, we are good.

Regardless of the division of torque, power, etc., an open differential is problematic in maintaining motion in situations of limited traction, as it wants to always reduce torque to the lowest common denominator (tire with least traction).

A locking differential, torque biasing differential, ABS-traction control, etc., are all devised to overcome the shortfalls of the open differential.

I will re-read the last few (long) posts. Maybe I missed something. I feel like I'm driving into a mud hole.:smiley_drive:

 

[Pskhaat]

So the goal for traveling is to move forward/backward and not get stuck. So if we make sure there is enough torque to any wheel with traction to keep the vehicle moving, we are good.

A few things, we need power to non-turning wheels; we need rotations associated with the applied torque and a bias or locked diff would do that. After that a focus on tire traction, but how often are we unlocked and glamper rigs flying wheels?

Save for a few years in an LC100, I have never had a locking differential in any of my 4WDs, yet I've managed to always find another line or some other throttle, braking, and steering modulation to assist, not to mention the right tire is a huge help, but maybe the extreme articulation from an LC80 helps greatly.

 

[michaelgroves]

A few things, we need power to non-turning wheels; we need rotations associated with the applied torque and a bias or locked diff would do that.

Right. But the power is derived from the torque - the reason we don't have power, is because we have insufficient torque to make the wheel rotate. Discussing torque is the simplifed version of analysing it in terms of power. The use of a locker or LSD does (may) indeed result in power, and hence movement, but it's because it allows more torque to be applied.

 

[Pskhaat]

OK, this is where I may need some clarification so let's walk through it (again we might be saying the same thing). Let's use the canonical example of a single axle, one tire aground the other flying with an open diff. We must include power though as we're talking about a dynamic system not a static one, the definition won't work without time as a variable.

The engine turns the input shaft and equal torque is maintained to both half-shafts. They flying tire turns, the grounded tire does not. You can apply torque to a wrench on a fast bolt with no rotation, yes? This is due to the ratio of resistance between the tire on the ground and that in the air: the one in the air has very little resistance and as such turns. It should not be receiving however any more torque than the grounded wheel. A differential always maintains torque.

The grounded wheel doesn't turn not because it has insufficient torque as you state but rather the power is diverted to the least resistive path to the flying wheel. In fact the applied torque may be perfectly sufficient to move the object.

Sidetrack IIRC from Uni: there is a case of average torque across a differential in that the applied torque is always equal to the least resistive path.

We are not lacking torque per se, rather we are simply lacking resistance (traction), and thus the power to the right wheel.

Now let's lincoln lock this open diff and you will indeed have unequal amounts of torque to each half-shaft. The one with the most traction will now get the most torque, or rather the torque ratios will now roughly equal the resistence/traction rations.

On your last statement concerning LSD/locker allowing for the application of more torque, I think I know what you're saying but I think it is a power discussion not torque: I believe you're saying that if you have a very low resistive path with the flying wheel, it takes very little torque to expend the power and thus the grounded tire would never have cause to move? If so then yeah maybe we're saying the same thing?

 

[Photog]

...........

On your last statement concerning LSD/locker allowing for the application of more torque, I think I know what you're saying but I think it is a power discussion not torque: I believe you're saying that if you have a very low resistive path with the flying wheel, it takes very little torque to expend the power and thus the grounded tire would never have cause to move? If so then yeah maybe we're saying the same thing?

Yep. That is what I was saying.

It also means the rig is stuck.

Choosing a different route, line, trail, stacking rocks, applying brakes, etc. always worked for me in the desert. It definitley does not work up here in the N-West trees, mud & rocks. Most of time, there is only one line. I had steering brakes in my VW Baja, and used them to get moving on many occasions up here. Just apply brake pressure to the tire that has very little traction, and on we go.:smiley_drive:

 

[michaelgroves]

I suspect we are saying the same thing, or at least reaching the same conclusions.
Where we differ is this:

The grounded wheel doesn't turn not because it has insufficient torque as you state but rather the power is diverted to the least resistive path to the flying wheel. In fact the applied torque may be perfectly sufficient to move the object.

We are not lacking torque per se, rather we are simply lacking resistance (traction), and thus the power to the right wheel.

Because torque is the result of the engine power working against the available traction, there is zero torque on the flying wheel, and therefore zero torque on the other wheel, due to the open diff. (OK, not quite zero, because there is the friction of the wheel-bearings, air resistance etc). How much torque the engine is capable of producing is irrelevant - it's unable to use its potential to generate actual torque.

As I said in a previous post, if you imagine a sensor on each half-shaft, showing a continuous readout of the torque, it would become clear that neither shaft is under torque.

So the reason the grounded wheel fails to turn is simply that there isn't any torque available. Although there is traction at that wheel, the engine meets no resistance from either side because the more the pinion tries to turn the ring gear, the more the diff just speeds up the the flying wheel. No resistance, no torque; no torque, no power!

The reason it gets complicated to talk about power, is that a diff doesn't share power between both sides evenly. If the torque really is zero, then so is power, and it's pretty simple. But as soon as the spinning wheel is no longer flying, but in mud, say, then we have some (equal) amount of torque on each shaft. But because they are turning at different speeds, they have very different amounts of power (power=torque x RPM). (If one is actually stationary, it will be is generating no power, even though it is generating torque and force!)

 

[milo12]

I have been following this thread with interest because I have the same issue.

No one makes a locker for the 11.5 inch rear on a newer Dodge diesel. It does come with a LSD.

I don't want to get bogged down in all the campfire discussion that while interesting won't get me unstuck.

The fiddle brake discussion did give me an idea. The rear parking brake is cable actuated. This is independent of the normal brakes. The parking brake is a single cable at the cab but then at the axle it splits to each rear tire. So I could just run 2 independent cables. I need to find and rig up a couple hand parking brakes but it should be relatively simple to do.

Of course that is just a theory. I'll let you know what happens in reality.

 

[Root Moose]

The fiddle brake discussion did give me an idea. The rear parking brake is cable actuated. This is independent of the normal brakes. The parking brake is a single cable at the cab but then at the axle it splits to each rear tire. So I could just run 2 independent cables. I need to find and rig up a couple hand parking brakes but it should be relatively simple to do.

Of course that is just a theory. I'll let you know what happens in reality.

This should work fine. I've seen it done on Jeeps and Suzukis in the past.

My only concern would be the torque of the diesel being able to over power the parking brake. If it is one of those drum inside the rotor deals it could be marginal. If it is an actuator on the caliper itself it should work fine regardless.

I'd still try it regardless of which type of brake it uses.

HTH