Increase Traction w/o a rear locker? How?

[adi]

I noted some problems with this explanation further back in this thread. Brakes hold all the wheels, so where does the additional traction come from? So far, I've yet to hear a convincing hypothesis about how brake-throttle modulation could work. Also, no explanation of why a controlled experiment of BTM fails.

My own hypothesis - pending a better one - is that traction is often better when a wheel turns slowly, and using the brakes makes the wheels turn more slowly. So does less throttle...

Traction comes from a momentary redistribution of torque. Its not a solution where you go up a hill with one foot on the brake, and the other on the throttle. Its a solution where you use the brakes to equalize torque between a wheel in the air and a wheel on the ground to "bump" over an obstacle. It creates (just making up numbers, no actual number crunching) lets say a 80/20 split of torque for a few milliseconds, and that might be enough for the tire(s) on the ground to get enough traction and momentum to get the tire in the air back on the ground.

The other way to look at it, which is almost impossible to reliably do in real life, there X amount of friction/rolling resistance for the tire on the ground that needs to hit it before it moves, and using the brakes you can get close to X on the wheel in the air to lessen the path of least resistance. Balancing the amount of brake is the issue of trying to rely on its success in reality.

Hopefully this might shed some light on the theory of the idea.

ETA: It will not get you out of a 2 foot hole. It's abilities are probably limited to level 3-4 trails (scale of 10).

 

[Cabrito]

I am really liking all the information in this thread.
I wish I had read it before I went to DV in April..

 

[michaelgroves]

Its a solution where you use the brakes to equalize torque between a wheel in the air and a wheel on the ground to "bump" over an obstacle. It creates (just making up numbers, no actual number crunching) lets say a 80/20 split of torque for a few milliseconds, ...

The question is how? How does using the brakes create any traction?

The brake acts on the spinning wheel to create a torque (momentary or otherwise) of x Newton.metres. Since the brake is working on both wheels, the force required to overcome the brake on the other wheel is also x N.m. So no additional force (traction) is generated at the stationary tyre.

Here's another question. Looking back at the experiment I described earlier, how would you reconcile my results with your understanding of how BTM works? Or, put it another way, what did I have wrong in my technique or the setup that made BTM fail to move the vehicle?

 

[KLF]

The question is how? How does using the brakes create any traction?

Or more precisely "Assuming an open differential, how does application of the brakes apply more (net) torque to a wheel with traction when the other is spinning?"
The concept has been referred to as "Left footed braking", "poor man's locker, "poor man's traction control, etc., and is oft repeated wisdom.
Yet a complete explanation of why it works (if it does) is not to be found. Most explanations incompletely point to an element that would make sense standing alone while conveniently ignoring the entire system.

This thread has the most extensive discussion I have been able to find on the web. But, for reasons MG pointed out here, a complete and accurate underlying explanation has not come out.

Anybody care to address this?

 

[SnowedIn]

The most complete explanation is going to involve how an open diff works to begin with. Most people go cross-eyed at that point, and your best bet is to look up a video or animated gif.

This is the brief version;

You have one wheel with zero traction and one with some degree of traction. On an open diff, the wheel with zero is going to spin freely and you go nowhere. Everyone is on the same page, up to that point.

Normally (i.e. not stuck and spinning), you can apply the brakes and still have forward progress. It's just slower. That is after all the point of the brakes.

Now let's say the wheel with no traction suddenly gained some; due to the nature of the open diff, both wheels would start turning and you'd be fine. Applying the brakes enough to artificially give resistance to the spinning wheel while not entirely halting the gripping wheel accomplishes a similar thing in the differential. The open diff neither knows nor cares about the difference between resistance from moving the vehicle and from braking.

 

[Webfoot]

If there is a perfect 1:1 torque bias and the brakes apply equal force to each wheel there can be no increased traction from brake application. But perhaps 1:1 bias is not perfect, or the brakes do not actually apply equal force to each wheel? There is some friction within the differential, and the momentum of the spider gears would have some effect, but I doubt these effects are perceptible. I therefore conjecture that if a traction increase is observed with brake application it is due to the brakes themselves.

The actual friction coefficient of pad on rotor is probably not invariant to surface velocity, and it is surely not invariant to temperature.
Many brake pads have a coefficient of friction that is positively correlated to pad temperature up to the point of pad failure. (Fig. 1 below) The pad on the side with the spinning wheel will heat up while the one on the stuck wheel will heat very slowly or not at all. This will create a torque bias within the brake system itself.

There are bound to be other nonlinearities in the brake system that work with or against this biasing. For example if the pads exhibit significant stiction a stationary wheel will be more firmly held in place, reversing the beneficial bias described above. These factors may be why some report success with braking open differentials while others have no success at all.


Figure 1 from from TireRack.com for Hawk pads. (Sorry for the gigantic image; the board software does not appear to respect image width tags in BBCode.)

 

[KLF]

This is the brief version;

[This is a repost. It appears the original was deleted as I tried to edit a typo]

The open differential and its constant 1:1 torque ratio is fully understood. And so is why braking ONLY the spinning wheel could be quite effective. That is the essence of modern "traction control."
But we have a a system with two wheels at opposite ends of the axle, both of which will be affected by the brakes.
(And can we assume one wheel has NO traction and is spinning, while and the other has some traction but is stationary due to lack of sufficient torque to drive it and propel the vehicle?)

Yes, applying braking (be it via the brake system or parking/emergency brake) to both the wheels does cause development of additional "drive" torque in the system. And, yes, that is applied to the stationary/stalled wheel. But we cannot simply ignore the whole system and stop our review at that point.
The act of braking also acts on the stationary wheel (through its brake) as drag - or "negative" torque.
So the axle/differential provides drive/positive torque and the brake provides braking/negative torque. Presumably in equal and opposite amounts due to the nature of the system.
Where is the NET torque that will take advantage of the unused traction of the stalled wheel to propel the otherwise motionless vehicle?
i.e. - How does the increased drive torque to the stalled wheel exceed the added drag/braking torque?

 

[proper4wd]

You guys are thinking about this all wrong.

Left foot braking works because what you are doing is equalizing the amount of force required to turn both wheels, exceeding the resistance generated y the single "stalled wheel".

Think of it this way - it turning the wheel with traction will require 1000 ft lbs, you apply the brakes to *both* wheels to create 1100 ft lbs of resistance at each. Each wheel now has equal load, and the torque is split evenly across the differential.

Now apply 2500ft lbs total torque to the differential, and each wheel will get 1250 ft lbs - overcoming the brake resistance of 1100 and delivering 150 ft lbs of driving torque to each wheel.

The "stalled wheel" traction load is irrelevant because the brake-generated load is greater.

 

[KLF]

You guys are thinking about this all wrong.

Left foot braking works because what you are doing is equalizing the amount of force required to turn both wheels, exceeding the resistance generated y the single "stalled wheel".

Think of it this way - it turning the wheel with traction will require 1000 ft lbs, you apply the brakes to *both* wheels to create 1100 ft lbs of resistance at each. Each wheel now has equal load, and the torque is split evenly across the differential.

Now apply 2500ft lbs total torque to the differential, and each wheel will get 1250 ft lbs - overcoming the brake resistance of 1100 and delivering 150 ft lbs of driving torque to each wheel.

The "stalled wheel" traction load is irrelevant because the brake-generated load is greater.

First, you cannot apply 2500 ft/lbs of torque to the differential if the greatest resisting torque is the 1100 ft/lbs provided by braking at the spinning wheel.
That is the inherent limitation of an open differential.
And, although it is oft repeated, an open differential does not "split" torque. In other words, it is not divided into fractional proportions.
The available (transferable) torque is identical at both wheels (1:1) and can be no greater than that of the wheel with the least resistance. Using your example, 1100 ft/lbs.
And with that maximum available 1100 ft/lbs, you are now trying to make the vehicle move - which requires turning the wheel with traction.
That means (1) overcoming the 1100 ft/lbs. of braking force you are applying to each wheel (oops, there nothing left) AND (in your example) also providing the 1000ft/lbs you have set as the amount needed to turn it.

Aside from the braking force, keep in mind that the stalled wheel is stalled, not because it requires some huge amount of torque to turn it, but because it is simply not getting any torque at all (if the opposite wheel is spinning).

 

[proper4wd]

Your logic is flawed Build yourself a model to illustrate.

 

[KLF]

Your logic is flawed Build yourself a model to illustrate.

If you believe that to be the case, please explain. I have pointed out your rather blatant erroneous assumptions.
What errors have I made and just what am I supposed to build a model of?

 

[proper4wd]

You are correct, torque can only be applied where there is resistance.

By applying resistance (by the brakes) evenly to both wheels, and then applying enough torque to overcome that resistance, both wheels will begin to turn at the same speed (as you noted, its how an open diff works).

This doesn't matter if one wheel is in the air and the other is on the ground. Equal application of the brakes generates equal resistance at each wheel.

By manipulating the brakes, you are essentially "tricking" the differential into reacting as if both wheels are on flat ground with an extremely high resistance.

 

[proper4wd]

Maybe think of it this way,

How do you ensure both axle shafts turn the same speed with an open differential? Equal load on both axle shafts

How do you apply equal load to both axle shafts?

 

[KLF]

You are correct, torque can only be applied where there is resistance.

By applying resistance (by the brakes) evenly to both wheels, and then applying enough torque to overcome that resistance, both wheels will begin to turn at the same speed (as you noted, its how an open diff works).

This doesn't matter if one wheel is in the air and the other is on the ground. Equal application of the brakes generates equal resistance at each wheel.

By manipulating the brakes, you are essentially "tricking" the differential into reacting as if both wheels are on flat ground with an extremely high resistance.

You are repeating the classic mantra which, I believe, relies upon a degree of misunderstanding and/or assumption of erroneous facts.
First, an open differential is a "constant torque" device. In no way does it ensure constant speed. Speed differentiation is why it exists.

Let's take the one wheel off the ground scenario. For that wheel to be spinning, it takes some very small non-zero torque (yes there are minor frictional issues to be overcome).
Let's just call this necessary small threshold amount +. So the minimal + is the torque being applied to the spinning wheel and is also the torque being applied to the stationary wheel.
If we brake the spinning wheel with braking force T, how much drive torque do we need to supply to keep the spinning wheel moving as before?
We need to overcome the braking torque with a drive torque of T, correct? At which point we have bunch more drive force but the wheel in the air is not moving.
But in that situation of balanced braking/drive how much more torque do we need to add to cause the wheel to spin again? It is only the threshold +.
That means when we brake with force T and provide drive force T, the wheel again spins at a drive force of T+.
And since the differential provides torque at 1:1, the stationary wheel is also receiving a drive force of T+.
Sounds great!
But at that point, we cannot add more drive torque because we have run out of resistance - the wheel is spinning again. You agreed that torque only exists with resistance.

So, the stationary wheel is now receiving drive torque of T+.
Sadly it is also affected by the braking resistance force T.
Which means that all that is left to turn the wheel and propel the vehicle is the puny threshold force, +.
That minimal net torque wasn't sufficient before we applied the brakes, why should it be so afterward?

 

[KLF]

Independent braking of the wheels, AKA "traction control" works.
No question.