Why doesn't increased normal force from ARB (antiroll bar) increase adhesion? 1

[NoahLKatz]

I've long been mystified by this aspect of how ARB's work.

Why doesn't transferring weight from inside to outside tires increase their lateral tractive capability?

Seems to me that's what would should happen when increasing normal force w/o increasing the lateral force.

I understand that increased slip angles may require steering angle correction, but that's a different issue.

 

[NoahLKatz]

> But you didn't define "lateral adhesion"

I believe I did - lateral tractive capability.

I guess my question wasn't clear.

I am NOT interested in oversteer/understeer.

It doesn't make sense to me that increased normal force doesn't increase it.

> 1. Cornering forces do not alter the TOTAL normal force at each end - just the L/R distribution.

I said why I don't believe that's the case; rather than repeat your statement, can you say why or what part of mine you disagree with?

 

[BrianPetersen]

Understeer or oversteer is going to put a limit to how much lateral acceleration you can actually use before it reaches the theoretical maximum (the coefficient of friction) - sometimes well before.

And, the average racetrack corner doesn't look like a skid pad, and the surface of the average piece of public road doesn't look like the surface of a racetrack, and keeping the average joe safe when he has to do an emergency lane change on a rain-soaked motorway doesn't look like much of any of that.

 

[NormPeterson]

I understand that the increase in lateral tractive capability is not commensurate with the increase in normal force, but it seems like there should be an increase nonetheless.

For the outside tire on a given axle taken in isolation, that is, of course, true. But it seems sort of academic to focus on the loading (and adhesion) at only one of the wheels on an axle while ignoring what's going on over on the other.


Norm

 

[GregLocock]

Sta bars change the left to right distribution of Fz, but they do not change the fore aft distribution. if they did one end of the car would rise up . Draw a free body disgram in side view.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[140Airpower]

"Sta bars change the left to right distribution of Fz, but they do not change the fore aft distribution. if they did one end of the car would rise up . Draw a free body disgram in side view."

Greg, you haven't seen racing cars where in a corner one end rises up?

Seen on FWD, forward weight biased cars where the rear rises and RWD, rearward weight biased cars where the front rises. This is a consequence of putting almost all roll stiffness at the light, non-driven end. I've even seen it on RWD, forward weight biassed cars -the heavy end rises up.
One could say that the bars connect to both ends through the chassis.

 

[140Airpower]

"Just about all farm tractors have zero front roll stiffness due to a center pivoting solid front axle. Rear suspension is the tires, which operate at low pressure. On ag tires, they roll quite a bit. Can-Am Spyder motorcycles with only 1 rear wheel also have, you guessed it."

True!

 

[NormPeterson]

I'd be careful about interpreting visual-only observations. Rising-rate springs could cause an overall increase in ride height at that end of the car (the outboard side drops less than the inboard side rises).

That said, I think there are a couple of mechanisms by which there could be some coupling between ride height and the roll stiffness provided by a sta-bar. Unequal arm lengths, maybe. Asymmetric geometry changes in roll putting the endlink on the outboard side at a slightly larger motion ratio than the one on the inboard side.


Norm

 

[GregLocock]

RE FWD lifting inside rear wheels. You have drawn an incorrect conclusion from a correct observation. Draw the FBD. Unless you were being ironic.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[NoahLKatz]

My apologies for spinning everyone's wheels.

As I lay in bed not falling asleep last night, I saw my error.

I realized that if zero roll stiffness at an axle means that latacc can't xfer weight onto that axle's outside wheel, it also means it can't remove it from its inside wheel.

That put the lie to my belief that a rear ARB xfers weight from the inside front to the outside rear wheel (which I had read in a car magazine when I was young and never questioned) is not true.

So now I get it; the axle with greater roll stiffness loses total adhesion because it gets a poorer wt xfer distribution, and adhesion is gained at the other because it gets a better wt xfer distribution.

Which is what everyone has been saying.

 

[gruntguru]

Pheeeew!!!

je suis charlie

 

[140Airpower]

"RE FWD lifting inside rear wheels. You have drawn an incorrect conclusion from a correct observation."

Greg, the observation is that ALL normal car suspensions raise the CG of the car upon cornering. The ones that have one end with roll stiffness disproportionately high compared to the weight at that end will raise that end the most.

 

[gruntguru]

140A.
That is not a correct assessment either. If the roll moment is purely moment, the chassis will rotate about the roll axis so no "jacking" will occur - unless a wheel lifts off the ground. OTOH "jacking" can also be caused by suspension design (eg swing-axle) where forces other than a "pure" roll moment come into play.

je suis charlie

 

[GregLocock]

Whatever happens, in side view, in steady state cornering, there is no front to rear weight transfer. Draw the FBD. Again. Which you obviously haven't. Again. JFDI.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[140Airpower]

Take pictures and measure what happens!

 

[GregLocock]

Ah, not ironic then. Incapable.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[gruntguru]

Norm mentioned rising rate springs which I agree may be a frequent cause on track cars. A common source of rising rate is the trusty bump-stop. If the rate was symmetric about the unloaded ride height (ie a droop limiter) you would see a wheel lifting.

je suis charlie

 

[GregLocock]

It's a common tuning condition on FWD. It is not ideal. The reason it happens is that to maximise traction you want the front roll rate to be soft, so to limit roll you have to use the back axle. But the rear axle is only lightly laden so the stiff rate makes it relatively easy to lift the inside rear wheel. The stiff rear roll rate also helps to reduce the understeer on the car, which is often helpful, it is otherwise quite easy to end up with a car that understeers if you accelerate, understeers if you brake, and understeers when you turn the wheel. Drivers whinge about that.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[140Airpower]

"Ah, not ironic then. Incapable."

Greg, I'm surprised you seem to believe "The" or "A" free body diagram can decide this question.
Of all the attitudes the suspension of a car on road or track can take for any condition of loading, side force and instantaneous ride height, the car would seek the lowest energy configuration at any instant, but this would be described by a family of free body diagrams. Vehicle dynamics software can provide this. If you have access to that I can be easily convinced if you would kindly post the curves for ride height and other interesting parameters.

 

[GregLocock]

Yes i spend every day running adams, or measuring cars on the track, or working things out with a pencil and paper. You are the one making the ludicrous proposition, it's up to you to show how it could work. A side view fbd of a car with weight and 2 axles shown is sufficient to demonstrate that in the absence of longitudinal acceleration there is no weight transfer.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[GregLocock]

obvioulsy m*g=F_f+F_r

F_f=b/(a+b)*m*g

Otherwise there must be accelerations in the plane of the diagram.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?