Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations KootK on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

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

Status
Not open for further replies.

NoahLKatz

Mechanical
Jun 24, 2016
47
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.
 
Replies continue below

Recommended for you

The total axle lateral force is the sum of the inside and outside tire forces at some slip angle. Because tires act as "softening" elements ( the more you add vertical load to them, the less incremental output force results and vice versa as a generalization. Remove load from them and they produce incrementally more force). So, fy(alpha,wlf + deltaw) + fy(alpha,wrf-deltaw) is reduced as deltaw is changed.

where fy = tire lateral force output function
alpha = slip angle
wlf = left wheel weight at 0 sideforce
wrf = right wheel weight at zero sideforce
deltaw = vertical load transferred side to side due to cornering, suspension spring and anti-roll bar forces.

To see this, just draw up a parabola of tire force with vertical load on the x axis and output force on the y axis. Slap a dot on the curve at some convenient point and go the same distance (deltaw) in each load direction. Then connect the dots with a straight line between the two new tire outputs. The line will lie below single point representing the tires with no load transfer.

Actually, tires are not "springs" as far as sideforce generating elements. They operate like dampers. You need forward velocity in order to generate lateral velocity. (no speed, no cornering).

And there are cases in which the tires on a vehicle having VERY large load reserve (They are way under their rated load capacity) do just the opposite. It these cases, you will sometimes see that adding an anti-rollbar WILL increase the grip of the pair of tires up to a point. Corvettes come to mind in this case. Their tires 'like' increased vertical load up to a limit. Pressure, rim width, and construction details influence this trait. And this phenomenon can easily be lost in some (most) forms of tire math models which can not accomodate a linear or increasing stiffness tire trait. It's there in the raw tire test data but not in the analysis. Oops. From there on out, it's lies, untruths, urban legends, dogma and blatant ignorance. In God we trust, all others bring data...

 
Thanks for the nice response.

Right now I'm too lazy to use your nice formulae, but is this it - in the case of the weight transfer being on the same axle, the nonlinearity of lateral tractive capability vs. normal load results in less being gained on the outside wheel than is lost on the inside?

Either way, what about the case where only the rear ARB is added or stiffened?

Then normal force is transferred from the inside front tire to outside rear tire.

As far as the rear is concerned, why doesn't this normal force "stolen" from the front increase the lateral adhesion of the outside rear tire (not to mention decreasing adhesion at the front)?
 
A stiffer rear ARB increases normal force at the outside rear AND the inside front. The normal force is "stolen" from the outside front and inside rear.

1. The total normal force at the front stays the same regardless of ARBs etc. Same for the rear.
2. The maximum lateral force available at the front for the given normal force will occur when each front tyre carries the same load. Any deviation from this will reduce the total lateral force available. (as explained by Cibachrome). Same for the rear. (This is why vehicles with a wide track and low CG corner faster.)

je suis charlie
 
Correction: Use a function like log(x-1) to model a tire. I was thinking parabola on its side. Brain fart. A sine function is more famous (as in Pacejakk).

The rear bar operates the same way. Because it also contributes to the total roll resistance moment, there is reduced load transfer at the front. The fractions of load transferred are the main factors in "TLLTD" (Tire Lateral Load Transfer Distribution) which is a major tuning setting and is specific to a tire construction, pressure, rim, etc. recipe.

Roll bars also affect other tire and chassis properties, some by intent and others unintentional. There are tire properties other than lateral force to deal with. And there are several types of "roll bars" (direct acting, etc).
 
> 1. The total normal force at the front stays the same regardless of ARBs etc. Same for the rear.

Incorrect and counter to long-established suspension tuning practice.
 
Noah said:
Incorrect and counter to long-established suspension tuning practice.

How so, when it's the transferred load rather than the static total Fz that results in changes to the total Y-axis grip and the slip angle(s) associated with generating it? For a quasi-static midcorner condition, anyway.


Norm
 
In a steady state corner the lateral force at each axle must be proportional to the static axle load. Ignoring aero and traction.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
> How so, when it's the transferred load rather than the static total Fz that results in changes to the total Y-axis grip and the slip angle(s) associated with generating it?

If the front has lower roll stiffness than the rear, then with latacc instead of normal force transferring proportionally from inside to outside wheel at both front and rear, weight will effectively transfer from inside front wheel to outside rear wheel.
 
"NoahLKatz (Mechanical)(OP)12 Sep 16 02:50
...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."

Noah, your last statement is the most important issue for car handling and can limit what you would like to do with front/rear roll stiffness. The most noticed effect of roll bar selection is on handling. You could end up with a car that makes more Gs on a skid pad, but is undriveable on the track.
That last statement also partially answers your question. Note that tires make a side force when not pointed in the direction of travel. The angle they make is called the slip angle and the force they make is not perpendicular to the direction of travel, but points behind. If the car is cornering about a turn center, then the side force of the tire does not point toward the center of turn, it points behind the center. This resolves into two components, a force toward the center and a force opposite the direction of travel, a drag force that slows the car. Therefore, the "cornering" force toward the center of turn is less than the side force. Also keep in mind that though increasing the normal force increases the tire to track friction (you hope), the compliance of the tire is not changed. It distorts more and requires more slip angle to make more side force.

Now suppose you had suspension with zero roll stiffness (Z-bar springing or beam axle with a single center spring). You have two tires always sharing equally the weight of that end of the car. For a certain cornering speed they will corner with a certain slip angle. Now with the same tires add the maximum roll stiffness that just begins to lift the inner wheel in a corner. The outer wheel now carries twice the weight as before and requires approximately twice the slip angle. You see that the side force points farther behind the turn center, etc, etc, etc.

Another aspect involves what happens in the contact patch. You know that a tire rolling with only a vertical force has its contact patch in static contact with the road. You also know that for rubber, static friction is much greater than sliding friction. Well, when a tire is cornering, braking or accelerating there is both static and sliding friction. Suffice it to say that under side force with the carcass and tread under high distortion the proportion of sliding vs static goes unfavorable. The tire with the higher loading and greater tread distortion loses G force due to an increase in sliding friction with a loss of static friction.
The tread tends to break away at the rear of the contact patch where the distortion is greatest and the weight is being lifted. But there is also sliding and gripping and sliding and gripping going on throughout the patch.

There are also changes due to heating and additional effects that are mysterious to me and some more effects that seem to be mysterious to everybody.
 
140A,

"... Therefore, the "cornering" force toward the center of turn is less than the side force."

Thanks, I hadn't read or thought of that aspect before.

I'm aware of the control and safety issues; I'm not proposing anything counter, just trying to understand what is still a conundrum for me.

I don't see how the case of zero roll stiffness at one end addresses my question; in that case there would be no lateral weight transfer at that end and all of the overturning moment reaction would appear as higher downward normal force on the outside tire of the other axle.

So I ask again, why doesn't that increase its lateral adhesion?

Or does it, but it's unusable because of the control issues?
 
The illustration about the zero roll stiffness with both tires sharing the vertical and cornering loads equally vs total weight transfer onto one tire shows the lessening in the geometric cornering power due to higher slip angle, as you acknowledge, and the poorer performance of the contact patch due to more distortion and a greater proportion of sliding.

 
Noah said:
I don't see how the case of zero roll stiffness at one end addresses my question; in that case there would be no lateral weight transfer at that end and all of the overturning moment reaction would appear as higher downward normal force on the outside tire of the other axle.

Zero roll stiffness does not mean that the geometric portion of LLT is zero. That would require locating the geo roll center at grade level - and even then, and together with a zero roll stiffness suspension, there'd still be a little LLT from the effective amounts of unsprung mass (wheels, tires, hub-mounted brake rotors, suspension uprights, etc.).


Norm
 
Noah said:
If the front has lower roll stiffness than the rear, then with latacc instead of normal force transferring proportionally from inside to outside wheel at both front and rear, weight will effectively transfer from inside front wheel to outside rear wheel.

Juggling roll stiffnesses alters the amounts of roll moment carried at the two ends of the car, and the load transfer remains proportional taking each axle and its roll moment loading by itself. In your case here, there would be less LLT up front covered by more LLT out back. What does change is the crossweight percentage (circle track terminology that I'm not at all sure how to work with).


Norm
 
The "zero" roll stiffness in the example is used for illustration of why and how load transfer diminishes cornering power. I don't know of a suspension with actually zero roll stiffness.
 
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.
 
Norm,

>In your case here, there would be less LLT up front covered by more LLT out back.

Isn't that more or less what I said?

Does anyone not agree that higher roll stiffness at an axle causes its outside wheel to have increased normal force transferred from the inside wheel of the other axle?

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.
 
"I don't see how the case of zero roll stiffness at one end addresses my question; in that case there would be no lateral weight transfer at that end and all of the overturning moment reaction would appear as higher downward normal force on the outside tire of the other axle.

So I ask again, why doesn't that increase its lateral adhesion?"


Depends whether you want to define "lateral adhesion" as cornering force (Fy) at that tyre or Fy/Fz (coefficient of friction). Fy increases but Fy/Fz decreases and therein lies your problem.

Repeat:
1. Cornering forces do not alter the TOTAL normal force at each end - just the L/R distribution.
2. Moving the L/R distribution away from 50:50 at one end REDUCES the total adhesion available at that end.

je suis charlie
 
> Fy increases but Fy/Fz decreases and therein lies your problem.

Finally, an answer, which is affirmative.

Why is it a problem?

I know coefficient of friction decreases, but I'm only asking about lateral adhesion.

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

That contradicts your answer above.
 
Why is it a problem?
You have posed us a question ie you have a "problem".

I know coefficient of friction decreases, but I'm only asking about lateral adhesion.
But you didn't define "lateral adhesion". If you are trying to predict understeer/oversteer from "lateral adhesion" the definition needs to be "effective coefficient of friction for that axle".
So if "that axle" is experiencing more weight transfer than the other, it will tend to have a lower "effective coefficient of friction" than the other.

That contradicts your answer above.
My answer above: "1. The total normal force at the front stays the same regardless of ARBs etc. Same for the rear."
My latest answer: "1. Cornering forces do not alter the TOTAL normal force at each end - just the L/R distribution."

Where is the contradiction?

je suis charlie
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor