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.

 

[140Airpower]

It just so happens that today in qualifying for Sunday’s F1 race in Singapore Sebastian Vettel’s car experienced some kind of suspension failure that had his front end lifting up in a corner with the inside wheel high off the ground. It was absolutely spectacular and it was a perfect example of what I had said. The failure was at the rear. A loss of roll stiffness at the rear made the front carry too much of the car’s roll stiffness, much beyond its share compared to the amount of weight it carries. The result was the front had extra lift. The rear then must have had less lift than normal.
Since the rear did not seem to drag, the failure most likely was only with the roll stiffness function of the rear suspension system. It was described as a broken rear sway bar. A perfect real world example of a sway bar at the rear affecting roll behavior at the front.

BTW "3-wheel racing" a term describing setting roll resistance front and rear to deliberately raise a wheel in cornering is well known and can be the fastest way around. It was used in F1 during the 2000s and on 911s in the 1980s.

 

[140Airpower]

"gruntguru (Mechanical)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."

What ultimately controls chassis lift is not the spring rates nor the suspension geometry. It is the fact that the tires exert centripetal force at ground level and the "centrifugal force" acts through the CG of the vehicle at some distance above the ground. As you well know, forces acting in opposite directions along vectors offset by a length produce a torque.
That torque can be shown, by drawing a line from the CG to the outside tire contact patch, to result in a lifting force at the CG. Ultimately, if the tire grip is sufficient combined with a CG high enough, it will actually lift both inside tires off the ground and tip the vehicle. ALL vehicles can be tipped with sufficient grip at the wheels and high enough cornering force.
I used to try to design (non-active) suspensions that would not lift the chassis in cornering. It can be done, but only up to a point and it requires compromising other characteristics like showing a kink in the curves of suspension behavior and may not be self restoring, especially not along the same characteristic curve.

With conventional suspension cornering force ALWAYS produces this lifting force at the CG.

 

[GregLocock]

blaa blaa blaa. Look at the FBD I posted. Criticise it.

Cheers

Greg Locock


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[gruntguru]

140A. Your "lifting force at the CG" does not exist. The couple due to centripetal and centrifugal forces is reacted by unequal normal forces at the inside vs outside tyres producing an equal and opposite couple. Yes, once the inside tyre normal force goes to zero, the wheel (and the CG) starts to lift. Until that point, there is no "jacking".

je suis charlie

 

[140Airpower]

"GregLocock (Automotive)
18 Sep 16 05:53
blaa blaa blaa. Look at the FBD I posted. Criticise it. "

Greg, your diagram is irrelevant to what I'm talking about. Since you seem not to know that, we must be talking past each other.

gruntguru, You imply that the inside wheels exert an equal force to the outside wheels in a corner. But, of course you know that is not true as you state in "Yes, once the inside tyre normal force goes to zero, the wheel (and the CG) starts to lift. Until that point, there is no "jacking"".

The inside tire normal force begins to go to zero (diminishes) just as soon as cornering begins. The suspension is usually designed to be compliant which means that as soon as the torque described exists, the CG begins to react to it by rising.

Originally I was going to give an example of a solid suspension vehicle to illustrate what it takes to avoid CG rising where what you describe can actually happen. And I was going to show what roll stiffness means with solid suspension and how you can get different roll stiffness front and rear with solid suspension and how this would also react in the manner I talked about with the end that has higher roll stiffness rising higher in roll (this time regardless of weight distribution), but that requires even a lot more mental flexibility and realizing that tipping involves roll.

Try to see if you can see what Sebastian Vettel experienced when his rear sway bar broke. I'd be glad to read your explanation of it that is different from mine.

 

[NormPeterson]

When his rear sta-bar broke, the boundary conditions for chassis attitude changed. Which in turn resulted in the inside front tire lifting (yet another change in boundary conditions - any/all further LLT had be taken at the rear for obvious reasons). IOW, it wasn't the same car any more, so an approach that implies linear behavior cannot be used to explain what was going on.


Norm

 

[GregLocock]

Greg: "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."

Airpower140: Greg, you haven't seen racing cars where in a corner one end rises up?

Ah I get it, you meant pitching, I meant literally rising up and up and up until the car was standing on end. Sorry.


Cheers

Greg Locock


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[140Airpower]

"IOW, it wasn't the same car any more, so an approach that implies linear behavior cannot be used to explain what was going on."

Norm, the crew knew what was going on because the reaction of the car was explained by a loss of rear roll stiffness. Note that the exaggerated lifting only occurred in cornering. They therefore knew the vertical stiffness was ok. They did not look at the front where the strange behavior was occurring. They went straight to the problem. That was also my guess.

 

[140Airpower]

"Greg: "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."

Airpower140: Greg, you haven't seen racing cars where in a corner one end rises up?

Ah I get it, you meant pitching, I meant literally rising up and up and up until the car was standing on end. Sorry..."

If you saw Sebastian Vettel's car and thought that was pitching then no, you didn't get it. But, I think you actually do get it, Greg, and are only being sarcastic as shown in your continuation.

 

[GregLocock]

No i don't watch F1, and i was serious, a longitudinal weight transfer in the absence of longitudinal acceleration s would result in the absurd posture I proposed, eventually.

Cheers

Greg Locock


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[140Airpower]

Noah,
I was going to add more to the discussion on the effects of slip angle, but the post would have been excessively long. However, I think you might find this interesting.
Remember that the slip angle points the side force generated by a tire in cornering actually behind the center of the turn which can be represented by a reduced force pointed toward the center along with a drag force pointed opposite the direction of travel. This describes a coasting wheel.

A driven wheel can generate a thrust force in the direction of travel that neutralizes the drag force resulting in the full side force being directed toward the turn center. So, a driven wheel can have better cornering power than a coasting wheel. However, for a 2-wheel drive car, one end is coasting and the driven end has to make extra thrust to make up for the drag at the coasting end. This kills the advantage the driven end has in cornering... UNLESS, you minimized it by concentrating as much weight as possible on the driven end. Then, there is always 4-wheel drive.

 

[NoahLKatz]

Very interesting, 140AP, thanks.

Greg, regarding your FBD, I'm a bit puzzled as to your choice of view; in side elevation the lateral force vector is invisible so deriving its force resultants and effects is problematic, no?

Front or rear elevation makes more sense to me.

 

[140Airpower]

Greg, if we are talking about weight, (measured on a force scale), and not mass then sway bars can transfer weight front to rear and vice-versa diagonally which in a corner means that the most critical outside pair of wheels can experience a direct dynamic transfer of weight from front to rear or the reverse. You can crank a twist into the bars to create a static stagger of weight. They do the equivalent of this in NASCAR. And yes, when you increase the weight on a tire without increasing the mass on it (in effect, the crank of the sway bar adds to the compression of the spring), the reaction force is an increase in lift that raises the ride height for that wheel.

 

[SwinnyGG]

Every single post of Greg's as far as I can tell either uses the term 'steady state', or is explaining the logic of a post that used the term 'steady state'.

By definition, 'steady state' means that there is no weight transfer occurring- any and all weight transfer that ever will occur has already occurred and the system is in equilibrium.

The point (I believe) Greg is trying to make is that once the system has reached equilibrium (in which case all forces are constant, are in equilibrium, and any transfer of forces laterally, longitudinally, or diagonally have already occurred and settled to a constant) the only way to transfer weight front or rear is longitudinal acceleration, which is either not present in steady state, or is present but constant so that there is no change in weight distribution occurring through any axis.

He's talking about steady state and the rest of you are making arguments about dynamic situations.

Greg, I apologize for putting words in your mouth, that isn't really my intent here, but watching you all talk circles around each other is exhausting.

The inside tire normal force begins to go to zero (diminishes) just as soon as cornering begins. The suspension is usually designed to be compliant which means that as soon as the torque described exists, the CG begins to react to it by rising.

It is true that cornering causes the normal force of the inside wheel to diminish- but that does not mean the CG is rising. Think about where the roll center is.

Until the inside rear wheel is no longer contacting the ground, it is still contributing normal force to the total for the axle it is attached to- this means two things: 1) the axle in question is still contributing to the roll resistance of the entire vehicle and 2) the spring of the outside wheel is not yet supporting 100% of the axle normal force, so as roll continues to accumulate that spring will continue to compress, and in almost all cases that means the CG is moving down, not up.

At the exact instant the inside wheel comes off of the ground, the axle ceases to contribute to the roll resistance of the vehicle as a whole, so the vehicle-level roll resistance plummets; the wheel also ceases to contribute any of the total axle normal force- this means that the outside wheel bears all of it, and will not compress further. This behavior means that the roll center of this axle ceases to be the geometric roll center of the suspension, and becomes a point somewhere in the contact patch of the outside tire. Any further added cornering force will cause this end of the car to try and rotate around this new roll center, which will cause the CG to move up away from the ground- but unless the suspension geometry (and thus roll center location) are very strange, the CG will not move up until this happens.

 

[NormPeterson]

the crew knew what was going on because the reaction of the car was explained by a loss of rear roll stiffness.

Sure - that explains the visible effect. But not whether load was being transferred diagonally across the chassis by the sta-bar, which is nominally what this thread is about.

It's been common knowledge for quite some time that lifting a front tire on a rear-heavy RWD car can happen either during cornering or more likely/noticeably when on the throttle exiting said corner when you have a combination of pitch and roll. I'm sure I can find decades-old head-on pictures of a Porsche 911 doing precisely this. courtesy of a front sta-bar chosen overly stiff to reduce oversteer tendencies - this being essentially similar to it having a normal front bar combined with a broken rear bar.


Norm

 

[gruntguru]

gruntguru said "The couple due to centripetal and centrifugal forces is reacted by unequal normal forces at the inside vs outside tyres producing an equal and opposite couple."

140A said "gruntguru, You imply that the inside wheels exert an equal force to the outside wheels in a corner."

gruntguru says "huh???"

je suis charlie

 

[NoahLKatz]

> if we are talking about weight, (measured on a force scale), and not mass then sway bars can transfer weight front to rear and vice-versa diagonally

Yes, upon further thought that seems right:

If both axles have roll stiffness, both will xfer wt w/lat acc.

Roll moment must be reacted, so taking the extreme w/zero roll stiffness at one axle, all the lat wt xfer must happen at the other.

IOW, wt will effectively xfer from inside wheel of one axle to outside wheel of the other.

So why the heck didn't anyone tell me I was wrong about being wrong?

Now we're full circle, and unless I'm missing something, my original question remains unanswered.

 

[gruntguru]

"Roll moment must be reacted, so taking the extreme w/zero roll stiffness at one axle, all the lat wt xfer must happen at the other.

IOW, wt will effectively xfer from inside wheel of one axle to outside wheel of the other."
No. The axle with zero stiffness will not change weight at either wheel (they remain equal). The other axle will see all the transfer ie from inside wheel to outside wheel - same axle - there is no diagonal transfer.

je suis charlie

 

[NormPeterson]

if we are talking about weight, (measured on a force scale), and not mass

So let's use the correct term here to avoid any such confusion - it's load transfer, not weight transfer. Regardless of how common the latter term seems to be in nontechnical conversation.

then sway bars can transfer weight front to rear and vice-versa diagonally

Unless what you're mean by "transfer weight" is really something like "transferring load transfer", no they can't. The resolution of front or rear roll moment resisted by each individual sta-bar into load transfers is statically determinate without looking at the other end of the car at all. Looking at both ends together was already done when the front and rear shares of the total roll moment picked up by each bar were determined, no need (not to mention misleading) to refer to both ends again.

You can crank a twist into the bars to create a static stagger of weight. They do the equivalent of this in NASCAR.

Not the same thing at all. Here, you're changing a force boundary condition (maybe calling it an applied displacement boundary condition would be more accurate) rather than a stiffness. I get that you'd do this to shift the resulting wheel loadings around a little to benefit behavior at some point during cornering, but it's a separate effect from anything you could do with sta-bar stiffnesses alone.


Norm

 

[GregLocock]

Seems odd that all the handwavers can't put a diagram together to present their case analytically..

Noah wrote "Greg, regarding your FBD, I'm a bit puzzled as to your choice of view; in side elevation the lateral force vector is invisible so deriving its force resultants and effects is problematic, no?

Front or rear elevation makes more sense to me."

Well, that's the power of using the side view. It demonstrates that in the absence of external longitudinal or vertical forces, or accelerations, no internal reactions to lateral loads can change the vertical force on each axle. You can get absolutely tiny changes by jacking one axle or the other, and hence moving the cg fore-aft, but if you do the calculation it is a small effect even with a production car, with a relatively high cg. You certainly can get jacking just from rolling the car, it is not likely to be more than 50mm at the axle centreline. So it'll tilt the sprung mass by 1 degree (late edit), oh dear I just lost interest entirely.





Cheers

Greg Locock


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