Roll center and effect on roll stiffness 3

[PAG24]

As we all know, the roll stiffness ratio effects the load transfers during steady state cornering (Why changing anti roll bars effect handling)

If we have a roll center on the ground, then the full lateral forces from corning acting through the CoG is 100% taken by the spring/dampers and the roll stiffness is calculated from the springs and anti roll bars.
Now if the roll center is at the same height as the CoG then all the cornering forces go through the suspension arms and the spring/dampers take nothing (Hence no roll)
Doesn't this mean the roll stiffness is now infinity? (taking the stiffness of the metal to be infinite?)

So doesn't this mean that a changing roll center height will change roll stiffness and hence change the handling?

I've always been taught that you take the stiffness ratio eg if the total stiffness at the front is 60kn/nm and the rear is the same then roll stiffness is 50/50.

But then this would only be true if the roll center is on the ground?

 

[BrianPetersen]

While it's true that you would have to use an antiroll bar of infinite stiffness to achieve zero body roll with low roll centers, the effects on the vehicle are nowhere near the same. Consider what happens when the vehicle goes over a one-wheel bump, for example. In both cases, the effects are bad. With a low roll center but infinite antiroll bar, there is no jacking or side-stepping but the wheel that didn't hit the bump is forced off the ground (and, that side of the vehicle now falls down by gravity while the vehicle as a whole gets driven upward by the compressed springs, leading to some nasty ride motions). With an excessively high roll center but weak springing, there is considerable side scrub on the tires and the vehicle body is kicked sideways.

The rule of thumb that you are using falls apart if you are examining the case of a roll center which is excessively high, as you have noted.

Real vehicles aren't designed with suspension having infinite roll stiffness and, for the most part, they aren't designed with excessively high roll centers.

 

[GregLocock]

OP You are confusing roll gain and roll stiffness. If you hold the body still, and move one wheel up and on wheel down, you are measuring that axles roll stiffness, which is obviously not infinite, whereever the RCH is.







Cheers

Greg Locock


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[PAG24]

I understand the difference in a straight line will obviously be different but I am not talking about that. Ill try to rephrase my question better as I've seem to missed the mark.

If we take a steady state corner condition. So corner of x radius with y constant speed (Not accelerating) and we have a vehicle that is completely symmetric, same track width front and rear and CG in the very middle of the wheelbase and the roll stiffness is 0.5.

Then the lateral load transfer for the front wheels will be the same for the rear wheels.

If we have the same tyres then the slip angles of the tyres will be the same and the car will be said to have neutral steering.

But now if we increase the roll stiffness at the front by increasing the stiffness of the anti roll bar or by increasing the stiffness of the springs at the front, then we will have more load transfer at the front, and less load transfer at the rear.

This will cause the front slip angle to be greater and rear slip angle to be less and will cause the car to understeer.

Now... My question is, if we raised the front roll center to the CG height then in this steady state corning condition the rear springs wont get to compress at all because the front suspension setup wont let the car roll at all.
This would cause a massive load transfer at the front and relatively little load transfer at the rear correct?

 

[NormPeterson]

Absence of roll due to making the roll moment arm equal to zero is not the same thing as making the roll zero via infinite roll stiffness. The first item addresses whether a roll moment is even going to exist, the second defines how much roll will be caused if the roll moment is nonzero.

If you only move the front RC up to CG height, while leaving the rear RC at whatever presumably lower height it's starting out at, you will still have some roll, which will divide its lateral load transfer effect according to the relative roll stiffnesses.

Your LLTD would still be heavily front-biased.


Norm

 

[GregLocock]

You need to set the roll 'axis' coincident with the cg to get no roll of the sprung body.

Cheers

Greg Locock


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[PAG24]

(assuming the tyres and chassis are rigid) and you set the roll axis coincident with the CG to get no roll.
Does the roll stiffness ratio still come into play when working out the lateral load transfer?
If it does how? as this I can't understand yet and if it doesn't then will the load transfer for front and rear be the same regardless of springs/ARBs used for a symmetric car?

 

[NormPeterson]

I've always been taught that you take the stiffness ratio eg if the total stiffness at the front is 60kn/nm and the rear is the same then roll stiffness is 50/50.

But then this would only be true if the roll center is on the ground

Assuming the vehicle to be a reasonable approximation of a rigid body, LLTD is the sum of at least three significant effects, two of which are purely static force resolutions not involving any sort of stiffness (i.e. fully statically determinate). Work out the load transfers from each of the three effects individually first before attempting to arrive at the total.


Norm

 

[KevinK2]

In the original hypothetical, there is no roll and the weight transfer is based on front and rear CG's and their distance from the ground. The wheel rates, and the roll stiffnesses, can be anything. Consider a cube with a shaft through the center ... subject it to lateral acceleration and there is no tendency for it to want to roll (ignore aero effects).

Yes, if front roll stiffness is 50% of total, Then it's a 50/50 distribution, front vs rear.

This is my way of answering the OP, sorry if repetitive.

Kevin

 

[WolfHR]

" You need to set the roll 'axis' coincident with the cg to get no roll of the sprung body."

Greg, this was always a bit unclear to me... I've seen Carroll Smith's explanation how to obtain inclination of the roll axis in side view, and it didn't 'feel right'; and I've seen explanations with different CoG heights for 'front' and 'rear' sprung mass and that RCs should be at equal distance from their respective CoG (I presumed something like the car was sliced at CoG and then CoG for front and rear part being calculated), but wasn't sure about it.

I've always wondered if this would mean that roll axis should be (in side view) parallel to the car's primary axis of inertia (or that of sprung mass)?

 

[KevinK2]

From my post:

"In the original hypothetical, there is no roll and the weight transfer is based on front and rear CG's and their distance from the ground."

Should be .."front and rear roll centers", not CG's

Wolf: The roll and CG axes do not need to be parallel to ecah other, or to the ground, for sideview. They typically never are. You need the front CG and rear CG, to establish the CG axis and the CG location on that axis. The other axis is based on F&R roll centers. The roll couple is based on the unsprung mass times the vertical distance from the CG to the roll axis. The roll couple will then be reacted at the front and rear, based on relative roll stiffnesses.

The chassis is assumed to be at least 10X any local rate.

Puhn fan, Kevin

 

[cibachrome]

The roll axis IS usually set parallel to the car's primary sprung mass inertial axis. And, ITS inclined because the masses and inertias of the dominant pieces that make up the vehicle's inertial axis tensor give it such a characteristic. That would be the motor and transmission to name 2. Forget about the windshield wipers...

 

[KevinK2]

Cibachrome, what is wrong with my descrption of how to set up these two independent axies, per the Fred Puhn method. I'm an engineer and car builder/racer, and saw no flaw in this approach to determining these axies.

 

[cibachrome]

Your statment about roll stiffness when the roll axis at the cg (actually the sprung mass cg) is incorrect. It just means that the moment arm is zero so that any or NO roll constraint is necessary to keep the sprung mass upright.

My bass boat anti-rolls during high speed cornering. Does that mean that the water provides negative roll stiffness?

The sprung mass is an inverted pendulum. It would fall over if there was no restraint. So why don't cars have very high roll centers? Because the structural reaction loads can be huge, the sensation of confidence is overwhelming, and the likelyhood of exceeding the friction capability of the surface is very likely. Plus roll kinematics are a great way to tweak insufficiencies in tire or suspension compliances and oror delta payload demands on handling. Even your race car has a full and empty gas tank load condition. Did you mount the tank at the roll axis centerline, I hope?

Keeping the roll axis aligned with the inertia ellipsoid (and near it) cuts down the mr^2 component of roll ineria. So the dynamic roll reaction dances better.

 

[WolfHR]

Plus the jacking effect, I would think (about high roll centers)... Throwing away lateral force to jack up the suspension rather than to generate centripetal force.

As for roll axis, I would think guys at Caterham know a bit or two about the cars, and their car with deDion and IRS have quite a bit different inclination (ISTR that IRS model has 30mm higher rear roll center, while deDion twice as much). Not that I'm making a point here- just a food for thought...

 

[GregLocock]

It is very misleading to compare roll centre heights for an IRS and a live axle.

Which rather reinforces my general opinion that RCH is damn close to being a busted concept.

Variations in GRC on a particular suspension seem to correlate to useful changes in measured parameters, but if you were to set an IRS up to a RCH = to the SLR of the tire, you would be an unhappy camper, yet it works perfectly well on a beam axle.

Cheers

Greg Locock


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[KevinK2]

"Your statment about roll stiffness when the roll axis at the cg (actually the sprung mass cg) is incorrect. It just means that the moment arm is zero so that any or NO roll constraint is necessary to keep the sprung mass upright."

I said: "The wheel rates, and the roll stiffnesses, can be anything."

What's wrong there?

"The sprung mass is an inverted pendulum. It would fall over if there was no restraint."

A rock supported by a needle would be unstable. If a typical car was stripped of all unsprung elements, and just supported by pins at the upper coil spring perches, it would be stable. It would be unstable if you tried to support it on a long, longitudinal 2x2 below it, if that's what you are implying.

"The roll axis IS usually set parallel to the car's primary sprung mass inertial axis"

Does not sound right. I think that at turn in, the mass polar momment of inertia of the sprung weight, relative to the roll axis set by roll centers, will initially resist body rotation. But for steady state cornering, that polar moment has no effect, and the roll couple will be based on the distance from the sprung weight to the roll axis based on roll centers.

Just trying to understand your comments.

.

 

[NormPeterson]

It would be unstable if you tried to support it on a long, longitudinal 2x2 below it . . .

Isn't that precisely what you are assuming with a LLTD model that sums the load transfers through the geo roll centers with the load transfers through the elastic elements and whatever contribution comes from the unsprung masses?

If a typical car was stripped of all unsprung elements, and just supported by pins at the upper coil spring perches, it would be stable.

Try to avoid confusing mathematical modeling or structural stability definitions with the absence of roll.

Commonly, a pin support is assumed to have high stiffness in its direction of constraint. But if you're trying to prove that roll stability exists in the absence of roll stiffness you have to construct your model such that in total it actually has zero (or at most, negligible) roll stiffness.

What do you suppose would happen to the sprung mass orientation in 3D space if you were to release those four pin stiffnesses to very low values but somewhat higher on one side than on the other? Mathematically, it would still meet the definition of being a stable structure (wouldn't rotate indefinitely about some axis), but you would see more than just heave motion.


Norm

 

[NormPeterson]

Edit last sentence to rear

It would still meet the definition of being a stable structure mathematically (you wouldn't be trying to divide by zero, and a computer solution shouldn't show it rotating indefinitely about some axis), but you would see more than just heave motion.

 

[NormPeterson]

read.