Front and rear roll angles

[Komodo86]

I have done a fair bit of mathematical and 3D modelling of independent automotive suspension systems, and have come to the conclusion that the in pure steady state cornering no matter what the combination of springs, ARB/sway bars, motion ratios or linkage geometry, the roll angle (ignoring the tyre flex component) is always determined by the deflection of the spring and that the load they see on each axle will directly correspond to the rate of the spring, meaning that they will deflect such that the wheel movement and thus roll angle, is the same at each end of the vehicle.

It has been suggested by an experienced engineer, that this is not the case and that the front and rear axles DO roll different amounts, however the mechanics of this were not explained. Surely if such a case was true, this means the chassis must be twisting to accommodate the difference in roll angle? In a typical modern chassis with say 15,000Nm/° torsional rigidity, surely the amount of differential roll, if any, must be negligible to the point of not being worth considering?

 

[GregLocock]

I haven't checked all of your maths, but I get a different answer for load transfer and hence wheel deflection.

Cheers

Greg Locock


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[Komodo86]

TC3000, that makes a lot of sense, thank you.

 

[NormPeterson]

What is the wheelbase of this car?


Norm

 

[Komodo86]

Sorry Greg there was an error somewhere.

Total load transfer is 781 lbs.

Geometric LT
Front 40.4lbs
Rear 58.99lbs

Unsprung LT
Front 21.17 lbs
Rear 19.76 lbs

Elastic load transfer is therefore 781 lbs minus the sum of the above (140.32 lbs) = 641.06 lbs.

The elastic load transfer is split according to the RSD, 36.14% front in this case

So front load transfer is 231.7lbs, rear is 409.4lbs.

Wheel deflection is then 231.7/284.37 = .815 at the front, and 409.4/502.55 = .815" at the rear.

 

[Komodo86]

Norm, the wheel base is 2560mm, however there is no input for it on the model.

 

[TC3000]

Komodo,

let's look at this another way round:
You have calculated a combined wheel rate (springs + ARB) of 284.5 lb/in front and 503.5 lb/in rear, if we combine the two, you get a total rate of 788 lb/in to resist roll of the suspended body/mass.

In other words you resist ~63.9% of the total roll moment at the rear of the car.
If you roll your body any given amount, what will be the torque experience by the body at the front and the rear?
The difference will try to "twist" your body.
If you assume a rigid body, it can't be twisted, but the reaction forces at the ground will be different --> your tyre "springs" will see a different load.
What is the twist angle of the body needed to make the reaction forces to the ground equal --> think about "cutting" your body in two parts.

On a different note, if you carry ~63% of the mass on the front, why is your front LT/OM lower then your rear?
What would be the overturning moment for each individual axle?
How is this overturning moment resisted/counter acted --> what is the resulting deflection/ roll angle of each axle?

 

[Komodo86]

So I make that 381.5lbs load transfer per degree of roll.

Of that 381.5lbs, for every degree of roll, the rear would take 247.7lbs and the front would take 133.4lbs.

The roll couple is 20" front and 17" rear. So 247.7 x 20 / 12 = 412lbft front and 133.4 x 17 / 12 = 188lbft at the rear.

The difference is 224lbft. Given the torsional rigidity of 5000Nm/° or 6779lbft/° the chassis twist is therefore .03°.

Is that on the right lines?

As for the wonky roll stiffness, it's a FWD car.

 

[Komodo86]

Got that the wrong way round.

133.4 x 20 / 12 = 222lbft front
247.7 x 17 / 12 = 350lbft rear

Difference is 128lbft and 0.018° twist.

 

[TC3000]

What would be the difference, if you assume a chassis stiffness of "0", as in the front and the rear of the car are linked via a revolute / hinge, or if the two systems would roll independently (as two systems)?

 

[Komodo86]

I don't really understand the question. That is what I just did was it not?

 

[NormPeterson]

Norm, the wheel base is 2560mm, however there is no input for it on the model.

Knowing that little piece of information better fits something I might use for sanity checking.

What the wonky roll stiffness distribution sounds like is FWD being adapted for autocrossing or track day use. In which case if you're running unusually high rear tire inflation pressures (relative to street specs), that will have an effect on all this.


Norm

 

[GregLocock]

I think you are still way out on wheel deflections. Does one inch sound reasonable at 0.8 g for a relatively softly sprung car with a conventional cgz?

Cheers

Greg Locock


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[Komodo86]

Norm.

Yes, it is a FWD car but not one set up for autocross. This car leaves the factory with that setup and is renowned for it's excellent handling. It's not the first time I have discussed the setup here and apparently it baffles the old school guys.


Greg.

You could well be right, I'll admit I am a little skeptical of such low numbers, however the maths all seems to check out. I'm open to suggestions on what you think might be more reasonable values for the given parameters?

 

[BrianPetersen]

"
Total load transfer - 744lbs
Front LT - 280lbs
Rear - 464lbs
"

How did you arrive at the proportions of front and rear load transfer?

If that was arrived at by proportioning the amount of wheel loadings based on their spring rates with the assumption of a rigid bodyshell between them (which appears to be the case) then of course you are going to get the same calculated wheel deflections, because it was a constraint that was put into your calculations - implied by assuming the load transfers are in proportion to total spring rates.

Do the calculations again by making the front and rear load transfers in proportion to the front and rear weight distribution.

I'm only doing rough calculations in my head but it's enough to indicate that the numbers are going to be different.

 

[Komodo86]

But that would be even more incorrect. The load transfer is only proportional to the weight distribution at 'full load transfer', that is when both inside wheels have fully unloaded. Up until this point, you have a partial load transfer, and the rates of transfer at each end are going to be dictated by the roll stiffness.

The stiffer axle transfers load faster than the softer one. This is a known fact.

The weight distribution at full load transfer (the car on two outside wheels) is the same as the static load distribution. Unless someone shows me test results that say otherwise, I consider this fact also.

My method satisfies both of these. Distributing the load according the weight distribution, for use at partial load transfers, does not. It suggests that A. the front axle will transfer the load faster, even if the rear is stiffer and B, that there is longitudinal load transfer at some point before full load transfer, which either instantly transfers back to the original axle at full load transfer OR, that the weight distribution at full load transfer is somehow different to the static distribution.

This, to me, sounds ill thought out.

 

[TC3000]

"I don't really understand the question. That is what I just did was it not?"

no
I had in mind to model the front and the rear as two independent systems, and I think Brian is hinting at the same thing.

Something along the lines of this:
Front sprung mass * lateral acceleration * moment arm (CG height - RC height) = Overturning Moment
Overturning Moment / half track width is a force acting at the wheel plane
Force @ wheel divided by 2*wheel rate (which is your lumped spring & ARB rate) = deflection at the wheel
sum of deflection at the wheel(s) / track width = roll angle

If you run this calculation, you will see that if unconstrained your front and rear would like to roll a different amount to reach equilibrium of moments.
The difference between the two individual roll angles is largest if the roll stiffness distribution is 100% at one axle.

 

[Komodo86]

OK, so lets run those numbers again. (Sorry, this is taking a while isn't it!)

SMf = 1465lbs
SMr = 855 lbs

MAf = 20"
MAr = 17"

OMf = (1465 * .8 * 20)/12 = 1953 lbft/°
OMr = (855 * .8 * 17)/12 = 969 lbft/°

Ff = OMf / (TRf/2) = 67lbs
Fr = OMr / (TRr/2) = 33lbs

Front roll = Ff / (2 * 284 lb/in) = 6.78°
Rear roll = Fr / (2 * 503 lb/in) = 1.9°.

OK, I see where you are coming from now.

 

[TC3000]

If you replace your rear springs with solid links, what does that mean to your model?
What happens in reality if the car can't roll around it's suspension, but does not have infinite stiff tyres?
If the rear rolls on it's tyres, but the front suspension is a lot softer then your tyre stiffness, will there be deflection in the front suspension or not?
If yes, but you still have no deflection in your rear suspension, what does this say about your suspension roll angle?

the front tyres and rear tyres have a individual plane, which can warp/twist against each other (front axle vs. rear axle), these plane(s) lay between the rigid body plane and the ground (which if flat is a plane too) plane. The rigid body can have only one overall angle in roll in relation to the ground plane (if the ground is flat), but this angle is made up of two angles: rigid body vs. wheel plane + wheel plane vs. ground plane.
The deflection of the tyre(s) (axle wise) in inverse proportional to the suspension deflection at the axle.
The stiffest axle will deflect it's tyres more/most.

 

[BrianPetersen]

The purpose of getting you to model the front and rear as two completely independent systems was not to imply that doing so had any semblance to reality, only to convince you that the bodyshell is indeed being subject to torsional loads. I believe that point has now sunk in.

The next step is to introduce a spring rate between the front and rear subassemblies, and find out where the equilibrium lies.

 

[Komodo86]

Yes I see how that works now, thankyou!

The spring rate being that 5000Nm/degree torsion stiffness?

Went and dug my copy of Millikens RCVD out of storage last night. I'm sure the answers are in there somewhere...