# Front and rear roll angles

1. Jul 9, 2013

### Kozy

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?

2. Jul 10, 2013

### jack action

Or one wheel will lift off the ground:

[Broken]​

Last edited by a moderator: May 6, 2017
3. Jul 11, 2013

### Ranger Mike

roll angle is not same on both end. This is because different forces are acting on each axel. Look at photos posted above by Jack..also you can not isolate reaction to weight transfer to spring rate only. ARB is a spring and you also have aero down force impacting on the event if speed is high enough. To discount the action of the tire is to miss the entire point of dealing with weight transfer. It is all about tires, Tires, TIRES.

4. Jul 13, 2013

### Kozy

Lifting a wheel is not what I am looking at. What has happened there is simply that one axle has transferred 100% of it's load and as a result the inside has lifted. I am concerned with the roll angle prior to wheel lift.

Mike, my calculatios for roll angles end up giving the same wheel deflection front and rear, no matter what combination of springs, roll bars, ride heights, motion ratios or track widths are used. If the wheel displacement is the same, the roll angle is the same. This just seems correct to me.

If the front and rear are to roll different angles, the difference must be taken in chassis flex. Now, assuming some 10,000N/m torsional rigidity, one would have to assume that given the loads in question, the degree of flex in chassis is going to be small, to the point of not being worth considering.

5. Jul 13, 2013

### jack action

Wheel lift is exactly a different roll angle at front and rear. Look closely at the picture, especially the one in black & white: With respect to the rear axle, the body has a certain roll angle, but with respect to the front axle, the body has a lesser roll angle, hence the inside wheel is off the ground.

Even if the wheel is not off the ground, the tire is also a spring and will «elongate» accordingly, until it can't anymore. That is why your body don't need to flex to absorb the different roll angles, the tires are the one «flexing».

6. Jul 13, 2013

### Kozy

I think we may being thinking of roll angle differently. I am thinking of the chassis roll angle, i.e. purely the displacements of the wheels in relation to the chassis (tyre effects ignored). You seem to be looking more at the wheels relative to each other and/or the ground, which would include the tyre displacement?

7. Jul 13, 2013

### Aero_UoP

That's one of the reasons they put anti-roll bars, right?
http://www.turnfast.com/tech_handling/handling_antiroll

8. Jul 14, 2013

### Ranger Mike

Lets look at what happens when an automobile goes into a high speed left hand turn. Sprung weight (SW) is transferred from the rear to the front of the car, SW is transferred from the left to the right side of the car, and SW is transferred from the left rear to the right front of the car. SW transferred depends on speed of corner entry and degree of banking of the track as well as capability of the tire to adhere to the track surface. Summation of these three events mean that the right front tire will be the heaviest and the left rear will be the “ lightest”. The left front tire would see an increase in weight followed by the right rear. You will see a high degree of camber change on both front tires because of this SW transfer. Depending on the configuration of the rear axel you would see a camber change as well since the unload condition of the left rear goes to droop.

I am not sure how you get the same wheel deflection..this could be a terminology miscommunication thing. I can tell you that the amount of SW is always going to be more on the front than the rear. Please clarify how the front and rear wheel deflection can be the same? If they are both running into the suspension bump stops this might be the case but you will be sliding the tires when this happens.

Anti Roll Bars or sway bars are a method to control body roll and these have a finite spring rate just like the front and rear springs. For a properly handling race car, these 3 rates should equal the total amount of SW transferred.

Last edited: Jul 14, 2013
9. Jul 14, 2013

### Kozy

Wait, what? In pure lateral acceleration, weight is transferred from front to rear? I don't think that's correct. One axle will transfer load faster than the rear, which might give the impression that it is sending load diagonally, but think about what would happen if the acceleration increased to the point that both inside wheels lifted. The weight distribution on the two outside wheels would match the static weight distribution, say in this case 62%.

My calculations will give (FRz + RLz)/Total vehicle mass = static front weight distribution, for all values of A.

Where FRz and RLz are the front right and front left tyre loads respectively.

Now, if weight is transferring diagonally as you say then (FRz + RLz)/Total mass would not be constant, that would suggest that at some point before both inside wheels are fully unloaded, all that weight that went forwards will transfer back, instantly, to the rear axle. This makes no sense what so ever.

No, this is no miscommunication, I mean what I say. When all is calculated to give the elastic load transfer going into the springs (i.e. total load transfer across an axle - geometric + unsprung) you are left with a load and a spring rate, which results in a deflection. In my model, that load is always proportional to the rate and thus the deflection of the spring is always such that when motion ratio is taken into account, the wheel deflection is always equal front and rear.

This just makes complete sense to me. All these other explanations seem to be wrong.

If you want to run the numbers with you method, here are my parameters.

Vehicle mass = 2575lbs.
Track front = 58"
Track Rear = 58"
Front weight distribution 62%
CGh = 22"
RChF = 2"
RChR = 5"
Axle height = 11.75" (F+R)

Spring rate
Front - 250lbs/in
Rear - 250lbs/in
Motion Ratio
Front - .7
Rear -.7

Gives a wheel rate of 123.5lbs/in front and rear.

Unsprung mass
Front - 66lbs
Rear - 62lbs

ARB (simple model)
Front
Diameter - 26mm
Length - 32"
Arm - 12"
Motion ratio - .55

Gives a wheel rate of 161lbs/in

Rear
Diameter - 22mm
Length - 32"
Arm - 6"
Motion ratio - .59

Gives a wheel rate of 379lbs/in

In total, we have a front wheel roll rate of 161+123.5 = 284.5lbs/in and a rear wheel roll rate of 379+123.5 = 503.5lbs/in.

Now, based on those parameters, at 0.8g lateral acceleration I get

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

Front unsprung LT - 22.17lbs
Front geometric LT - 40.40lbs
Front elastic LT = 290lbs - 22.17lbs - 40.40lbs = 218.7lbs.

Rear unsprung LT - 19.76lbs
Rear geometric LT - 58.99lbs
Rear elastic LT = 464lbs - 19.76lbs - 58.99lbs = 385.39lbs.

218.7lbs into 284.5lbs/in = .77" wheel deflection at the front.

385.4lbs into 503.5lbs/in = .77" wheel deflection at the rear.

There are small small rounding errors written here as I've lifted numbers straight off the spreadsheet. The spreadsheet gives exactly the same wheel deflection front and rear, which is what I would expect otherwise the chassis will be twisting to accommodate different angles at each end.

10. Jul 14, 2013

### Ranger Mike

Kozt...you said "Wait, what? In pure lateral acceleration, weight is transferred from front to rear? I don't think that's correct. " I never said that.

Kozy - your statement "One axle will transfer load faster than the rear, which might give the impression that it is sending load diagonally, but think about what would happen if the acceleration increased to the point that both inside wheels lifted. The weight distribution on the two outside wheels would match the static weight distribution, say in this case 62%. " I think there are cases when the rear transfers load faster is called over steer and is quite common.
One huge factor is the tires interaction with the pavement. This is the critical thing regarding wheel deflection, weight transfer and the like..And this will impact on total wheel deflection. BTW,,wheel deflection does not mean a thing to me..it is the ability of the tire to adhere to the track surface.
see bottom corrected post

Last edited: Jul 14, 2013
11. Jul 14, 2013

### Kozy

Where did you get 515lbs total load transfer?

How do you figure 75% goes to the front?

12. Jul 14, 2013

### Ranger Mike

kozy- I am at the track and again..do not have access to my notes on calculating spring rates so I looked on Racer car suspension class post on page 2 to see how I calculate our spring rates..like I said I had the calc for 1.3 Gs . If I remember correctly..you need the Gs your tires are capable of, car weight, Center of Gravity and track width..so .8 G x 2575# car x 22" CG = 45320
45320 / track width of 58" is 781 total weight transfer..right??
75% of 781 = 586 pounds coming to the front end.. so you need spring and arb rate of 195#

Last edited: Jul 14, 2013
13. Jul 14, 2013

### Kozy

No disrespect Mike, but your math makes no sense.

First, how do you 'know' 75% of the 781lb* load transfer goes to the front?

Secondly, how do you get from 586lbs load transfer to needing a spring AND bar rate of 195lbs? What do I need 195lbs for? Keeping the car off the bump stops? surely you'd need to know how much travel there was before they came into play?

The current car has a total front rate of 284lbs/in. On that roll stiffness, I figure out how much load transfer goes to the front, it's almost directly proportional to the roll stiffness distribution, which in this case is around 36%. The remaining 64% transfers across the back axle.

Trying to work it backwards simply does not work, how can you possibly know the load on the outside front without knowing the roll rate?

*This applies on the general equation you have listed. My method splits up the unsprung and sprung mass load transfer, which results in a different (lower) figure. The sprung mass in this case is 2320lbs and has a roll moment of 18.9" (22"-3.1"RCh at 62% wheelbase) giving 603.5lbs elastic load transfer + 40.94 unsprung + 99.4lbs geometric load transfer = 744lbs total.

Last edited: Jul 14, 2013
14. Jul 14, 2013

### Ranger Mike

Am at disadvantage being away from my notes..aka data base of old binders full of chassis set ups..so can not remember the math regarding the % right rear weight,,i can tell you that practical experience confirms the right rear spring rate works out to 25% more weight..ergo the 75 - 25 rule of thumb.
regarding the spring rate of 195...you have two front springs and one ARB so the 585 # front end weight divided by 3 = 195# spring rate.
Kozy..if you are happy with your method of calculating load transfer...great...
I got a race to run

Last edited: Jul 14, 2013
15. Jul 14, 2013

### Kozy

Very well Mike, crack on!

Meanwhile, does anyone else have any thoughts on my theory? That is that in pure roll, the wheel rates are proportional to the wheel loads meaning the wheel displacements are the same, and therefore there is only one roll angle and no twisting moment acting between the axles.

16. Jul 14, 2013

### jack action

The torsional stiffness of the car is what it is. You could easily modeled a torsional spring between the two axles and find out the actual difference between the rear and front roll angles, which will be zero if the frame and the tires are infinitely stiff. I guess that makes you right on that point.

Some note about what you stated in the OP and your calculations:

But in your calculations, your wheel rates does include the anti-roll bars and not the spring alone. That was kind of shocking statement in your OP.

In an other of your statement, you said:

I think it could make sense, as when both inside wheels are off the ground, it could be due (maybe always?) to one axle hitting the bump stop. Thought experiment:

Imagine you have an solid rear axle with no suspension at all, and a front axle solidly fixed as well, but with a pivot point, like this tractor:

[Broken]

The rear axle has an infinite roll rate and the front axle has a zero roll rate.

While cornering, all the weight transfer goes to the rear axle. At one point, the rear wheel will lift such that the rear outside wheel will have the rear weight plus all the weight transfer. If the vehicle continues to roll, the front axle will reach its bump stop, lifting the front inside wheel as well. At this point, the front axle will also have an infinite roll rate and weight transfer will shift instantly to the front.

Last edited by a moderator: May 6, 2017
17. Jul 15, 2013

### Kozy

No, I stand by what I said. The deflection of the springs is what causes the suspension to move. The ARB rate combines with the spring rate to add total roll rate. In doing this, the bar effectively takes load away that would go to the springs meaning the car rolls less. Take the example of inifinate bar rate and zero spring rate, the bar would take all of the load away from the springs, meaning no roll. The springs and the load upon them are what allows the car to roll. Everything else just acts to take away spring load resulting in reduced roll.

Interesting one! OK, I ran a similar setup though my model, I'll tell you what it tells me and lets see what we make of it?

I set the rear track at 60", front at 40", CGh at 25". Rear spring at 10000lbs/in front at .1lb/in. (Can't have 0) Weight distribution 30% front.

No arbs, but the FRC is 11" (I believe it's at the axle centerline on a solid axle?) and rear at 15".

This tells me, again, that for all values of A, the front weight distribution is 30%. The load transfer at the front cannot be zero, due to the unsprung effects and the non zero RCh. The load transfer is 17% at the front, and 83% at the rear. This acts to ensure that the axle weight distribution always remains the same. This is because the weight going across the rear can only ever be half the static rear weight. Once that's gone, that axle is done.

The wheel lifts at .89G (it's a high performance racing tractor?) At this point all the remaining load transfer goes across the front axle, and with very little roll resistance, it rolls quickly. Crucially though, the weight distribution F/R is still 30/70. It's just that 70% of the load is on the RR tyre and the fronts are splitting their 30% between them. Up to the point that the inside front comes unloaded at 1.04G all that front axle is doing is shifting its 30% share of the weight over to the outside wheel.

If it hit a bump stop somwhere on it's way to maximum roll, then yes the spring rate would go infinate, and all that would happen is the remaining load transfer across that axle would happen instantly.

No weight went forwards, and none gets sent back.

Again, this just makes sense to me displayed mathematically.

Where some differential roll will stem from is the tyres. A tyre with the massive sidewall of the rear tyre, carrying 70% of the static load, is going to compress more than the smaller front tyre carrying less load.

Is this the source of chassis twist?

Last edited by a moderator: May 6, 2017
18. Jul 15, 2013

### Ranger Mike

you can teach an old dog new tricks

Kozy...my apologies. I was in a less than favorable mood yesterday...qualifying was the pits...and race did not go good...anyway I got back to home base and dug up the notes. The weight transfer figures I had posted were from an old class by Duke Southard of New Smyrna Beach Florida. Short Track Chassis set-up. It was from a class on proper spring selection. The ball park figure assumed ( death to an engineer..never assume nothing) 25% of the transfer
weight goes to the outside rear tire and 75% comes forward in a turn UNDER BRAKING CONDITIONS. Your original post said in pure state acceleration and I agree whole heartedly this is the case...until braking comes in the picture. Dukes formula gets a typical week end warrior in the ball park spring rate wise but it was a bad assumption on my part to interject it in a detailed discussion of weight transfer. I will amend my post in Race Car Suspension Class to discuss this in detail with applicable formulas.
Thanks again for getting this old dog back on the beam.
rm

19. Jul 15, 2013

### Kozy

No problem, always good to get the old grey matter sparking!

What I'm doing is putting a load of my models online and I want to make sure that I've got stuff as accurate as possible as I do not want to be touting incorrect information. Combined lat./long. acceleration is a bit beyond what I am gunning for (at the moment at least, I have models for each individually) so if it turns out that in pure roll the amount of chassis twist can be neglected then I am happy I do not need to incorporate it into the model.

I'll certainly have to work on adding the tyre deflection in, that has become apparent!

20. Jul 15, 2013

### Ranger Mike

I would not worry about twist..
question- In all my research the formula for total weight transfer is pretty easy to calculate. All the relative discussions I have seen uses the wheel base and track width and CoG height and CoG location...and has provision for Gs,,,is the percentage front to rear weight transfer based on wheel base to CoG location value?
the easy way out is to buy a weight transfer calculator of the web and roll with it but I have to understand the math..
any ideas?
also I have been getting a figure of 20% weight transfer rear to front under 1 G braking ..over several examples..I hate assumptions but is 20% a solid number?

21. Jul 15, 2013

### Kozy

The 'total' load transfer would be, if it were ever achieved. As I mentioned, the weight distribution with both inside wheels unloaded will be the same as static load distribution. Since cars rarely reach this state and will almost always have at least one of the inside wheels loaded, we are left with a partial percentage of load transfer. Under these conditions, percentage of load transfer allocated to front and rear depends on the roll stiffness distribution. Obviously for that one is required to know the spring rate, bar rate, RCh etc etc.

You can have 80% of the weight over the front axle but if the rear takes 80% of the roll stiffness, that's the end that's going to see the faster rate of load transfer up until the point that all the load has been transferred. I underlined that as I feel it is significant. In total, the front will see more load transfer, because there is more load there to transfer in the first place. What is different though is that while the inside rear has load on it, the rate of load transfer at the front will be low as the rear is doing the work.

If the load transfer rate is 100lbs per .1g then of that 100lbs, 80lbs will go across from inside to outside rear, and 20lbs will go from inside to outside front. If that rear axle only has 500lbs on it at rest, then it can only transfer up to 250lbs off the inside wheel. This would suggest that the inside rear wheel would lift at .32G. From this point, all further load transfer takes place at the front, to the point we reach total load transfer.

At the point of lift, the front axle will have transferred 64lbs, verses to the rears 250lbs. Now increase that acceleration from .32g to .52g. All of that 100lbs per .1G now goes across the front axle. So now at .52g we have that 64lbs + 200lbs additional = 264lbs total vs the rears locked out 250lbs. Small change sees the bias shift! The front would continue transfer load until the inside front became unloaded and we reach the point of total load transfer.

Note: At all points, the weight distribution is still 80% front.

Now, this is an extreme example to illustrate the point, and most cars won't be lifting a wheel until north of .7G lateral, so we can see that for a lot of the range of expected accelerations, we have all four wheels planted and some value of partial load transfer which is being directed according to the roll stiffness distribution.

My one is in work and it will be free to use. A basic version is available currently. http://blackartracing.zxq.net/Load Transfer 2.php. I am happy for you to review the maths. :)

The short answer is... it depends! (Don't you just hate that answer!)

Longitudinal load transfer is subject to the same rules as lateral, only using the wheelbase instead of the track. I would say that 20% is a pretty solid assumption though. You can effectively take vehicle weight out of the equation and call it a relationship between the wheelbase and CGh.

Forwards weight transfer % = (CGh * A)/Wb

Generally speaking the cars with higher CGs (SUVs) have longer wheel bases, so yes it would generally end up being about 20% in most cases. I'd say 15-25% would cover the majority of passenger vehicles as they all end up falling within the same narrow window for both parameters, say 2.25m - 2.75m for wheel base and 0.4m - 0.7m for CGh.

Last edited: Jul 15, 2013
22. Jul 15, 2013

### Ranger Mike

calculator looks pretty good!! question

Front lateral load transfer distribution ..how is this calculated?
At some point for total calculations.
..the tire performance curve has to be added to the mix. Once you start adding down force via the weight transfer this adds more traction to the front tires to the point they skid.
This added traction will increase the Gs and even more weight can come forward...
which means the ARB and spring rates have to be increased.
completing the full circle
,,,,right?

23. Jul 15, 2013

### Kozy

FLLTD is the sum of the outside front and inside rear tyre loads, divided by the vehicle weight. If the front transfers more weight, the % is higher and the car will under steer and visa versa.

Yes you could run it as a circular model. I have not yet done this for lateral acceleration, however I have done for braking and it is, excuse the french, a total mindf*ck.

In that model, maximum braking occurs just before the front wheels lock, so a simple task to solve the equation "Front Grip - Front brake force = 0" was done. Sounds simple, and it is until, as you have with the lateral load transfer, you realise the two components are interlinked. Obviously the grip is dependant on the forward load transfer, which is dependant on the brake force, but the brake force you can deploy is dependant on the grip...

It took a little while, but I figured it out. That model will go on the website too at some point. Way down the 'to do' list at the moment though...

24. Jul 15, 2013

### Ranger Mike

you just nailed what being a crew chief is all about...holy cow..why wont it handle on this track??
Look forward to the new package ...present one if awesome...

25. Jul 15, 2013

### jack action

I have to say that your reasoning makes sense. Thanks for an insightful discussion.

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