Overview

Topic 1 - Roll stiffness parameters

Previously, we calculated the desired spring rates that we want for our suspension, but there is more to the vehicle’s handling characteristics that we need to consider.  For example, the amount of body roll that we experience as the vehicle goes around a corner is important to us.  Generally speaking, for a high-performance vehicle like the one we have been examining, we would want to minimize body roll by having a reasonably high roll stiffness.

Roll is controlled, in part, by a suspension member referred to as the ‘sway bar’ or ‘anti-roll bar.’  This bar is mounted to the chassis and attaches at the ends to the assemblies containing the wheel spindles. 

Anti-roll (sway) bar

Text version

As the car experiences body roll, while going around a corner, the suspension on the side furthest from the centre of the corner will compress (move downward) and the suspension on the side nearest the centre of the corner will unload and extend (move upwards). 

Effect of an anti-roll bar vehicle body roll (Doody 2013:4) / CC BY-NC-ND 4.0

This will twist the anti-roll bar, and thus the stiffness of the anti-roll bar will resist the roll of the body.  A stiffer bar will allow less body roll.

To do this, we need to know the height, h, of the centre of gravity of the chassis, body and drivetrain, which excludes the wheel, tyre, hub, spindle, brakes, and any other components that are separated from the chassis by the suspension spring.  These are referred to as the ‘sprung mass’ and the ‘unsprung mass’ respectively.  What we normally know is the height of the centre of gravity (CG) of the whole car, including the unsprung mass.  So, we need to find the CG of just the sprung mass.

Topic 1 - Application

The following table gives us more of the vehicle’s parameters.  

Basic data table for the vehicle

Wheelbase (distance between the front and rear wheel centres): 261.6 cm, vehicle centre of mass is 38.1 cm above the ground.

What we know is the height of the CG of the whole car, including the unsprung mass - we need to find the CG of just the sprung mass.

${\text{W}}_{\text{TOTAL}}{\text{h}}_{\text{TOTAL}}={\text{W}}_{\text{SPRUNG}}{\text{h}}_{\text{SPRUNG}}+{\text{W}}_{\text{UNSPRUNG}}{\text{h}}_{\text{WHEELCENTER}}$  

${\text{h}}_{\text{SPRUNG}}=\frac{({\text{W}}_{\text{TOTAL}}{\text{h}}_{\text{TOTAL}}-{\text{W}}_{\text{FRONTUNSPRUNG}}{\text{h}}_{\text{WHEELCENTER}}-{\text{W}}_{\text{FRONTUNSPRUNG}}{\text{h}}_{\text{WHEELCENTER}})}{{\text{W}}_{\text{SPRUNG}}}$ 

${\text{h}}_{\text{SPRUNG}}=\left[\right(11401\text{ N}\left)\right(38.1\text{ cm})-(756\text{ N}\left)\right(28.6\text{ cm})-(1112\text{ N}\left)\right(32.4\text{ cm}\left)\right]/9533\text{ N}$  

$=39.5\text{ cm}$  

We can now calculate the sprung weight distribution and the total weight distribution:

Topic 2 - Roll centre

There is a point known as the ‘roll centre’, about which the vehicle’s sprung weight tends to roll or rotate.  The roll centre may be different for each axle, which means the car does not roll evenly, front to rear.

Roll centre (Milliken and Milliken, 1995:611)

At either axle, the roll centre can be located as follows.

1. Project a line from the centre of the right tyre’s contact patch with the road, through the instantaneous centre for the right side. The instantaneous centre is the point where the arc of the car’s roll has its centre.
2. Repeat for the left side.
3. The roll centre is where these two lines intersect.

The lateral acceleration forces on the car can be assumed to be applied at its CG. Therefore, the moment arm for the rolling moment at either axle of the car is the height difference between the CG (where the force is applied) and the rolling centre (i.e. the point about which the car is rolling). 

We should know approximately the roll gradient of that we would like to have. Failing any previous experience from our development program, the Millikens give a target range of 1.0 – 1.8 degrees/G for sedans and .25-.50 degrees/G for aerodynamically designed pure race cars.

Topic 2 - Application

Let us approximate the rolling moment arm, d, for the vehicle by using a weighted average of the front and rear roll centres (from the data table), weighted by the percentages of sprung weight distribution that we calculated in the previous step.

$\text{d = (height of CG) – (weighted average of roll centre heights)}$ 

$\text{d = 39.5 cm – [ (7.9 cm)(45.7\%) + (17.3 cm)(54.3\%) ] = 26.5 cm}$ 

The lateral force on the car during cornering is the sprung weight, W, times the lateral acceleration, A_Y.  The roll moment is therefore equal to this lateral force times the moment arm, d, which we just found.

Assume the car can corner at one G of lateral acceleration (A_Y = 1)

$\text{M =}{\text{dWA}}_{\text{y}}\text{ = (26.5 cm)(9533 N) (1) = 2526 Nm}$ 

Keep in mind that this is the moment if the car is cornering at 1G.  If the car cornered at 2G, the moment would double, etc.  So we can think of the moment as being

$\text{M = 2526 Nm / G}$ 

If we use a target roll gradient (RG) of 1.2 degrees/G (again following a Milliken recommendation) for our example car, we can calculate a desirable roll spring rate.

${\text{K}}_{\text{Rotational}}=\text{M}/\text{RG}=\frac{2526\text{ Nm/G}}{1.2\text{ deg/G}}=2105\text{ Nm/degree}$ 

Topic 3 - Finding our desired roll springs

There is some roll resistance provided by the springs which we previously sized at each corner.  The anti-roll bars must provide the rest of what we need.

The front of our car has Independent Suspension and can be represented as:

${\text{K}}_{\text{Rotational}}=\frac{\text{Moment causing roll}}{\text{Angle of roll}}=\frac{\text{M}}{\text{θ}}$  

${\text{K}}_{\text{Rotational}}=\frac{\text{(Force in spring) (moment arm to center) (2 springs)}}{\text{(linear deflection of spring) / (distance to spring from center)}}$ 

${\text{K}}_{\text{Rotational}}=\frac{\text{M}}{\text{θ}}=\frac{\text{F(T/2)2}}{\text{x / (T/2)}}=\text{(2F/x)}{\text{(T/2)}}^2=\frac{{\text{2K T}}^2}{4}=\frac{{\text{K T}}^2}{2}$ 

Topic 3 - Application

F/x is the spring rate we found previously, as the front suspension target, 63400 Newton/metre.

${\text{K}}_{\text{Rotational}}=\text{(2F/x)}{\text{(T/2)}}^2=\frac{{\text{2K T}}^2}{4}=\frac{{\text{K T}}^2}{2}$ 

$=\frac{\left[\right(31700\text{ N/m}\left){\left(165\text{ cm}\right)}^2\right]}{2}\text{(π rad/180 degree)}=753\text{ Nm/degree}$  

The rear axle has Solid Axle Suspension and can be represented as:

${\text{K}}_{\text{rotational-Tires}}=\frac{700000\text{N/m}{\left(165\text{ cm}\right)}^2}{2}\text{(π rad/180 degrees)}=16630\text{ Nm/degree}$ 

${\text{K}}_{\text{rotational-Springs}}=\frac{31450\text{N/m}{\left(119.6\text{ cm}\right)}^2}{2}\text{(π rad/180 degrees)}=393\text{ Nm/degree}$ 

$\text{Combine: }\text{1/k = 1/16630 + 1/393}$       $\text{yields }\text{k = 384 Nm/degree}$  

We previously determined that we wanted 2105 Nm/degree of roll resistance. So anti-roll bars need to provide whatever is not provided by the springs, which would be

2105 – 753 – 384 = 968 Nm/degree

The Total Load Transfer (TLT) is a measure of how much of the car’s weight shifts to the heavily loaded side as it experiences body roll in a corner.  TLT is related to the lateral acceleration, A_Y, the total weight of the car, W, the CG height, h, and the track, T, by

$\text{TLT}/{\text{A}}_{\text{Y}}=\left({\text{W}}_{\text{TOTAL}}\text{h}\right)/\text{T}$ 

For a car cornering at 1 G,

$\text{TLT}=\frac{\left(11401\text{ N}\right)\left(38.1\text{ cm}\right)}{165\text{ cm}}=2632\text{ N per G}$ 

For our car, let us assume that the actual load transfer distribution is  equal front to rear, i.e. 50% at each.

$\text{Front: (50\%) (2632) = 1316 N/G}$ 

$\text{Rear: (50\%) (2632) = 1316 N/G}$ 

We can now calculate the stiffness required for each sway bar by summing moments about a point on the ground below the vehicle centreline for the front end when cornering at one G:

$\text{2(1316 N)(165 cm/2)}$ 

$=\text{2(378 N)(28.6 cm) + (4360 N)(7.9 cm) +}{\text{K}}_{\text{bar\&springs}}\text{(1.2 deg/G)(1G)}$  

${\text{K}}_{\text{bar\&springs}}=\text{1342 Nm/degree}$  

So the front suspension has to supply 1342 Nm/degree and we already showed that the springs supplied 753 Nm/degree, so the anti-roll bar must supply the remainder and have a stiffness of 571 Nm/degree.

We previously determined the total roll resistance that the anti-roll bars need to provide to be 968 Nm/degree, so if the front anti-roll bar supplies 571 Nm/degree, then the rear bar must supply the reminder, or

$\text{968 – 571 = 397 Nm/degree}$ 

Summary

We have now accomplished the sizing of the suspension springs and anti-roll bars for our vehicle.  Having developed all of the relationships, it is possible to see how they could be turned into a computer programme which could be menu-driven and calculate the impact of an independent parameter on its associated dependent parameters.  Or, alternatively, determine values of an independent parameter which would yield a target-dependent parameter.

Reference and bibliography

Doody, M. (2013). ‘Design and development of a composite automotive anti-roll bar’ [online]. Electronic Theses and Dissertations. 5040. Available from <https://scholar.uwindsor.ca/etd/5040 > [25 March 2020]

Gillespie, T. (2000). Fundamentals of Vehicle Dynamics.  Detroit USA: SAE International.

Milliken, W. & Milliken, D. (1995).  Race Car Vehicle Dynamics. Detroit, USA: SAE International.

Milliken, W. (2006). Equations of Motion. Cambridge, USA: Bently

Seward, D. (2014). Race Car Design.  London: MacMillan.