New discussion point: Sway bars!
Roll center (according to SAE) - The point in the transverse vertical plane through any pair of wheel centers at which lateral forces may be applied to the sprung mass without producing suspension roll.
Layman's definition - This is a neutral point for your suspension. Applying lateral force at this height will generate no roll (vertical movement at either corner).
Let's consider a 1.0 g turn in a 2645 lbs car. That means ~2645 lbs force will be applied to the center of mass/gravity. Here's an illustration:
4 possible options:
1)
Roll center height = center of gravity
There is no roll. If there is sufficient grip in the tires, your car will turn like a go-kart or a door-hinge (flat). This does sent alot of force through the control arms, and the spring/damper are not used at all.
2)
0 < roll center height < center of gravity
There will be a moment (torque) generated, since the lateral force is applied at a different height than the reaction force.
Torque = Force * distance
Roll Torque = 2645 * abs(center of gravity height - roll center height)
Since there is a torque, there will be reaction forces. Typically this duty falls on the springs and sway bars. Most race cars try to keep the roll center height at 15-30% of the center of gravity height.
3)
roll center height = ground height
The control arms won't be loaded, and all forces will be sent through the spring/damper. Not horrible, just sub-optimal.
4)
roll center height < ground height
The spring/damper will see an amplified force, and can cause the control arms to see the wrong type of force (compression vs tension). This isn't necessarily bad, but I can't recommend ever having an underground roll center, unless you overbuild the spring/damper to compensate for it.
TL;DR - Try to keep the roll center between the center of gravity and the ground. Lower is better, but don't go underground.
Load Transfer
Let's assume that we stay between #2 and #3 (because that's how a properly designed suspension should be). In a cornering maneuver, the reaction forces are generated when the outside spring is compressed and the inside spring is extended. For simplicity, let the roll center height be 10 in. Let's initially assume that there is no sway bar and the reaction force at both ends is half of the total reaction force:
Reaction force = 1/2 * roll torque / half track
Reaction force = 1/2 * [2645 lbs * 0.53] * abs(18.1-10 in) / 29.9 in
Reaction force = 189.3 lbs
Deflection = reaction force / spring rate
Deflection = 189.3 lbs / 131 lbs/in
Deflection = 1.445 in
Roll angle = arctan(deflection / half track)
Roll angle = 2.767 deg
Sway Bars
Sway bars add coupling between the wheels. Any difference in height will create a torque that will "lift" the outside wheel in an attempt to equalize the wheel heights again.
Situations:
1) In a single wheel bump, the full length of the bar is used to control one end, so the bump stiffness is halved from the numbers calculated.
2) In a two wheel bump, the whole bar has no effect.
3) In a turning maneuver, it should follow the formula:
K = pi*G*d^4*(MR)² / (16*R²*L)
where
pi = 3.141592653
G = elastic modulus
d = bar diameter
MR = motion ratio of control arm swaybar link
R = radius arm of sway bar
L = length of sway bar
basic data:
G = 8.14 x 10^10 Pa for spring steel
d = 0.018 m (front) 0.014 m (rear)
MR = not sure... but it's 0.6 on a Miata
R = not sure... but it's 0.225 m (front) and 0.121 m (rear) on a Miata
L = not sure... but it's 0.830 m (front) and 0.850 m (rear) on a Miata
Therefore:
K (front) = 14375 N/m
K (rear) = 17761 N/m
From this, we see that the FR-S is a sports car built with soft springs and stiff roll bars. This means that it will feel silky smooth on the highway, but any difference in wheel height will be heavily resisted. Strictly speaking, this is not good race car dynamics... but it works great on street cars.