Total Lateral Load Transfer Distribution

[Sleeper68]

When you add roll rate from either more spring or the ARB, you load the outside tire......and I think the inside front to some degree.

That doesn't happen when you raise the roll center

Raising a RC will load the outside tire of that end of the car, as will an increase in roll rate (ARB or ride springs). The difference is mainly that an increase in roll rate of an end increases the total roll rate of the system (i.e. the car). A RC change will merely "move around" the weight transfer you already have. A rear RC moved up will load the outside rear more, the inside rear less, and will reduce front weight transfer which unloads the outside front and loads the inside front. Just as you said.

From Dennis Grant:

"Weight transfer has two components:

Unsprung Weight Transfer: This is the contribution to weight transfer from the unsprung mass of the car.
Sprung Weight Transfer: This is the contribution to weight transfer from the sprung mass of the car, which itself is broken into two sub-components:

Geometric Weight Transfer: This is the contribution to weight transfer from the lever that comes of the difference between where the suspension naturally wants the sprung mass to roll (the roll centre) and where the sprung mass naturally wants to roll on its own (the projection of thecentre of gravity onto the ground plane)
Elastic Weight Transfer: This is the contribution to weight transfer from the reaction force of the springs and bars as they attempt to resist the spring mass roll.
These three components are additive, meaning that total weight transfer is the sum of all three components.

As a first guess, one might assume that the sprung mass works effectively the same as the unsprung mass, so weight transfer would be the sprung mass (Ms) times the lateral acceleration (LatA) times the height of the CG of the sprung mass (CGhs) divided by the track - and you'd almost be right.

The problem is determining the length of the lever through which the cornering force acts. For the unsprung mass, it's the projection of the CG on the ground (the ground on the tires being the centre of rotation). But for an articulated suspension, the point around which the suspension wants to roll is controlled by the interaction of the suspension links and pivots with each other.

Think of it this way - if you push a door on the handle, it will pivot around the hinges. If you take the door off the hinges and stand it up on an ice rink and then push the handle, it will (assuming it doesn't fall over) pivot in place, somewhere near the centre (where the projection of the CG onto the ice surface is). The suspension works the same way as the door hinges.

The trick then becomes figuring out where that pivot point - the roll centre lies. And that's where things start to get REALLY complicated.

Firstly, because locating the roll centre is really not a straightforward process. Secondly, because there is an intrinsic amount of weight transfer that is a function of the difference in height between the roll centre and the ground (the pivot point for the unsprung mass) and a second component based on the difference in height between the height of the roll centre (RCh) and the height of the CG of the sprung mass.

So to help calculate these sub-components, we call the first the Geometric Weight Transfer and the second the Elastic Weight Transfer.

Some textbooks refer to the Geometric Weight Transfer as "jacking force" and it can be helpful to keep that term in the back of your mind - because based on where the roll centre lies, that jacking force can be positive, negative, or neutral. And to further muddy the waters, the roll centre can (probably will!) move as the suspension articulates!

Breaking these down into equations, we get:

WTg = Ms * LatA * RCh / T

WTe = Ms * LatA * (CGhs - RCh) / T

Now if you pay attention to these equations, you can see that the location of the roll centre is really very important:



If the roll centre height is on the ground (0) then the geometric weight transfer is zero (no jacking force) and the elastic weight transfer is maximum;
If the roll centre height is on the CG, then jacking force is maximum and the sprung mass won't roll (the car will try and flip over the outside tire);
In between these two extremes, you get an intermediate result; and
If the roll centre is below the ground, you get "anti-jacking";"