Where is leverage calculated from in jointed system (car wishbone/spring rate vs wheel rate) Please let me know if this is on the right SE site.
In a car suspension, the spring typically acts partway along the wishbone:

As such, there's formulae for working out the leverage ratio that exists in the suspension. The method I've been using involves the ratio between the length of the lower wishbone (d1+d2 in this picture) and the length between the inner pivot and the spring mount (d2).
However, in my research I've found a number of different ways of measuring the leverage ratio. The point of contention seems to be where the outer end of the lever is. The second method I've seen includes the additional length between the end of the wishbone and the centrepoint of the tyre. They measure the lever arm as b+c in this picture:

In terms of the physics, is the lever measured as 'b' or 'b+c' in the above mechanism?
Sorry if I haven't explained that very well, feels like a messy description. I've asked the question on a number of automotive forums, but they seem to rely on 'I've been told it works like this', without explaining any of the underlying principles as to why.
Edit: Or is it more accurate to view it as a compound series of levers?
Lever 1: the ratio between c and d
Lever 2: the ratio between a and b

 A: This is basically a simple Torgue $\tau$ equation. The Pivot point is correctly defined, and then we have the forces $F$ doing things in certain distance $r$, and the equation is;
$$Fr=\tau$$
But the as Force is a vector, and the distance $r$ must be tangential (perpendicular) to this vector, the situation is not so simple.
So first of all, even the $a$ is not correctly defined in your pictures. I have drawn it in my picture with red; the true value "a" is thus slightly less than the one in your definition.
This same goes to the tire forces. If we simplify this case and consider the wheel alignment angles Caster, Camper and Toe to be zero. Then in this rare case, when driving on straight line on even road the $r$ for the Tire force would be $b+c$, but if the car is turning left or right, this would immediately vary and might be approximately $b$ These are shown with blue lines on the Drawing.
This concludes that both ways to calculate this Leverage is just an approximation of the reality which also is variable in away that even both approximations are acceptable.
It also should be noted, that even the load of the car influences to the length of $a$, and also causes the Camber Angle to vary in away that either the inner or outer edge of the wheel is more loaded, which moves the force vector position accordingly.

A: Leverage ratio
The leverage ratio $i_L$ is the relation between the  spring deflection $d_S$ and
wheel deflection $d_W$ .
$$i_L=\frac{d_S}{d_W}$$
The leverage ratio is depending which wheel suspension you have and of the geometry of the wheel suspension.
Example: Mac Pherson strut

you start with the static equilibrium, thus we just look for the dynamic deflections.
Point B is hinge joint thus the suspension  can rotate about the x  axis .
Point w Wheel will move on a circle to w' and also point c to c'
The components of the vector $\vec{R_{w'}}$ are:
$$\vec{R_{w'}}=\left[ \begin {array}{c} 0\\ -\cos \left( \varphi 
 \right)  \left( d_{{1}}+d_{{2}} \right) \\ \sin
 \left( \varphi  \right)  \left( d_{{1}}+d_{{2}} \right) \end {array}
 \right] 
$$
thus the deflection is:
$$d_W=\parallel\vec{R_{w'}}\parallel=d_1+d_2$$
The components of the vector $\vec{R_{c'}}$ are:
$$\vec{R_{c'}}=\left[ \begin {array}{c} 0\\ -\cos \left( \varphi 
 \right) d_{{2}}\\ \sin \left( \varphi  \right) d_{{
2}}\end {array} \right] 
$$
the spring deflection is $$d_S=\parallel\vec{R}_{CA}\parallel-\parallel\vec{R}_{c'}\parallel$$
but because our calculation is from the static equilibrium, the spring deflection  is:
$$d_S\mapsto -\parallel\vec{R}_{c'}\parallel=-d_2$$
The leverage ratio for this suspension is:
$$i_L=\frac{d_S}{d_W}=-\frac{d_2}{d_1+d_2}$$
and the spring force is:
$$\vec{F}_S\approx-\frac{d_2}{d_1+d_2}\,d_w\,\begin{bmatrix}
  \cos(\theta) \\
  \sin(\theta)\\
\end{bmatrix}$$
