Progressive rate suspension using a simple (non-progressive) spring and a modified bracket
The typical setup for inboard spring/shock has a 90 angle bracket used to transfer force between the pushrod and the spring/shock.
The bracket has legs of lengths Lp (leg connected to the pushrod) and Ls (leg connected to the shock).
At normal suspension loading the bracket will be at a certain angle.
Let's define this position to be 'zero' degrees.
To make the math easier we will assume that the pushrod and shock are pushing 90 degrees to the bracket at this position. We can derive the math for the case where the pushrod or spring do not act at 90 degrees to the bracket, but it doesn't help illustrate the idea.
Fp is the linear force of the pushrod resulting from the weight of the vehicle.
The torque on the bracket due to the pushrod is: Tp = Fp*Lp*cos(angle).
Since we are at the resting position, the spring must be pushing back with exactly the amount of force to cancel the weight of the vehicle.
The torque on the bracket due to the spring is: Ts = Fs*Ls*cos(angle).
Note these two torques must be equal.
The cos() term cancels out and we find that the 'suspending' force on the pushrod is directly proportional to the force from the spring.
The bracket acts as a simple lever to increase or decrease the force of the spring.
Now consider a bracket where the two arms do not form a 90 degree angle.
Let's explore the case where the bracket is a simple bar.
One end of the bar is a fixed pivot point, and both the pushrod and the spring are connected to the other end of the rod.
(Please ignore the thin line shown at 45 degrees, this is a bug in my CAD program)
Again we assume the pushrod and the spring are 90 apart from each other.
Let's imagine what happens if we load the suspension so that the bracket rotates through various angles.
If we totally unload the suspension, the spring will push the the bar to it's point farthest away from the spring.
Let's call this angle zero.
The torque on the bar resulting from the spring is: Ts = Fs*L*sin(angle).
The torque on the bar resulting from the pushrod is: Tp = Fp*L*cos(angle).
For any given suspension loading these two torques must be equal.
We can derive the equation: Fp = Fs * sin(angle) / cos(angle)
There are some interesting results from this setup:
The resultant force on the pushrod goes from zero to infinite as the bar travels from zero degrees to 90 degrees.
The force on the pushrod is not linear with respect to position. It is in fact progressive (the more the suspension is compressed, the stronger the effective spring force)
The length of the bar will set the amount of suspension travel.
The bar at zero degrees should be wheels at full extension (wheels off ground).
The bar at 90 degrees should be wheels at full compression (undercarriage almost touching ground??).
If you graph the pushrod force vs. position you will notice the effective rate (slope of force vs. position) of the spring is not actually controlled by the real rate of the installed spring.
Rather is it controlled by the position of the bar. The amount of pre-load applied to the spring will control the position of the bar at normal suspension load. Therefore the preload controls the effective rate of the spring.
This is the beautiful effect that lets us buy only one spring, and use simple adjustments to get the desired effective spring rate.
Sure this sounds good on paper, but would it actually be useful?
As the suspension gets fully compressed the effective spring rate gets very big very quickly. Will this cause handling problems? I assume it's better than 'bottoming out'.
How does the damping force of the shock get modified by this setup?
Will the damping force be correct to be 'critically damped'. (Critically damped is the condition where the suspension will take the minimum possible time to settle after being displaced. Too little damping will cause the suspension to oscillate. Oddly enough, too much damping can cause the suspension to take longer to settle)