Stiff Shocks

[hardingfv32-1]

When going through a turn does a stiffer shock setting apply more force to a given wheel/tire?

ASSUMPTION: We are not concerned with geometry changes do to chassis roll, etc.

Brian

[jpetillo]

Brian, excellent question. Assuming a zero roll, and assuming a given static ride height, a stiffer suspension (not talking about damping, and not talking about hitting a limiter so preload doesn't matter) will limit the amount the car with jack up under cornering. That will limit the CG height, where a stiffer suspension will have a lower height. Since load transfer on a zero roll resistance is only a function of CG height, the stiffer suspension will have less load transfer.

[Just to cover my fanny, if we use a variable stiffness suspension, such as running up against a droop limiter (soft or solid), we can stop the CG height wherever we want (like we do), and then we can decouple stiffness from load transfer since we can dial in whatever CG height we want while cornering]

Is this the direction you were thinking about for this discussion - am I on or off base?

This may be a subject for another thread later, but perhaps another question should be how a stiffer suspension affects cornering over various disturbances, but then I think damping needs to be included. John

[Lynn]

In autocross, with its rapid transitions, stiff shocks are often used to simulate stiffer springs in classes which are not allowed to change springs. I can't see this working as well in racing with the slower transitions.

[CitationFV21]

When going through a turn does a stiffer shock setting apply more force to a given wheel/tire?

ASSUMPTION: We are not concerned with geometry changes do to chassis roll, etc.

Brian

Too many variables.

Front or rear.

Bump or rebound (or both)

ChrisZ

[hardingfv32-1]

I guess I'm thinking about how the forces are different when using a soft "low speed" shock setting versers a stiff setting. I would presume the forces come from weight transfer, but how do those forces get to the contact patch? How is these force divided between the shock and spring?

Brian

[jpetillo]

Brian, I agree with Chris - this is too wide open, still. We need to know front or rear, since it will be quite different for zero roll resistance setups than for our front suspension.

I would suggest looking at a standard setup and not zero roll to start - so maybe the front. Let me start off stating what's obvious to get us onto the same page and then you can tell us if this is the direction you intended the thread to go.

The spring force increases roughly linearly with roll for the outer tire. It starts with the static weight for that corner of the car and ends with the additional force needed to compensate for the weight transferred at full roll.

For the damping force, let's assume that the damping rate is chosen to be critically damped for the system. In that case, the shock's damping will allow the car to roll over and smoothly come to a stop a full roll. The shock's force is only velocity dependent, so it starts off providing zero force. As the car rolls, the shock will increase its force to some peak, and then relax back down to zero force as the car comes to rest at full roll.

About how the force is divided between the spring and shock depends on your damping and spring rates. It also depends on how the wheel is turned and how the driver is transitioning the car into a particular turn, etc. Maybe you can define more constraints to keep it simple. John

[hardingfv32-1]

Yes, simple std setup, one coil-over per wheel BUT no sway bar. Smooth constant radius turn with full throttle and no brake input.

"About how the force is divided between the spring and shock depends on your damping and spring rates." Do we have a fixed amount of force that must be divided between the shock and spring at any moment in time? If so, what does it matter what the if the shock is stiff or soft?

Brian

[jpetillo]

"About how the force is divided between the spring and shock depends on your damping and spring rates." Do we have a fixed amount of force that must be divided between the shock and spring at any moment in time? If so, what does it matter what the if the shock is stiff or soft?

No, we don't have a fixed amount of force that must be divided between the shock and spring at any moment in time. That's because some of the load weight transfer force can go into accelerating the car in roll, and only the force provided by the damper and spring is transferred to the tire. The specific stiffness of the shock and the spring will affect that balance.

If you have a very, very stiff shock, as the car enters the turn and wants to roll, almost all the weight-to-be-transferred force is provided by the shock and immediately transferred to the tire. The car will continue to roll ever so slowly until the full roll is reached at which time the shock will be providing no force. The point here is that the any weight transfer force is immediately, on starting the corner, transferred to the tire with a very stiff shock.

If you have a very, very weak shock, the weight to be transferred is provided by the spring only and again will be transferred to the tire. The force to the tire will linearly rise as the car rolls until full roll is reached, and it can roll over much faster than with a shock to slow it. So, instead of the full weight transfer force being applied immediately to the tire (as with the stiff shock case), the force is taken up partially by the spring (which transfers to the tire) and the rest goes into accelerating the car in the roll direction and that does not go to the tire (until later). The roll will stop as the spring force increases with roll angle to stop it. Since the springs have to stop the inertia of a possibly fast rolling car, they have to provide the standard weight transfer force, and then the springs have to provide the additional force to stop it. In contrast to the stiff shock case, the no shock case initially transfers little force to the tires, and then as the fast(?) roll is stopped a much larger force is transferred to the tires until the roll rebounds (now less force to the tires) and settles back to the static roll angle that it would have when the roll stops (regular, static roll force, but less force than the first time when it had to stop the fast roll). Depending on the damping, this could oscillate more than once around the static roll angle with the tire force going up and down with the oscillations.

So, a nominal shock rate (perhaps the critical damping rate) gives you something in between where the initial force to the tires is reasonable and then it is allowed to grow quickly, but the roll is damped so that it can't roll fast and gain roll speed and it settles to the static roll amount without overshooting (too little damping) or undershooting (too much damping). Of course this changes with every car, driver, corner, track conditions, tires, etc. Did that make sense?

John

[hardingfv32-1]

John

"That's because some of the load weight transfer force can go into accelerating the car in roll, and only the force provided by the damper and spring is transferred to the tire. The specific stiffness of the shock and the spring will affect that balance."

I can't find anything that describes the acceleration of the car in roll. Can you provide a rough formula to demonstrate your idea.

I don't understand why we are not working with a fixed amount of force at any moment in time. I assume it is probably derived from weight transfer.

The best I can find on the subject is that force is transfered can be temporarily proportioned based on the shocks until the spring platform stabilizes.

Brian

[jpetillo]

Brian, it's really no different than the force balance equation F=m*a, where an unbalanced force will accelerate a body. For vertical forces on a suspension, the force balance equation is m*a = F_external - k_spring*x - k_damping*v. So, when the spring and damper (the last two terms) don't balance the external force, the car will move (accelerate) up and down in relation to the tires. Think of shoving down on a car with no dampers and then letting go. In that case the car will bounce up and down. That bouncing motion is the car accelerating up and down from the force of pushing down on it or the restoring force from the springs.

This is exactly what would happen in roll if you removed all damping (and friction) and shoved on the roll bar. It would rock (roll) back and forth - that's the roll acceleration I was referring to. I can write the equation for the roll if you want.

You said "I don't understand why we are not working with a fixed amount of force at any moment in time. I assume it is probably derived from weight transfer." So, let's say there is a fixed amount of force, it is balanced between the spring, damper and roll acceleration, and not just the spring and damper.

[hardingfv32-1]

Searching through Milliken for some rays of light in there section on wheel loads. They clearly state that all their discussions will be in a steady state condition. All of the formulas they discuss are familiar. It occurred to me that we are not discussing steady state if we are talking about the effects of shock valving.

Is this the source of my confusing?

Brian

[jpetillo]

Brian, yes, this is a dynamic situation, not steady-state.