Basics Of Shock Absorber Tuning- 3

In the first post of this series I talked about transmissibility, but I never really explained it. Let’s back up and finish that thought while showing why I think double-digressive shock valving is the key to good grip. This is going to be fairly technical, but if you slog through it to the end I think you’ll feel rewarded.

In Figure 1 below I’ve reproduced the standard one degree of freedom (1-DOF) transmissibility vs. frequency ratio chart that’s in all textbooks (and many papers) about vibration isolation. A 1-DOF model means one spring and one mass. Most of you probably already know that a standard car suspension is at least 2-DOF at each corner, i.e. it has two masses (sprung and unsprung) which can both move. This was illustrated in the previous blog post. Therefore it takes two numbers to describe the state of the system, i.e. two positions, one for each mass, thus 2-DOF. So, the figure below is for a model that’s even simpler than a real suspension, but I think we can learn a basic truth or two from it nonetheless.

Figure 1

So, what does “frequency ratio” mean? That means the ratio of any particular input frequency to the natural frequency of the system, i.e. the Fn we talked about previously.

So, a frequency ratio of 1.0 on the horizontal axis means 1.0 x Fn which is Fn. A ratio of 2 means a frequency of 2Fn. Remember, a typical Fn for a sporty car might be 1.5Hz, so a ratio of 2 would mean 2 x 1.5 = 3Hz for that particular system. This is the ratio of the frequency, the cyclic rate, of a disturbance to the system as compared to the system natural frequency.

The transmissibility ratio, let’s call it T, are the numbers listed vertically on the left side of the chart. They mean the ratio of response to input. So if the T is 1, it means that if the tire (in our case) encounters a bump 1 inch high then the mass of the car (the sprung mass in our case) will also move up 1″. The ratio of output to input (T) is then 1.

This is exactly the case at zero frequency ratio (the left, vertical axis) and it might represent, for instance, going over bumps very slowly. Imagine slowly creeping over a series of 4″ high speed bumps. How much does the car lift over each one? Right. 4 inches. If you go slow enough the spring won’t compress at all and the sprung mass won’t bounce upward.

Now please notice the five curves that are labeled with the values 0, 0.1, 0.2, 0.5 and 1. These numbers indicate percent of critical damping or damping ratio in the system associated with each curve. 0.1 means 10% of critical damping, 0.5 means 50% of critical damping, etc.

Critical damping is the exact amount of damping that will make the disturbed mass return to it’s original position in the least amount of time and not overshoot. Anything less and there will be some overshoot, i.e. some amount of bouncing around the zero point. Any more damping and it will take longer for the mass, once disturbed, to get back to the zero point. It might take 0.5 seconds, or 5 seconds, to get back to zero, depending on the damping level. If we have excess rebound damping in our shock absorbers as compared to bump damping the system responds too slowly after a compressive impact and doesn’t have time to get back to zero (the static ride height) before the next input, i.e. the spring can’t force the shock to extend fast enough so gravity drops that corner of the car a little bit. After a series of inputs near Fn the entire car may “jack down” onto the bump stops. On the other hand, if we provide more bump damping than rebound the car will tend to jack itself upward, increasing the ride height until the bumps (or driver inputs) cease. Fine tuning the shocks with shaft velocity histograms, something most professional race teams do, is at least partially based on the idea of equalizing the damping (energy dissipation) between bump and rebound, in which case the car will not tend to jack in either direction.

An aside: Do you think you could fine tune a shock that has, say, four times as much rebound as bump, as many linear-valved shocks do have, to produce equal energy dissipation between bump and rebound? Not bloody likely.

If there’s no damping at all in the system, then theoretically, if the disturbance is at Fn = 1, the system will resonate out of control. This is why the upper portion of the damping ratio = 0 line can’t be seen. It’s off the top of the chart. It goes upward to infinity.

Please note that as the damping is increased the resonant peak we see at a frequency ratio near 1 becomes less pronounced. When the damping is critical (1.0) then T is just slightly more than 1.0 at a frequency ratio of 1 (Fn). Even though there’s no line on the chart for more than 100% critical damping, nothing prevents us from using more, but the T-curve will never drop below 1.0.

One more key thing about this chart: notice that all the lines converge back to exactly T = 1 at a certain point, about Fn = 1.4, and then to the right of this point they all descend below 1, but they are not all equal after Fn = 1.4. The 100% critically damped line transmits about 60% of a disturbance at a frequency ratio of 3. At the same frequency ratio the 10% damped line transmits only about 15% of the disturbance. This is a huge difference, a factor of 4. That would be the difference (400%) in the vertical movement of the sprung mass of a car in our case.

Again, I need to remind you that this is a very simple system model, but this chart says that no matter how much damping you add, T equals 1 at Fn x the square root of 2. No matter how much damping we add it won’t ever go below 1 from zero to 1.414Fn frequency ratio. (At least we’ve obtained control over the resonance.) That means that for any spring-mass system there’s a range of input frequencies that will always significantly affect the mass. This is a basic limitation of this type of passive vibration isolation.

Alternatively, to the right of 1.414Fn damping hurts transmissibility. The more damping we have the higher the T at all frequencies above 1.414Fn. If we don’t have any damping out there then T can get very small. This basic relationship has been shown to mostly carry through to 2-DOF systems (what we have at each corner of our cars) as well. (See part 2 of this series where a 2-DOF model showed this result.)

Why do we care? Basic Truth #1: It turns out that we can basically equate low Transmissibility to both good ride quality and low tire force variation. We then equate low tire force variation to high cornering grip.

I know, that’s a lot of equating, but each has been shown to be basically true by many different people.

What’s the real-world effect? Here’s an example. I used to work at a place where I traveled over a very poorly paved country road on my way home each day. It was not a series of isolated bumps, but a consistently bad, quite rough surface. At the speed limit the ride was really bad. But, if I slowed down (reduced the bump input frequency) the ride got even worse. About 20% below the speed limit was the worst… in fact, it could get downright dangerous. The car would begin to bound and yaw even though the road was straight. On the other hand, once I sped up to 20% over the speed limit the ride began to smooth out. At 200% of the speed limit it was downright smooth, though it had become dangerous for other reasons. (Many of my coworkers were ticketed for speeding on this stretch of road.) Near the speed limit the bump frequency was probably exciting either the Fn of the unsprung or sprung masses of the car, perhaps sending the wheels off the pavement and causing the corners of the car to pitch and jump.

The classic example of an input with a consistent frequency is a road paved with cobblestones. At the right (relatively slow) speed driving over cobblestones can produce an input of around 10Hz. Guess what has an Fn of 10Hz? The unsprung mass at each corner of many cars. The tires can be seen lifting off the surface of cars with “normal” amounts of damping in their shocks. The tires move vertically up and down more than the vertical gap between the stones, i.e. T is greater than 1.0. The unsprung mass is resonating at it’s natural frequency and it can feel like you’re driving on ice.

Remember, this is a theoretical result for a simple system. There is no system in nature that actually has zero damping, but there are plenty of systems that have so little damping that resonance is a real problem. Resonance can destroy crankshafts, for instance, if they are run at the natural rotational frequency and the harmonic “balancer” (damper) is incorrect, non-functional or has been removed. This is also why the Porsche flat-12 engine developed in the 1960’s for the 917 has the power take-off from a gear in the middle of the engine block. This divides the crankshaft into two shorter, stiffer halves, each half having an Fn above the maximum engine speed just like the flat-six of the 911. No harmonic damper was necessary and the crankshafts didn’t break (too often.) Most V8 motors have crankshafts that are not stiff enough to push the Fn above the RPM limit, so they must have harmonic dampers to prevent failure.

Now we have arrived at Basic Truth #2 that I think we can take from this standard Transmissibility chart: Since, to produce the least T and therefore produce the best grip (if you accept all the “equates”) we want lots of damping to the left of 1.414Fn and little to no damping to the right, that means we want digressive (or even regressive) shock valving.

If you don’t quite see how I got to this point it may be because on a shock dyno chart we don’t see a frequency ratio, we see shock shaft velocity. So, there’s one more “equate” I must bring in, namely:

High shaft velocity equates to high input frequency and vice-versus.

When it comes to actual roads this relationship has been shown to be basically true. High shaft velocities are typically caused by short, steep bumps where the peaks are close together, producing a high frequency input. Low shaft velocities are caused by longer, smoother bumps where the peaks are farther apart, producing a lower frequency input. This is simply a fact given the nature of paved roads. We know from data gathering in cars that driver inputs, like corner turn-in, produce a low-frequency, low shaft speed input as well.

Where is the dividing line between high and low frequency input? It’s at Fn of the sprung mass times the square root of two: 1.414 x Fn as stated previously. In the automobiles we drive Fn ranges from about 1.0Hz for a softly sprung passenger car up to 3Hz for a very stiff, non-aero race car, which gives us dividing lines that range from about 1.5Hz to 4Hz. (The Fn of the unsprung mass is, in all cases, quite a bit higher than 4Hz, which is an interesting twist and will probably be discussed in the future.)

Ah, but what shock shaft velocities are associated with this range of frequencies? That’s not so clear. The general relationship of higher equals higher and lower equals lower holds, but there is no hard and fast precise relationship, I don’t think. (That would make this too easy!) Dennis Grant decided to use 3in/s as the dividing line for a car set up with 2.2/2.5Hz front to back Fn numbers. I’ve seen everything from this number down to 0.5in/s on actual shocks. The ones I’ve been running have knees at about 1in/s when set stiff, though this was done by a shock rebuilder long before I knew what to ask for and they are, in general, too stiff, i.e. they have too much total damping due, I think, to the linear rebound curve and the aggressive but digressive bump curve. (I think I told them “somewhere to the left of 3in/s” and I have to run them at full soft in compression to get best grip, which is not optimal for shock performance.)

This post has presented the core analytical support for having the damping force vs. shaft speed curve rise quickly to a “knee” and then change to a lesser slope after that. This is the real reason for double-digressive shock curves. This is also why linear shock curves, either on bump or rebound, are generally not the best for grip. They tend to be incapable of producing enough damping below the dividing frequency (1.414Fn) without producing too much damping above it, especially if you must run on a poor surface. Given this limitation, and the fact that too much low speed bump damping throws a car off-line when hitting a sharp bump, the shock builder has little choice but to crank in a lot of rebound, often the 4 to 1 ratio mentioned earlier, to get the total damping up high enough for good body control and sufficiently fast transient response.

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