Bang for Buck: ek9 springs/shocks on Em1/Ek4

[Alvis]

^ You know what man, you absolutely read my mind. I was literally just doing a search comparing Spoon v KONI Yellow shock data to see the difference.

I have the Spoon data below, but I have no idea how to interpret this - does anyone know what this means in 'very simple' language?

COM/TEN
0.05m/sec:
F85/-50kgf
R60/-36kgf

0.1m/sec:
F120/-84kgf
R90/-55kgf

0.3m/sec:
F218/-147kgf
R148/-83kgf

My spring rates are:
lbs (pounds) = 320F/200R.
kg/cm = 6.5F~ 5.5R

If I was to combine these shock rates with the spring rate would this = medium/ firm / hard / harsh ride quality?

OK guys, I'm pretty sure I've worked out what those shock figures mean - now correct me if I'm wrong - but:

1. The Spoon fixed damper numbers above are showing compression (com) and rebound (ten) (note the KONIs are rebound adjustable valved only - which is not necessarily a bad thing)

2. 0.05, 0.1 and 0.3/ sec refers to the shaft velocity of the damper in metres per second


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SO, if we look at a shaft velocity of 0.1m/sec this means the compression force applied to the front is 120kg and 90kg at the rear.

Essentially, the higher the velocity = the higher the damping.

This is also a non-linear relationship. This means if you double the velocity it doesn't necessarily mean the damping is doubled, it may be as much as four times as much for example/

For the below data I have drawn on some of the info from a motorcycle tech article, but regardless car or bike, the theory is the same.

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So know that we know this, the next question is what does this really mean and how can we understand the data?

I've tried to replicate the Spoon fixed damper rebound and compression data on a simple chart to show a curve with rebound and compression - again if someone can confirm if I've got this right, that would be great.

Note: the rebound graph is inverted because we call compressive forces positive and tensile forces are considered negative. To get a more complete picture of the entire damping curve, simply put the two curves together: Each time the shock goes through a complete cycle, it passes through zero twice-once at the bottom of the stroke and once at the top of the stroke. When the wheel hits a bump, the shock accelerates to a maximum velocity, slows down and then stops compressing.

At this point, the vertical velocity of the shock is zero. It then changes direction, accelerates in the negative direction, slows down and stops again when it's done extending:




NOW, the next question obviously is: what part of the curve is low-speed and what part is high-speed - what's a harsh ride and what's an acceptable ride? There are not specific terms defining specific velocities, they are relative terms so I guess I have to compare it to other shock data if I can find it to find this answer.

Source: http://www.sportrider.com/tech/146_9612_tech/index.html
The point is, even though the terms high- and low-speed damping are not specific velocities, they do allow suspension tuners and designers to communicate. Dyno tests have shown that velocities as low as 0.05 m/s are very important when making changes to a damping curve.

Remember that compression (or bump-damping) occurs when the wheel contacts a bump and compresses, and rebound or tension damping occurs as the spring forces the shock or fork to extend. The fact that the compression velocity is forced by the size and shape of the bump means it sees a very wide range of velocities.

Rebound, on the other hand, is largely controlled by the spring force and therefore sees a much smaller range of velocities. Typically, rebound may see one third of the velocity of compression.

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Let me know what you guys think and if I've got this right or I'm talking out of my arse lol