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Old 02-02-2001, 07:52 AM   #1
Memnoch
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Post Question about spring rates and dampening rates

What exactly do the numbers mean? I did a search on the old technical forum and this one but I didn't find anything except that old thread about weight transfer (which rocks btw).
I don't want to know that 300lb/in is stiffer than 200lb/in or whatever. But what that designation means itself.

Trying to be as clear as I can
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Old 02-02-2001, 08:51 AM   #2
Nat
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Lets see if I can put some of my ME schooling to work.

Springs, the spring rate (lb/in or kg/mm) is just a measurement of how much force it requires to compress the spring. If a spring is rated at 300lb/in, it will require 300 pounds of force on the spring in the axial direction to deflect it one inch. The rate is dependent on a few things, wire diameter (d), mean coil diameter (D), number of coils (n), and material constant the spring is made of (G). The formula ends up being spring constant k=Gd^4/8nD^3. The basic spring force equation is F=kx, where (F) is the force on the spring, (k) is the constant from the equation above, and (x) is how far the spring is compressed. Springs can get fun when you add "tender" or "helper" springs to the main spring. These are usually softer and will decrease the spring rate untill fully compressed. Springs can also be wound so the rate is progressive, where the rate changes as the spring is compressed to different heights.

The damping rate is a measure of how much energy is dissapated by a fluid from the system. It is determined by the piston hieght and diameter, distance between the piston and the clinder wall,and viscous coefficient of the fluid. Units are in force*distance. The damping ratio is good to know, since it can tell if the system is over/under/or critically damped. The formula for that is =natural damping coefficient/calculated coefficient. An underdamped system will bounce around a lot until it settles down, while an overdamped system will immediately go to the rest state. A critically damped system will allow small amounts of system movement, but return to the resting state quickly. Damping is a lot harder to understand, and the problems I have don't really relate to cars.

Hope this helps somewhat, hopefully someone can explain damping a little better for you.
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Old 02-02-2001, 08:58 AM   #3
Memnoch
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Gotcha. I've always loved physics so even though I never studied it, it's easier for me to understand using physics. That way you can use the same basis for other applications too.

I didn't think just one reply would do it for me but it did! Thanks

PS. Damn you're good
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Old 02-02-2001, 09:00 AM   #4
Scottie
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Well if they are linear springs, then 300lbs/in would mean it would take 300 lbs to deflect the spring 1 inch. To deflect it another inch (2 inches total) it would take another 300 lbs for a total of 600 lbs.

Progressive springs are a different story. Their spring rates start out soft, say 150 lbs per inch, and progressively get stiffer. Now, this does not mean that the spring would deflect an inch if you apply 150 lbs. No, as the spring deflects its rate increases, so its rate would likely be higher than 150 lbs before it finishes deflecting that first inch.

It would really take a graph to show this correctly. You would want to graph the force (lbf) on the vertical (y) axis and deflect on the horizontal (x) axis. As the name suggests, a linear spring's force vs deflection curve would be a straight line with its slope being the spring rate. A progressive spring's curve would not be a straight line, and as the name suggests its slope would progressively increase (steeper and steeper slope).

Many suspension manufacturers offering progressive springs, such DMS, will list the spring rate at the static ride height.

Dampening rates are the force required to move the dampener's shaft(piston) at certain velocity. The three basic types of curves that you see when plotting force vs velocity are linear, digressive, and progressive.

Agian, similar to how progressive springs are spec'd, dampener manufacturers only tend to give the dampening force at one specific shaft velocity.

These single numbers make it hard to compare how the ride and handling of one suspension may compare to another. A set-up with linear springs and linear dampening rates may have the same single number spec's, say 225lbs/in for the springs and 310/150 for the rebound and bump dampening, as a set-up with progressive springs and digressive dampening, but their ride and handling will be different.

[This message has been edited by Scottie (edited February 02, 2001).]
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Old 02-02-2001, 09:03 AM   #5
Tony
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Spring rates: Force per unit length. So with a 300 lb/in spring it would take 300 lbs or force to compress the spring 1 inch. A 200 lb/in spring would only take 200 lbs of force to do the same thing. In metric units it is given as Kg/mm (so many Kilograms to compress the spring 1 milimeter). Technically these force measurements are actually weight measurements but I don't want to confuse the issue.

Damping rates: Aren't usually given but they are in force per unit speed. Rearranging the units gets you lb*s/in. Don't worry about the units.

So to simplify a spring will apply more force the further it gets away from it's equilibrium point and a shock (or damper) will apply more force the faster you try to move it.

Hope this helps.

[Edit]: Sheesh you guys are fast. We'll just call my response the simple answer

Tony

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Old 02-02-2001, 09:10 AM   #6
Scottie
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Damn Nat you beat me to it. I guess I should have taken typing.

Nat refers to the spring constant(k) which in the case of a linear spring is its slope on the force vs deflection plot.

[This message has been edited by Scottie (edited February 02, 2001).]
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Old 02-02-2001, 10:17 AM   #7
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So with a progressive spring with an initial static rate of 150lbs/in, when compressed that initial inch, the next inch (depending on the slope) could either be the same 150lb/in or maybe 250lb/in.
If it was 250lb/in for that second inch we'd have 400lbs to force the spring to compress two inches. Just clarifying

Thanks everyone for the prompt (damn ) responses!
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Old 02-02-2001, 12:30 PM   #8
Scottie
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The change in spring rate with progressive spring will tend to be smooth. In other words there will be on steps or jumps in the spring rate.

If it takes 225lbs to compress a progressive spring 1 inch, it does not mean its spring rate at one inch of deflect is 225lbs/in. Likely the starting spring rate was lower, say 150lbs/in, thus the spring rate at 1 inch of deflect would be higher than 225lbs/in. This may be a little confusing, but it would be hard for me to explain any better without graphs and such.

In my use of the term "static ride height", static refers to the car being at rest. In such case, each spring is compressed by the portion of the car's weight that it must bear. For a perfect 50/50 weight distribution that would be 1/4 of the car's weight.

Thus, for a progressive spring, the spring rate of 225lbs/in at the "static ride height" only tells you what the the slope of the force vs deflection curve would be at that point. It doesn't tell you how much deflection there is, nor could you calculate how much deflection there would be with just that number. In other words, it does tell you a lot. I think the most you could extract from such a number for progressive springs is the relative relationship of front to rear stiffness. Which would be helpful in the area of oversteer and understeer.

As far as analyzing a particular suspension set-up, determining if its underdamped, overdamped, or critically damped and what not, I would have to say its beyond my present level of expertise. And probably beyond what you would learn in achieving your average BSME. Must undergrad text books cover simple spring, dampener (dash pot), and mass systems, but I wouldn't necessarily call a car's system simple. For one think, the dampening rate is different for bump and rebound, even when you're talking strictly about linear dampening rates. Then you have the two different masses to consider, sprung and unsprung weight. Throwing in progressive springs, and dampening rates that can vary with both velocity and position, and your head will start to throb. At least my does.

[This message has been edited by Scottie (edited February 02, 2001).]

[This message has been edited by Scottie (edited February 02, 2001).]
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