One of the common questions asked by those of us new to suspension tuning is “what is coilover preload, and what is its effect on the properties of the spring, particularly spring rate?”.

The first part of this question is easy. Many coilover units come with a threaded strut body. The lower spring perch is threaded onto the body. This perch is turned to raise or lower the ride height of the vehicle. Preload is when the spring is compressed between its upper and lower perches on the strut body before the assembly is placed in the car.

The more important and contentious part of the question concerns the effect preload has on the properties of the spring. The answer to this question is contentious because the principles behind it are, at least for many people, counter-intuitive the first time around.

Many people intuitively associate spring rate with the amount of resistance a spring provides when it is deformed (stretched or compressed). The resistance provided by a spring against deformation is known as restoring force, and is variable depending on how much the spring is deformed. The spring rate is the amount of force (weight) needed to compress a spring a given distance. For example, a 400lb/in spring rate means that it takes 400 pounds of force to compress the spring one inch. Here is why preload has no effect on spring rate:

## Hooke’s Law

Hooke’s Law: F = -kx

There are tons of resources to find out exactly what Hooke’s Law is, but for our purposes it’s enough to know that this equation tells us what is going on with a spring when we know two of these three variables:

F = Force, aka Restoring Force (Weight)

k = Spring Constant (Spring Rate)

x = Distance Spring is Displaced From Equilibrium (Compressed or Stretched)

I’ve created a crude example below to illustrate a few of these variables. Spring A is at equilibrium. This means that there is no force being applied to it, and therefore also has not been displaced (x = 0). Spring B has force applied to it, illustrated by the arrow facing downwards with the F above it. The distance the spring has moved downward from equilibrium by the force is called the displacement (x), illustrated by the red line.

Notice that there is also an arrow facing up with an F below it. This is here to illustrate an important property of the spring. All force that is applied to the spring will cause the spring to exert the exact same amount of force in the opposite direction, pressing back against the force being applied to it. This is known as restoring force.

## A Simple Example Without Preload

Let’s start with two variables we know, and use the equation to find the third:

I recently got my Mazdaspeed3 corner balanced. The front-left corner of the car came out to a weight of 1066lb. We have to remember that this weight includes all weight on the front-left corner, including unsprung weight (wheel, tire, brake assembly, wheel hub, drive shaft, control arm, ect.), so we need to remove all that weight from the figure. I estimated the unsprung weight of that corner to be around 143lb (I got the odd number of 143 because I knew the exact weight of my wheel/tire combo was 43, and I estimated the remainder of the weight to be 100lb), so:

*1066lb – 143lb = 923lb*

*F = 923lb*

The other variable we will use is K, which is the spring constant (spring rate). We will use the spring rate from the Mazdaspeed Coilovers, which reports the front-left spring rate to be 400lb/in. What this number means is that it takes 400 pounds of force to compress this spring one inch. So:

*k = 400lb/in*

Now let’s setup our equation:

*F = -kx*

923lb = 400lb/in(x)

Solve for x:

x = -2.31in

Remember that x equals displacement, or distance the spring was moved by the force. The negative indicates that the spring was compressed (as opposed to stretched). So x = -2.31in means that a force of 923lb placed on a 400lb/in spring compressed that spring 2.31 inches!

## A Simple Example With Preload

Let’s recreate the last example, but this time with a spring preloaded one inch. First what we are going to do is use Hooke’s Law to find the amount of restoring force (F) the 400lb/in (k) spring has with one inch of preload (x):

*F = -kx*

*F = (-400lb/in)(-1in)*

*F = 400lb*

What this F value tells us is that our preloaded spring is exerting 400 pounds of restoring force on it’s perches before it is even installed in the car. Now we need to install this preloaded coilover into the car. This is the point at which many people’s intuition leads them astray. When you factor in the weight of the car pressing down on the coilover you have to SUBTRACT its weight from the force already stored in the spring! So since our spring already had 400 pounds of force stored in it we subtract that amount of the 923 pound weight of the car:

*938lb – 400lb = 523lb*

Since our spring already had 400 pounds of force preloaded, the spring only has to absorb another 523 pounds (F). Let’s find how much additional compression will take place (x):

*F = -kx*

*523lb = 400lb/in(x)*

Solve for x:

*x = -1.3in*

So when we place the car’s 923 pounds on a spring preloaded one inch (400 pounds of stored force) it compresses the spring an additional 1.3 inches. Add this to the preload compression:

*1.3in + 1in = 2.3in*

Look back at our first example with no preload. You end up with the exact same amount of spring compression (2.3 inches) regardless of whether or not there was any preload!

Remember! Any force already stored in a spring acts against any force that is being added to it! This goes for all linear and progressive rate springs!

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Your theory on preload is flawed. One inch of Preload does NOT shorten the shock or lower the car 1″ or even at all. So a preloaded coilover will not lower a car the same amount as a non-preloaded coilover.

Sorry, I see I read it wrong and you are talking spring length, not ride height in a car, as where my head was at when reading this.

No problem! Do you know which section gave you that impression? Maybe I could make the point more clearly.

So when setting preload on my car. I have a single turbo a gt42 to be exact to get an idea of weight, how can i preload the coil overs on my car to compensate for that weight so that when i sit in the car the ride height and balance is equal?

Preload is not used to compensate for weight nor is it used to set the ride height or weight balance. Additionally, equal ride height in each corner does not necessarily equal proper balance (when you say balance I assume you mean weight distribution). Proper balance with coilovers is set using 4 scales, one under each wheel, which is a process known as corner balancing. Once the weight distribution is set, the ride heights in each corner may or may not be equal. Moreover, when corner balancing the vehicle preload may or may not be added, but this is simply a by-product of adjusting the spring perch is order to shift the vehicles weight around.

I bought a k sport coilovers for my Nissan 240sx 1992. When I installed them, they were so hard that the car didnt moved from its position, I put the stock back again. When I bought them they said “36 levels of damping adjustment allow fine tuning of the damper to suit both handling and comfort needs” I dont know what to do with them to make them a LOT softer. Could you give me a hand?

I’m sorry, I’m not familiar with your vehicle or the k-sport line of coilovers, so I don’t think I can help you. Good luck!

Something I am missing here. To compress the spring to 400lb/in of preload and the system is static or balanced, then as you said the spring pushing up with 400lb but the ring has to be pushing down with 400lb. Then when you add 938lb to the system is not the total weight now 1328lb? I know my thinking has to be flawed, it just been a while since my physic. And it was physic I at that

My only way of seeing this is some sort of potential energy cause by the preload, but my poor mind is not seeing it yet.