How To Calculate Spring Rate–and How to Understand Cutting Coils | Handling Basics

By Staff Writer
Sep 17, 2022 | suspension, springs | Posted in Shop Work , Suspension & Handling | From the May 1997 issue | Never miss an article

[Editor's Note: This article originally ran in the May/June 1997 issue of Grassroots Motorsports]

Story by John Hagerman

For the last four decades, serious race cars have sported fully independent suspensions. Street cars, including some of the most notorious sports models, have tradition-ally used independent suspension at the front end only; at the rear, designers usually opted for the simplicity and low expense of leaf springs hooked to a live rear axle.

However, the import invasion of the '70s introduced the average American to the pleasure of good handling vehicles, and all automobile manufacturers were forced to respond to the resultant consumer demand. Thanks in particular to the ubiquitous McPherson strut, fully independent suspensions have become commonplace.

Central to most independent suspensions is a simple device: the coil spring. Although alternative springs (such as the torsion bar) are available to suspension designers, coil springs have been the usual choice.

Coil springs offer many advantages when used for racing and performance car suspensions: they are compact, light in weight, inexpensive to manufacture, and nearly infinitely durable. But one feature of coil springs that is perhaps most important to the performance enthusiast is that they allow for simple modification of spring rates and ride heights.

Unfortunately, many enthusiasts unfamiliar with the details of spring design do not take advantage of this easily modified area of the suspension. They hesitate to modify or replace their springs for fear of making an expensive mistake. For these people (as well as for those at the other extreme who ruthlessly attack their suspensions), we offer the key to understanding coil springs.

A wealth of knowledge can be gained simply by studying the formula for the spring rate (or stiffness) of a coil spring. The formula is:

Where k=spring rate; d=diameter of the spring wire; G=shear modulus of elasticity of the spring material (11,500,000 for a steel spring); D=mean diameter of the spring's coils; and N=number of active coils.

The spring rate (k) is normally measured in pounds per inch of deflection; for example, 100 pounds per inch. A spring having this rate will compress one inch under a 100-pound load. Two hundred pounds will compress the spring two inches, three hundred pounds will compress the spring three inches, and soon. This relationship will also be exhibited when the spring is extended rather than compressed.

In studying the spring rate formula, notice that the length of the spring has nothing to do with the spring rate. Therefore, the spring can be stretched or compressed without changing its rate. The spring is said to be "linear." The length of the spring affects only the ride height of the car; on a given suspension with non-adjustable spring perches, a shorter spring will lower the car, while a longer spring will raise the car. This fact also demonstrates that old, sagging springs still have the same spring rate as when they were new (unless coil-binding occurs; more on this later). Sagging springs do, however, affect suspension performance by limiting bump travel, and by causing the suspension to operate about a different segment of the camber curve. This al geometry may in turn induce undesirable handling characteristic

Next, refer back to the formula and notice the effect of the n of coils (N). The number of coils is the number of times the spring is wrapped around. If a spring has 10 coils as opposed to 20, then all else being equal, the 10-coil spring will be stiffer. Notice that relationship is also linear: the 10-coil spring will be twice as stiff as the 20-coil spring. Therefore, when coils are cut off a suspension spring to lower a car, the spring will also be stiffened.

It is also important to notice that the value used for N in the spring constant equation refers to the number of active coils. If two coils are in contact with one another when in compression, they are not active; the contact (coil binding) prevents the coils from acting as spring elements during spring compression. However, if both ends of the spring are secured and the spring is extended (rather than compressed), the bound coils will become active, since the coils become unbound during spring extension. Suspension springs commonly have two bound coils at each end of the spring. These coils would not be counts when determining N.

Normally, coil binding is highly undesirable, since suspension loads suddenly skyrocket when the binding occurs. This sudden increase in load occurs because all of the coils on the spring bind almost simultaneously, causing the spring to become, in effect, a block of steal. However, with proper design, coil binding can be used to an advantage. A rising-rate spring can be created by manufacturing the spring so that spacing between the coil layers varies along the length of the spring. As the spring is progressively compressed, this difference in spacing causes coil binding to occur first between the coils that are closely spaced. This initial coil binding reduces the number of actin coils (N), thereby causing the spring rate to rise. This type of rising-rate spring is very common in motorcycle suspensions.

The overall diameter (D) of the coil spring is a very important factor in determining the spring rate. From the equation, it can be noted that a larger diameter will yield a weaker spring. Furthermore, the effect of the diameter change is to the third power. Therefore, a spring that is twice the diameter of another spring will have '/z3, or one-eighth, the stiffness of the smaller diameter spring. The most important factor in determining spring stiffness is the diameter of the spring wire (d). Referring to the equation, note that the spring stiffness will increase with increasing wire diameter. A doubling of the spring wire causes 24, or 16, times the spring stiffness. Therefore, an increase of as little as 0.200 of an inch in spring wire diameter can stiffen a spring by over 20 percent (based on a 0.333-inch wire diameter spring).

Though a smaller coil diameter (D) will provide a stiffer spring rate, smaller coil diameters cause springs to tend to buckle during compression. Buckling is the phenomenon that occurs when a column (in this case, a spring) is compressed by the ends. The mid-point of the column begins to bend, and the column bows into an arc. A spring having a small coil diameter and a substantial length may buckle under compression, causing the spring to contact and rub the shock absorber body (assuming a concentric or coil-over shock design). An alternative spring design that would eliminate this problem would incorporate a larger coil diameter (which would re-duce spring stiffness), along with a larger diameter spring wire to restore stiffness.

The G term is the final variable in the spring constant equation. This term is the shear modulus of elasticity (also called the modulus of rigidity) of the spring material. The shear of modulus of elasticity is a measure of the ability of a material to resist deformation when a load is applied. Under a given load, a higher-shear-modulus mate-rial deforms less than does a low-shear-modulus material. The shear modulus of all types of steel is virtually identical.

If the shear modulus of all steels is approximately the same, why do "spring steels" exist? Spring steels are simply high-strength steel alloys that have the ability to withstand high loads without permanent deformation. Low-strength steel springs would initially behave in the same manner as high-strength steel springs; however, the low-strength steel would exceed its strength limitations at a much lower load. Low-strength steel can be used as a spring material if the loads seen by the spring in service remain sufficiently low.

Now that the spring constant equation has been analyzed, how is this information to be used? As an example, assume that the owner of a Lotus Europa wishes to lower the front end of his car by cutting the springs. Cutting the springs will raise the spring rate, but by how much? A stock Europa front spring has a spring rate of 116 pounds per inch and has 11.5 active coils. These two factors are the only variables needed to solve this problem. (In fact, if only the percentage change in spring rate is desired, only the number of active coils need be known.) Restating the spring constant equation:

Since d, G and D will remain constant, these terms can be ignored. Assuming the Europa owner desires to cut two coils from the springs, the spring constant will be 1/9.5/11.5, or 1.21 times higher than the stock spring. This equates to 11.5/9.5 multiplied by 116 pounds per inch, which equals a spring rate of 140 pounds per inch. This spring rate represents an increase of 21 percent over the stock spring rate. With these figures, the Lotus owner can make an informed decision regarding how many coils he needs to cut from the spring before he does the actual cutting, thereby avoiding any unpleasant surprises.

As has been outlined here, a great deal of information can be derived from the simple equation for the spring rate of a coil spring. If the spring constant equation is used as a reference when making decisions regarding coil spring modification, specification, or substitution, the likelihood of a successful decision will be considerably enhanced.

So get out those pocket calculators and start modifying your suspension the logical, scientific way. Your competitors may not notice your improved mathematical abilities, but they will be sure to notice your n quicker times.

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View comments on the GRM forums
noddaz GRM+ Memberand UltraDork
1/29/21 11:33 a.m.


I will have to read this several more times to get it.  lol


David S. Wallens
David S. Wallens Editorial Director
2/5/21 12:59 p.m.
noddaz said:


I will have to read this several more times to get it.  lol


Cool. We recently posted a few other suspension fundamental articles from the same series:

Understanding anti-roll bars

Understanding alignments

slantsix Reader
4/3/21 7:42 p.m.

Love it Good stuff from GRM.... back from the historical times when I was young, dumb and did not know about GRM.



Thanks for (re)sharing!



fearlesfil New Reader
7/19/21 3:38 p.m.

Back in '88 I put all this in Excel. I added fields for current and desired ride heights, length of control arm, where the shock/ strut mount intersected that arm (so it could calculate the lever arm length), and spring compressed and free heights. From these plus the items in the article, the current spring's stiffness and the weight on that corner are calculated.

Then I use Excel's "goal seek" Tool for a "number of coils to cut" field, which calculates to 1/10th of a coil to get to the desired ride height. As it iteratively cuts 1/10th of a coil at a time, it recalculates the increase in spring rate (due to fewer free coils/ reduced free height) and the resulting drop in ride height. When the difference between the desired and calculated ride heights goes negative, it stops and reports the most recent iteration.

The "incremental steps for reference" are just there as a sanity check that the calculated number of coils to cut are reasonable. 

I've used it successfully on several Mustangs (street and track), a Toyota AE86, a Mitsubishi Mighty Max (Dodge D-50) oval track truck, and a Formula Ford. 

8valve Reader
7/21/21 1:52 p.m.

Interesting "spring steel" and regular steel act the same way.  Had no idea.

" increase of as little as 0.200 of an inch in spring wire diameter can stiffen a spring by over 20 percent (based on a 0.333-inch wire diameter spring)."

Does not compute! *smoke from ears* 

Also steal vs steel is peeving my pet off.

twentyover Dork
1/23/22 3:04 p.m.

"Therefore, an increase of as little as 0.200 of an inch in spring wire diameter can stiffen a spring by over 20 percent (based on a 0.333-inch wire diameter spring)."

Can't be right

0.333e4 is  .01229

.533e4 is  0.0807,  or 6 1/2 times the spring rate. Now if you were to say

"Therefore, an increase of as little as 0.0200 of an inch in spring wire diameter can stiffen a spring by over 20 percent (based on a 0.333-inch wire diameter spring).",

0.353e4 is 0.01552, now you're talking 20%


Or am I missing something?

randyracer New Reader
6/8/22 2:30 p.m.

I miss John Hagerman, he really nails the engineering.  Is he still around, David Wallens?


I do think he left out an important consideration in cutting springs, however.  Most stock spring have a lot of preload, and that raises the effective spring rate.  Like the adjustable lower spring perch on some motorcycle shocks, especially vintage Japanese.  

Thus, when we cut coils from a stock spring, we do stiffen the spring as in article, but we reduce preload, so the change is not as much as it might be.



rslifkin UberDork
6/8/22 3:42 p.m.

Preload doesn't necessarily have the same stiffening effect on a typical car suspension.  Unless you're running so much preload that the weight of the car is less than the preload force (and therefore the suspension is fully extended at rest), reducing preload won't have any effect on the effective spring rate except for when the suspension is fully drooped.  

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