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    suspension formulas

    Coil spring rate

    The formula for calculating the stiffness or rate of a coil spring.

    Spring rate (K)=wire dia (d) to 4th power times Spring modulus 12.000.000 all divided by 8 times the number of active wraps(N)times Diameter of the coil(D) cubed
    . t
    K = Spring rate in pounds per inch
    W = Diameter of the spring wire in inches
    G = 12,000,000 for steel springs (a constant)
    N = Number of active coils (number of coils that are free to move + 1/2 coil)
    D = Diameter of the coils measured to the center of the wire, in inches

    Note the relationship between the number of active coils (N) and the spring rate (K).
    If you cut off half the active coils the spring rate is doubled.

    Last edited by DuceAnAHalf; 04-17-2007, 05:36 PM.

    R.I.P. Jason P Harrill 6-12-06

    Sway Bar Calculations

    Provided by KLA Industries to answer the question of how the drop-link length affects the rate of a sway bar.

    Here is the angle of the dangle formulas for calculating the change in sway bar stiffness. First of all there are only 2 ways to change the stiffness of a bar. The first is to change the bar size. For that change you would use the formula below.

    If you're substituting a larger diameter bar that otherwise has the same dimensions, the rate increase is simply:

    (new diameter/old diameter) to fourth power

    So when you change your bar from 18mm to 19mm, the stiffness increased 24%. (19/18)^4 = 1.24

    Now, if you want to calculate the actual rate or have adjustable sway bars (several adjustment holes to change length of lever arm) and want to figure out the rate increase, you'll need to measure the bar and use a formula (from Fred Puhn's "How to Make Your Car Handle").

    Formula for sway bar stiffness of a solid steel bar

    500,000 D^4
    K (lbs/in) = --------------------------------------------------
    (0.4244 x A^2 x B) + (0.2264 x C^3)

    A| / \ C
    | / \

    A - Length of end perpendicular to B (torque arm - inches)

    B - Length of center section (inches)

    C - Length of end (inches)

    D - Diameter bar (inches)

    On a bar where the bars have nearly a 90 degree angle between the arms and center section, A = C for the calculations.

    A : : C (= A)
    : :

    Example calculation for a 25mm front bar D = 25mm = 0.984 inch

    A = 11.25 inch
    B = 31.5 inch
    C= 11.25 inch

    500,00 x 0.984^4
    K = --------------------------------------------------------------------- = 233 lb/in

    ((0.4244 x 11.25^2 x 31.5) + (0.2264 x 11.25^3))

    Now for the second way to change the bar rate is to lengthen or shorten the A measurement, which is the Length of end perpendicular to B (torque arm - inches)

    As you know the sway bar ends move up and down in an arch motion. Raising and lowering the bar will effectively change the torque arm inches, which is the a measurement in the formula. Changing the torque arm inches will either change the rate softer or harder. Changing the drop link length makes this change. To find the rate change you would measure the A Length of end perpendicular to B (torque arm - inches) with the original setting and calculate the rate. You would then make the change and re-measure Length of end perpendicular to B (torque arm - inches) and calculate the rate. By dividing the new rate into the original rate you can determine the change in percent.

    Now to the answer, 6mm is equal to .236. The effect on the torque arm inches by lowering the sway mar will be less than .1. In the final analysis, the effect of 6mm is so small that it would be hart to even measure.

    The greater problem we need to be concerned about is improper adjustment causing a pre-load on the bar. The pre-load is caused by not having the bars ends neutrally adjusted when making the change. Pre-load in sway bars generates extremely differing handling characteristics between left or right turns (especially at corner entry phase). Depending on which way the bar is out of adjustment, it (the bar AND the car) will think it is already leaning into the turn so it will twist the bar more, or it will be starting with a "twist", then go to neutral bar load, then go to twist. You can see how the amount of load transfer for a left vs. right turn of the same speed, radius, degree of lean, conditions, etc. would be very different. The solution is simply to be sure the bar(s) have no pre-load on level ground, with the driver

    (or someone/something of same weight) in the seat.

    Credits: Ken Arnold, KLA Industries - Some information derived from published sources such as Fred Puhn's "How to make your car handle".


    R.I.P. Jason P Harrill 6-12-06


      Suspension, Chassis, Drivetrain, Brakes. Everything to do with handling. Also includes transmission discussions.

      R.I.P. Jason P Harrill 6-12-06