//How to select and use Hyperco Bump Springs

How to select and use Hyperco Bump Springs

Hyperco Steel Bump SpringsHere is some additional information regarding the Hyperco line of precision bump springs…

The design load is solid load for all rates from 800#/in through 3500#/in.  For rates 4000#/in to 5000#/in, the design load is the maximum load that can be applied before the spring takes a set.

You can find the total deflection available by taking the design load and dividing it by the rate.  For instance on the 800#/in spring, the design load of 887# divided by the 800#/in spring rate yields 1.09” of deflection.  887/800=1.09.

All of the springs through the 3500#/in rate are understressed and can be compressed to solid thousands of times without affecting free length, rate or load.  For any given rate, the springs do not need to be deflected to solid or design load.  You can travel them any length you need within that constraint.

To start out, you need to know the approximate load that your vehicle puts on a spring at its maximum through the corner.  Then take the initial rate spring you want to use and take it to the maximum deflection point that you see and determine the spring load at that deflection.  Subtract that from the total load you are going to need through the corner and that will be the minimum design load from a bump spring that you will need.

For instance: let’s say that you see a maximum of 2340# of load on the right front through a corner and you want to use a 12B0275UHT as your main spring on the right front.  Let’s also assume that you have 650# of sprung weight on the right front and that you want to travel the main spring 3.0” additional before the spring rate starts to rise.  650# of sprung load compresses a 275lb/in spring 2.364”.  The additional 3” of travel requires another 825lb (3.0” x 275#/in) of load on the spring.  650# + 825# =1475# of force contributed by your main spring.  This is now the point where you want the rate to increase.  A bump spring is not a stacked spring.  It is a nested spring and acts like the inner spring in a dual valve spring set.  Therefore the bump spring rate adds to the main spring rate from this point on.  Now let’s say you will travel the suspension another 0.750”.  0.750” x  275lb/in spring rate =206# of load from the main spring.  Add that to the 1475# and you have 1681#.  Subtract the 1681# of load from the required 2340# maximum load that you see and you get 659# of load required in an additional 0.75” of travel.  Divide 659lb by 0.75” and you get 878lb/in minimum spring rate needed to achieve this load.

You usually want to round up to the next highest rate and that would be 1000#/in.  Divide the 659# load by 1000#/in of bump spring rate and you will find the bump spring deflects 0.659.”  That is not too far from the 0.750” of travel we are talking about.  Additionally the bump springs can be shimmed to change the “pick-up point” where they come into action.  Also, if you want a slightly higher rate feel, there is nothing wrong with trying a 1200#/in rate.

The key point to remember is that using a bump spring creates a “nested spring” configuration where the rates are additive.  With stacked springs, the combined rate is always less than the lowest rate spring.  With nested springs the combined rate is always greater than each of the single spring rates.  The combined rate in our example above would be 1275#/in.; 275#/in from the main spring and 1000#/in from the bump spring.

2017-05-03T12:29:56+00:00 September 8th, 2015|Tech Tips|