One of the guiding principles of spring design is understanding that rate and deflection go hand-in-hand, inversely so. The higher the spring rale, the less deflection required for a given force. This basic concept makes it very easy to know what you’re up against as far as load tolerances called out on a spring print.

Let’s say the print calls out a load with a tolerance of ±20 lb. How much deflection is needed to produce that 20 lb? If you know the spring rate, you are in luck. This information is typically already called out on the part drawing. The rate of any spring is calculated with following formula:

**Rate = Load/deflection or R = P/f**

As with any formula, this can be algebraically rearranged to produce other results. Therefore,

**Deflection = Load/Rate or f = P/R**

Now, we already know the required load tolerance is ±20 lb. But, how far does a spring need to move to produce that force? Let’s assume the Rate shown Is 250 lb/In. With this data, we can now calculate the travel required to produce 20 lb.

**f = 20 lb/250 lb/in**

**f = 0.080″**

Now comes the important part. What if the rate of the spring is 2500 lb/in? We have a totally different situation. This is because the deflection needed is now

**f = 20 lb / 2500 lb/in**

This equates to 0.008′”! The load tolerance is likely not achievable with standard methods and Engineering will need to speak with the customer’s Engineering to explain and get some input.

Now, for those who know exactly how much distance/movement 0.008″ represents, this principle is already in your bag of tricks. But many times, engineers must talk with nonengineering counterparts such as salespeople or buyers who do not have the math background to understand these concepts. It’s not their bailiwick—they are involved with a different math base.

I find it very important to be able to communicate concepts to those involved. So, if I were to tell the voice on the other end of the phone that the load tolerance is ±0.008″ and their reply is “I don’t know what that means,” I have an out.

My go-to response is to explain that a standard sheet of paper is 0.004” thick. This is something that nearly anyone in any field can understand. If I tell them that this distance is the same as the thickness of two sheets of standard paper, it provides them with a measure that they can understand quickly and plainly.

It works for me.

It is very important to be capable of explaining to others the importance of tolerances and why some tolerances are reasonable and some are not. When tolerances are too small, it becomes a processing issue.

Once this is understood, a sincere effort can be made to see whether the tolerance is a true requirement or if it was assigned at some point in time with standards and practices that do not apply to spring tolerancing and conventions.

By: Randy DeFord, Engineering Manager Mid-West Spring & Stamping