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Ask Anything - February 2014


John Wade; Stafford, England: Some years ago you published a formula for calculating the spring rate of a spring using wire diameter and number of coils. This was in a reply to a question from a reader. I meant to but never did make a note of the formula. Do you have easy access to it in your back issues/tech notes, and if so could I please ask for it again?

Jeff Smith: It's always fun to get tech letters from outside the U.S., this time from England! This is a relatively easy one, and we just got this formula off the Internet from someone with engineering expertise. The formula is somewhat simple but requires some specific engineering information—namely something called torsional modulus of the spring wire. Don't worry, we had to look up the definition so we'll save you the time. Basically, this relates to torsional rigidity, which is a material's resistance to twisting, or more accurately the ratio of the amount of force (torque) required to twist a given bar to generate a given amount of angle or twist at the end of the bar. So the variables would be the properties of the steel, the diameter of the bar, and its length. One property of carbon steel relates to its elasticity—or its ability to bend without breaking and also to return to its original shape. This is a very important factor for a spring that's constantly twisted. Now let's take a steel bar and bend it around a mandrel to create a coil spring. The length of the bar relates to the number of active coils in the spring. An active coil is defined as one that does not touch an adjacent coil. The overall spring diameter plays into this equation as does, of course, the diameter of the wire that makes up the spring. A larger wire diameter increases the spring's resistance to being compressed. The formula uses what is called "mean coil diameter," which is basically the outside diameter (OD) plus the inside diameter (ID) divided by two. For a 5.5-inch OD spring with a 0.625-inch diameter wire, this number is 5.185 inches.

A spring's ability to resist compression is called the spring rate. This is expressed in terms of pounds per inch. As an example, if we stand a coil spring on end and apply 500 pounds of load to the top of the spring and it compresses exactly 1 inch, then this spring has a rate of 500 pounds per inch. Note that this is not 500 pounds per square inch. That is a pressure reading (such as air pressure inside a container) where force is applied equally in all directions. A spring has a rate of force that is applied in a linear direction.

The one engineering concept we need to include in our spring equation is the torsional modulus of a steel spring wire. This is given as 11.25x10^6 (10 to the sixth power), or 11,250,000. Now that we have this number, we can apply the formula:

Spring Rate = G * D^4 / 8 * N * D^3
G = torsional modulus of steel: 11,250,000
D = wire diameter: 0.625
N = number of active coils: 6
d = mean coil spring diameter: 5.185 inches

Now we can plug in the variables. Let's solve D^4 and d^3 first:
D^4 = (0.625 * 0.625 * 0.625 * 0.625) = 0.1525
d^3 = (5.185 * 5.185 * 5.185) = 139.39
Spring Rate = 11,250,000 * 0.1525 / 8 * 6 * 139.34
Spring Rate = 1,715,625 / 6688.3
Spring Rate = 256.5 pounds per inch (lbs/in)

Now if we reduce the number of active coils from 6 to 4, the formula tells us that the rate will increase. This means the new denominator (the number below the line in simple division) will be a smaller number: 8 * 4 * 139.39 = 4,460

Spring Rate = 1,715,625 / 4,460
Spring Rate = 384 pounds per inch

A simpler way to measure coil spring rate is to use a scale to measure the actual load created by the spring. Let's place a coil spring on a scale and compress the spring 1 inch. Let's assume the scale reads 500 pounds—this isn't the rate, this is just the initial load. We need to preload the spring to get a more accurate reading. Now compress the spring another 2 inches and the scale now reads 1,800 pounds. Subtract the initial load of 500 pounds from the total (1,800 – 500 = 1,300 pounds). Now divide that 1,300 pounds by 2 because we want to know the rate in pounds per inch, so 1,300 / 2 = 650 lbs/in.

These calculations can be used for any type of coil spring, such as a valvespring. Let's say you want to reduce the installed height of a valvespring by 0.100 inch and you'd like to know what the new closed and open valve spring loads will be. If you know the spring rate, the calculation is easy. Let's say the installed height was 1.850 and the original load was 90 pounds. With a spring rate of 415 lbs/in, decreasing the installed height to 1.750 means a change of 0.100 inch. The math will be 415 x 0.1 = 41.5 pounds of load increase on the valve all the way through its entire lift curve—assuming it's a constant-rate spring. Now let's toss a variable into the works. A beehive spring is tapered at the top, which means the overall spring diameter is not constant, which makes this a variable-rate spring. This means the above calculation will not be accurate since the rate will change as the spring is compressed. This also assumes that no coils touch, which will affect the spring rate because now we have fewer active coils. In terms of a quickie eyeball evaluation of any coil spring, shorter springs offer a higher rate as does a thicker wire diameter. Longer springs with more active coils will reduce the spring rate. I hope this helps you with your spring calculations.

CC Quickies

Art Carr is the man behind California Performance Transmissions, and by the date on this NHRA Competition license, he's been doing this stuff for a long time. This license was issued to him in 1966 to drive a AA/Altered to speeds of up to 170 mph. That was bad fast back in the day.

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