Input your coil's measurements to find it's spring rate.
Q -Mike,
Finally, I can ask an expert. I’m a member of a Classic Mustang forum where the “popular wisdom” is that if one cuts a spring, say a fraction of a coil (like 1/4 to 1/2), the spring *rate* increases.
I’m an engineer (but alas an EE!) with no practical knowledge on the matter — BUT, this “popular wisdom” tends to run contrary to my intuition.
OK,
within elastic limits of a spring in compression, does cutting a spring’s free length change the rate (i.e., the “k” in F=k*x)??.I think it changes the rated load (as a percentage of the reduction in overall length) as the reduced overall length also reduced the total “F,” or load, the spring can handle (max deflection/compression but I just don’t see how the metal knows that part of it has been lopped-off or not.
So, is *k* a function of spring length (and number
of coils)?I would really appreciate a bit more of information on what factors drive the *k* value for a given spring design (mostly I’m focused on coils for front springs here).
I’ve read your web sites Spring Tech 101 and it doesn’t address this specific question in detail.
Also, I know lots of restomod folks cut springs to achieve a certain ride height. But is this really the best practice if limited to less than, say, 10% of a spring’s original length??.
I think I already know the answer
to this, but I want an expert opinion please.And, do you by chance know what the factory ride height of a ’67 Mustang GT at front and rear as measured from the spindle/axle centers to the fender lip center are??.
I cannot find this data anywhere.
Scott
A – Scott,
“Popular wisdom” rules. Cutting coils does increase the spring rate. Let me explain why.
The strength of a spring, leaf or coil is a function of the cube of the steel used. Keeping with the subject of your question, coil springs, the diameter of the wire and the length of the wire will give us the amount of steel used.
For this whole discussion we will be talking about springs with the same wire diameter and the same inside diameter. The only thing that will change will be the length of the wire used to wind the spring.
The longer the wire is the lower the spring rate. As the wire get shorter, such as when cutting the coil, the spring rate increases.
So everyone has a clear understanding lets describe what “rate” is.
Rate is the amount of weight it takes to deflect a spring one Inch.
A very common mistake is to think that spring rate is how much a spring supports. How much weight a spring is designed to support is called “Load” or “Designed Load” or”Load Rate”.
Rate and Load Rate are two totally different animals.
The calculation to find the rate of a coil spring is:
11,250,000 times the wire diameter to the 4th power divided by 8 times the active number of turns times the mean diameter cubed.
Active turns are the number of turns of the spring that do not touch anything. Any part of the coil which makes contact with anything becomes inactive, that is it no longer functions as part of the spring.
The mean diameter is the inside coil diameter plus one wire thickness. Or the outside coil diameter less one wire thickness.
Let’s say for example a 1967 Mustang GT front spring is made from .610 wire and has an inside diameter of 3.875″ and has a free height of16.145″ (not installed) and is deflected down to 10.5″ (load height) when loaded to 1,519 Lbs. (load rate) This spring has a spring rate of 269 Lbs.
This spring has 9.33 total coils but 1.33 coils touch the spring seat so they are inactive leaving 8 active turns. (I know this from the Ford blue print).
The mean diameter is 3.875 + .610 (The inside is the important diameter because it is the inside of the spring which is used to locate the spring on the corresponding suspension parts. The outside diameter is not considered because it will change with a change of wire diameter)
Do the math-
11,250,000 x (.610 x .610 x .610 x .610) / 8 x 8 active turns x (4.485 x 4.485 x 4.485) = 269 Lbs.
Double check the math – 16.145 – 10.5 = 5.645 deflection. 1,519/5.645 = 269
Now if we cut say 1/2 turn off this spring the active turns become 7.5.
So 11,250,000 x (.610 x .610 x .610 x .610) / 8 x 7.5 x (4.485 x 4.485 x 4.485) = 287 Lbs.
While the rate is increased the load is unchanged. Rate is the amount of weight required to deflect the spring one Inch while load is the amount of weight the spring will support at a given height.
Cutting coils is limited to those types which have tangential ends. Tangential ends are those which spiral off into space. If you tried to stand the spring on end it would fall over.
Square ends and pigtail ends, both will stand up, and can not be cut because the finished product will not mount correctly in the suspension.
When altering ride height one must be aware of much more than just the springs. Brake lines, steering, shock length and other areas of interference. We do not offer coil springs which will alter any ride height more than 2 Inches. Nor do we recommend anyone alter the ride height more than 2 Inches.
While we have all sorts of springs which will vary ride height, spring rates and ride quality on the shelf, cutting coils maybe, in some cases, the only way to achieve the desired stance one is looking for.
I do not know the distance from the spindle to the fender lip, but we do have other measurements which may help. Send me your fax number.
I hope this answers your question. – Mike
