I'm tired of having to post up and argue with complete noobs over spring options and everyone always asking the same questions. I did a whopping 5 minutes of research and found a wealth of information that can be understood by pretty much anyone. Some of it seems complex, but I'm compiling it to help not onlythe technically inclined, but alsothose that see formulas and have brain-friction.

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Spring height and rate (lb/in) formulas and calculations
Spring tech:
The formula for calculating spring rate, or stiffness;
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.



List of Spring information; (SN-95 including new edge)
Very close estimates, variables are dependent on wether they are mounted on a coupe, vert, gt, v6, etc.


So take any one of these springs, and simply by doing the proper formula, you can calculate a very accurate ride height and spring rate for whatever setup you choose to run, regardless of wether or not the springs are cut. When you cut your springs you can also acurately calculate your spring rate, and possibly modify the spring to work with the inteded performance. Higher front rates for auto-x, lower front rate for drag, etc.

A simple way to look at the formula is to say whatever percentage of the spring section removed, double that added to the old spring rate is the new spring rate.

Ex.
K = unknown spring rate
W = 0.5" diameter
G = 12million (constant)
N = 9.5 active coils + 1/2 coil = 10
D = 5"

Stock spring rate = 341lb/in
With one spring removed = 380lb/in (Variable N = 8.5 active coils + 1/2 coil = 9)

Removing 1 coil effectively raised the spring rate 40lb/in

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This information was obtained directly from a spring manufacturer:


The strength of a spring, leaf or coilis a function of the cube of the steel used.Keeping with the subject of your question, coil springs, thediameter 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 aboutsprings with the same wire diameter and the same inside diameter. The only thing that will change will be thelength 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.Avery 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". [/align]
The calculation to find the rate of a coil spring is:
[blockquote]
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. [/blockquote][/align]
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 springis made from .610 wire and has an inside diameter of 3.875" andhas a free height of16.145" (not installed) andis deflected down to 10.5" (load height) when loaded to 1,519 lbs. (load rate) This spring hasa spring rate of 269lbs.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) [/align]
[blockquote]
Do the math-
[blockquote]
[blockquote]
[blockquote]
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. [/blockquote][/blockquote][/blockquote][/blockquote][/align]
Double checkthe math - 16.145 - 10.5 = 5.645 deflection. 1,519/5.645 = 269 [/align]
Now if we cut say 1/2 turn off this spring the active turns become 7.5. [/align]
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 [/align]
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.

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Hope this helps a bit, and by all means please discuss

I pretty much figured this crap out a long time ago when i researching it all for my own car, so ask away.

And before it gets asked 100,000,000 times;

Replace your struts and get an alignment if you alter your ride height.



Ryan[/align]

 

Yeah but according to "simple physics"you are wrong..

 

I wasn't aware that the laws of physics changed based on level of difficulty lol

You have to make sure to factor in the difference between dead coils and active coils. Dead coils won't affect anything more than ride height. Figuring out where the dead coil ends and the active coil begins becomes more difficult, I think.

 

Who do you think you are posting real tech info in the V6 section?

 

I know, I know, but I try to bring the section up a wee bit.

I don't want us to get dumbed down like the S-197 section.

 

I like how you are trying to mock me by using "simple physics", truth is, what ryan just posted proves what I said. When you change the spring, you change the behavior (ie spring rate)...and when I said simple physics I just meant that through one semester of a college phyics class you learn about Hooke's Law. Which basilcy said the force a spring exerts is based on a spring constant K (which is explaned in Ryan's OP) and the distance the spring is stretched/compressed...Now I will say I didn't knowmuch about cutting springs before, so I assumed the spring would be considerably altered...Continuting on topic though...Ryan, you are correct about saying if you only remove "dead coils" you won't affect the spring behavior...but in the truth of this, it must be said that lowering your car by cutting your stock spring will not change spring rate, so the main purpose for doing so should be for looks mostly. I would assume the only performance advantage is the lowering of center of gravity which will give you slightly better handling...So, if you want performance (I think it would be ok to assume people could all agree now with all the information), you should look into getting an aftermarket spring with higher spring rate.

Oh, and +1 on new shocks/struts and getting an alignment.

 

Yes I was mocking you and i'm glad you enjoyed it.

Ryan what exactly is brain-friction?

 

Mike: Brain friction is where something is too complex for a simple mind to handle and the left and right side start to violently shake against each other causing "brain friction." Too much and they'll lock up.
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Now, back on topic. Cutting the stock springs PAST the dead coils will affect spring rate. it won't be tremendous, but it will increase it. Cutting dead coils will only affect the ride height by the height of the dead coil removed. Cutting out pieces of an active coil are what will increase your spring rate.

Cutting the topp off of our rear coils effectively lowers the car, without changing spring rate, since it's a dead coil anyway. I haven't looked at a set of stock front coils in over a year to recall if there are any dead coils in it or not, but I don't think there are any.


Changing the LOAD is what makes the height change not equal to the height of active coil removed. This is where we determine the final ride height after cutting, and settling etc. This is why a spring from a cobra will msot likely lift a v6 front end, the load is different, but the rate has not changed at all.

 

I like the spring chart, very useful![8D]