Chad05TL

Unless there is a specialist out there, I bet nobody knows. But here goes;

If you have a spring that is "yay long". And you cut 1/4" off the spring as measured when it is uncompressed, will the ride height (or compressed height) be 1/4" lower too? or more than 1/4" lower?

BROlando

Spring rate rises as coils are cut off.

And most cars don't have a 1:1 motion ratio.

So its not possible to answer without knowing the car and spring.

Ossman14

Chad05TL

I think it will be more than a 1/4" if I cut off 1/4"...

Chad05TL

The reason I want to trim a little, is because I have Type-S springs (for automatic) and my 05 is an automatic... so I may get a slight bump from OEM. And I dont really want to do that.
So how much weight difference is there? How much is the effect?
Well, according to my picture, the TypeS is 63# heavier.. But you have to divide that by 2 because there are 2 springs.. So each spring is has 31.5# more lift when mounted my 05TL than a Type-S. (and it will be slighter stiffer - bonus)
So how much travel is 31.5#? Well, all I know is that I can stand the spring up on the floor and squat down and apply most of my weight, and it barely moves..
So I think if I cut off 1/4" off the spring it will over compensate for the added 31.5 to 63# of lift(however you want to figure it) and then some!! So I should get a small drop from the TypeS OEM ride height.
The reason I choose 1/4" and not more, is because I think I can only safely cutoff 1/2 a loop off the bottom before the coil starts to spread out. I may be able to cut off 65% of the bottom loop, but I can't tell yet until I mount it on the shock.

Chad05TL

Here is another guy who says if you cut off 1 coil, it will get you 2" of drop. But his coils are about 1" thick. That is like twice as thick as these TL's.. but you get the Idea!
so;
Cut 1" and you get 2" drop.
Cut 1/4" and you get 1/2" drop.
etc..

but remember, on these TL's, you cannot cut the top of the spring. And I see only about 1/2 (50%) of 1 loop that is safe to cut off the bottom. So, that is about 1/4". You may be able to cut 60-65%, maybe 70% of 1 loop.. but I have not done it yet. And I have to just cut a piece at a time to see how fast the spring widens from the bottom. So, trial and error

Chad05TL

correction. It wont be any stiffer. I am cutting off 1/4" off the narrow end of the spring. The narrow end is the firmer side. So it may actually get softer... haha As a spring spreads out wider, it becomes softer. Plus it's only a 1/4", it wont effect firmness.
Even if you consider a constant rate spring like in the video above, the spring rate is not changing.. Only the length is changing. Now, it may bottom out easier, but that won't change the spring rate.

BROlando

How does one shorten a spring without making it stiffer?

IRL, you *may* not notice the increase from cutting off a small part of the pigtail, maybe.

But physics dictates that altering length also alters rate. Shortening a spring makes it stiffer.

I'm not following the rest of your calculations. And I'm not sure you understand what motion ratios are (no offense).

Chad05TL

physics dictates? Show me the math from a credible source.. And maybe I will believe. I would LOVE to learn more.

I do think a shorter spring will bottom out sooner. And therefore it will firm up quicker if you approach the max compressed length.
However I do not think the spring rate changes.

Think about eibach progressive rate springs. It rides soft, but only firms up at an increased rate because of the progressive rate..
That is why they have a progressive rate spring in a lot of cases because it makes it feel softer. But since they lower it also, they add a section of the spring that is higher rate to keep it from bottoming out.
Without that higher rate section of the spring, it would collapse sooner because it is an inch lower.

BROlando

Idk what eibach springs have to do with this. And I can't quite follow the above logic.

But taking an existing spring and then cutting it will make it stiffer. That is what I mean.

The spring rate is baked into the spring height. The spring flexes "x" amount per "y" length. Shortening it by cutting it makes it stiffer.

I realize I'm not a trustworthy enough source for this very established information.

Hooke's law can be referenced for this. Try referencing that on google or something. Lmk if you still disagree.

Chad05TL

I posted this a long time ago.
K (spring rate) is not determined by the length of the spring.
F is determine by 2 variables. #1 Spring Rate , #2 the length of distance that a spring is pulled or compressed from its resting position. Which is totally separate from spring rate.
F is always negative, presumably because the force is always opposite to the direction the spring is being compressed or stretched.


where;
F is the force the spring exerts
k is the spring rate of the spring.
x is the displacement from equilibrium length i.e. the length at which the spring is neither compressed or stretched
The spring rate of a coil spring may be calculated by a simple algebraic equation or it may be measured in a spring testing machine. The spring constant k can be calculated as follows:
where;
d is the wire diameter,
G is the spring's shear modulus (e.g., about 12,000,000 lbf/in² or 80 GPa for steel), and N is the number of wraps
D is the diameter of the coil.

This video only addresses the Force the spring exerts when it is stretched or compressed. Which is VARIABLE. And always negative because it opposes Change.

BROlando

Ok.

As shown by your equations above...

Spring rate is baked into the ORIGINAL length of the spring.

If you take a 10" long spring that is rated at 500LB/IN and cut it in half...you now have two 1000LB/IN springs.

Get what I'm saying?

If you take an existing spring and then cut it...both of the portions you are now holding in your hands are a higher rate than the original spring.

Why?

Because each winding/coil flexes a certain amount in order to give a spring its rate.

If a 500LB/IN spring has 10 windings, each winding must flex 0.1" when you put a 500LB weight on the spring.

Inversely meaning that the only way it can be a 500LB/IN spring is if there are 10 windings that each flex 0.1" with a 500LB load.

If you cut off 5 windings, each one will still only flex 0.1". Because the steel's modulus has not changed. Each winding has the same stiffness.

So...a 500LB weight will now compress the spring only 0.5". Which means you now have a 1000LB/IN spring.

Its stated in the fancy equation you posted. Number of windings/working coils is a factor.

Yes?
Yes.