So I guess my question is, Do the coilover kits come with more then just coilovers, and they're selling only parts of what you need?
Or is this just some crappy product to make a quick buck?
Unless you plan on doing some serious racing I wouldn't suggest coilovers to anyone. The coilovers you have shown are just junk and I would slap myself if I had ever ordered those. If you do choose to get coilovers, stick with the ones that have a known good reputation. Can't go wrong there with a good quality set as compared to something that you will be kicking yourself in the head over.
This has been about a year now. Are the BOMZ coilovers still considered junk or are there a few people out there who have tried and like them. I have attached the link to the web site for BOMZ. I found that they are going for about $60 something to about $90 on different web sites. I am also attaching a web site that has these and some others at a pretty low price.
You really want to get a "matched set" of springs AND shocks or know what springs and shocks by differing manufacturers work well together. Mixing and matching can be a crap shoot if you don't know what you are doing. You need a good shock to dampen the stiffer rates of shorter springs and a shorter shock to maintain suspension travel. Some good kits are made by Eibach, H&R's, and Ford Racing.
Posted via FF MobileTo quote a prev. comment, the name is descriptive (bomz) You could take a torch to your springs to lower them like some of the 1950`s rodders did.... It`s the cheapest lowering kit.....& the results match the price.....
The old adage you get what you pay for definitely applies to suspension.
And yes 05 to 07 use a different steering knuckle, there is an adapter out there that lets you sue 001 to 05 kits, or you can swap the knuckles to SVT one and your brake calipers will work with them.
is this for coilovers?
for just a normal drop 00-05 (even svt) will all work excluding wagons. then you have the change in 06/07. at least that's what i know..
i know absolutely nothing about coilovers..
Yes that the case for coilovers or regular Shocks, the struts on the 05 to 07 are the same diameter from top to bottom, the 00 to 05 taper to a smaller diameter where the strut fits into the knuckle.
I have the solution.. I have been riding on cut springs for 45k and upgraded my shocks and it rides and sits nice.. My car is dropped about 2 3/4 to 3" all around.. I took 2 coils off the front and 2.5 in the back.. It gave me a little stiffer spring rate and sits even.
I really doublet cutting your springs gave you a stiffer spring the majority of springs are linier in the spring rate for example100lbs/in springs will need 100lbs for 1 inch of compression 2inches will take 200lbs and so on. Cutting them will not change that rate.
Incorrect, it will stiffen them. A spring is a torsion bar, albeit a wound up one. As you shorten a torsion bar it takes more force to twist it. For example, on a linear rated spring that has five coils, if you remove one equal length coil, you effectively increase the spring rate 20%.
You’re correct that a torsion bar will get stiffer if shortened, and your also correct that is you shorten a spring that it will become stiffer in torsion but not for the same reason http://en.wikipedia.org/wiki/Torsion_spring
However that is not how they work in compression or tension as you can see in the like I posted for Hooks Law the force is equal to the spring constant times the distance if you look at the units of force are lbs, kg or whatever you like the spring constant units are in lbs/in and distance is in inches. When you multiply lbs/in X in = lbs (i.e. you get (lbs X in)/in= lbs the distance units cancel out), so from that you can see that it the spring will not stiffen when cut and actually will not be able to take as much load because you have shortened its free length. The spring constant does not chance that’s why it is a constant.
"The equation below can be used to estimate the spring rate of a coil spring:
Spring Rate = G x d4
8 x n x D³
G = Shear modulus (11,385,500)
d = coil wire diameter (inches)
n = number of active coils
D = mean coil diameter (inches)
For example, we can calculate the spring rate for a coil spring with a coil diameter of 0.625 inches, a mean diameter of 5 inches, and 10 active coils. Using the equation above gives us a spring rate of about 174 lb-ft/in.
Shear modulus (G) is the same for all steel springs. It changes only when the spring material changes. Coil wire diameter (d) is a constant number unless the spring is designed as a progressive spring using a varying spring wire size. For that case, this equation is no longer valid. Number of active coils is the number of free coils (complete circle) free to move with the spring action. Mean coil diameter (D) is the distance through the centerline of the coil as shown above.
Using the equation can help you compare coil springs or help you design a set of custom coil springs. Note that coil height specifically does not affect the spring rate significantly, only the active number of coils does. Note also that coil spring rates remain nearly constant even if the coil sags over time."
So......per these examples, the adding or subtracting coils do have a dramatic effect on the spring rate.
Cool I learned something today I didn’t realize how the spring constant was calculated. Guess I sould have looked into that. I see how the number of coils changes the equation k=(Gd^4)/(8nD^3)
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