What does this mean regarding to suspension:

10k front 6k rear for the street setup
12k front 8k rear for the sport setup

Which setup is good for my car?

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by bibong &raquo;</TD></TR><TR><TD CLASS="quote">What does this mean regarding to suspension:

10k front 6k rear for the street setup
12k front 8k rear for the sport setup

Which setup is good for my car?</TD></TR></TABLE>

Where did you see it or who said that to you?

 

umm, i was doing a search on omnipower coilovers and that came along.

 

it refers to the spring rate (spring constant). basically means how stiff its going to be or how much force its going to take to compress the spring for a given unit of length. hope that helps

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by xhatchracerx &raquo;</TD></TR><TR><TD CLASS="quote">it refers to the spring rate (spring constant). basically means how stiff its going to be or how much force its going to take to compress the spring for a given unit of length. hope that helps </TD></TR></TABLE>

Well said.

Been awhile since I took physics but I believe it's the "K" in the spring rate equation.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Syndacate &raquo;</TD></TR><TR><TD CLASS="quote">

Well said.

Been awhile since I took physics but I believe it's the "K" in the spring rate equation.

</TD></TR></TABLE>

Yeah, it is... Hooke's Law of Elasticity states that the force required to compress a spring (f) is equal to the spring constant (k) times the distance the spring is elongated or compressed (x):

F = kx

basically, the numbers are the stiffness of the springs. The larger the number, the stiffer the springs.

And, yes, I'm currently taking physics, so yeah...

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by reaper2022 &raquo;</TD></TR><TR><TD CLASS="quote">

Yeah, it is... Hooke's Law of Elasticity states that the force required to compress a spring (f) is equal to the spring constant (k) times the distance the spring is elongated or compressed (x):

F = kx

basically, the numbers are the stiffness of the springs. The larger the number, the stiffer the springs.

And, yes, I'm currently taking physics, so yeah...
</TD></TR></TABLE>

wow, good ****.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by reaper2022 &raquo;</TD></TR><TR><TD CLASS="quote">

Yeah, it is... Hooke's Law of Elasticity states that the force required to compress a spring (f) is equal to the spring constant (k) times the distance the spring is elongated or compressed (x):

F = kx

basically, the numbers are the stiffness of the springs. The larger the number, the stiffer the springs.

And, yes, I'm currently taking physics, so yeah...
</TD></TR></TABLE>

And H-T's best noob of the year award goes to &gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&g t; Reaper2022!!!!!!

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by bibong &raquo;</TD></TR><TR><TD CLASS="quote">What does this mean regarding to suspension:

10k front 6k rear for the street setup
12k front 8k rear for the sport setup

Which setup is good for my car?</TD></TR></TABLE>

What car do you have and what goals do you want to take it to?

The way I read these is 8k would be 8kg per millimeter. or 448lbs per inch. so it would take 448 lbs to compress the spring one inch.

FYI-
12kg per mm = 672 pounds per inch
10kg per mm = 560 pounds per inch
8kg per mm = 448 pounds per inch


Modified by LowSlow93Sol at 10:04 AM 1/29/2007

 

i have a 2000 civic HX coupe. I just want a setup that would ride smooth and wouldnt hurt the car if i push it once in a while.

 

Honestly go with the softer springs. (10k/6k)
But these are still pretty stiff so I would expect a pretty firm ride.

I have 450#(8k) front and 350#(6k) rear and they are plenty stiff for most people.

 

k means kg/mm which is the pressure needed to compress the spring, so 10k would mean it takes 10kg to compress the spring 1mm

i'm daily driving on 12k/8k and it's perfect for the way i drive, anything below 10k/8k feels wimpy soft to me

 

Millimeters and Kilograms are great, but for the Americans ...

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Eran &raquo;</TD></TR><TR><TD CLASS="quote">Here's how spring rates work:

Say you have a front rate of 200lb/in. (ie, 200F/xxxR). That means that each front spring takes 200lbs of pressure to compress one inch. The higher your spring rate, the more force it takes to compress it, the stiffer your ride.

Linear vs Progressive:

Linear springs have the same rate from top to bottom. No matter where pressure is applied, it's going to take the same ammount of pressure to compress one inch. Progressive springs use "inactive" coils to have two completely separate spring rates. The bottom portion of the spring will take less pressure to compress, thus giving a softer, smoother ride, where the top portion will take more pressure to compress, thus improving handling in more agressive cornering.</TD></TR></TABLE>

 

to all you guys for good info!

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ek forever guy &raquo;</TD></TR><TR><TD CLASS="quote">

And H-T's best noob of the year award goes to &gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&g t; Reaper2022!!!!!!</TD></TR></TABLE>

Is that good or bad? lol