Lowering early 944 front end without lowering springs
07-28-2006, 10:00 PM
#16
Burning Brakes
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Originally Posted by M758
You CAN'T!
The only 924/944/968 that can lower the front without new springs is the 944 Turbo S or 968/944S2 with M030. These cars have factory adjustable spring perches.
The only 924/944/968 that can lower the front without new springs is the 944 Turbo S or 968/944S2 with M030. These cars have factory adjustable spring perches.
07-28-2006, 10:19 PM
#17
Race Car
"All a spring is is a coiled torsion bar. By cutting it youre reducing the leverage and therefore increasing the spring rate. How do you think the spring is made initially."
A spring is a coiled torsion bar but cutting it does not reduce the "leverage". Leverage is torque applied in the torsion bar (circular torsion bar in a coiled spring), which is determined by the twist (Omega in degrees radian since is a circular spring integrated over the y axis in a vertical spring) times the force applied (force times length = torque). The leverage length in this case is fixed (diameter of the coil) so the only thing you can change is the force (weight of the car or bumps you hit).
How does cutting the spring affect the spring constant? It doesn't. Simply put it, if you straightened a spring and make it into a straight torsion bar, cutting the torsion bar will absolutely not change the spring constant, which is why they call it a constant. It will change the amount of force it can handle (how much it will twist before it breaks or "bottoms out"). The ONLY WAY to change spring constant is by changing the material or increasing the diameter of the spring, which is how "stiffer" springs are made with the same height. This is provided you don't do anything like changing the annealing process of the steel changing the material property of the metal, which is not what we are talking aobut here.
"What happens if you put two 200 lb springs in series? The result is a 100 lb spring. Spring rates are added like capacitors total = 1/(1/R1 + 1/R2....)."
This is not true. If you put 2 200lb springs on top of the other (200lb is a force, or rate of the spring), you changed the distance it can travel and the amount of force it can absorb through spring travel. The spring constant is still the same, the distance it will travel is doubled. Its like a spring scale when you go fishing. A 20 pound fish will stretch the spring twice the distance whereas the 10 pound fish will stretch the spring once. If you place 2 200 pound springs side by side, you effectively doubled the spring constant (instead of twisting one spring with X bar thickness with 1x spring constant, now you are twisting 2 bars with X thickness, therefore 2x spring constant.
Unfortunately I got to spend hours figuring out the spring constant of a spring using multivariable calculus 20 years ago and spent time in a material sciences lab with an Indian guy who can barely speak English explaining atomic dislocation energies of body or face center cubic lattice structures so I get cursed with remembering crap like this.
A spring is a coiled torsion bar but cutting it does not reduce the "leverage". Leverage is torque applied in the torsion bar (circular torsion bar in a coiled spring), which is determined by the twist (Omega in degrees radian since is a circular spring integrated over the y axis in a vertical spring) times the force applied (force times length = torque). The leverage length in this case is fixed (diameter of the coil) so the only thing you can change is the force (weight of the car or bumps you hit).
How does cutting the spring affect the spring constant? It doesn't. Simply put it, if you straightened a spring and make it into a straight torsion bar, cutting the torsion bar will absolutely not change the spring constant, which is why they call it a constant. It will change the amount of force it can handle (how much it will twist before it breaks or "bottoms out"). The ONLY WAY to change spring constant is by changing the material or increasing the diameter of the spring, which is how "stiffer" springs are made with the same height. This is provided you don't do anything like changing the annealing process of the steel changing the material property of the metal, which is not what we are talking aobut here.
"What happens if you put two 200 lb springs in series? The result is a 100 lb spring. Spring rates are added like capacitors total = 1/(1/R1 + 1/R2....)."
This is not true. If you put 2 200lb springs on top of the other (200lb is a force, or rate of the spring), you changed the distance it can travel and the amount of force it can absorb through spring travel. The spring constant is still the same, the distance it will travel is doubled. Its like a spring scale when you go fishing. A 20 pound fish will stretch the spring twice the distance whereas the 10 pound fish will stretch the spring once. If you place 2 200 pound springs side by side, you effectively doubled the spring constant (instead of twisting one spring with X bar thickness with 1x spring constant, now you are twisting 2 bars with X thickness, therefore 2x spring constant.
Unfortunately I got to spend hours figuring out the spring constant of a spring using multivariable calculus 20 years ago and spent time in a material sciences lab with an Indian guy who can barely speak English explaining atomic dislocation energies of body or face center cubic lattice structures so I get cursed with remembering crap like this.
07-28-2006, 10:32 PM
#19
Race Car
Spring constant is constant. The distance travelled depends on number of coils. Nothing in your equation uses spring constant so you must be confused with the rate.
07-28-2006, 10:33 PM
#20
Lazer Beam Shooter
Rennlist Member
Rennlist Member
Ahoi mates, but the distrant traveled is negativized by the distance in theory. So therefore if he drives over a boulder, the spring rate of the cut vs stock spring will spring the car to do a backflip and land perfectly on its 4 wheel axis. This is furthermore aided by the fact that the 944 has 50/50 weight distrubution.
07-28-2006, 10:36 PM
#21
Race Car
Rock you are funny. BTW, spring constant is measured in Kg/Seconds squared. If you see that in the equation, let me know.
Lets throw in a magnetic containment field, inject antimatter, extract the quarks and tachions byproduct and put it in a flux capacitor to make Mr. Fusion.
Lets throw in a magnetic containment field, inject antimatter, extract the quarks and tachions byproduct and put it in a flux capacitor to make Mr. Fusion.
07-28-2006, 10:36 PM
#22
Rennlist Member
Of course nothing in my equation uses spring constant. Thats the equation for the spring constant (or rate).
I'm not going to post on this matter anymore. Do your own research and come to the correct conclusion on your own.
Regardless, my point stands - reducing the # of coils (cutting the spring) will increase the spring rate (lb/in, kg/m, whatever).
I'm not going to post on this matter anymore. Do your own research and come to the correct conclusion on your own.
Regardless, my point stands - reducing the # of coils (cutting the spring) will increase the spring rate (lb/in, kg/m, whatever).
07-28-2006, 10:40 PM
#23
Race Car
Serge, I am not trying to be a pain. I am just trying to keep this guy from doing something stupid by cutting the spring and bottom out when he hits a bump sending the car out of control. We have a difference in terminology but in engineering, the phrase "spring constant" is used differently. Peace.
