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many of you know I have eibach pro kit on my car, its one of the slightest lowering springs, but i have heard good things about people cutting these springs. How much did you guys cut off to get wheel to fender on the front and the back. My tires are 215/45/17 on a i4 car and i4 springs.
I dont want to cut much off to ruin the performance of it, i was thinking 1/2 coil each would do it?
Keep in mind that if you over-lower you'll increase your roll couple to the point where the lower center of gravity is no longer worth-while. Most 'slammed' cars are likely hitting their bump-stops 80% of the time, unless they have coilovers.
cutting 1/2 a coil will get you like... a 3 MILLIMETER DROP.
1" is going to need ~3 coils cut.
remember when you cut coils you soften the spring.
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125 front wheel horsepower with major retard issues between 4500-5200RPM -
OD switched off, even when not in 3rd results in major power loss/rpm drop.
remember when you cut coils you soften the spring.
Opposite actually. Cutting springs increases spring rate. And it will do it unevenly (because the front and rear springs are very different) and may unbalance your car, this is especially true for progressive coils. You may also run into problems with the spring seating correctly which may be hazerdous.
If the drop is not good enough... go with another spring. Do it right or don't do it at all.
cutting half of a coil does more than 3 millimeter. I wouldn't recommend any more than 3/4 of a coil though if you want to keep the same "feel". I cut my stock springs about a coil and half with the tokico shocks and they've been perfect for 4 years now but i can't guarantee it's going to be the same for you.
cutting a coil needs to be done properly. if you cut it from the wrong end, you lose spring-rate, if you cut it in the right place, it'll eliminate preload, giving you an overall stiffer ride. My car has v6 whitelines with 1 coil cut from each of the fronts. it feels stiff as hell, and if you sit on the front end of my car, you won't even budge it.
i have the same rims as you do, and i have cut springs
sprint springs and my fronts are cut 1.5 coils. i did it with a die grinder...
it looks better, but rides worse, slightly too low.
Opposite actually. Cutting springs increases spring rate. And it will do it unevenly (because the front and rear springs are very different) and may unbalance your car, this is especially true for progressive coils. You may also run into problems with the spring seating correctly which may be hazerdous.
If the drop is not good enough... go with another spring. Do it right or don't do it at all.
WRONG WRONG WRONG WRONG
linear springs are a certain rate (lets say 4kg/mm) thats the value for teh WHOLE SPRING
i fyou remove a coil you now have LESS COILS. less coils means less spring rate at the same preload.
im prety sure the eibachs are progessive rate so cutting the bottom of them will amke it A LOT SOFTER.
the reason most say it is stiffer is because the final stroke of dampers is very stiff.
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cutting loweres springs results in A LOT less drop then cutting stock springs, as you already have less coils.
we cut 1/2 the spring out of our 97lancer springs (performacne lowered linear rate springs) to get an extra 1" drop. by 1/2 i mean 3.5 coils remaining of hte spring.
i have 1 whole coil cut out of my camry rear springs (prog rate lowered) which gave me about 5mm.
if you cut 1/2 of a stock spring out youll end up on the bump stops.
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dz coil over sleeves are $180 for camrys ifyou really must. (ebay)
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125 front wheel horsepower with major retard issues between 4500-5200RPM -
OD switched off, even when not in 3rd results in major power loss/rpm drop.
As you shorten a spring, the spring rate increases. Its physics, you cannot argue this.
Hooke's law for an ideal spring: F = kx
Suppose we apply a force that compresses a 10-coil spring by 1 cm; then each coil is compressed by 0.1 cm. Now compress a 5-coil spring by 1 cm; then each coil will be compressed by 0.2 cm. According to the ideal spring equation, the force of compression for a single coil is directly proportional to the displacement of the coil. The force needed to compress a single coil by 0.2 cm must be twice as large as the force needed for a 0.1 cm compression. Thus, the spring constant of the 5-coil spring must be twice that of the 10-coil spring. In general, the spring constant is inversely proportional to the number of coils in a spring so shorter springs are stiffer springs.
I don't care if it is progressive or linear or what. Less coils = stiffer. More coils = softer.
Just like if you stack a 200lb spring on a 300 lb spring (like helper springs in coil-overs). It should act like a 200lb spring until you crush the helper spring, right? Wrong. The intial rate is 120lb. Don't believe me, go ask a physics professor... or an F1 crew chief.
Hell... reverse it. If you have a 300lb spring, you can make it softer by stacking a stiffer 500lb spring on top of it. The initial rate with that combo would be 194lb/in.
It may be counter-intuitive, but its fact.
Last edited by Hawkeye-SE; 09-16-2005 at 12:59 AM.
The thing to be careful about in cutting a spring is this: I believe part of a spring is softer than another, the reason for this is to add comfort to the ride, the other part of the spring is rigid and contains the actual 'spring-rate.' Where you want to cut is from the softer part, this will actually increase stiffness significantly.
As you shorten a spring, the spring rate increases. Its physics, you cannot argue this.
Hooke's law for an ideal spring: F = kx
Suppose we apply a force that compresses a 10-coil spring by 1 cm; then each coil is compressed by 0.1 cm. Now compress a 5-coil spring by 1 cm; then each coil will be compressed by 0.2 cm. According to the ideal spring equation, the force of compression for a single coil is directly proportional to the displacement of the coil. The force needed to compress a single coil by 0.2 cm must be twice as large as the force needed for a 0.1 cm compression. Thus, the spring constant of the 5-coil spring must be twice that of the 10-coil spring. In general, the spring constant is inversely proportional to the number of coils in a spring so shorter springs are stiffer springs.
I don't care if it is progressive or linear or what. Less coils = stiffer. More coils = softer.
Just like if you stack a 200lb spring on a 300 lb spring (like helper springs in coil-overs). It should act like a 200lb spring until you crush the helper spring, right? Wrong. The intial rate is 120lb. Don't believe me, go ask a physics professor... or an F1 crew chief.
Hell... reverse it. If you have a 300lb spring, you can make it softer by stacking a stiffer 500lb spring on top of it. The initial rate with that combo would be 194lb/in.
It may be counter-intuitive, but its fact.
x2, ap physics w0t
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98 V6 XLE
Dell 5100|SB Extigy|PPI DCX 730|NEXT 4.00|Monitor 2.70|ED Nine.1|LPG 26||Alpine Type R|ED 11ov2's and some other junk
eibachs are wide, soft on the bottom, hard and more coils on top. by cutting the bottom coil i will loose some comfort and make it more stiff, right?
if i cut the top, i will loose the progressive spring rate? how would I even cut the top if the spring bends to be completely flat on the top for the strut mount?
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Last edited by RedPhoenix; 09-16-2005 at 05:05 AM.
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