Lowering Megaladon
Burning Brakes
Joined: Oct 2004
Posts: 904
Likes: 0
From: Murphy NC
Maybe I can help. By cutting 1/2 a coil off a spring you increase its rate by 10%. Thats right. Increase. Cutting 1 coil makes it 20% and so on. Ever notice aftermarket lowering springs? They are shorter with less coils. Factory are taller with more coils. You can increase or decrease rates by the thickness of the wire, diameter of the coils, and number of coils. The "softness" in your ride is all mental as it is really stiffer. Hope this helps. Justin
Melting Slicks
Joined: Feb 2000
Posts: 3,052
Likes: 0
From: Vancouver BC, Canada
Originally Posted by Gordonm
Grade 5 is good but a Grade 8 is better. Stick with a Grade 8 you would hate to have a failure due to a weak bolt.
grade 5: 85 ksi
grade 8: 120 ksi
1/2-20 has tensile area of .1599 in^2
so proof load is:
grade 5: 13600 lbf
grade 8: 19200 lbf
I’d estimate the load fluctuates between 1000-2000 lbs
fatigue is not of too much concern as the load is always in tension. (reversing loads are the worst ie alternating tension-compression as in bending)
for comparison, when you torque your connecting rod bolts (3/8-24 x 2” ??),
stretching them .0065” results in 8500 lb of force
Melting Slicks
Joined: Feb 2000
Posts: 3,052
Likes: 0
From: Vancouver BC, Canada
Originally Posted by hosspowerinc
Maybe I can help. By cutting 1/2 a coil off a spring you increase its rate by 10%. Thats right. Increase. Cutting 1 coil makes it 20% and so on. Ever notice aftermarket lowering springs? They are shorter with less coils. Factory are taller with more coils. You can increase or decrease rates by the thickness of the wire, diameter of the coils, and number of coils. The "softness" in your ride is all mental as it is really stiffer. Hope this helps. Justin
here is a handy formula (for linear rate springs)
everyone knows Hooke's law F=Ky
F=force applied
K=spring constant, rate
y=deflection
using Castigliano’s theorem of strain energy
(insert arm-waving here)
K=G*d^4/(8*N*D^3)
G=modulus of rigidity (carbon steel= 11.5E6 psi)
d=wire diameter
N=number of active coils
D=mean diameter of helix
Everything else is constant so as you can see, if N decreases, K will increase
Hence cutting out coils increases the spring rate
Burning Brakes
Joined: Oct 2004
Posts: 904
Likes: 0
From: Murphy NC
here is a handy formula (for linear rate springs)
everyone knows Hooke's law F=Ky
F=force applied
K=spring constant, rate
y=deflection
using Castigliano’s theorem of strain energy
(insert arm-waving here)
K=G*d^4/(8*N*D^3)
G=modulus of rigidity (carbon steel= 11.5E6 psi)
d=wire diameter
N=number of active coils
D=mean diameter of helix
Everything else is constant so as you can see, if N decreases, K will increase
Hence cutting out coils increases the spring rate
Hey thanks. I was looking for that formula but I miss placed the book. Justin
everyone knows Hooke's law F=Ky
F=force applied
K=spring constant, rate
y=deflection
using Castigliano’s theorem of strain energy
(insert arm-waving here)
K=G*d^4/(8*N*D^3)
G=modulus of rigidity (carbon steel= 11.5E6 psi)
d=wire diameter
N=number of active coils
D=mean diameter of helix
Everything else is constant so as you can see, if N decreases, K will increase
Hence cutting out coils increases the spring rate
Hey thanks. I was looking for that formula but I miss placed the book. Justin
Racer
Joined: Jun 2002
Posts: 254
Likes: 0
From: Lonoke Arkansas
"Very confused here... they shouldn't get stiffer ...???... you reduce the pre-stress when you install the shortened spring."
The amount a spring compresses when installed depends on the weight it must support and the spring's rate (lbs/inch).
Shorten the spring (reduce the number of active coils) and the spring's rate will increase (more lbs/inch).
The shortened spring will have to compress less to support the same weight.
Originally Posted by 69autoXr
I'm not familiar with pre-stress. Is it a material property?
Shorten the spring (reduce the number of active coils) and the spring's rate will increase (more lbs/inch).
The shortened spring will have to compress less to support the same weight.
Le Mans Master
Joined: Oct 2004
Posts: 5,269
Likes: 54
From: Orlando Florida
St. Jude Donor '05
By 'pre-stress' I meant the amount by that the spring is compressed when installed between the upper and lower pocket.
The uncut spring is compressed by a decent amount - I'm guessing roundabout 3" when installed. After cutting one coil off the spring is only compressed 1-1/2" when installed.
Then, when you get the car off the jacks the spring compresses even more when supporting the car's weight.
I agree with the posts above, the formula is correct, but I am 100% sure the front feels softer than before cutting the spring.
Maybe VBP has an answer to this ???
The uncut spring is compressed by a decent amount - I'm guessing roundabout 3" when installed. After cutting one coil off the spring is only compressed 1-1/2" when installed.
Then, when you get the car off the jacks the spring compresses even more when supporting the car's weight.
I agree with the posts above, the formula is correct, but I am 100% sure the front feels softer than before cutting the spring.
Maybe VBP has an answer to this ???
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