|
A -
Scott,
"Popular
wisdom" rules. Cutting coils does increase the spring
rate. Let me explain why.
The
strength of a spring, leaf or coil is a function of the cube of the
steel used. Keeping with the subject of your question, coil
springs, the diameter of the wire and the length of the wire will
give us the amount of steel used.
For this
whole discussion we will be talking about springs with the same
wire diameter and the same inside diameter. The only thing that will
change will be the length of the wire used to wind the spring.
The
longer the wire is the lower the spring rate. As the wire get shorter,
such as when cutting the coil, the spring rate increases.
So
everyone has a clear understanding lets describe what "rate"
is. Rate is the amount of weight it takes to deflect a spring one-inch.
A very
common mistake is to think that spring rate is how much a spring
supports. How much weight a spring is designed to support is called
"Load" or "Designed Load" or "Load
Rate". This is cover in Spring
Tech 101.
Rate and
Load Rate are two totally different animals.
The
calculation to find the rate of a coil spring is:
11,250,000
times the
wire
diameter to the 4th power
divided by 8
times
the
active
number of turns times
the mean diameter cubed.
Active
turns are the number of turns of the spring that do not touch anything.
Any part of the coil which makes contact with anything becomes inactive,
that is it no longer functions as part of the spring.
The mean
diameter is the inside coil diameter plus one wire thickness. Or the
outside coil diameter less one wire thickness.
Let's say
for example a 1967 Mustang GT front spring is made from .610 wire
and has an inside diameter of 3.875" and has a free height of 16.145"
(not installed) and is deflected down to 10.5" (load height)
when loaded to 1,519 lbs. (load rate) This spring has a spring rate
of 269lbs.
This
spring has 9.33 total coils but 1.33 coils touch the spring seat so they
are inactive leaving 8 active turns. (I know this from the Ford blue
print).
The mean
diameter is 3.875 + .610 (The inside is the important diameter because
it is the inside of the spring which is used to locate the spring on the
corresponding suspension parts. The outside diameter is not considered
because it will change with a change of wire diameter)
Do the
math -
11,250,000 x (.610 x .610 x .610 x .610) / 8 x 8 active turns
x (4.485 x 4.485 x 4.485) = 269 lbs.
Double
check the math - 16.145 - 10.5 = 5.645 deflection. 1,519/5.645 = 269
Now if we
cut say 1/2 turn off this spring the active turns become 7.5.
So
11,250,000 x (.610 x .610 x .610 x .610) / 8 x 7.5 x (4.485 x 4.485 x 4.485) =
287 lbs
While the
rate is increased the load is unchanged. Rate is the amount of weight
required to deflect the spring one-inch while load is the amount of
weight the spring will support at a given height.
Cutting
coils is limited to those types which have tangential ends. Tangential
ends are those which spiral off into space. If you tried to stand the
spring on end it would fall over.
Square
ends and pigtail ends, both will stand up, and can not be cut because
the finished product will not mount correctly in the suspension.
When
altering ride height one must be aware of much more than just
the springs. Brake lines, steering, shock length and other areas of
interference. We do not offer coil springs which will alter any ride
height more than 2-inches. Nor do we recommend anyone alter the ride
height more than 2-inches.
While we
have all sorts of springs which will vary ride height, spring rates and
ride quality on the shelf, cutting coils may be, in some cases, the
only way to achieve the desired stance one is looking for.
I do not
know the distance from the spindle to the fender lip, but we do have
other measurements which may help. Send me your fax number.
I hope
this answers your question and enjoy your t-shirt.
- Mike
|
