No, it doesn't, and you are an idiot.
Shortening the spring means the spring bottoms out quicker since there is less travel, and the rate goes infinite, since metal isn't compressible. So if that's what you mean by increasing the rate, you're still an idiot.
Seriously? It's not the number of coils, you numbnuts, it's the DISTANCE BETWEEN THEM along with other factors, like the width of the spring wire and metallurgy of the material. Cutting springs doesn't change any of those, other than giving you less suspension travel. You can't just compare two springs' stiffness by what they look like.
Good grief.
So, OP, don't cut your springs, unless you're a flat-brimmed ricer douchebag who drives Subarus or Volkswagens. There are plenty of decent lowering springs available for this platform and lots of options to buy used in the classifieds section, and zero legitimate reason to cut springs.
Vince, you are wrong, so very wrong.
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 take a coil spring 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 269 Lbs.
This spring has 9.33 total coils but 1.33 coils touch the spring seat so they are inactive leaving 8 active turns.
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 lbs
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
Vince, are you sure you want to stand by the statement that the rate doesn't increase with cutting coils or portions thereof off?