Springs 101 and cutting

[Guest 510kamikazifreak]

Only for the noobs :lol: :P

 

Found this on another site, thought may be the answer to many questions that seem to be out there.this will help figure spring rates etc for Coil Springs

 

 

 

 

 

 

 

"Popular wisdom" rules. Cutting coils does increase the spring rate.

 

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 front spring is made from .610 wire and has an inside diameter of 3.875" and has a free height of16.145" (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. (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.

 

See this tech question on Cutting Coil Springs for a more complete explanation.

 

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.

[Guest 510kamikazifreak]

In a lot of cases it is fine to cut the coils. However this is one case where it can not be done.

 

Your springs have what is called "pigtail" ends. The ends have a smaller diameter then the body of the springs.

 

The only type of coil springs that can be safely cut are those with "tangential" ends.

 

Tangential ends look as though they just twist off into space. The spring would fall over if you tried to stand the spring up.

 

Now that I have disappointed Frank, let me tell those with tangential end springs how to cut the RIGHT way.

 

DO NOT USE HEAT. Did you hear me......NO HEAT

 

Cut the coil with a hacksaw or a rubber cutoff wheel. NOT A TORCH!!!! And be sure to use safety glasses.

 

Don't cut off the whole amount you need to lower your ride. Cut off about 3/4 of the amount you wish to drop your ride. Then re-install the springs and drive the car like you stole it, do not baby it.

 

Then recheck the ride height. If you are lucky you'll be at the right height. If not, take the springs out and cut off a bit more.

 

If they are front springs you better be using a spring compressor.

 

Remember, there ain't no way of making them longer, so be careful.

 

 

 

 

 

 

 

 

 

 

 

 

http://www.eatonsprings.com/springtech101.pdf

^^^

All the info you could need to help along the way

 

 

 

Spring Rate - Spring Load - Load Rate

The most Misunderstood and the most Misapplied spring terms

 

What you are about to read is the gospel and anyone who says otherwise is a know-it-all who knows nothing.

 

RATE - Half the difference between the loads 1 Inch above and 1 Inch below a specified position.

 

Or put an easier way, it is the amount of weight required to deflect a spring 1 Inch.

 

The lower the rate, the softer the spring.

 

LOAD- The amount of weight on the spring.

 

LOAD RATE- The amount of weight a spring is designed to be supporting at a certain height.

 

Load Rate and Rate are not interchangeable terms.

 

Now let's put what you just learned to use.

 

A spring which is designed with a free arch of 6 Inches and a loaded height of 1.25 Inches when supporting 600 lbs would have a Load Rate of 600 Lbs. and a Rate of 126 Lbs.

 

6 Inches free height minus 1.25 Inches loaded height = 4.75 Inches of deflection

 

600 Lbs. divided by 4.75 Inches of deflection = 126 Lb. rate

[RedBanner]

much needed info, thanks

[Guest 510kamikazifreak]

Hope it helps

[RedBanner]

Bump

[Stupid_fast]

Some applications with the "pigtail" ends is possible, I did it in my toyota cressida. Both front and rear require spring compressors, the front spring pearches have a lip, so you can cut the lower coils off. The rears are tapered towards the lower end so you cant cut the bottom, have to cut some off the top and its good springs are secure.

Have to use your own judgement when deciding to cut or not to cut, safety first!

 

Only con is the front ones now groan a little bit! Not loudly, but noticeable.

 

IMO chopping for a small drop with NEW shocks is OK, a new set of KYB's where only $120 for my application. Agressive stance should always be on coilovers.

 

My toyota handles much nicer after dropping it 1", it should have come this way from the factory!!

[Xander_42]

There is one small error in this: spring rate is measured in lbs/in, not straight lbs (correct units make an engineer's life way easier, because if you have the correct units, you basically never need to remember an equation).

[RedBanner]

Isnt Spring rate is measured in inch pounds?

[Xander_42]

Isnt Spring rate is measured in inch pounds?

 

No. Inch pounds (actually pound-inches) is a measurement of torque. It is usually used for smaller torque values. Equivalent, obviously, to 1/12 of 1 lb-ft (pound foot, not ft-lb/foot-pound which is a measure of work/energy).

 

Pounds/inch is spring rate because if you divide a force by it, the result will be the number of inches that force will compress the spring. You can also multiply the spring rate by a distance (in inches) to figure out the force it takes to compress a spring by that distance.

 

N/m is the metric way of expressing spring rate.