spring rate ?

[theking]

if you stack two 300 pound/inch spings together of equal length, like in a coilover, would that halve the rate to 150?

[CurleyMan]

No it would still take 300 lb's per inch to compress the spring. In other words no-matter how many you stack together you would need the same amount of weight on each spring to come to full compression. If infact you stacked two coils on top of each other you would need double the listed weight to compress the combo..... two 300 lb springs would need 600 lbs to reach full compression. Think of it as if they where leaf springs.... more leafs of any given spring rate = more capacity...
that's my:twocents: , i'm sure someone may disagree but i'm pretty sure i'm correct in my explanation.

[Bill Potter]

No it would still take 300 lb's per inch to compress the spring. In other words no-matter how many you stack together you would need the same amount of weight on each spring to come to full compression. If infact you stacked two coils on top of each other you would need double the listed weight to compress the combo..... two 300 lb springs would need 600 lbs to reach full compression

I think you are correct...

but would 600lbs compress 2 inches, or full compression?

[elementalvoid]

CurleyMan is correct. Essentially it is like you bought a longer spring.

To answer your question broncb2b: 2 inches. Full compression depends on how long the spring is.

[Bill Potter]

CurleyMan is correct. Essentially it is like you bought a longer spring.

To answer your question broncb2b: 2 inches. Full compression depends on how long the spring is.

Yes, that is what I thought Ben intended I just got confused when he said full compression.

Another question...in the real world does a single rate spring really keep the same rate until it is fully compressed? Or is it just an average used to rate the spring? If not does the rate climb or drop off?

[theking]

Take this for example. Take a single 300# spring and place it under a 3000 weight. It compresses 10 inches. Then take another 300# spring and place it under the first 300# spring and 3000# weight. The second spring will subsequently compress 10" under the load. thus, 10" + 10" = 20" for both springs. 3000#/20" = 150 #/inch.

Additionally, coils springs are analogous to a long torsion bar in a coil. If you double the length of a torsion bar (or double two coil springs) it make it easier to twist. ie. strain is force over area. double the area should halve the force required.

Am i wrong?

[Bill Potter]

I have a headache!!!

I'm not sure.

[CurleyMan]

I still think your wrong because if you stretch a 300 lb per inch torsion bar from 12 inches to 24 inches it will then become a 150 lb per inch bar..... the metal in the coil is specifically set up with the three hundred lb per inch rate.... a 6 foot 300 lb per inch spring should compress the same as a one foot spring under the same load.... However the springs would be made from very different diameter metal... the 6 footer would have very thick metal and the one footer would have thinner metal...
At least i think this is true
I too am getting confused
maybe Allen can chime in on this one.

Or you could buy two cheap little springs from the hardware store and do a mini experiment with a paper weight

[tmorgan4]

If you run 2 springs, you've got 2 different rates. Primary and Secondary. Assuming each spring is the same length and 300#, the primary rate is 150#, secondary rate is #300.

There are some good calculators out there. Just have to look around.

http://www.swayaway.com/TechRoom.php

Go to "Online Calcuators" and there's a Dual-rate calcuator in there.

[tmorgan4]

Great, this has got me confused. This calculator is telling me different. :rolleyes:

http://members.aol.com/Stvns/springcalculationR3b.xls

EDIT: Nevermind the last comment. I was putting in the wrong numbers in the second calculator. Use the second calculator as it is a lot more detailed. Put in spring rate and travel for primary and secondary in the top left, and leave the helper rate and travel blank.

[CurleyMan]

ok i'm lost And totally confused let me just stick to the leaf spring questions

[Reserector]

ok i'm lost And totally confused let me just stick to the leaf spring questions

Don't get all wound up about it. :crazy:

[theking]

lol. Sorry guys.
I don't get what is meant by secondary rate? the units are in just inches.?
i think no matter what combo you use the springs will compress a given amount per unit load, one linear rate, that is until the coils bottom out on the softer spring. just like each leaf in a leaf spring is different but add up to a single rate.

So if you have a 300# and 100# stacked supporting a 300 pound weight, the 100# spring will compress 3" (300/100) and the 300# spring 1" (300/300) = a total of 4" of compression.
300/4"gives you a rate of 75 pounds per inch.

I brought this up cause we're doing coilovers in a frontier and i want to estamate how much they'll compress with dual 300# springs.

Ben, you're right. a 300# spring no matter what shape, lenght, or whateever will behave the same because it's rated at 300#/inch. But make one out of two of these and the you have to change the rating. Just like cutting a coil in half would double it's spring rate. I think???? I've said enough for today. lol

[tmorgan4]

You got it. You've got a constant spring rate until one of the springs bottoms out. That's the joy of coilovers..based on spring rates and lengths you can adjust when the second "stiffer" rate kicks in.

[captinamazin]

when two springs are stacked end to end they are said to be "in series" (as opposed to "in parallel"). Not going to derive this here but the sort answer is that the effective spring rate is { (1/k_1) + (1/k_2) }^-1
read: inverse of k_1 + inverse of k_2, all to the negative 1st power.

http://scienceworld.wolfram.com/physics/SpringsTwoSpringsinSeries.html
look at equation (6)

NOTE: if both springs went from frame to axle, as opposed to being stacked, then the new effective spring rate is k_1 + k_2, assuming both must deflect the same amount which is not perfect but kinda close.

same force = series
same displacement = parallel

hope this avoids costly miscalculations

[theking]

Damn. :duh:

Not bad for a first post! :bowdown:

:wave:

[Ryan Gee]

(AxB)/(A+B) is the simple way... if I remember right.