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Nicely done.

I found a site called Eng-tips.com which is all types of engineers asking and answering questions. From there I pulled this quote:

"So far as the original post goes, the spring rate will be unaffected so long as the number of active coils remains the same. The rate of a spring is inversely proportional to the number of active coils."

Read the whole discussion here
 

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Discussion Starter #42
rate of a spring is inversely proportional to the number of active coils

Urethane said:
Nicely done.
Thanks, but I'm certain that gtdream will have reservations and find shortcomings.

Urethane said:
I found a site called Eng-tips.com which is all types of engineers asking and answering questions. From there I pulled this quote:

"So far as the original post goes, the spring rate will be unaffected so long as the number of active coils remains the same. The rate of a spring is inversely proportional to the number of active coils ."

Read the whole discussion here
I particularly enjoyed reading the post that asked, " The real point is do you really want better handling, or do you want the psudo racer look."

We know what Iggy's answer was. More than a few others feel the same way that he does about "that 4x4 gap".:D For myself, my concern is more in the area of performance than apprearance.

Chacun à son goût.

I also noted that these engineers were able to easily separate the issues of strength and spring rate, something that gtdream seems unwilling or unable to do.
 

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Sleepy:
Thanx for taking the time & effort to carry out the experiment in such a clear and concise way. I think this subject can be put to bed now unless gtdream cares to add anything........................................................well?
 

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hold up.

How did you calculate the spring rate??

The spring rate is calculated my manipulating the equation:

F=-k(x)

force equals negative k times delta x.


I guess im not used to seeing stuff done in the non metric system. My numbers that I got are not the same as your but lead to the same conclusions. Using your data.

quote:
"It ain't rocket science. I taught myself that stuff in a couple of weeks, not all that sinusoidal/moment of force mumbo jumbo that gtdream and TiZBaD99 keep trying to bring into the discussion, but how wire diameter and coil pitch would affect spring rate."

to you what i said may be mubo jumbo but to me coil pitch is mumbo jumbo, i guess its all subjective.

Also the fellow who is an engineering major (like myself) should agree with you after this. This is a very well done, and well presented experiment which should put this matter to rest. You did a very comendable job.
 

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Discussion Starter #45
faint prize

TiZBaD99 said:
hold up.

How did you calculate the spring rate??
Load / diff

diff is (freeLength-compressedLength)

So I guess it's SpringRate = Load / (freeLength-compressedLength)

TiZBaD99 said:
The spring rate is calculated my manipulating the equation:

F=-k(x)

force equals negative k times delta x.
Sorry that stuff that seems so obvious to everyone else still seems to escape you. It can all be proven without any k, without any delta. Only an engineer would try to complicate such a simple concept. You can't do it using k. k changes when you remove a coil.

TiZBaD99 said:
I guess im not used to seeing stuff done in the non metric system. My numbers that I got are not the same as your but lead to the same conclusions. Using your data.
Let it never be said that I'm not sympathetic to the metric point of view. Translated using 1 inch = 25.4 mm and 1 lb = 454 g. Go nuts.



Seems like, even in metric, the spring rate increases as the length (number of coils) decreases.

TiZBaD99 said:
Also the fellow who is an engineering major (like myself) should agree with you after this.
Most should; engineers? Not always.

TiZBaD99 said:
This is a very well done, and well presented experiment which should put this matter to rest. You did a very comendable job.
Aw shucks, yer just saying that because it's true:eek:

We'll see if it's put to rest after gtdream has checked in. I think he's got a little of that engineering-viscosity stuff flowing in his veins, too.
 

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well sleepcounter, i have this odd and wholly unnatural urdge to concede defeat, though i believe it is due to the fact that i have been proven wrong, with photographic evidence using a coil spring, of course im going to attempt to find something wrong, but that is because i love to argue, anywho, i shall not burden the other readers of this forum with further argument. the spring rate does indeed change with the number of coils. there i said it, it still feels wrong, but i have no way to refute the evidence, though i do intend to run your experiment myself at a more convinient time.. i think the lab is open on thurs.....
 

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Discussion Starter #47
almost everyone onside

gtdreamdxbudget said:
well sleepcounter, i have this odd and wholly unnatural urdge to concede defeat, though i believe it is due to the fact that i have been proven wrong, with photographic evidence using a coil spring, of course im going to attempt to find something wrong, but that is because i love to argue, anywho, i shall not burden the other readers of this forum with further argument. the spring rate does indeed change with the number of coils. there i said it, it still feels wrong, but i have no way to refute the evidence, though i do intend to run your experiment myself at a more convinient time.. i think the lab is open on thurs.....
Between now and thursday, do a little research on punctuation, sentences and paragraphs, okay? I run out of breath trying to read your run-on sentences.

Grab yerself a fresh ball-point (my cat toy was a high-miler that had been stepped on a time or two) with consistent spring pitch.

I was hoping to evince some proof one way or the other on your contention about spring rate increasing as the (same number of coils) spring is further compressed. But if you evaluate the spring rate numbers, you'll see they're not consistently in either direction. Maybe my measuring tools are inadequate to resolve such differences. Maybe your lab has better stuff for that.

But I'm most confident that you'll find (no matter how counter-intuitive you might think it is) that the spring rate will increase as you remove coils.

As far as burdening the other members of this forum, I think we've lost most and those who remain are really and truly interested in something more insightful and mature than name-calling and thread-closing (thanks, Panther, for correcting that re-opening message).

Keep burdening us.
 

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just so you know you did use the same formula that i mentioned, and you calculated the new K values

f=-k(x)

is same as force/load difference

rewritten as k=-f/(x)

the reason there is a negative is because it is convention to use the non compressed spring as 0 length (displacement) and compressed as - and streched as +.

what you did should be and I am sure is, a lab that is done by physics students around the world
 

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Discussion Starter #49
ex-honorability

TiZBaD99 said:
you did use the same formula that i mentioned
Ya don't suppose I'm turning into an [shiver] engineer, do ya?

TiZBaD99 said:
what you did should be and I am sure is, a lab that is done by physics students around the world
Well, if I'm so friggin' smart, where's my diploma?:eek:

[Cue Wizard of Oz music]

'goon needs to see my diploma before I'm allowed to post here. Last week, in his "jacka$$" message, he said:

imagoonmx6 said:
When you produce your PhD in mechanical Physics, come back and argue.
BTW, speaking of physics student around the world, there was a veiled hint by gtdream that he was gonna audit my results (subsequently confirmed in pmail), so that "lab that is done by physics students around the world" is presently being done, once more, as we speak.

Return tomorrow for the next exciting episode of, "Spring Rate Revisited" [organ crescendo/glissando goes here] :eek:
 

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don't let this all go to your head sleepcounter, but your analogy with the 2X4 was correct. It really does not matter if its a leaf spring or a coil. If you are trying to bend a material it will resist. the more you have of the material the more you actually have to bend. even if the material is in a coil you are still bending it.

hope that makes some sense.
 

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Discussion Starter #51
SJ weighs in

el norm said:
don't let this all go to your head sleepcounter, but your analogy with the 2X4 was correct. It really does not matter if its a leaf spring or a coil. If you are trying to bend a material it will resist. the more you have of the material the more you actually have to bend. even if the material is in a coil you are still bending it.
Thanks, norm. I was actually saying pretty much that back in August when I said, "When you remove some length of a coil (or leaf) spring (or torsion bar, for that matter -- torsion bar is just a coil spring straightened out), you INCREASE the spring rate. It compresses less under the same load.)

Find the original gory stuff where I was saying that back last August here

el norm said:
hope that makes some sense.
Made perfect sense for me from the outset. It's been the engineers who've been the hard sell (and who want a coil spring to be subject to different rules than those which govern a leaf spring).
 

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Discussion Starter #52
lab results

gtdreamdxbudget said:
of course im going to attempt to find something wrong,
Still no response.

gtdreamdxbudget said:
i do intend to run your experiment myself at a more convinient time.. i think the lab is open on thurs.....
Can we take your lack of response as confirmation of my findings?
 

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Re: lab results

Nope, that is just circumstantial.

No rebuff does not equal the correctness on the theory.

The burden of proof does not rest on the non-believer.

~Chris

SleepCounter said:
Still no response.



Can we take your lack of response as confirmation of my findings?
 

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Acctually no, I was unable to get into the lab before the end of the semester. Additionally, two limiting factors to my time has been the onslaught fo the holidays ( whoever said they were happy should be shot ) and also my wife and I have been house hunting and have found one. Hopefully we will close on the house at the end of February. Another difficulty I have had was the death of our tercel which was brought about due to the junk they call a carburetor. My solution to the last difficulty has been custom making an adaptor plate to install a weber 2 barrel, and turn the SU unit into a boat anchor. Dont be discouraged, I will eventually get to doing your experiment, it has mearly taken a back burner to the other projects in my life, esspecially after being sent home with my tail between my legs:tup:. Oh, and BTW, great write up on the holland boards for the camshaft and volumetric effeciency.
 

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Discussion Starter #55
gtdreamdxbudget said:
I will eventually get to doing your experiment, it has mearly taken a back burner to the other projects
Good luck on the house and the tercel -- looking forward to your next lab.

gtdreamdxbudget said:
Oh, and BTW, great write up on the holland boards for the camshaft and volumetric effeciency.
Glad to know someone is reading.:)
 

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I read both threads, all 6 pages, didn't find any of the info I was looking for, found some interesting info and got a headache.
So I thought I would share the Thread Ache with the rest of the forum.

It is hard to believe it took so many pages to resolve such a pointless argument.

Anyone ever use a coil spring or valve spring compressor, ever notice the more spring coils you have in the compressor the less torque (pressure) it takes to compress the spring enough to install it. The lower the number of spring coils in the compressor the closer each coil has to be compressed to the next to shorten the spring enough to install it.
So the further the coil travel and more torsion the metal is under the harder it gets to compress.

Because spring rate is based on a fixed travel distance from a free position cutting a coil spring will always increase it's resistance to that travel proportionate to the % removed.

Based on the resistance properties of metal coil springs the 1st inch of travel will have less resistance than the 2nd inch of travel.

But does that fact actually improve handling or mean anything? what if the compression test was done for a certain vehicle at a certain suspension travel point, so the springs get measured from 10" to 8" no matter the springs length (full strut extension to bump stop, what it's a performance car). If two manufactures bring in springs one brings in a 16" coil with a spring rate of 200lbs per inch and the other brings a 14" spring with a rate of 220lbs. In the operational range of the spring test 10 to 8" what are the resulting spring rates and in that operational height 10-8" what weight are the the coils reacting to. (Can anyone answer this question pretending spring rate is the same for every additional inch of travel).

At what travel distance do they test the spring?
Why does that mater, preload: (for this example I will use 2 coils with the same spring rate, the 10" coil would have smaller diameter metal and less coils, I will also mention increased resistance with increased compression but will pretend it's linear/equal to keep the math simple. )

If you have a 10" coil perch (strut perch to mount) and you install a 200lb per inch 12" coil and compress it 2" to install.
Then install a 200lb per inch 10" coil in the another 10" coil perch.
Now measure 1" travel on both coils, the 12" coil is on it's 3rd inch of compression and the 10 inch will be on it's 1st inch of compression. Based on the fact both springs have 200lb / inch spring rate the and the more the spring is compressed the more it resists the 12" spring will require more pressure to move it. (If every inch of travel had = resistance it would take 600lb to move the 12" spring 1inch and 200lb to move the 10".

However there is a second preload on a coil spring and that is the weight of the car.
Lets say the front coil springs carry 600 lb each, the 12" springs would drop 1" (if each inch was equal) but the 10coil would drop 3inches.

Despite the fact both springs have the same spring rate when measured on a press and under the weight of the car. they will perform very differently in the this example. The 12" inch spring will have 1" of upward travel above it's car stance position and will operate that travel between 400-600lb, the 10" spring will have 3" of upward travel and operate that travel between 0 to 600lb giving it a bouncier ride.

Using the same example but this time a 12" spring is cut to make the 10" spring, the 10" spring would have a spring rate of 240lb per inch of travel, so a stiffer spring rate. Would only drop 2.5" when preloaded by the cars weight it will still operate between 0-600 lb above vehicle stance, another bouncy ride.

If you cut the 12" spring to 11" it will have a rate of 218lb per inch, sit with 1inch of perch preload and drop 1.75" on the car. that 1.75" and will operate that travel between 218lb and 600lb.

From ride height downward the 11" coil would demonstrate a higher spring rate, but suspension travels in two directions and what goes up must come down.

So it would seem spring length is as or more important than spring rate when it comes to performance.
The preload in the strut perch (or cross member to frame) decide at what pressure of travel the cars weight puts on the coils between the strut fully extended and the coil spring at ride height. 0lb to cars weight, 300lb to cars weight, 450lb to cars weight...

So how do you lower your car, increase spring rate and decrease wheel travel and bounce?

You shorten the coil spring perch space, shorten the strut shaft length. So if you installed the 11" cut spring in a 9" perch it would require 436lb to compress it 2" to fit the perch and would drop 0.75" inches with the car weight on them, they would operate that 3/4" inch of travel between 436lb and 600lb ( in their 3rd inch of travel).
Have less bounce and travel than the 12" coil in 10" perch.

It is Funny that Larken who set this bullshit ball rolling (OP) for the dung beetles to play with was simply asking an honest question and posted only three times in all six pages and seemed to ignore all the arguing.
Larken actually is a machinist, programmer and engineer, he designs, builds, assembles, makes his own control computer (assembled on mother boards) and writes his own programs to control his CNC machine tables.

He realized the above stance bounce issue shortly after cutting his springs so he lathed down the strut stem and threaded the struts one inch farther, shortening the perch height and stiffening the coils above and bellow ride height, he posted a thread about that and the dung beetles crawled in criticizing the dangers of milling and threading a solid shaft and because a front strut shaft spins in the strut to turn the front wheels it should not be in a lathe, but if the strut shaft spun in the strut as the lathe turned he wouldn't have been able to do any milling (apply resistance when carving off material).
To be honest however as smart as he was Larken never had his cars aligned and they drove like shit, anyone want to know how to make your car run slower 1/4 mile times with a car under the same traction and acceleration, run more than a 1/4 mile, think of the 1/4 track as a straight line 1320 feet long (well it is) think of Larken's cars as an analog pulse on an oscilloscope, he ran 1450-1600 feet of track on 1320 feet of road.

How coil steel works is if two springs are the same length, made of the same tensile material, coil 1 has 2x the number of coils with half the diameter, coil 2 has half the coils and twice the diameter, coil 2 will have a slightly higher spring rate because of the pitch (angle) of the coils, the closser to straight the harder to compress, but coil 1 will have a smother transition of torsion load.
 

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Is no one good at math. The question is:

If two manufactures bring in springs one brings in a 16" coil with a spring rate of 200lbs per inch and the other brings a 14" spring with a rate of 220lbs. In the operational range of the spring test 10" to 8" what are the resulting spring rates and in that operational height 10"-8" what weight are the the coils reacting to?
Can anyone answer this question pretending spring rate is the same for every additional inch of travel.
 

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Is no one good at math. The question is:

If two manufactures bring in springs one brings in a 16" coil with a spring rate of 200lbs per inch and the other brings a 14" spring with a rate of 220lbs. In the operational range of the spring test 10" to 8" what are the resulting spring rates and in that operational height 10"-8" what weight are the the coils reacting to?
Can anyone answer this question pretending spring rate is the same for every additional inch of travel.
Hmm...Id forgotten about this thread (to be fair, it is more than 15yrs old now....amazing how time flies)

A lot of the confusion of this thread is from folks attempting to place accurate values on stock or "stock type" aftermarket lowering springs. These springs are variable rate springs. That variability makes things really hard to lock down accurate values for them. Its a constantly moving target based on coil thickness, winds per inch, taper of the spiral, length...its a headache really.

Anyway...to your question..."pretending spring rate is the same for every additional inch of travel". This is easy..you are describing a simple linear rate spring. Its simply "X-weight per inch of compression", and that ratio doesnt change regardless of how much you compress the spring or what starting height is (hence linear).

For your example....a 16" tall spring of 200lb/in spring rate is compressed to 8" height. So...8(200lb/1")=1600lb/8". It took 1600lbs to compress this spring 8". Its uncompressed height started at 16" and now, 1600lbs later, stands at 8" tall (16"-8"=8"). To compress it 9" takes 1800lbs (now at 7"tall)....10" takes 2000lb(at 6" tall), etc, etc until the spring coil binds(all the coils have compressed to the point where they all touch).

Same math for your example of the 14" tall spring with a 220/in spring rate. To compress it 8" takes 1760lbs (spring now sits at 6" tall)...9" of compression takes 1980lbs (at 5" tall), etc, etc.

Gavin
 

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Hmm...Id forgotten about this thread (to be fair, it is more than 15yrs old now....amazing how time flies)

A lot of the confusion of this thread is from folks attempting to place accurate values on stock or "stock type" aftermarket lowering springs. These springs are variable rate springs. That variability makes things really hard to lock down accurate values for them. Its a constantly moving target based on coil thickness, winds per inch, taper of the spiral, length...its a headache really.

Anyway...to your question..."pretending spring rate is the same for every additional inch of travel". This is easy..you are describing a simple linear rate spring. Its simply "X-weight per inch of compression", and that ratio doesnt change regardless of how much you compress the spring or what starting height is (hence linear).

For your example....a 16" tall spring of 200lb/in spring rate is compressed to 8" height. So...8(200lb/1")=1600lb/8". It took 1600lbs to compress this spring 8". Its uncompressed height started at 16" and now, 1600lbs later, stands at 8" tall (16"-8"=8"). To compress it 9" takes 1800lbs (now at 7"tall)....10" takes 2000lb(at 6" tall), etc, etc until the spring coil binds(all the coils have compressed to the point where they all touch).

Same math for your example of the 14" tall spring with a 220/in spring rate. To compress it 8" takes 1760lbs (spring now sits at 6" tall)...9" of compression takes 1980lbs (at 5" tall), etc, etc.

Gavin
Nice to have you back, even if it ends up being temporary.

I think in both threads a few members associated spring rate with spring tensile strength and couldn't understand how cutting a spring gives the coil winding a higher tensile strength because in their mind the coil must be stronger(thicker/harder grade steel) for the spring rate to increase.
Because spring rate is only measure in fixed distance traveled from a free standing position across the full coil, with the same spring steel and coil distance a shorter spring will always have a higher spring rate.
In one members mind it would seem spring rate is tensile strength, either the yield strength of the coil steel or the coil spring rate was tested between only 2 coil windings over a given distance, so take a factory spring and compress the 2 middle windings X distance. Then cut the top coil winding off the coil and measure the resistance between the same 2 middle windings at X distance. They would both have the same resistance and the same tensile strength = same spring rate.

But the real problem is that more important details where not addressed because this was a thread about egos and not information.

The application, vehicle, suspension travel, free standing perch length, standing position under vehicle weight... all make a big difference in how X spring rate will perform on a vehicle.


The question in my first post limited the total suspension travel from 8" to 10", At 10" the strut is fully extended and at 8" it's on the bump stop.

what if the compression test was done for a certain vehicle at a certain suspension travel point, so the springs get measured from 10" to 8" no matter the springs length (full strut extension to bump stop, what it's a performance car). If two manufactures bring in springs one brings in a 16" coil with a spring rate of 200lbs per inch and the other brings a 14" spring with a rate of 220lbs. In the operational range of the spring test 10 to 8" what are the resulting spring rates and in that operational height 10-8" what weight are the the coils reacting to. (Can anyone answer this question pretending spring rate is the same for every additional inch of travel).
In this scenario the 14" spring can't compress to 6", the 16" can't compress to 7"....

The 16 inch spring like you say would operate at spring rates between 10"= 1200lb to 8"= 1600lb (higher load bearing/operational weight).
The 14 inch spring would operate at spring rates between 10"= 880lb to 8"= 1320lb (lower load capacity/operational weight but still a higher spring rate).

However using more details, number of coils per spring, diameter, type of metal and it's properties complex math and engineering to calculate travel to resistance the 16" spring would likely have an equal or higher spring rate than the 14" spring from 10" to 8" travel.

If the both had the same space between the coil windings and where made of the same coil material the 14" spring would have a 225 lbs per inch spring rate.



The idea behind the question was to have people figure out the difference between manufactere coil spring rates and application spring rates. Coil springs are rated/measured in a fixed value of travel (1") from a free standing position despite length or application. A vehicle operates at a fix point of travel bump stop to fully extended strut.
If a coil spring rate is measured withing the operational range of the vehicle it is going on instead of free standing on a press the numbers turn out very different. The 14" coils have a 10% higher spring rate than the 16" but the 16" operate at 36% to 21% higher weight.

I am guessing variable spring rate labeling is due to a difference in coil diameter thickness of the metal or tempering, usually at the top of the coil resulting in the first X distance of travel as the weaker section of coil compresses enough to match the resistance of the thicker coil steel giving sections of the coil different spring rates.
 
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