how to convert kN/m and kNs/m to kg/mm or lbs/in ?

[xagna]

LFS shows spring rate and damping force in newtons.
Does any one know how to convert kN/m and kNs/m to kg/mm or lbs/in which is used in the U.S?

[tristancliffe]

From memory...

1 kN/m = 0.001 kN/mm
0.001kN/mm = 1N/mm
1N/mm = 1/9.80665 kg/mm (or 0.10197kg/mm)

1kN/m = 2.590079 kg/inch
2.590079 kg/inch = 5.70819 lb/inch

[Gunn]

This (http://www.onlineconversion.com/force.htm) might be handy at times too.

[Hallen]

This (http://www.onlineconversion.com/force.htm) might be handy at times too.

lol, that is one of my favorites too :)

Xagna,
But why do you want to convert it? One of the things I wanted to know was spring/dampening rates for real cars, but it seems that suspension manufacturers don't like to publish that data.

[xagna]

I was curious about how real car's spring rate would translate into LFS's newton. Wanted to recreate my car in LFS and use it as a training tool.

[tristancliffe]

You could remove the springs off your car, and use something to compress them a known amount. You then calculate from that the 'spring-rate' in whatever units you want.

So rather than ask the spring manufacturers, you 'ask' the springs themselves. Might be a bit harder with dampers (you can't push them with your arms and measure anything usable), and could potentially be difficult with anything other than LFS's suspension geometries (unless you're good at maths).

[Bob Smith]

It'll never work perfectly as no spring is truly linear.

[JTbo]

What is formula to calculate spring value from spring dimensions?
I know it is not simple, but if metal is know it should be just the math to calculate stiffness. Any engineers near by?

[tristancliffe]

The ones in LFS are, aren't they? So an approx value of spring rate would work in LFS. It would be close enough?

Sprint Rate (N/mm) = Applied Force(N)/Change in Length(mm)

Spring Rate(N/mm) = G d^4/ ( 8 N D^3)

where
G = Modulus Of Rigidity (N/mm^2)
d = diameter of wire (mm)
D = Major external diamter of spring (mm)
N = Number of coils

I think - it's quite late, and I might have misread/misinterpreted it.

[JTbo]

The ones in LFS are, aren't they? So an approx value of spring rate would work in LFS. It would be close enough?

Sprint Rate (N/mm) = Applied Force(N)/Change in Length(mm)

Spring Rate(N/mm) = G d^4/ ( 8 N D^3)

where
G = Modulus Of Rigidity (N/mm^2)
d = diameter of wire (mm)
D = Major external diamter of spring (mm)
N = Number of coils

I think - it's quite late, and I might have misread/misinterpreted it.

Thanks Tristan :)

I did give a go for it:
G=1080
d=12
D=127
N=6,5
End result is 0,210249 N/mm ?

I did made calculations with excel spreadsheet that is available in here (http://www.janiervast.com/tiedostot/Spring_stiffness.xls)
G is actually bit of guess, 1080-2010 N/mm2 is best data I have come across, so that I need to research more, but it should be in right range.

[tristancliffe]

Typical Values for The modulus of Rigidity for different Spring materials are listed below

Material Modulus of Rigity = G (N/mm^2)
Carbon Steel 78,600
316 Stainless 68,900
Brass 34,500
Berylium copper 50,000

So, with your attempt and my 78600N/mm, I get 15.3N/mm.

Now the next thing is - does that sound about right... Well 15N is about the weight of 2 apples. And 2 apples on a car spring won't make it deflect 1mm. So something is wrong - what spring is it your basing it on?

[JTbo]

Typical Values for The modulus of Rigidity for different Spring materials are listed below

Material Modulus of Rigity = G (N/mm^2)
Carbon Steel 78,600
316 Stainless 68,900
Brass 34,500
Berylium copper 50,000

So, with your attempt and my 78600N/mm, I get 15.3N/mm.

Now the next thing is - does that sound about right... Well 15N is about the weight of 2 apples. And 2 apples on a car spring won't make it deflect 1mm. So something is wrong - what spring is it your basing it on?

My car's front springs, suspension is quite soft, but then again car weights 1100kg, weight distribution is 53% at front 47% at rear.

15.3kN/mm would sound much closer?

That modulus of rigity I took from one company's page that makes car springs, maybe I won't buy from them, lol.

Steel used for springs is bit different from normal carbon steel, it has some features from stainless steel, I think that it has also bit different modulus of rigity, but I think that carbon steel is close enough to get formula sorted out.

Total length of my front spring is 312mm, 80kg did compress it around 1,5-2cm but that is huge guess so can't really do much calculation with that.

One thing, in lfs it is kN/m so N/mm should be yet converted to be able to compare to LFS values, that is x1000 to both sides, right?

[tristancliffe]

For the time being the value of G will do - sure the exact spec of steel will make a difference, but not much.

80kg is 785N, so at 15N/mm it shoudlc compress 52mm!!! Something is still wrong :S Not sure what yet....

1kN/m = 1000N/m = 1000N/1000mm = 1N/mm
so
1kN/m = 1N/mm - a nice conversion isn't it :D

[JTbo]

Surely could be that it compresses 5cm with 80kg, only testing I have done is lean quickly to spring and estimate amount of compression, not exactly accurate, eh :D

Would just need something heavy that I know what is weight so I could perform better test, just can't think anything like that at the moment.

I return at evening to this subject, maybe I find something :)

[JTbo]

I think that result should be somewhere near 30-40kN/m area, but I'm not too good in math, should have been in those lessons :(

Anyway front springs carry approx 583kg so that will be 291,5kg per each spring, if I rise my car up for tire change spring will be uncompressed around 15cm and it is still 2-3cm compressed then, bit difficult to get exact measure as there is snow out there and also that grandma would come again to beat me with a stick if I start to take front suspension out in parking lot :schwitz:

I did change front shocks and springs almost year ago so that how much spring is compressed when car is lifted is from memory, don't know if that makes lot of difference though?

Anyway I get around 19N/mm with those numbers and it can't be right either, too low value.

[tristancliffe]

Okay, using those values you just gave...

The front spring supports 219kg = 2855N, and compresses approximately 180mm from it's free length. Assuming that the spring is 100% linear (they never are quite, but we'll have to make do with this assumption for now, and LFS's springs ARE linear, so even better) that gives us

2855/180 = 15.9N/mm

My calculated value above with a rough value of G, your values of everything else, gave 15.3N/mm, so perhaps we are not wrong after all. I think it's fair to say that the spring rate can be considered somewhere between 15 and 16N/mm.

[JTbo]

I did read from here http://www.whiteline.com.au/default.asp?page=/faqsprings.htm

I make calculations and according to that my car's front springs indeed are 19kN/m rate, maybe there is something wrong in LFS or then my springs just are getting stiffer when they are compressed more as in LFS car is bottoming out in garage if I put so low spring rate.

If springs have feature that makes them stiffer when they are more compressed then I think that there is some correction factor for that, just need to find that.

[JTbo]

Could be as you say, as spring is not fully uncompressed when installed to car and that can make bit of confusion I guess.

But in LFS 17 is minium and 1100kg car is not using springs at all in that value, it is using only bumbstops.

Hmm, we did crosspost.

I just played with my new VB so it put out this (http://www.janiervast.com/tiedostot/Spring_Calculator_installer_beta1.zip)
Maybe some use, maybe not, just did it to refresh my memories from VB.

[tristancliffe]

It's a number of things.

1. Our value of G may be be perfect - I just used a generic one.
2. The coil spacing is not linear - the coils become closer at the ends on most car springs.
3. Some are even slighly tapered overall, which as the effect of changing D throughout the compression, and a varying N as the distance between each coil changes too.
4. The modulus of rigidity is not uniform - surface treatments (hardening, shot peening, painting, heat treatment), surface finish and spring age all act to alter G across the diameter of the wire.

With so many variable it becomes quite hard to make a spring calculator that takes it all into account. In real life, when building a car, the spring calculations only give you a 'ball-park' figure. Further improvments to the spring rate (either for handling, comfort or whatever) are mostly done on a trial and error basis. That's why car manufacturers do many many miles of testing and improving - the mathematical and computer models they use can only get them so close to reality.

Edit:

Okay, another attempt to be sure...

1100kg car = 10791N (lets assume whole mass of car is sprung). Supported by 4 springs, with a 50:50 weight distribution front:rear and side:side (makes my life easier).

I'd tend to think that a car spring would compress about 80mm - 100mm when the car is on the ground relative to the spring being on the bench and free. This would give (assuming 100mm for nicer numbers) 27N/mm.

From memory (and looking at a rule) I'd estimate spring wire to be about 15mm diameter, the coil diameter to be about 125mm, the number of coils to be about 9ish. With a value of G at 78,600N/mm^2, this would give about 28.3N/mm.

Until someone does a few proper measurements on car springs we won't know for sure. I'm back to work tomorrow, and I could measure the spring rate (roughly) of a Toyota Sera spring that is used in a 'Handling Pack' (don't think it gives more grip, but it makes it less rubbery to drive, and boy racers always assume lower and stiffer = better).

[JTbo]

Sounds very good indeed. Of course there are numerous things to think about, but these may give good estimate, so one can be in correct range when putting his/her car data to LFS suspension settings.

d, D and N are very close, inside 1mm even with D as I measured those back then when springs were in table, so if other springs give ok results then we just must accept that my car does have somewhat soft springs and there is possibility that in LFS has something odd as even 19kN/mm springs won't support 1100kg car.

I wonder if dev's have accurate data from springs or have they used some general math to come to current rates and weight carrying capability?

[JTbo]

Funny, if I calculate with 6,5 coils I get that 15,3 but if I 'cut' 2 coils I get 22,1.

I'm not sure if I understand this totally, but every coil carries certain amount of load, if there is less coils each coil need to carry more load. Now if we cut 2 coils then spring has 30% less coils and less stiffness or same stiffness, but certainly not more stiffness?

Too complicated for me, lol.

[tristancliffe]

I'd expect LFS's springs are linear, and the value of spring rate is simple a case of force/compression.

I'd quite like to hear Bob's comments on this, as he has done a lot of work with making road car setups already. I think he calculated the spring rate from an ideal spring frequency, but I would to hear his reason why our 'calculated'/guessed spring rates are quite a lot lower than his Easy Road setups.

A coil spring is bascially a wound up torsion bar. Removing coils is the same as shorting a torsion bar. Take a plastic rules and grip the ends. Twist -> easy to twist. Now hold one end and the middle so the effective length of the ruler is halved, and it will be harder to twist, hence the spring rate increases.

[JTbo]

I'd expect LFS's springs are linear, and the value of spring rate is simple a case of force/compression.

I'd quite like to hear Bob's comments on this, as he has done a lot of work with making road car setups already. I think he calculated the spring rate from an ideal spring frequency, but I would to hear his reason why our 'calculated'/guessed spring rates are quite a lot lower than his Easy Road setups.

A coil spring is bascially a wound up torsion bar. Removing coils is the same as shorting a torsion bar. Take a plastic rules and grip the ends. Twist -> easy to twist. Now hold one end and the middle so the effective length of the ruler is halved, and it will be harder to twist, hence the spring rate increases.

Thanks, that explained a lot again. Eventually I may learn something here :D

[Bob Smith]

I notice that the formula you presented (Tristan) doesn't seem to take enough into account. I've found the formula elsewhere online so I'm not doubting it's accuracy. However this means that the tightness of the coils has no impact on the stiffness?

I also found this: http://home.earthlink.net/~bazillion/design.html

You are quite correct though, I picked what I thought to be likely spring frequencies of a road car, and used the weight of the car to determine the required spring rate.

[JTbo]

I found one site that has term 'dead coils' also it shows how to measure active coils that are coils that affect to spring rate, so after counting coils again I got 4 instead of 6,5. So that makes spring rate of 24,86 kN/m.

There was also other perhaps useful information, I have not yet read/understood all.

Here is link http://www.off-road.com/dirtbike/tech/2002spring/

Also found another link http://www.acewirespring.com/ace-wire-spring/coil-springs/coil-springs.htm

Here is all info that I measured from my front springs:
Length 312mm
Coil outer diameter 127mm
Coil inner diameter 104mm
Wire width 12mm, there was black plastic surface that I guessed to be 0,2mm actual measure was 12,2mm
length from Dead coil to active coil top 53mm bottom 52mm Measured from end of wire.
Active coil to active coil 61mm
Total coils 6,5 I did measure to wire tip to wire tip (end point of wire, is that tip?)

I do have older springs in warehouse, but those are different spec, as these values are from GLT version (sportier, 20mm lower).

[axus]

Well 15N is about the weight of 2 apples. And 2 apples on a car spring won't make it deflect 1mm. So something is wrong - what spring is it your basing it on?

15N the weight of 2 apples? 15N = ~1.5kg with the Earth's gravitational acceleration. That is 750 grams per apple... not sure where your apples grow but I'm looking at a 1.5kg pack of apples that has 13 in it...

How about working in reverse?

1100/4 = 275kg/spring
275*9.8 = 2695N

Assuming spring stiffness of 19N/mm

2695/19 = 141mm

Assuming that the springs still have 3cm (30mm) of travel left when unloaded.

Anyway front springs carry approx 583kg so that will be 291,5kg per each spring, if I rise my car up for tire change spring will be uncompressed around 15cm and it is still 2-3cm compressed then

Based on this, your car should have a shock absorber travel of 15cm (I will take it as 14 as that is what my measurments said it should be) + 2cm + another 2-3cm further compression after the car is loaded for diving and squatting. So that makes a motion range of 180-190mm (please tell me if I am talking utter rubbish, but unless you want bottoming out I believe this should be roughly accurate). I put these values (19kN/m [=19N/mm] spring rate and 190mm suspension travel) into the XR GT - it has roughly the same weight as your car - 1167kg and practically the same weight distribution so it should be about right (I am way too lazy to use mecanik to tweak it now but if you want me to I will). Then I used a damping of 3.8kNs/m at the back and a damping of 4.5kNs/m at the front because with these values the car's springs compress and uncompress when dropped (with any smaller value they would compress again afterwards - this is the way most road cars should behave when dropped). It looks perfectly sensible to me... I will assume the differential is open. With no roll bars it will hit 0.9 lateral g's on the BL car park and with 10kN/m at the back and 20kN/m at the front it will do 1 lateral g. You will have to tell me your car's camber values, toe values (these two should be on your wheel-alignment invoice or a document that you got when you did your wheel alignment), tyre pressures, roll bar stiffnesses (if there are roll bars) and gearing (this isn't too important but you should find it somewhere)... and, of course the driving wheels and then I could recreate your car pretty closely in LFS... If you give me this information I will also do a mecanik profile for the car to get the power, torque, mass, etc correct. The values you gave don't seem unrealistic to me though... just out of interest, WHAT CAR IS IT? :thumb: :D

[JTbo]

I have made setup that has almost all data correct for my car, springs, shocks and arb settings are guesses also steering angle.

This is what repairmanual tells.
Front suspension:
camber -60' to 0'
Caster 7 deg 35' +/- 45'
KPI(Inclination) 9 deg 30' +/- 30'
Toe in 4.0 +/- 1.0mm

Rear suspension(not adjustable, de-dion rear axle, leaf springs not possible to get accurate in LFS):
Camber -2 deg +/- 30'
Toe in 0 +/- 3mm

Steering wheel turns, lock to lock 4.4

Tires are 185/60/14
Pressures on track are 2,5-2,8bar on road 0,4bar less.

Ground clearance is 141mm
Here is data that I have measured from front shock pic1 (http://www.janiervast.com/kuvat/iskarit/etuiskari_volvo_360_ver2.jpg) pic2 (http://www.janiervast.com/kuvat/iskarit/ylapaa_mitta.jpg)

Gearing I have in LFS, but only with LFS Tweak I can change wheels to proper size and therefore gearing is not working correctly because of too big wheels.

ARB data I don't have, my poor car has ARB only at front, it is only 12-15mm can't remember how it was.

My grandma car see here (http://www.janiervast.com/projekti/)
6,5MB video of granny drive (http://www.janiervast.com/videot/summer_2005/8_5_2005_nopein_2nd.AVI)

If we can find a way to have realiable close results from measuring springs, it would help to get some car data to get into lfs setups but also it could help car enthusiast like me to check different springs when getting used parts so one knows approx of stiffness.
Anyway I find this theory stuff quite interesting :)

[tristancliffe]

However this means that the tightness of the coils has no impact on the stiffness?Well my equations don't take into account the length of the spring, so more coils could mean a longer spring or tighter coils. Either way, same effect.
I also found this: http://home.earthlink.net/~bazillion/design.html (http://home.earthlink.net/%7Ebazillion/design.html) Arrrgg, Wahl stress factors - now you're getting in complex real life design rather than quick approximations...

15N the weight of 2 apples? 15N = ~1.5kg with the Earth's gravitational acceleration. That is 750 grams per apple... not sure where your apples grow but I'm looking at a 1.5kg pack of apples that has 13 in it...Don't know what happened to my brain there. 1 apple is roughly 1 Newton (which is where the apples falling on his head story came from), and for some reason I have applied that so 1 apple ~ 1 kg :x Ooops, well spotted. This will be where I got confused with my 'is it reasonable' checking!

Good to know your calcs end up with similar results to mine too :)

JTbo, I'm gonna have a look at your figures and see what it looks like. :up:

Okay, with the average coil diameter taken as 120 I got a spring rate of 29.5N/mm which seems good. Tried the formula that Bob provided, but it's not clear what units to use, especially for the Maximum Fibre Stress, so I didn't do it, and I'm too lazy right now to read up it and Wahl factors etc. Maybe later. For now, with your values for the suspension, plus the above answers for spring rate you'd be somewhere close.

[JTbo]

How spring is measured and how to use that information seems to matter mostly in this one, I must write some instructions how to measure so your formula will be useful to all.

This thread has now quite lot of information of springs so I better save this somehow.

Thank you from your work with this :)

[axus]

Quick question...
How does one work out the force applied on the springs with a given acceleration (let's say just longitudinal to make it simple) - I can imagine how the rest would work from there...

[JTbo]

Quick question...
How does one work out the force applied on the springs with a given acceleration (let's say just longitudinal to make it simple) - I can imagine how the rest would work from there...

I don't perhaps understand 100% but I think that you are looking weight transfer now, if acceleration is 0.5g then certain % of weight is sifted towards rear, springs, shocks up/downhill etc. affect to this and I think it is way too complicated for me :schwitz:

I try to find location for my car's mechanic setting and setup to go along with and start new thread there about my car in LFS.

[axus]

Yes, I am refering to weight transfer - I presume the center of gravity has a part to play here and acts like a pivot point around which the car dives and squats... but from there onwards I can't think of much.

[tristancliffe]

Well the force that causes the acceleration acts through the wheels, yet the center of gravity will be somewhere higher. This creates a moment about the CoG, which, basically, adds on the rear spring load, and subtracts from the front spring load. Opposite with brakes.

Suspension geometry, anti-dive/squat, damping effects etc all make their mark in this scenario (it's usually varying and transient), so doing a sum to work it out won't tell you much. I think.

At it's most simple, a 500N force acting the wheels, which a CoG 100mm from the road surface will create a moment of 50Nm. Assuming the wheels are 1.5m from the CoG (in both directions i.e. the CoG is in the middle) then additional force on the rear wheels will be 75N, and the front wheels will 'lose' 75N. This change of load on the springs will change the extension of them.

[Bob Smith]

Yes, I am refering to weight transfer - I presume the center of gravity has a part to play here and acts like a pivot point around which the car dives and squats... but from there onwards I can't think of much.
If I've understood Colcob's posts in the past correctly, the roll center and CoG height are independant. The roll center is usually way below the ground and is due to the suspension geometry, whereas the CoG just affects how much weight transfer and subsequently how much body roll there is.

Edit: Tristan, changed your sig I see, no more Bob & meat-juice comments ;)

[axus]

I see. Thanks for the explaination, guys. :thumb:
From what I understand of your post, Bob, the CoG moves with suspension travel...? Jeesh - this is enough to make someone throw up! I would have assumed it moves as fuel in the tank moves around and things like that but as I understand it the CoG also rotates around the roll center. If that is bellow the ground, the CoG changes as the car dives and squatts and then the moment changes even if the same force is maintained constant, and so does the distance of the CoG from the wheels. :pillepall Simulating all this must be hell. I'd love it! :D

[tristancliffe]

Simulating all this must be hell. I'd love it! :D

Thats why GTR, GTL and 'sims' like that don't bother - they just have a nice table of values. LFS is great because not only does it simulate it, but it simulates it well!

[the_angry_angel]

Totally off topic (sorry): If I've understood Colcob's posts in the past correctlyHas anyone seen colcob recently? He seems to have dropeed off the face of the earth.

Tristan, changed your sig I see, no more Bob & meat-juice comments ;)Now thats just asking for someone else to add it into theirs ;)

[JTbo]

How would be possible to have spring length to formula?

I'm just thinking if we have two springs, both are equal every other aspect but one is shorter than another, I guess they have not equal stiffness then?

That is because adding more coils make spring less stiff and if you have two different length springs with same number of coils, then other have to be stiffer than other, because if I cut longer one so that it is equal length with shorter one they will have different number of coils, or is only number of coils that matters not spring length?

For example 400mm and 800mm spring, both have 4 active coils, cut 800mm to half and you will have 2 coils vs 4 coils.

Maybe total length should be divided with number of coils, that would give relation between those, but how to use that value in formula instead of number of coils. Eh, I can't think this out :(

Maybe I just can't think of number of coils as torsion spring, I'm thinking length of spring. However length of wire is perhaps what counts here, as if spring is cut that is shorter, no matter how long is spring's external length. Still it feels funny. Grrr, my brain is not working anymore, must get some sleep ;)

[Bob Smith]

Well this formula does not have anything to do with length, so therefore if everything is identical except length, the stiffness of the two springs will also be identical.

If you cut a spring in half, it gets stiffer, as simple as that.

Basically ignore the length when thinking about the stiffness, unless of course you shorten a spring, as you will also be changing the number of coils.

[JTbo]

Well this formula does not have anything to do with length, so therefore if everything is identical except length, the stiffness of the two springs will also be identical.

If you cut a spring in half, it gets stiffer, as simple as that.

Basically ignore the length when thinking about the stiffness, unless of course you shorten a spring, as you will also be changing the number of coils.

Hehe, I better leave thinking to others for a while, all things that involve round shapes are so hard to understand (like a woman for example) ;)

[axus]

Could on calculate the roll center? Even if a programme is used to analyse an RAF file to find it? Would this be of any help to anyone?

[JTbo]

Little miniature spring, but hey, it is a spring and not even one of those evil progressive ones ;)

1,6mm wirediameter
17mm outer coil diameter
15,4 inner coil diameter
That makes 16,2 average that is used in calculation
3mm coil to coil
3 coils

with 1kg load compression = 7,7mm

Using formula Tristan provided "Spring Rate (N/mm) = Applied Force(N)/Change in Length(mm)"

Spring rate = 1,27 N/mm

Ok now important thing, I did cut one coil off and look:

with 1kg load compression = 4,4mm

Spring rate 2,23 N/mm !

One tiny little problem here, I don't exactly know what material this spring is made of, it won't rust and it is shiny bit like stainless, but I think that it is not stainless, but just with special coating or something.

As Material Modulus of Rigity = G (N/mm^2)
Carbon Steel 78,600
316 Stainless 68,900
Brass 34,500
Berylium copper 50,000

There is no match, also 19800 gives correct spring rate for case 1, but when I cut one coils leaving only 2 coils it won't match anymore.

Any ideas what is wrong here?

[tristancliffe]

So in case 1, G appears to equal -> 19773N/mm^2

In case 2, G 'changes' to -> 23146N/mm^2

Well, this could be for a number of reasons. As the spring is quite small, the wire length is quite short, so every 'section' of metal is having to twist a bit further than it'd really like to - exceeding the elastic limit across parts of it's cross section.

Also, some spring steels don't have constant elasticity, and so (as the wire bends more per unit length in case 2) appears to have a different value of G (I know G is rigidity, but elasticity and rigidity are closely linked).

Also, it could be measurement errors :p

But without playing with tiny springs here, I can't really double check, and I just so happen to be fresh out of tiny springs.

The material, by the way, is likely to be a either a high-carbon spring steel allow or, perhaps, a stainless steel spring alloy. If you wanted to have something new to do, you could try working out the volume of metal within the spring (area x wire length) (don't cheat by straightening it :p), and also weighing it accurately (that'll test you). Then you can get a value of it's density, and that might help you find out what it is. Is it magnetic by the way (not that that tells us much as some stainless steels are magnetic). Either way, here's a few samples of spring material...

* High Carbon Spring Steel - These spring steels are the most commonly used of all spring materials because they are the least expense, are easily worked, and are readily available. They are not suitable for springs operating at high or low temperature or for shock or impact loading.

* Alloy Spring Steel - These spring steels are used for conditions of high stress, and shock or impact loadings. They can withstand a wider temperature variation than high carbon spring steel and are available in either the annealed or pre-tempered conditions.

* Stainless Spring Steel - The use of stainless spring steels has increased and there are compositions available that may be used for temperatures up to 288°C. They are all corrosion resistant but only the stainless 18-8 compositions should be used at sub-zero temperatures.

* Copper Base Spring Alloys - Copper base alloys are more expensive than high carbon and alloy steels spring material. However they are frequently used in electrical components because of their good electrical properties and resistance to corrosion. They are suitable to use in sub-zero temperatures.

* Nickel Base Spring Alloys - Nickel base alloys are corrosion resistant, and they can withstand a wide temperature fluctuation. The material is suitable to use in precise instruments because of their non-magnetic characteristic, but they also poses a high electrical resistance and therefore should not be used as an electrical conductor.

[JTbo]

Measurement error it sure can be, working with mauser and kitchen scale may not be most accurate method :D

I did other measurements too and in one spring came up as stainless steel, very nicely I could calculate from spring dimensions and also test spring rate and it was same, but when I cut spring situation changed, calculated rate was not enough stiff anymore.

Must find bit bigger springs to test :P

This material seem to attach very well to magnet.

Impossible to measure weight of spring, my scale is not enough precise, spring weights under 10 grams or so.

I got several of this kind of springs when cargo box falled from truck and one truck drove over that box, then there was hundred of thousand springs all over the road, so I took few to safe home :nod:
I have no idea where these could be used.

Need to dig more of my stuff, I think that somewhere I had valvesprings, but could be that this 5kg scale is not enough for those.

[Bob Smith]

I got several of this kind of springs when cargo box falled from truck...
:D should have stopped there.

[JTbo]

:D should have stopped there.

Life is sometimes much more amusing than dreams ;)

My car's valve springs are stiffer than suspension springs, funny but could be true as valve spring need to be very stiff to prevent valve bouncing.

Wire diameter 4,4mm
Coil diameter 32,4mm outer
Coil diameter 23,7mm inner
Makes around 28,05mm avg
3 coils

[JTbo]

I did made yet more findings.

This is page of spring designing software:
http://www.hexagon.de/fed1_e.htm

Specially this image is interesting as there is all data we need to make calculation and also end result:
http://www.hexagon.de/gif/fed1_2e.gif

What is interesting is that G is 80000, but G400 is 71556 and with this G400 I get to same result as in that picture, 64 N/mm.

I don't understand much from that page, but perhaps you will get more out from it?
Anyway that G400 seems to be something to do with temperature of +400C.