Lotus Elan

Spring length

PostPost by: SADLOTUS » Sun Aug 28, 2005 9:46 pm

Just a query at the moment. I've had a set of Spyder rear 2[size=7] 1/4[/size] inch adjustable spring seats suspension made up. But if the car still sits too high after full adjustment, is it acceptable to grind half an inch or so off the spring (equally both sides of course) - what will it affect apart from spring length? or do I have to find a supplier of shorter springs? - I thought they all came in set sizes anyway.
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PostPost by: types26/36 » Sun Aug 28, 2005 10:21 pm

Cant tell you about grinding springs but I have heard of heat been applied to springs with an oxy torch untill it lowered the car :? haven't done it myself so dont know the effects :roll:
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PostPost by: SADLOTUS » Mon Aug 29, 2005 6:33 am

Had a think in my favourite place (bed) and decided there must be loads of different spring lengths made, so checked in the Demon Tweeks catalogue and found lengths from 4 to 14 inches in many poundages, even some with fewer windings, ie less weight/more space.

http://www.demon-tweeks.co.uk/catalogue ... rom=search

http://www.demon-tweeks.co.uk/catalogue ... rom=search

Still curious about the grinding off though.
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PostPost by: rgh0 » Mon Aug 29, 2005 11:15 am

A coil spring is just a torsion bar wrapped into a spiral shape. Grinding or cutting off coils shortens the spring free length but also stiffens the spring by shortening the effective length of the torsion bar making up the spring.

Cutting an inch off will not reduce the static ride height by an inch as the shorter but stiffer spring will now compress less under the static load. The stiffer spring will also affect the handling and ride quality to some degree as it will deflect less under dynamic bump and corner loads.

Resetting a coil spring by heating and then compressing or stretching to reduce or increase the free length will affect the static ride height but not change the spring stiffness because the torsion bar length has not been change.

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PostPost by: brassrngfm » Mon Aug 29, 2005 7:25 pm

Hmmm - Now I give up. I had always thought that springs generated a particular force proportional to the degree compressed and that it was fairly linear - e.g. 80 pounds/inch over the useful spring length. Of course progressive springs with unique changes in diameter as the spring compresses are a different matter. So, why WOULDN'T grinding off an inch change ride height? The torch option scares the crap out of me for fear of altering the spring rate in an unpredictable way along the length of the spring. When I picture the spring compressing, I see some torsion, but don't picture that much torsion proportionately. (Also bending moments). I am an engineer - but admittedly a Chemical variety.
I'm seeking knowledge, not embarrassment. (remember the Chevy Chase/Jane Curtin skit on Saturday Night Live?)
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PostPost by: bill308 » Mon Aug 29, 2005 10:58 pm

Paul,

I believe Rohan is correct in this instance. The classic algorithm for a spring is:

F=kx

where:

F - is the applied force
x - is the spring deflection
k - is the spring constant or spring rate

A change in the spring rate will be reflected by a change in the spring constant such that:

k=F/x

If a linear spring is shortened, the deflection x, for a given force F, will be reduced. Consider a spring with a 10-inch free length and a stiffness k. A force F will compress the spring a distance x.

If one now shortens the spring to 1/2 it's free length (5-inches) and then applies F, the spring will compress x/2 resulting in:

k=F/(x/2)

Since the denominator is smaller by 1/2, the resulting spring constant (k) will be twice as large, or the rate is doubled, if the spring is cut in half.

Good job Rohan. It's not intuitive, but you are correct.

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PostPost by: bill308 » Mon Aug 29, 2005 11:16 pm

Paul,

I believe Rohan is correct in this instance. The classic algorithm for a spring is:

F=kx

where:

F - is the applied force
x - is the spring deflection
k - is the spring constant or spring rate

A change in the spring rate will be reflected by a change in the spring constant such that:

k=F/x

If a linear spring is shortened, the deflection x, for a given force F, will be reduced. Consider a spring with a 10-inch free length and a stiffness k. A force F will compress the spring a distance x.

If one now shortens the spring to 1/2 it's free length (5-inches) and then applies F, the spring will compress x/2 resulting in:

k=F/(x/2)

Since the denominator is smaller by 1/2, the resulting spring constant (k) will be twice as large, or the rate is doubled, if the spring is cut in half.

Good job Rohan. It's not intuitive, but you are correct.

Bill
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PostPost by: rgh0 » Tue Aug 30, 2005 10:20 am

Paul

Cutting an inch off a springs free length will change and reduce the ride height determined by the springs static compressed length it just will not be a 1 to 1 relationship. I tried to describe the effect in non techical / mathematical terms. Given a chemical engineering training If you dig up the design equations for a coil spring and apply them to your spring you will see the effects in mathematical terms and can determine what the static ride height change will be for any length cut of your springs. From the same equations you can also determine the impact of cutting the free length on the spring rate in tems of lbs/inch.

While it is possible to design rising rate coil springs via a collapsing coil design that effectively shortens the number of working coils progressively as the coil compresses this approach is not normal or necessary on an Elan.

Heating and resetting coil springs is not a job for an amateur and needs to be done by a professional spring manufacturing or reconditoning company to ensure the end result is what was desired and that the springs metallurgy has not been detrimentally affected by the process.

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( Both Mechanical and Chemical Engineer )
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PostPost by: SADLOTUS » Tue Aug 30, 2005 10:39 am

CHAPS!!!
I'm impressed!
Things are never as simple as they might appear. I imagined a spring manufacturer making a mile or so of say 110lb spring and cutting it up into various lengths of 110lb springs!

I also have enough aggravation in my life without having to check the poundage of a spring I've cut up, heated and mucked about with!

Thanks for your enlightenment, methinks I'll stick with known manufactured springs.
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PostPost by: M100 » Tue Aug 30, 2005 11:20 am

If the springs are from a known source as they appear to be then i'd suspect something else is wrong.

Are you tightening up the wishbone pivot bolts AFTER the car is on the ground?
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PostPost by: SADLOTUS » Tue Aug 30, 2005 8:41 pm

It's a bit of a mix and match setup; spyder wishbones, Spyder 2.25" adjustable shock tube, koni inserts, 100lb TTR springs, TTR poly bushes. Wishbones are able to turn easily with bushes done up tight. still a while 'til I'm able to drive it and settle the suspension. Car looks just about OK at the moment but if I ever wanted to lower it any more the adjustment is used up - suppose I need to check that Tony Thompson and Spyder springs are the same length.
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PostPost by: mark030358 » Tue Aug 30, 2005 10:16 pm

Hi there,
I noticed you said poly bushes. These WILL make your car ride higher, as you cannot pinch them up. I have fitted these and the rear end sits up. I took them off the front for the same reason. I have just taken off my NEW springs which are correct rate and length and put back the 32 year old ones after powder coating as they a a fraction smaller and softer.

cheers
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PostPost by: brassrngfm » Thu Sep 01, 2005 7:28 pm

I found the spring formula I was looking for: Take a look here and the formula goes back to the basics and clearly explains why cutting a spring lowers height, but because it reduces the number of turns slightly - ups the spring constant a tad. So to get the same spring rate at a lower height - you have to change modulus or the diameter (larger=bad for Lotus) or the wire diameter (ever so slightly smaller).
Now it's all logical!

http://www.insideracingtechnology.com/eibach1.htm

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