Graham CS

'it is intended to incorporate both the torsion and coilover springs'

Your right, I was being lazy on my typing but thats what I meant when I said

'The .56 rate on the paragon site is for calculations with torsion bars.'

I really should have said helper springs and torsion bars.

How are the KLA's doing at the back?

I think you chose the best route in sorting out the springs rates by ditching the torsion bars.

Still havn't sorted mine.

My house is taking all of my time at the moment (nightmare!)

Nearly there and I will be able to concentrate on the car.

Have you done any more work on the banana arm replacements?

Graham

P.S. I read an article in track and race cars magazine that the back should be set up significatantly softer than the front on a front engine, rear wheel drive car. I assume this is a weight issue and due to our transaxle and weight distribution that I should ignore this.

968cs_red

Graham,

Think you are right about the softer rate at the back not being totally applicable for our application due to the transaxle layout. I seem to remember reading that the weight distribution is around 52/48 front/rear.
Obviously though as you are aware you get a significant amount of weight transfer to the front during braking so thats one reason the front should be stiffer really... but then too stiff compared to the rear and you'll get too much understeer..
It's an artform all this suspension tweaking....

Thanks for the info about the helper spring being 50lb/in....... this certainly seems to make more sense when working out the effective rate at the rear...seemed too high before.
New calculation:
effective rear rate = (0.70 x 50) + 175 = 210 lb/in.

Is that really true about the helper spring being for towing trailers????? surely not?!!! The helpers were only there for the M030 setup.... who would get an M030 CS and then tow stuff?!.....

cheers guys,
Matt.

Graham CS

I know it sounds weird

I read it on here.

It was from a seminar with the guy who did the chassis I think.

Graham

flash968

may have been that part of the towing package included the rear spring setup and he just put 2 and 2 together and came up with 5 - happens

to answer your question - love my kla setup - much nicer than stock - now need to play with the front a bit - the eibach/rtech art progressive springs have an initial rate that is too low for what i want on hard cornering - very nice on the open road though and for daily driving - i will have to compromise my ride quality a bit and go to the coilover conversion on my koinis and install a linear spring to get the rate and ride height i want - the h&r is also too tall for me by a long shot

Damian in NJ

This was said at Hershey, although many things the gentleman said either didn't make sense or were untrue. Has anyone seen a non mo30 with helper springs for towing? Not me. And why would Porsche add them to every mo30, considering they knew these were the most sporting of all? Makes even less sense. So although it was quoted by an ex Porsche wrench, I think he got it wrong.

flash968

no clue on this end - they did so many odd things, i thought it was a possibility - more likley the M030 was fitted as a part of the towing package - i was only speculating that an abbreviated version may have existed - i've never seen a towing package, so i don't know about this one

anybody out there have one?

968cs_red

Thanks for the info Damian… it does sound slightly illogical (as Mr Spock would say! :-)

Anyway guys – thought you may be interested in some calculations I have done. I thought I’d better try and support/prove/disprove all these figures flying about with a scientific experiment!...... so I did some measuring of my rear springs…both the free length and the length while under static load with the car at rest.

================================================================

Weight of the car:
For this we will take the weight to be 3000lb. I think this is a good enough figure.

Weight distribution:
This we will take to be 52/48 front/rear. If anyone has any other figures or real world measurements please let me know.

Rear torsion bar:
This is 25.5mm diameter for the 968 and has an effective spring rate of 175lb/in. I have yet to confirm whether this is definitely the effective rate at the wheel or a calculation needs to be done using the length between the torsion bar and where the force acts. Anyone?

My measurement (rear helpers M030):
Free length = 170mm.
Length under static load = 104mm
Therefore length change = 170mm – 104mm = 66mm = 2.6 inches.

===============================================================

Calculation 1: Weight at each rear corner

Weight at each corner = (3000lb x 0.48) / 2
= 1440lb / 2
= 720lb

Calculation 2: Effective rate at each rear wheel given the measured spring compression

Effective rate = 720lb / 2.6 inches = 277 lb/in

Calculation 3: Using the above, calculate the helper spring (hs) effective rate

Total effective rate = (0.7 x hs) + effective rate of (25.5mm dia.) torsion bar

Therefore:

277lb/in = (0.7 x hs) + 175 lb/in

So:

hs = (277lb/in - 175 lb/in) / 0.7

hs = 145 lb/in

Well there you go. Allowing for the damper taking some of the load which we haven’t taken into account, this seems to agree with the 120lb/in figure I have seen floating around.

Anyone spot any mistakes or have any thoughts?

Cheers,
Matt.

flash968

since you measured the free length, can you get the wire diameter, number of coils, and coil diameters?

968cs_red

Hi Flash,

Well I know the wire diameter is 9.0mm - this is in the 968 workshop manual.
As for the number of coils..I can do. Won't be able to for a bit though..maybe we can see from this picture?

http://dspace.dial.pipex.com/prod/di...es/968M030.jpg

Comes from this page, which is pretty good I think:
http://dspace.dial.pipex.com/prod/di...suspension.htm

How many coils you reckon?

As for the coil diameter. The inside diameter of the smaller top and bottom coils is 2 1/4 inches. As for the larger diameter - don't know but can measure at some point...

cheers,
Matt.

flash968

cool - yes, i will be able to calculate the rate fairly closely once i have the outside diameter of the large coil

thanks

968cs_red

Hi Flash,

How is the calculation done? Do you have the formulas?

Matt.

flash968

the formula is:

K=(W to the 4th power times G) divided by (8 times N times D cubed)

K is the stiffness of the spring
W is the diameter of the spring wire in inches
G is 12,000,000 for steel springs
N is the number of active coils (number of free coils + 1/2)
D is the diameter of the coil measured to the center of the wire

we would have to calculate 2 springs with the different coil sizes and then proportionally calculate the spring

there are 3.5 active coils, but one of them is the smaller diameter - therefore we will use 4 as the value for N

when calculating the proportions, and based on 3.5 active coils, we will use the factor of .285 for the multiplyer of the small coil rate and .715 as the multiplyer for the large coil rate and then divide the sum by 2

therefore that spring should be about (sort of guessing at the large diameter at 4.5): about 70#

anybody see any flaws in the math here? i'm dizzy now

968cs_red

Hi Flash,

Yeah dug up the formula last night in my Chassis Engineering book (Herb Adams)

K = G (d^4)
-----------
8 N (D^3)

G = 11.25 x 10^6 = 11 250 000

I think there are 4 active coils... I used this as a guide http://www.bannerspring.com/quote.htm

Anyway, my calculation:

wire diameter = 9.0mm = 0.35 inches
active coils = 4
mean diameter - I measured the inside diameter of the larger coils last night - this came out to be 106mm
smallest mean diameter = 2.25 inches + (0.5 x 9.0mm) = 2.43 in
largest mean diameter = 106 mm + (0.5 x 9.0mm) = 110.5mm = 4.35 in

K = (11.25 x 10^6) x (0.35^4)
------------------------------------
8 x 4 x [ (( 2.43 + 4.35) / 2)^3 ]

K = 168820.3
--------------
32 x (3.39^3)

K = 168820.3
------------
1247

K = 135 lb/in

mmmmm well I reckon adjusting it slightly as there is more of the larger diameter, if you see what I mean..(thats why you suggested those multipliers I presume?). which would give a larger number on the bottom of the equation, and so a smaller K... I reckon 120# is plausible..

???

Matt.

968cs_red

p.s why the rate came up too high in the previous calculation - using the weight of the car etc.. was because I forgot to think about unpsprung weight!... so the actual weight on the springs should've been less and so the rate would've worked out to be less...

flash968

the small diameter is actually the inner plus one full diameter to reach the wire center (half a dia on each side) so it would be 2.6 dia and not 2.43 - that also lowers the rate

yes, that's why i had the multipliers

yup - actual weight on the spring is a good 50# less per corner, according to paragon - i haven't weighted the stuff yet, but sounds plausible

i was pretty surprised when my math showed about the same number that i have been told the spring is - sure wish somebody would just step up and give us a spec - can't use a spring tester either, because it's progressive

i think i've figured out what happened with the paragon chart - i think the conversion rate should be .65 and not .56 for a stock orientation - it looks like simple juxtoposition - i get a higher number because my upper shock mounting point is not at as much of a slant, so i get a better advantage - i did some quick geometry yesterday and iif i start with .65 and add my advantage percentage i get REALLY close to the physical compression measurements

here's a simple test - place blocks under the front frame so the springs can't compress, but don't lift the car - throw a known weight in the trunk and measure the deflection (it will need to be at least 300 lbs) - you should get pretty darned close