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  #1  
Old 07-05-2009, 04:36 PM
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W123 and W140 suspension data

Also posted in Tech Help, but getting no response.

I need spring rate, wire diameter, overall diameter, and overall height for Non-SLS W140 springs. Both the Heavy-Duty versions and the Normal ones.

This information is available for W123s in the shop manual, for W124s in microfiche, not sure where for W140s. Anyone here work on W140s?

This is preliminary "research" for building a replacement for the SLS system in a W123 300TD. Basically, I am using the overall diameter and overall height numbers to make sure the springs will fit (comparing to W123 springs) and trying to get a spring rate that is good for about 500lbs more weight on the rears than a W123 sedan. Spring constant can then be calculated with F=kx and ride height at any weight can be determined.

Also helpful would be unloaded front and rear axle weights for W123 wagon and W123 sedan. Please post here if this information is available!

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  #2  
Old 07-05-2009, 05:43 PM
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Location: Carson City, NV
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From when I weighed my 240D:

Steer axle 1800 lbs

Drive axle 1560 lbs

Total 3360 lbs

This was a manual-everything 240D with a full tank of fuel and otherwise empty. I'm not sure if they recorded the weight before or after I got out to get my scale ticket.
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Whoever said there's nothing more expensive than a cheap Mercedes never had a cheap Jaguar.

83 300D Turbo with manual conversion, early W126 vented front rotors and H4 headlights 400,xxx miles
08 Suzuki GSX-R600 M4 Slip-on 22,xxx miles
88 Jaguar XJS V12 94,xxx miles. Work in progress.
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  #3  
Old 07-05-2009, 06:41 PM
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Location: Carson City, NV
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You can calculate spring rate according to the dimensions of the spring.

K=(d4G)/(8D3Na) (the four and the three are supposed to be superscripts)

Where d is the wire diameter, D is the coil diameter, Na is the number of active coils, and G is the shear modulus of the material. The shear modulus for spring steel should be about 11.5 million psi if I'm doing my math right. The number of active coils depends on the end treatment of the spring. I forget what suspension springs usually have for end treatments.

Na=Nt-1 for plain-ground ends

Na=Nt-2 for squared or squared-ground ends

Nt is the total number of coils in the spring.
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Whoever said there's nothing more expensive than a cheap Mercedes never had a cheap Jaguar.

83 300D Turbo with manual conversion, early W126 vented front rotors and H4 headlights 400,xxx miles
08 Suzuki GSX-R600 M4 Slip-on 22,xxx miles
88 Jaguar XJS V12 94,xxx miles. Work in progress.
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  #4  
Old 07-06-2009, 07:07 AM
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Join Date: Apr 2005
Location: D.C.
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Quote:
Originally Posted by Skippy View Post
You can calculate spring rate according to the dimensions of the spring.

K=(d4G)/(8D3Na) (the four and the three are supposed to be superscripts)

Where d is the wire diameter, D is the coil diameter, Na is the number of active coils, and G is the shear modulus of the material. The shear modulus for spring steel should be about 11.5 million psi if I'm doing my math right. The number of active coils depends on the end treatment of the spring. I forget what suspension springs usually have for end treatments.

Na=Nt-1 for plain-ground ends

Na=Nt-2 for squared or squared-ground ends

Nt is the total number of coils in the spring.
Thanks Skippy. The W123 shop manual gives a number in units of mm/1000N for spring rate for each spring, from there I just plug into F=kx to get k, then use it to solve for any (reasonable) x.

Thanks also for the weights. I'll now assume the drive axle of the wagon has 500lbs more on it (fuel tank slung out behind the rear wheels, extra body, extra glass, and I want to make sure it can still carry weight in the back. So figure 1050lbs on each wheel means 4670N.

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Last edited by GREASY_BEAST; 07-06-2009 at 07:13 AM.
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