Mercedes spring rate deciphering - PeachParts Mercedes-Benz Forum

rwd4evr · #1 03-02-2019, 12:27 PM

Does anyone know what the correct conversion process would be to get a usable spring rate number from the Mercedes books
" Mm of travel per 1000 newtons "
To pounds per inch or kg per mm. I can't find anything and I can't quite wrap my head around how you would go about it since it's a set weight and not distance.

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Roncallo · #2 03-03-2019, 08:36 AM

Quote:

Originally Posted by rwd4evr

Does anyone know what the correct conversion process would be to get a usable spring rate number from the Mercedes books
" Mm of travel per 1000 newtons "
To pounds per inch or kg per mm. I can't find anything and I can't quite wrap my head around how you would go about it since it's a set weight and not distance.

Sent from my SM-J737A using Tapatalk

Does the manual actually say mm per Newton.

If it does divide mm/KN by 5710.1

if it says 1000 Newtons per mm travel then multiply KN/mm by 5710.1

rwd4evr · #3 03-03-2019, 09:24 PM

Here's a page. I'm not certain off hand what my 78 450slc is though. No epc at the moment.

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Frank Reiner · #4 03-03-2019, 11:09 PM

rwd:

Use these #s to do your conversion:
1000N = 225 lb.

225/deflection (in mm) x 25.4 (mm/in) [or, 5715/deflection] = lbs. per inch

Example: 225/21.2 x 25.4 = 269.8 lb/in

rwd4evr · #5 03-03-2019, 11:26 PM

Awesome. Thank you!! Gotta keep. That formula. I'll never keep. That straight.

rwd4evr · #6 03-06-2019, 12:38 AM

So. With your help I've determined the rear spring is 359.43 lbs/inch. The problem I have now is that I have two coils cut off it. So how can I figure that out? Coil count and spring diameter I'll have to check.

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rwd4evr · #7 03-06-2019, 02:01 PM

348.47 lbs/in front springs.

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Roncallo · #8 03-10-2019, 11:56 PM

Quote:

Originally Posted by rwd4evr

So. With your help I've determined the rear spring is 359.43 lbs/inch. The problem I have now is that I have two coils cut off it. So how can I figure that out? Coil count and spring diameter I'll have to check.

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The answer to that problem is a little more complex. The linear portion of the spring stiffness will be inversely proportional to the length. Unfortunately some portion of the part of the spring you cut off is not in the linear portion of the spring in fact ~3/4 of the first coil are against the control or spring saddle and not acting as a spring at all.

The best you can do is get the range of stiffness percent which will be somewhere between the original stiffness and the original stiffness*(original length/cut length). My guess is about 1/2 way in between for the first two coils.

Roncallo · #9 03-11-2019, 12:02 AM

Quote:

Originally Posted by rwd4evr

Here's a page. I'm not certain off hand what my 78 450slc is though. No epc at the moment.

Sent from my SM-J737A using Tapatalk

A 450SLC has a very stiff front spring. I use 450SLC springs in my V12 powered 560SL to return the car to stock height due to the increased weight of the engine. The extra 200LB up front of engine and mods dropped the nose about 1" with the stock 560SL springs.

rwd4evr · #10 03-11-2019, 01:19 AM

Yeah. I'm working on a 450slc. I've got two coils removed in the front. Spring rate isn't bad but it's a little too low. Iike it where it is but suspension geometry is screwed at this point. Rear is a hair low too. I'm putting adjustable dampers on and I'm stiffening it up more. I need about 800 front 600 rear.
I'm planning on removing a 3rd coil up front and a second in the rear and adding spacers under the spring to get the height back. I'm going to put adjustable perches in the front with proper rated springs(soon I hope) but the rear is a bit tricky to do that with the shock inside. The upper mount doesn't allow for a true coilover without some seriuos hacking either. It's a drift only car at this point and it needs to be way stiffer than it is. Body roll is pretty bad.

The springs in the car are 115 324 2204 and
114 321 0704

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rwd4evr · #11 03-11-2019, 01:35 AM

Quote:

Originally Posted by Roncallo

The answer to that problem is a little more complex. The linear portion of the spring stiffness will be inversely proportional to the length. Unfortunately some portion of the part of the spring you cut off is not in the linear portion of the spring in fact ~3/4 of the first coil are against the control or spring saddle and not acting as a spring at all.

The best you can do is get the range of stiffness percent which will be somewhere between the original stiffness and the original stiffness*(original length/cut length). My guess is about 1/2 way in between for the first two coils.

So the actual spring rate would be more accurately figured minus one coil to begin with then determined with the removed coils from there? So the front is 11 coils with a 4" ID. Say one is dead. So 40" of spring. So if I removed 30% of it how does that affect it?

I wish I could go back and tell me to go to school for mechanical engineering.

I'm pretty intelligent but under educated. That's the kind of stuff that's a bit harder to teach yourself. I've been learning more and more geometry building this car but the actual engineering aspect, forces and loads.......

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Roncallo · #12 03-11-2019, 10:50 PM

Quote:

Originally Posted by rwd4evr

So the actual spring rate would be more accurately figured minus one coil to begin with then determined with the removed coils from there? So the front is 11 coils with a 4" ID. Say one is dead. So 40" of spring. So if I removed 30% of it how does that affect it?

I wish I could go back and tell me to go to school for mechanical engineering.

I'm pretty intelligent but under educated. That's the kind of stuff that's a bit harder to teach yourself. I've been learning more and more geometry building this car but the actual engineering aspect, forces and loads.......

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Divide the spring rate of your 40" section buy 0.7. That will give you the new spring rate of your now 28" section.

Frank Reiner · #13 03-12-2019, 12:12 PM

Quote:

Originally Posted by rwd4evr

So the actual spring rate would be more accurately figured minus one coil to begin with then determined with the removed coils from there? So the front is 11 coils with a 4" ID. Say one is dead.

If there are 11 coils and 1 is dead, then 10 coils are active.
If 2 coils are removed, then 8 are active.
The active wire length is therefor 80% of the original.
Divide the original spring rate by .8 to calculate the new rate.

rwd4evr · #14 03-12-2019, 04:19 PM

Quote:

Originally Posted by Roncallo

Divide the spring rate of your 40" section buy 0.7. That will give you the new spring rate of your now 28" section.

Thanks. I figured that could work but I just didn't know if it would be as simple and linear as that.

Roncallo · #15 03-12-2019, 11:13 PM

Quote:

Originally Posted by rwd4evr

Thanks. I figured that could work but I just didn't know if it would be as simple and linear as that.

Its not that linear. Where the dead coils end changes with load. But its as best an estimate as you will get.

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