Springs, explained...again... (archive)

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You have a valid point here, but I must make something very clear from the very beginning before I go on and explain some things.

535i, 535iS, and M5 springs are THE SAME RATE!!! PERIOD. No iffs, ands, or buts. I have measured them personally, I have BMW's own technical info, I have seen the ride heights, I have done all the research - as have all the officers in the M535i SIG may I add. Of course people like Ed (Advisor to 'insane-sick-mods') like to use 500 lb in springs and have their kidneys bleed, but I digress.

Anyway, The 528e USA version came with a variety of different springs during the course of production. They roughly varied between 150 and 165 lb-in, and in fact later ones came with the exact same springs as on 535is, 535ises, and M5s. Of course, build date and specs will account for the use of slightly different spring rates for the 528es during their 82-88 USA production run.

Also, let me clarify the website posted before. It has been OFTEN cited as a source, but is wrong on the account of the actual spring rate. The comparisons between 528e, 535i and M5 are close to correct, but the actual calculated spring rate number is wrong. But I digress.

As for your conclusion, in short, yes, the less weight you have pressing down on a spring, the less it well compress. Simple physics. So the 528e, at roughly 200lbs lighter overall than the 535i, will compress their springs less (if fitted with the same spring rate) than the 535i.

You should also notice how BMW made an effort to point out that the M5/M6 stood 0.4" lower to the ground than the 535i. Think about it. M5/M6 weighs more, will compress same spring rate more, thus will be lower to the ground than the 535i. It does NOT have lowering springs - it is all fancy BMW PR. (If we assume roughly 165 lb-in spring rate, and an approx. additional 250 lb in weight for the US M5 over the 535i, with an even distribution over all the wheels, we come out with 62.5 lbs additional weight on each spring. Let's say about 70 lbs to estimate due to weight distribution, etc. Plug in the numbers into F=kx (F = Force (weight), k = spring constant, x = distance of compression in our case), and since it's linear we don't even need to figure out the pre-existing weight, so plugging in F=70lbs, k=165lb-in, x=????. X=0.43 inches. HEY, look at that! We can do the same calculation for the rear and get a similar result, but I don't have the spring constants readily available.)

Likewise, since the 528e is lighter, and some use the same spring rate, then it will sit at a higher ride height than the equivalent 535i.

Oh, if you want to use that website's numbers, then the M5 will be sitting approx. 0.52 inches lower than stock.

Reading a post, a few threads down, from someone asking about what to do to make a 528e suspension better stirred up a question I've had for a while. The answer might be "it doesn't matter," which is actually what I'm expecting, but still. . .

The baby six is lighter than the big six, right? Therefore, there's going to be less weight in the front.

The website given -- http://www.cyberflunk.com/m5/tech/suspension/mf172.html -- says that 528e and 535i front springs have the same number of coils (7) and are the same length (14 inches). Even the coil diameter is the same. What makes the 535i springs a higher rate (and, therefore, stiffer) is the wire diameter.

That being said, since the 528e springs are softer, the engine pushes down on it more. If you put 535i springs that are the same length but stiffer in there, wouldn't the springs hold up more and keep the engine from pressing down on it as much, hence making the front end sit a little higher?

Someone please shed some light onto this for me. I'm sure I'm royally confused.

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