OK, this is truly Off Topic, but...

I am struggling with an analysis of a double spring problem.

The issue at hand is the Morgan front suspension. It has a sliding stub axle on a fixed kingpin (see this drawing). So the kingpin is fixed to the frame at both the upper and lower ends, and the stub axle holding the wheel slides up and down on the kingpin.

There are 2 springs, the upper main spring carrying the weight of the vehicle, and a shorter, stiffer rebound spring at the bottom. Now comes the hard part - there is a pre-compression load on the springs. So for the first part of motion, the springs interact (pushing against each other), until the rebound spring is fully extended. From then on the rebound spring is out of the picture and the wheel is riding the main spring.

What I have been trying to figure out is what the effective spring rate is for the initial part where the springs interact. There seems to be two opinions - that it is stiffer than the "only main spring" scenario because of the preload, and that it is less because the springs partially cancel each other.

I would imagine there is a simple standard formula for this, but I haven't ben able to find anything...

http://en.wikipedia.org/wiki/Hooke%27s_law

The preload is a red herring. The additional force per unit compression is linear, so over the whole range where both springs act (where neither is released nor bottomed-out), the resultant force is the same whatever the preload force. And that resultant force is just like that of a single, less stiff, mainspring.

A suspension system that suddenly got less stiff at a certain excursion would be a really alarming and dangerous thing!

Peter

I disagree - the *amount* of pre-load is not significant, but the fact that the pre-load has pre-compressed the bottom (rebound) spring *is* significant. The simple applications of Hooke's law on the wikipedia page only work for straight-forward parallel and serial situations, this is more complicated. And yes, there is definitely a discontinuity - where the lower spring is fully extended and starts to float.

Well, maybe I've misunderstood the mechanism, then. My answer only applies in this situation
where the kingpin with the two springs on is held in a constant-sized bracket, and the wheel is attached at bearing <O>, which can freely move along the pin. Because of the linearity, isn't this just the same as Wikipedia's parallel-springs picture
except with one of the springs having a negative spring constant?

Peter

Yes, except the springs aren't attached at ends, so once the shorter spring is fully extended, it "disconnects", thus providing the non-linear transition.

I'm just not quite sure that it is equivalent to the "two springs in parallel" scenario even in the initial phase.

I agree with Peter's first post.

The only way the spring constant will change is if the coil spring is wound in such a way as to allow certain coils to collapse before others (commonly called a 'cargo coil' over here in the US).

The way to prove it is to consider the forces on the slider when in position x. By Hooke's law, the mainspring supplies a force k0(x-x0), where k0 is the characteristic spring constant of the mainspring, and x0 is the rest position if there were no rebound spring. The rebound spring supplies a force -k1(x-x1), where k1 is the spring constant of the rebound spring, x1 is the rest position if there were no mainspring, and the minus sign is because (over the whole travel where both springs are acting) the springs push in opposite directions.

So the total spring force is k0(x-x0) - k1(x-x1).

This can be rearranged to (k0-k1)(x - (k0x0 - k1x1)/(k0-k1))

Which is the same force as would be applied by a weaker spring (strength k0-k1) with a different offset (note that the term on the right is a constant, independent of x).

Peter

But what if k0 = k1?

I think what we have is case 4 of this:
http://www.madsci.org/posts/archives/2001-02/981476550.Ph.r.html

Urk. Yes, you're quite right, I lost a minus sign somewhere.


Yes, OK, yes we do. Both springs are acting so that the further up the kingpin the slider is, the greater the downwards force on it. (In the case of the rebound spring, the lesser the upwards force, but that's the same thing.)

So you really do have a suspension that gets less stiff once excursion reaches a certain amount. Presumably that excursion isn't very large? i.e. the rebound spring isn't much longer unstressed than it is installed with the car at rest?

Peter

Originally, yes, but it seems over the years small changes have accumulated making the preload pressure higher, putting more stress on the rebound spring. Most of the attempts to solve the issue have focused on the shocks or the slider bushes, instead of just shortening the spring...

Yes, it's 2 springs in parallel, so F=k1x1 + k2x2. Once the bottom spring (s2) disconnects, it's just one spring with F=k1x1. Note that F=0 on the bottom spring when it disconnects -- that is your x=0 point for that spring.

The restorative force as a function of distance will look like a ramp that suddenly becomes steeper when the second spring is acting.

[quote=TigerJimmyThe restorative force as a function of distance will look like a ramp that suddenly becomes steeper when the second spring is acting. [/quote]

That's the desirable part - stiffer rebound. Unfortunately it also makes the initial bounce response stiffer (and anti-progressive).

You could have some custom upper coils wound for you with something like this in mind.

Yeah, dampers are supposed to handle some of this. It depends -- do you really want stiffer rebound or do you want to smooth out really rapid transients?

Springs give a force proportional to displacement (x). Dampers give a force proportional to velocity (dx/dt).

And of course on a Morgan suspension you have a fair bit of the third alternative, static friction (bushings against the kingpin). So not entirely trivial to analyze the whole combination.

Well, as my fantastic Controls professor used to say, "friction always acts to stabilize the system, so there's no need to model it." Particularly given the magnitude of the forces involved in a car suspension, you can leave those terms out.

My bigger question is -- exactly what are you trying to accomplish?

Well, with the Morgan that stabilizing is actually significant and of interest - some people report managing surprisingly well without any shocks/dampers at all just because there is so much friction

Basically just understand the system, and how different tweaks affect it. A good example is that a lot of people seem to shorten the springs in order to get a softer suspension - and of course accomplish exactly the opposite, except when the shortening drives the suspension in a mode where the rebound spring is fully extended and floating free.