Talk Morgan Forums Morgan Sports Cars Classic Morgans (pre-CX) Function of rebound spring

The bit I don’t get is that he states the two springs (main spring and rebound spring) are in series so the initial total stiffness is the addition of both spring rates. As I see it, since only one spring is compressed when the axel block is moved by a bump this is not the case? Surely the main spring compression is initially aided by the (partly compressed) rebound spring so it effectively reduces the stiffness of the main spring and “softening” the bump effect?

I don't need to cut down my rebound springs, (happy with SSL) but on a technical point surely this would leave a nasty end on the wire which would not sit flat on the adjacent mating faces top & bottom

When you look at the existing springs the last turn is brought in a little tighter so that when its cut off and ground flat you're left with a continuous circular face on each end to spread the load

Not easy to describe, but if you look at one you'll see what I mean

The answer to this is best understood by thinking about preload adjusters on the rear springs of motorcycles.

How do they work? By compressing the spring using a stepped or threaded adjuster so you can apply some force to the spring when the bike is at normal static ride height. What is the effect? It makes the spring harder and less compliant to road bumps and undulations.

Now think about a 140lb/inch main spring on a Morgan. How much force does it need to compress it by 1 inch? 140lb. How much force does it need to compress it by 2 inches? 280lb.

Now think about the effect the rebound spring has when it is pushing up against the bottom of the stub axle when the car is at static ride height. What is the effect? The rebound spring, being much stronger than the main spring, is going to compress the main spring so the stub axle is at the point where the forces balance out. The rebound spring won't be free as it is still pushing against the stub axle, and the main spring will be squashed too, being shorter than it would otherwise be. Remember the far ends of both springs are constrained by the same bit of metal, the Crosshead.

So we now have a situation where the main spring is preloaded by the force of the rebound spring, exactly like the motorcycle rear suspension is compressed by the adjuster. How much is it going to be compressed by depends on the relative poundage of the two springs. Let's pretend just to make the illustration easier, that the main spring is compressed by 1 inch. We'll ignore the weight of the car and the occupants as it's the same in any scenario.

Now let's go for a drive. We hit a bump. That bump is going to impart a force to the suspension. How much force do we need to compress the suspension by 1 inch so the car moves smoothly over the bump rather than jolting up into the air? 280 lb. We're using the second inch of the spring's movement as the first inch is already been taken up by the rebound spring.

So our main spring is behaving not as if it was a 140lb/inch spring, but rather as if it was a 280lb/inch spring. That's why having a rebound spring under compression makes your front suspension harder than it should be and causes the 'nodding dog' effect when you go down the road.

You're absolutely right, cutting springs is not to be advised.

Really it's only a theoretical solution and not a good practical solution.

Either get the correct length and rate spring or use SSL..

Yep. And then the problem becomes getting hold of the correct length spring even once you've found what the length should be.

One other option may be to consider re-engineering the bottom plate of the crosshead with spacers to allow the standard rebound spring to be used.

The SSL solution is definitely the right solution in my view.

Difficult to get your head around but although Peter Ballard mentions springs in series, he then explains that Morgan suspension has the compression and rebound springs effectively in parallel when both are in play due to them being above and below stub axle - same deflection, different loads, rates added.

Quote:
So although there is in ‘our’ case one spring above and one spring below the hub, these springs are actually in parallel, not series.

Consequently its this pre-load effect that causes an initially higher (rebound + compression) spring rate which then reverts to the lesser compression spring only rate when rebound spring completely unloads.

Peter Ballard says:

"So although there is in ‘our’ case one spring above and one spring below the hub, these springs are actually in parallel, not series"

The springs are similar to 2 springs in parallel. For having 2 springs in serie, they must be on the same side of the axle block.

When the axel block is move up by a force F:
the main spring (k1) is compressed,movement +dx mm
the rebound spring (k2) is released, movement -dx mm

When you write the equation of equilibrium of forces (sum of forces=0):
F-k1.dx+ k2.(-dx)=0
F=(k1+k2).dx

When both springs are in contact, the stiffness is equal to k1+k2 and you get the maximun stiffness of the suspension. It is counter intuitive.

Oh.....sorry you are faster than me, but anyway I wrote it and now I post it in addition of what you say regarding parallel forces.

I would like to add a point that is often misinterpreted.

First of all:

It is easy to say that if two springs are connected in succession, it is a series connection. This applies as long as you measure at the end points of both springs. The compliance adds up, and the total spring force decreases, regardless of whether the springs are pulled or compressed.

And it is just as easy to see that two springs that are arranged next to each other and on which a total pressure or tension is exerted represent a parallel connection. Again, it doesn't matter whether you pull or push the springs, the compliance is lower than with just one spring and the total force of the two springs is higher.

And now we come to our Morgan.
There is a special arrangement of springs which look like a series connection, but in reality they are a parallel connection.
This is the case when the measuring point (or in other words the point of attack for the common forces), i.e. our wheel hub, is located exactly in the middle between the two springs.

If we now have a slight compressive preload, then all spring forces from the main spring and the rebound spring add up in a linear way.
This means that not only a possible preload of the main spring is effective, as described here in the post with the motorcycle example, but that In Addition! the force of the strong short rebound spring must be added. Remember, it is a parallel circuit.

This condition makes driving over light bumps very hard and uncomfortable as long as this compressive preload adds both spring forces.

If you now drive over a more violent bump...or even worse...if you drive fast through a curve, then this preload is interrupted e.g. at the curve outer front wheel...only the main spring works! That means, it is a sudden omission of the high spring force of the rebound spring in the whole system. The car really buckles. This is not comfortable and can even be dangerous.

For this reason, it is important that the rebound spring is not allowed to "press" on the entire system in the case of static charge.

Total springforce in the example below is 2xc1 + c2 regarding m

Exactly what I was thinking!

Something else less well understood complicates this issue.
Gyroscopic Precession generates a force on spinning objects.
A sudden force on the axis of spin in our case will firstly cause the wheels to suddenly react by their tending to move in towards each other at the leading edge (toe-in). As the axle assembly moves the other way and against the lower rebound spring the opposite forces generate the opposite force so that the wheels then try to toe-out.
Any slight free play in the tie rod bushing will result in sudden ocillation between toe-in and toe-out.
For a motorcyclist the effect is noticed when pushing the handlebar (left or right): the bike lays over to the same side and NOT the other side as you might expect .