M900 suspension question

[stopintime]

for the non-engineery types, springs go with hooke's law, like bears go with candybars:

http://en.wikipedia.org/wiki/Hooke's_law

Maybe I'm reading things into that wiki content, but it does says that the force with which the spring "wishes" to return, is greater the further away from it's unstretched/uncompressed state it has been forced, doesn't it? Meaning some progressive behaviour even for linear springs? Sounds obvious to me....

[battlecry]

My spring wishes it were a pony.   

Like Ato says, ideal springs obey Hooke's law.  Their wishes are irrelevant.  If Kyle's springs misbehaved, they were compressed beyond their operating range, which is a bad practice. 

What the Wiki means is that the force is proportional to a constant times the length of compression or extension, so the more it is compressed or extended, the more the force.  The constant is the spring rate and it should not change unless the material reaches a yield state and its properties change. 

[junior varsity]

+1

[stopintime]

My spring wishes it were a pony.   

Like Ato says, ideal springs obey Hooke's law.  Their wishes are irrelevant.  If Kyle's springs misbehaved, they were compressed beyond their operating range, which is a bad practice. 

What the Wiki means is that the force is proportional to a constant times the length of compression or extension, so the more it is compressed or extended, the more the force.  The constant is the spring rate and it should not change unless the material reaches a yield state and its properties change. 

There are some things here that I don't expect to fully master, but I'd like to establish a simple fact in my brain....
Please help me out - even a linear spring will resist compression with greater force when it's fully compressed than when it's just a little compressed, right? Meaning it's easier to compress a linear spring the first inch than the last inch?

[junior varsity]

Well, its a focus on the extremes in which it no longer behaves spring-like

Think of a spring stretched out as far as it will go. What does it begin to look like? A rod/rope/etc. No longer behaving spring like.

Think of a spring compressed as far as it will go. What does it look like? A hollow tube. No longer behaving spring like.

[stopintime]

That much I understand, but I'm asking what it's behaviour is between the extremes. First and last inch different?

[battlecry]

"Meaning it's easier to compress a linear spring the first inch than the last inch?"

Shouldn't be, Stopintime.  But separate the concept of the force required to compress the spring an inch from the concept of the total force the spring is holding up.    The first inch you compress a spring, the force to compress it that one inch, call it X, is the same as the total force the spring is holding up.  If you keep compressing the spring, the force required to compress it from inch 3 to inch 4 will still be X, but it will be X on top of 3X and now the spring will be carrying 4X force.

[stopintime]

Now it's beginning to sink in.

My reasoning makes sense to me, after reading the latest post.
Allow me - if I try to compress a spring by hand force, it will be easier the first inch than the second a.s.o.
Live bike action on the other hand, is working as described with the 3x and 4x story.

Am I getting closer?

[junior varsity]

I think you are there - what you mean by "easy" should be defined:

Its just as "easy" to go from 3-to-4, in that you are adding the same amount of force. No, you can't think "go in further, spring!" and not push any harder, you have to push the same you have been pushing for the existing compression, plus the unit amount to go further.

If it takes 50 lbs to push spring from a to b, then to go from a to c, you would need the force for a to b (50) plus b to c (50), net result :100.

This is different than non-linear springs.

In those "progressive rate" springs, you won't simply add another 50 lbs to compress another 1 inch (or whatever the rates might be). You would progressively have to add more and more force to compress the same distance. This would be like,

a to b = 50 lbs.

a to c = ab (50) + bc (55) = 105

a to d = ab (50) + bc (55) + cd (60) = 165

So to move from a to d, it would take 15 more lbs than a linear rate spring.

[koko64]

This is intriguing. (Never too old to learn)

[stopintime]

Ato - what I mean by "easy" is that if a person is only strong enough, by hand, to compress a spring by an inch, the same person won't be able to compress it the next inch.

If that's true, I will see it as a sign of a spring being progressive no matter what kind of spring it is. It might not be a correct technical term though.

[junior varsity]

No. That would be incorrect terminology.

If a person was only capable of compressing a spring 1 inch, regardless of spring type, they would be unable to compress the spring any further.

I think you are missing that the spring "stops" compressing until more force is applied. Whether its x more lbs to go z more inches, or if its x^y to go z more inches is the difference in naming.

[stopintime]

Thanks, that's confirming my understanding. I'm far from understanding the physics, but that's ok.

Thread jack over, I guess

[junior varsity]

i think it is a helpful threadjack section for those interested in progressive rate springs. More readings (more of a how-to, really): http://www.webbikeworld.com/motorcycle-shocks-suspension/progressive-fork-springs/

[Speeddog]

How about this:

'Linear' or 'straight rate' spring

For example, it's a 50 lb/inch spring.
It takes 50 lb to compress it the first inch.
It takes 50 lb *more* to compress it the second inch.
So, a total of 100 lb to compress it 2 inches.
150 lb to compress it 3 inches.
And so on...

'Progressive' spring

50 lb to compress it the first inch.
60 lbs *more* to compress it the second inch.
So, a total of 110 lb to compress it 2 inches.
70 lbs *more* to compress it the third inch.
So, 180 lb to compress it 3 inches.