[kidwoo]

eh? not following you.

I hate the c-clip holding my springs in place on my 888.

I stripped one off the damper rod last month.


look everybody! The canadian said "eh"

 

[Transcend]

I hate the c-clip holding my springs in place on my 888.

I stripped one off the damper rod last month.


look everybody! The canadian said "eh"

I am now throroughly confused, but I'll just nod and smile.

 

[kidwoo]

I am now throroughly confused, but I'll just nod and smile.

You've never seen the way the springs are preloaded on an old 888?

There's an aluminum top cap that rests on the top of the spring and that's held in place by a little c-clip (seriously....pubic hair diameter). All this is concentric to the damper rod which has little grooves that seat that c-clip. Think lower spring collar on an old vanilla dh rear shock.....same thing just smaller. Those grooves flare out over time and actually get stripped on a hard hit......essentially making seating the clip impossible.

Dumbest design ever.

 

[Transcend]

You've never seen the way the springs are preloaded on an old 888?

There's an aluminum top cap that rests on the top of the spring and that's held in place by a little c-clip (seriously....pubic hair diameter). All this is concentric to the damper rod which has little grooves that seat that c-clip. Think lower spring collar on an old vanilla dh rear shock.....same thing just smaller. Those grooves flare out over time and actually get stripped on a hard hit......essentially making seating the clip impossible.

Dumbest design ever.

wow.5

 

[gemini2k]

The curve of the graph sliding left makes a lot of sense. No offense though, but I think your math may still be a little off. 40% of 180 is 72 lbs and sag for situation one is therefore 2.4 inches, excatly 30%, or are you factoring in the fact that your fork doesn't recieve the force from exactly 90 degrees. If so, with a 64 degree head angle even though the front wheel gets 72 lbs of force, the spring is actually only recieving 51.2 lbs. of pressure, that results in a sag of 1.7 inches. To bottom out that, according to my calculations, you'd need a force of 337.5 lbs pushing perfectly perpendicular to the ground, assuming the head angle remains 64 degrees throughout compression. And that's all without any preload factored into the spring at all.

You've described it pretty well with the sliding graph and I think I get it, but don't bikes use progressive springs though? if a 30 lbs spring takes 30lbs to move one inch that doens't mean it takes 60lbs to move it two inches or 90 to move 3 does it? That's only with a linear spring rate, right? Would a progressive spring affect the way the curve moves to the left with preload or just the actual shape of the curve? I'm sorry, but none of this really adds up for me.

Are there any engineers around? I'm basing this all on some high school physics and calculus classes I took 5 years ago. Where is Weagle?

i dont believe theres any "definition" of progressive and linear springs. They are simply ordinal words, not cardinal (oooh fancy terms kids). It just refers to the exponent of the x in ure trusty hookes law equation f=kx. thus ure 30lbs=1inch, 60lbs=2inch is correct for an ideal spring. but who knows what kind of springs are in these forks. i don't

 

[vitox]

yea yea i didnt want to get into any trig because i suck at it, but main thing is i think you got the notion about what preload is.
normally bike springs are thought of as linear, and yes, if they were progressive, preload would use the lighter part of the spring rate.

oh and i took a look at the figures and hehe yep all wrong...

 

[davep]

As Vitox said, the common bicycle springs are 'linear'.

A spring with the same geometry the entire length (coil pitch and wire diameter etc) and made of homogenous material will be essentially linear. Proogressive springs have varying coil pitch or varying wire thickness.

Pre-load on a spring will do exactly the 'sliding graph' idea. A progressive spring would have a curve on that same graph, not a straight line. The curve would be concave up. Air springs behave like this.

You can also see why air springs seem to 'wallow'. If the endpoints of a progressive spring curve are at the same end points of a linear spring (both start at zero, and bottom at the same force), the 'mid stroke (mid graph) of the progressive spring will be less than the progressive spring.

 

[dhkid]

Correct. Nitrogen is inert, dry and is not affected by temperature changes (big deal in suspension). This is why it is used. Atmospheric pressure changes will still affect it.

umm, no. it still expands and contracts with change in temperature.(it is still a gas after all) it just does it in a linear fasion. air on the other hand will have water vapour and all sorts of other stuff in it making it change pressure in relation with temperature slightly eratically.

Transcend,
the weight of the fox 40 is with a ti spring, even for 07. with a steel spring it comes to 7.1-7.2lbs.

the stock spring in the 07 40 is still ti, its just one weight heavier.

the reason you feel your 40 is much smoother now is because the main danmper runs on bushings now. the older models just used orings. thus reducing the stiction even more.

and the bottom out cone is not user servicable. its still needs to be sent to fox. although it now has more adjustments, amount of danmping, low, medium, hard. and where it will engage in the travel, sooner later.

as for air shoxs, the main problem was heat. they heat up too much over the course of a dh run, changing the spring rate, ect.

hope this clears things up a bit.
adam.