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Greetings- Thinking of a 160 lyrik coil u- turn for my Knolly Endo. I do alot of climbing here in Northern AZ , so want reduced travel for those long climbs. Do the fork's characteristics change when wound down to 130 to 140? . Not really sold on the two step reliability, so want to stick with the uturn coil
thanks
 

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denjen said:
The fork will get stiffer the more you lower it. All the U turn basically does is compress/preload the spring. Its not a big deal if you are just using it for climbing.
Are you sure? The U-Turn assemblies that I have seen do not work like that. What happens is as you turn the knob at the top of the assembly, the bottom of the spring threads into the lower stop. All that is happening here is the spring is getting shorter. It is not getting more compressed.
 

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ride_nw said:
Are you sure? The U-Turn assemblies that I have seen do not work like that. What happens is as you turn the knob at the top of the assembly, the bottom of the spring threads into the lower stop. All that is happening here is the spring is getting shorter. It is not getting more compressed.
This is correct, the spring does not get compressed, just shorter.
As for the feel at lower travel, it's not a drastic change. I can ride mine at a lower travel and the only thing I notice is the HA is different.
 

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As has been mentioned, the u-turn just shortens the spring which is supposed to make it sag slightly less when at lower travel settings. The spring rate doesn't change that much.

Also keep in mind that there is no preload adjustment on the coil u-turn so there is no way to fine tune your sag.
 

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~Disc~Golf~
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denjen said:
The fork will get stiffer the more you lower it. All the U turn basically does is compress/preload the spring. Its not a big deal if you are just using it for climbing.
Wrong, the spring acts the same

*Edit - I see this has been addressed by posters previous - it's just that when I saw a glaringly bad bit of info on the second post, I had to reply! ;) )
 

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denjen said:
Oops!!! Sorry about the wrong info. Thats how it was explained to me, and I hardly ever use mine so I have not had the chance to notice.
The spring rate does become a little firmer as it is lowered. But it is not by preloading, it's because the there are fewer coils to compress. The different in firmness while lowered is not very much.
 

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derby said:
The spring rate does become a little firmer as it is lowered. But it is not by preloading, it's because the there are fewer coils to compress....
huh? :confused:
Unless it's a variable rate/progressive spring (which it isn't) the rate does not change - 100 feet or one inch...it's the same
 

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highdelll said:
huh? :confused:
Unless it's a variable rate/progressive spring (which it isn't) the rate does not change - 100 feet or one inch...it's the same
I'm no mechanical engineer (but I play one on mtbr), but gotta agree with this. With a standard spring, the rate remains constant whether it be over 5 coils or 30 coils. That's why so many love coil springs...linearity through the travel rather than harsh ramp up like an air fork:thumbsup:
 

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I believe derby is right on this one. I recall learning about this in some class I took a long time ago. I found it discussed in another forum and here is one guys technical explanation.
http://www.lotuselan.net/forums/elan-f15/spring-length-t1935.html

F=kx

where:

F - is the applied force
x - is the spring deflection
k - is the spring constant or spring rate

A change in the spring rate will be reflected by a change in the spring constant such that:

k=F/x

If a linear spring is shortened, the deflection x, for a given force F, will be reduced. Consider a spring with a 10-inch free length and a stiffness k. A force F will compress the spring a distance x.

If one now shortens the spring to 1/2 it's free length (5-inches) and then applies F, the spring will compress x/2 resulting in:

k=F/(x/2)

Since the denominator is smaller by 1/2, the resulting spring constant (k) will be twice as large, or the rate is doubled, if the spring is cut in half.
edit, nvm, he mentions the rate becoming firmer, which it does not. But the shorter spring will not compress the same amount under the same force if it's shorter.
 

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d-bug said:
...
edit, nvm, he mentions the rate becoming firmer, which it does not. But the shorter spring will not compress the same amount under the same force if it's shorter.
Your last sentence is evidence the rate changes when the same coil is shorter or longer.

A coil is a wound torsion bar. The same bar cut shorter will not twist as easily. The rate changes.
 

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derby said:
Your last sentence is evidence the rate changes when the same coil is shorter or longer.

A coil is a wound torsion bar. The same bar cut shorter will not twist as easily. The rate changes.
haha! I was just in the shower thinking about this and that's the exact same conclusion I came to. What threw me off was thinking per coil was going to be stiffer, which it can't be without physical changes to it.
Yes, you are right! I over analyzed it. :blush:
 

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hmmm, well, no
a spring it "a torsion-bar" in a way, yes. But how it 'reacts' is linear.
say it takes 1 pound (lb) to move a spring coil 1" - ah snap! that's the rate!
the rating doesn't change!
these coils are extruded out at 'bunches' of feet and then chopped up - this is all across the board - not just mtb -
How many pounds to compress one inch? - 1 inch coil? 20" coil?.. same
 

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highdelll said:
hmmm, well, no
a spring it "a torsion-bar" in a way, yes. But how it 'reacts' is linear.
say it takes 1 pound (lb) to move a spring coil 1" - ah snap! that's the rate!
the rating doesn't change!
these coils are extruded out at 'bunches' of feet and then chopped up - this is all across the board - not just mtb -
How many pounds to compress one inch? - 1 inch coil? 20" coil?.. same
Think of it like this.
Take a spring with 20 coils, put 200lbs on top and it compresses 20mm. In that scenario each coil compresses 1mm. Cut 5 coils off that same spring, put 200lbs on top and each coil still compresses 1mm, but the total length of the spring will compress by 15mm, not 20mm like the longer spring. So to compress the spring with 15 coils 20mm, you'd need more than 200lbs. That equals increased spring rate. The rate per coil is no different, but for the overall spring it is.
 

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derby said:
The spring rate does become a little firmer as it is lowered. But it is not by preloading, it's because the there are fewer coils to compress. The different in firmness while lowered is not very much.
This is correct. Highdell, you're correct about how springs work, but you've got a misconception about how u-turn works.

When you lower a u-turn fork, what you're doing is reducing the number of coils in between the upper spring perch and the lower spring perch. It has the same effect as cutting the spring, not compressing the spring.

When you cut a spring in half, the remaining half-spring does not have the same rate as the whole spring had. If 20 pounds was enough to compress the original spring by one inch, then it will compress each half-spring by a half-inch - the spring rate for the cut spring is now twice what the original spring rate was.

If you want to use the torsion bar analogy, imagine clamping the middle of the torsion bar instead of the far end. The torsional stiffness of the material is unchanged, but since you're now operating on a shorter length of bar, the resulting torsional stiffness is increased (doubled).

The lower spring perch in a u-turn fork is threaded, so that rotating the spring causes the spring perch to move up and down the spring. It's kind of hard to put into words, but I just swapped springs in my Pike and operating the mechanism in my hands made it really obvious how it works. It's pretty clever. It does not compress the spring but it does shorten the effective portion of the spring, just like cutting would.
 

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Ride-NW's image just loaded. Nice. Allow me to elaborate....

That shows the spring in the "high travel" setting.
The black thing he's pointed out is the power spring perch.

When you rotate the spring, the perch travels upward, so in the "low travel" setting, the effective portion of the spring is the part between the upper perch and that short section where the coils are all bound together. The lower segment of the spring is now BELOW the spring perch.

Compressing the fork brings the upper and lower perches closer together, but that section of spring that's now hanging below the lower spring perch is not part of the equation anymore - it's as if you had cut off that section of the spring. So, in that setting, the fork is both lower and stiffer.
 
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