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  1. #1
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    Shorten the top out spring on DLR2 100mm lefty?

    Anybody ever do this to get full travel with low air pressures instead of putting in the softest spring?

  2. #2
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    I meant negative spring.

  3. #3
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    If you shorten a spring, the springrate will go up, so I guess that's the opposite of what you want.

  4. #4
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    Shouldn't need to, they are nowhere near as aggressive as the DLR2 ones, and are only about an inch and a half long anyway, and soft on top of it.
    This is a Pugs not some carbon wannabee pretzel wagon!!

    - FrostyStruthers



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  5. #5
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    Quote Originally Posted by Boozzz
    If you shorten a spring, the springrate will go up, so I guess that's the opposite of what you want.
    The spring rate will not go up if you shorten the spring. That is a common misconception.

  6. #6
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    Quote Originally Posted by MendonCycleSmith
    Shouldn't need to, they are nowhere near as aggressive as the DLR2 ones, and are only about an inch and a half long anyway, and soft on top of it.
    The DLR2 was what I was talking about. Seems everyone is *****in (me too!) about less travel with these springs in. Another question, what does it do? I'm guessing it makes the intial travel (small bumps) more plush. What happens if you take it out completely?

  7. #7
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    Quote Originally Posted by shift96
    The spring rate will not go up if you shorten the spring. That is a common misconception.
    Oh yes it will. It's a misconception that it won't. You calculate spring rate by taking some variables (wire diameter D, number of active coils N, mean coil diameter C) and some constants (modulus of spring steel M, and a constant for all springs K), and putting that all in a formula. You get: Spring rate = (M * D^4) / (K * N * C^3). As you can see, you devide by active coils. This means that less coils for the same spring means a higher spring rate. You dig?

  8. #8
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    Quote Originally Posted by Boozzz
    Oh yes it will. It's a misconception that it won't. You calculate spring rate by taking some variables (wire diameter D, number of active coils N, mean coil diameter C) and some constants (modulus of spring steel M, and a constant for all springs K), and putting that all in a formula. You get: Spring rate = (M * D^4) / (K * N * C^3). As you can see, you devide by active coils. This means that less coils for the same spring means a higher spring rate. You dig?
    You are correct. I read too quick this morning and posted. A common misconception is more preload will raise the rate. I hear that all the time.


    K = Spring rate in pounds per inch
    W = Diameter of the spring wire in inches
    G = 12,000,000 for steel springs (a constant)
    N = Number of active coils (number of coils that are free to move + 1/2 coil)
    D = Diameter of the coils measured to the center of the wire, in inches

    example:
    K = unknown spring rate
    W = 0.5" diameter
    G = 12million (constant)
    N = 9.5 active coils + 1/2 coil = 10
    D = 5"

    Stock spring rate = 341lb/in
    With one spring removed = 380lb/in (Variable N = 8.5 active coils + 1/2 coil = 9)

    Removing 1 coil effectively raised the spring rate 40lb/in

  9. #9
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    Quote Originally Posted by shift96
    You are correct. I read too quick this morning and posted. A common misconception is more preload will raise the rate. I hear that all the time.
    No worries. Yes, the preload story is something else and indeed a widespread "myth". Anyway, can't help you with your original question unfortunately.

  10. #10
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    Is it true that whatever you cutoff percentage wise, double it and add it to the spring rate of the spring you cut and the sum of the 2 will be your new ballpark spring rate? I also hear that if you introduce heat into the spring while cutting it all bets are off. Bandsaw or cutoff whell with coolant only

    Thanks for the catch!

  11. #11
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    Quote Originally Posted by shift96
    Is it true that whatever you cutoff percentage wise, double it and add it to the spring rate of the spring you cut and the sum of the 2 will be your new ballpark spring rate? I also hear that if you introduce heat into the spring while cutting it all bets are off. Bandsaw or cutoff whell with coolant only

    Thanks for the catch!
    No. Just use the formula, it's easy enough.

    Easy to illustrate: say you've got a 1125 lbs spring. It has a wire diameter of 1, 10 coils, and a mean coil diameter of 5.

    Springrate = (11,250,000 * 1^4) / (8 * 10 * 5^3) = 1125

    You cut off 20% of the spring. All remains constant, but the fact that it now has 8 coils. You fill in the formula:

    Springrate = (11,250,000 * 1^4) / (8 * 8 * 5^3) = 1406

    As you can see, that isn't 20% of 1125 times 2 plus 1125 (which would be 1575).

    As for your other point, I don't think introducing heat will change anything for a steel spring. The only thing that could be affected here is the spring modulus of the metal. But steel returns to its original shape and properties after heating as far as I'm aware, so I don't see this having any effect.

  12. #12
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    Quote Originally Posted by shift96
    The DLR2 was what I was talking about. Seems everyone is *****in (me too!) about less travel with these springs in. Another question, what does it do? I'm guessing it makes the intial travel (small bumps) more plush. What happens if you take it out completely?
    It's about 4 inches long, if you remove it completely, you'll create a situation where the piston inside the cartridge, COULD, potentially move up too high, and damage internals. Just get a super light one if you don't want the "plushening" effect of the proper one.
    This is a Pugs not some carbon wannabee pretzel wagon!!

    - FrostyStruthers



    www.mendoncyclesmith.com

  13. #13
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    Thanks for the info.

  14. #14
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    Quote Originally Posted by shift96
    Thanks for the info.
    You're welcome

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