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  1. #1
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    used this to work out the length of my new bike chain before fitting, seemed okay a bit of quick maths and voila!
    the info was on websites i navigated to from here, cant remember which ones!

    is doing it this way wrong or okay - doesn't seem to be mentioned in my maintenance book or on other websites to do with maintaining bikes. this is the first time i have fitted a chain and just want to check its okay to do it this way!

    i like the idea of it being so prescise! you can trust maths, cant you?!

    what do others think?

    Working out the exact lenght of chain required

    L= Length in inches
    CSL= Chain Stay Lenght-
    from centre of crank bolt to centre of rear axle in inches
    (CSL)=17 1/4 use thousands of inch=(17.25)
    F= number of teeth on largest front chain ring=52 teeth
    R= number of teeth on largest rear cog=28 teeth
    /= divide

    L = 2x(CSL)+F/4+R/4 + [(F–R squared)x(0.01275)x(1/CSL)]+ 1

    L= (2x17.25)+52/4+28/4+[(52-28 squared)x0.01275x1/17.25]+1

    L= 34.5+13+7+[576x0.01275x0.057971]+ 1

    L= 34.5+13+7+0.425739+1

    L= 55.925739 (round up from 1/2 inch, down from less than 1/2 inch)

    L= 56 inches

    There is this too (but i didn't understand this one)
    LK = 0,157a + 1/2 Z1 + 1/2 Z2 + 2

    LK = chain length in links (number of pins)
    a = chainstay length in mm (quick release skewer to bottom bracket center)
    Z1 = number of teeth on the largest chainring
    Z2 = number of teeth on the largest sprocket

    Example:
    Chainstay length a=420mm, Z1=44 teeth, Z2=28 teeth
    LK=0,157 x 420 + 44/2 + 28/2 + 2
    LK = 108,94 which means 108 links

    The result is always rounded up or down to the next even number.
    This is due to the fact that one end of the chain has to have the opposite type
    of link to the that of the first link in order to allow the chain to be rejoined.

  2. #2
    la vache fantôme phantomcow2's Avatar
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    Why bother going through the math? There is a simple way to get chain length, start by putting your gearing in "small small" and look to see if the derailleur cage is backing into the actual derailleur becuase of the spring, I usually start by removing two links. your derailleurs cage should be more or less parallel to the floor, well so the cage does not back up into the derailleur.

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  4. #3
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    Park tools website has a couple of different methods which all give the same result.

  5. #4
    Senior Member sydney's Avatar
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    Quote Originally Posted by phantomcow2
    Why bother going through the math? There is a simple way to get chain length, start by putting your gearing in "small small" and look to see if the derailleur cage is backing into the actual derailleur becuase of the spring, I usually start by removing two links. your derailleurs cage should be more or less parallel to the floor, well so the cage does not back up into the derailleur.
    That only works if you are using the right derailer for the application. Don't do this if you run a short or medium cage with a triple.

  6. #5
    Senior Member sydney's Avatar
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    Quote Originally Posted by hooplethemott
    used this to work out the length of my new bike chain before fitting, seemed okay a bit of quick maths and voila!
    the info was on websites i navigated to from here, cant remember which ones!

    is doing it this way wrong or okay - doesn't seem to be mentioned in my maintenance book or on other websites to do with maintaining bikes. this is the first time i have fitted a chain and just want to check its okay to do it this way!

    i like the idea of it being so prescise! you can trust maths, cant you?!

    what do others think?

    Working out the exact lenght of chain required

    L= Length in inches
    CSL= Chain Stay Lenght-
    from centre of crank bolt to centre of rear axle in inches
    (CSL)=17 1/4 use thousands of inch=(17.25)
    F= number of teeth on largest front chain ring=52 teeth
    R= number of teeth on largest rear cog=28 teeth
    /= divide

    L = 2x(CSL)+F/4+R/4 + [(F–R squared)x(0.01275)x(1/CSL)]+ 1

    L= (2x17.25)+52/4+28/4+[(52-28 squared)x0.01275x1/17.25]+1

    L= 34.5+13+7+[576x0.01275x0.057971]+ 1

    L= 34.5+13+7+0.425739+1

    L= 55.925739 (round up from 1/2 inch, down from less than 1/2 inch)

    L= 56 inches

    There is this too (but i didn't understand this one)
    LK = 0,157a + 1/2 Z1 + 1/2 Z2 + 2

    LK = chain length in links (number of pins)
    a = chainstay length in mm (quick release skewer to bottom bracket center)
    Z1 = number of teeth on the largest chainring
    Z2 = number of teeth on the largest sprocket

    Example:
    Chainstay length a=420mm, Z1=44 teeth, Z2=28 teeth
    LK=0,157 x 420 + 44/2 + 28/2 + 2
    LK = 108,94 which means 108 links

    The result is always rounded up or down to the next even number.
    This is due to the fact that one end of the chain has to have the opposite type
    of link to the that of the first link in order to allow the chain to be rejoined.
    geeeeeezzzzzzzzzzzzzzzee, why bother if you are using a specific cassette, unless it just gives you a nice warm fuzie and enjoy wasting time. Just use the bi/big plus 1" with out going thru the derailer and it works every time. Otherwise,if you are using the correct derailer for the application just use the small/ small, no drag method and you will be good for any combination that derailer is speced for.

  7. #6
    Junior Member
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    seems as though certain methods cant be used under certain circumstances and not others! so can the mathematical option be used under all circumstances? if it can i'll stick with the maths!

    oh and i did feel warm and fuzzy knowing i had done it right
    as for wasting time.... well the maths took a grand total of 30 seconds - a decent pay off between time spent and result achieved

  8. #7
    Senior Member sydney's Avatar
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    Quote Originally Posted by hooplethemott
    seems as though certain methods cant be used under certain circumstances and not others! so can the mathematical option be used under all circumstances? if it can i'll stick with the maths!

    oh and i did feel warm and fuzzy knowing i had done it right
    as for wasting time.... well the maths took a grand total of 30 seconds - a decent pay off between time spent and result achieved
    Whatever.....Who cares what you do as long is it gives you get the right answer and a warm fuzzie.

  9. #8
    Senior Member Retro Grouch's Avatar
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    St Peters, Missouri
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    16,521
    Quote Originally Posted by hooplethemott
    seems as though certain methods cant be used under certain circumstances and not others! so can the mathematical option be used under all circumstances? if it can i'll stick with the maths!

    oh and i did feel warm and fuzzy knowing i had done it right
    as for wasting time.... well the maths took a grand total of 30 seconds - a decent pay off between time spent and result achieved
    The mathematical method has exactly the same drawbacks as the trial fit methods. I hate to tell you this, but the answer that you get is going to be the same as trial fitting using the big/big method. It has the same drawback too in that if your derailleur doesn't have adequate slack take up, your chain is going to go slack in the little/little combination. Now personally, I don't think that's a serious problem, I use the big/big trial fit method for sizeing chains because I think that it's safer than the little/little method. It just goes to show, however, that it's good to understand the underlying concepts of whatever you're doing.

  10. #9
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    Feb 2005
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    Yes that was what i was attempting to do 'achieve an understanding of the underlying concepts' - attempting anything, least of all putting on a new chain, without that approach can usually lead to poor results!

    As for 'who cares' - well i posted this message after needing information and finding it! sharing information and making a job easier for the next person seems like a good thing to do! so perhaps caring is not so bad, its good to have the discussion and pass on information, isn't it?

    i could have used the information, completed the task and left it at that. but i figured that i got something from this forum and it seemed right that i give something back (some information).

    bye everyone!

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