I've seen formulas to calculate chain length for multi-speed bikes, but I need something for single speed.

Example: 42T Chainring 17T Cog 406mm (16") Chainstay Length

The answer in this case should come to ~94L.

  • 1
    I have made a strikingly question before, although I wanted to find chainstay length while you want to find number of chain links. Should this count as a duplicate? bicycles.stackexchange.com/q/8608/2355 May 7, 2013 at 3:28
  • And if you want a still more technical explanation: math.stackexchange.com/q/123361/27435 May 7, 2013 at 3:30
  • And, just in case you want to find the "magic gear" for a vertical dropout bike, I think it's risky, since minor variations on the parts themselves can make the chain become too tight or too slacky... Not to mention the ideal length will quickly vanish as the chain wears out. (well, just a thought) May 7, 2013 at 3:34
  • I did see the other similar question but as you mentioned it would have to be re-worked to solve for chain length. I'm not competent enough to understand how to do that unfortunately!
    – Jason Wood
    May 7, 2013 at 4:13
  • The formula I would use, to get the practical length (i.e. the one that is known to work) - wrap a chain around the cogs and match up one end to the link that it meets when tight, count the links.... Or were you wanting the theoretical value.
    – mattnz
    May 8, 2013 at 4:32

1 Answer 1



Scroll to the bottom, equation is there, you want the "Rigorous Equation".

Better solution: start with heltonbiker's question on the math stack and ask how to convert to chain length.

Happy Riding.

  • Actuallly it's just move things to the right side of the equation, leaving "L" (chain length) isolated on one side. May 7, 2013 at 12:16
  • (well, actually you have to get the radius from number of teeth, but that is "natural" too...) May 7, 2013 at 12:19
  • The park tool rigorous equasion is perfect. Just have to remember to take off the +1 for a single speed bike.
    – Jason Wood
    May 7, 2013 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.