i made an error. arctan((r2-r1)/d) is only half the angle; alpha is 2*arctan((r2-r1)/d). the chain is tangent to the cog and chainring; lines that are orthogonal to the chain at each point of tangency are parallel to each other, go through the centers of the cog and the chainring, and make the same angle (alpha) with the vertical that the chain makes with the horizontal ("similar triangles").

i also need to correct the equation for the bottom, angled length of chain. here is the correct set of equations:

alpha = 2*arctan((r2-r1)/d)
arc around chainring = r2*(pi + alpha)
arc around cog = r1*(pi - alpha)
length on top = d
length on bottom = [d^2 + (r1 - r2)^2*(2 + 2*cos(alpha)) - 2*d*(r2-r1)*sin(alpha)]^(1/2)

you can derive these results by drawing the picture in cartesian coordinates with top dead center of the chainring as the origin. then:

upper point of tangency for the cog is [-d,0];
lower point of tangency for cog is [-d - r1*sin(alpha),-r1*(1+cos(alpha))];
lower point of tangency for chainring is [-r2*sin(alpha),-r2*(1+cos(alpha))].

i don't have access to a good, to-scale picture-drawing tool here, or else i would be happy to draw a picture. you are probably more likely to believe it, however, if you draw it yourself!

I'm still not convinced that the "d" term you have in the denominator of the alpha equation doesn't need to be adjusted for the gear ratio.

I'll think about this this evening. I'm almost done with this spreadsheet and want to finish it before I leave work.

i.e., I think implicit in your derivation you assume that you can alter the frame of reference so that the top of the cog and chainring are parrallel and that this line is parrallel to the line denoted by "d". But I'm not sure if that's what you've done.

wait a minute. by similarity, the "top" and "bottom" lengths of chain should be the same, correct? you are right that "d", as i am using it, is not the actual distance between the centers of the bb and the hub; it is slightly less. sometime later i will look at this figure carefully and determine how to calculate the straight lengths of chain from "d", "r1" and "r2". that is all you need.

okay okay okay - forget everything i wrote before. here's the deal:

let "d" be the actual distance between the bb and hub centers, "r1" be the cog diameter and "r2" be the chainring diameter. than:

alpha = arcsin((r2-r1)/d)
straight lengths = [d^2 - (r2 - r1)^2]^(1/2)
cog arc = r1*(pi - 2*alpha)
chainring arc = r2*(pi + 2*alpha)

i finally realized the significance of the symmetry. sorry for all the confusion.

Start with long chain. Subtract links until it's the proper length. Leave the math alone, that stuff'll kill you.

for my bike, that's what i do. but i also like a math problem, in the abstract.

This is an improvement and more than I could come up with.

However, I'm not entirely sure, but I still have the same objections to your "d" term in your new derivation. I think putting that in and not accounting for the gear ratio is tantamount to accepting the 180 degrees assumption.

Anyway, the spreadsheet will be done on monday. It will show the chainlength and gear inches of a gear choice, determine the number of gearings that would work with vertical dropouts of a given chainstay length, and provide all feasible 'flop' combinations with a given horizontal drop range and primary gearing choice. what fun!

if all goes well i can scan my sketch and post it on monday as well. the 180 ° assumption is out the window. "d" is the hypotenuse of a right triangle, whose long side is parallel with the straight part of the chain, and equal in length. that is where the adjustments to the arc lengths come from: chainring arc greater than 180, cog arc less than 180. also, why the straight lengths of chain are less than "d", by an amount that accounts for gear ratio. cheers!