Technical question - please help

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  • I struggled to find the words to put this into a Google search term that would throw up what I’m looking for...

    On a fixed gear bike, how much further along the dropout will the rear axle move if I up the number of teeth on the sprocket by one tooth (chain length, tyre and everything else remaining the same)? I’m wondering how much closer to the seat tube my tyre will get if I get a bigger sprocket but keep the same number of links in my chain.

    Thank you in advance to any nerd geniuses in the house : )

  • It will be the difference in the radius of the sprocket of the old and new. You could calculate this. The difference I guess will be miniscule. How many teeth on the current sprocket?

    Edit: shouldn’t you still be able to put the axle in the same position regardless of the sprocket?!! Unless you’ve got no slack.

  • Ask tester.. I remember him answering it already somewhere

  • Good thinking! I feel very unimaginative now. It’s a 17t - does that make any difference? Thank you

  • Tooth pitch is 12.7mm (0.5"). You can work out a cog's main dimensions from there.

  • On that basis I’d guess 2.02 mm. But I’m totally not qualified to answer this.

  • Thanks all. That makes things a little clearer...from all this, can a concrete distance be deduced that the axle will move along the dropout with every increase in tooth on the sprocket?

  • No, the difference would be less with every additional tooth

  • Ok, right, mind-bending stuff

  • http://www.machinehead-software.co.uk/bi­ke/chain_length/sprocket_radius_table.ht­ml

    Cog Teeth Radius (Inches) Difference (Inches)
    10 0.8090
    11 0.8874 0.0784
    12 0.9659 0.0786
    13 1.0447 0.0787
    14 1.1235 0.0788
    15 1.2025 0.0789
    16 1.2815 0.0790
    17 1.3606 0.0791
    18 1.4397 0.0791
    19 1.5189 0.0792
    20 1.5981 0.0792
    21 1.6774 0.0793
    22 1.7567 0.0793
    23 1.8360 0.0793
    24 1.9154 0.0793
    25 1.9947 0.0794
    26 2.0741 0.0794
    27 2.1535 0.0794
    28 2.2329 0.0794
    29 2.3123 0.0794
    30 2.3918 0.0794
    31 2.4712 0.0794
    32 2.5506 0.0794
    33 2.6301 0.0795
    34 2.7096 0.0795
    35 2.7890 0.0795
    36 2.8685 0.0795
    37 2.9480 0.0795
    38 3.0275 0.0795
    39 3.1070 0.0795
    40 3.1865 0.0795
    41 3.2660 0.0795
    42 3.3455 0.0795
    43 3.4250 0.0795
    44 3.5045 0.0795
    45 3.5840 0.0795
    46 3.6635 0.0795
    47 3.7430 0.0795
    48 3.8225 0.0795
    49 3.9021 0.0795
    50 3.9816 0.0795
    51 4.0611 0.0795
    52 4.1407 0.0795
    53 4.2202 0.0795
    54 4.2997 0.0795
    55 4.3793 0.0795

  • Lol. So I was right-ish! 0.0791 inches = 2.01 mm.

    Where I was wrong though, is that the difference gets larger with every additional tooth. I don't understand that.

  • I think that is because as the number of teeth increase the polygon we're describing more closely approximates to a circle. A circle has a constant ratio of circumference to diameter - pi. Whereas, for the regular decagon of a 10 tooth sprocket, the ratio of the diameter and perimeter is 3.09, for a hexagon it is 3, and the ratio of the diagonal of a square to the perimeter is 2 time square root of 2 or 2.828. The difference between each difference will get smaller and smaller.

  • Had to read that about ten times, but I see what you're saying.

    I just thought, the size of each additional tooth is constant - whereas the circle is constantly getting bigger. In my head, that means the bigger the 'circle' gets, the less one additional tooth would stretch it. But obviously not.

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Technical question - please help

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