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  • Road gearing
  • billytinkle
    Free Member

    If I were to have 50/34 chainrings and a 105 short cage rear mech are there any restrictions on which cassette I can go for?

    Also am I right in thinking that a 10 speed shimano cassette will fit in place of an 8 speed shimano cassette?

    mr_average
    Free Member

    You need to check the ‘capacity’ and ‘maximum sprocket’ from the Shimano website and then compare with your proposed setup. Capacity is the front teeth difference (yours is 16) plus rear difference (e.g. A 12-26 would be 14). If you are at or under the spec all will be well, if it’s over it may not be possible to set the chain length so all gears can be used. You also shouldn’t exceed the max sprocket spec (it’s probably 28T), as the mech may hit the sprocket.

    Neither of these values are absolute and in the case of capacity you can always set the chain length based on the big to big combination and accept that some small to small gears shouldn’t be used.

    My understanding is that a 10 speed cassette will fit an 8 speed hub, but my experience only goes as far as 9 speed.

    thisisnotaspoon
    Free Member

    10s will fit a 9s hub, 10s road cassetts come with a 1mm spacer as 10s is actualy slightly narrower.

    You’ll be limited by the capactity (the ammount of slack chain the short cage can deal with) of a short cage mech with a compact chainset (especialy one with a 34t), IIRC mines on the limit with a 36-50 chainset and 13-27 cassette (i.e. 14+14 range = 28), so the only cassette you could use on a 50-34 with a short cage mech would be a 11-23, which kinda defeats the point of using such a low geared compact chainset.

    billytinkle
    Free Member

    Thanks gents. The Shimano spec for the short cage says total capacity of 33T, max front difference of 16T, max sprocket 28T and min sprocket 11T.

    So my current 50/34 difference is 16T and should be fine. And if I went for a 11-28 cassette the difference would be 17T. 16 + 17 = 33T total capacity and all should be well?

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