Originally Posted by 13ollocks

Clearly I'm bored - the 2012 London Olympic track was 250m with 85m "curves" - this equates to a 27.06m radius. Assuming a bike cleaving to this radius, and assuming a 10cm difference between the RHS drivetrain and a putative LHS drivetrain (I'm assuming 10 cm across the BB), the LHS drivetrain would describe a 26.96m radius (84.69m curve distance) in the time the RHS chainring travelled 85m - 99.6% of the RHS drivetrain speed through the air. Given that drag is a cube function of speed, this would mean that drag on the LHS drivetrain would be 98.9% that of the RHS drivetrain. A potentially significant marginal gain?