*Question*: What distance will be recorded on a bicycle computer on a straight, flat 1 km path if you cycled in a zig-zag fashion, swerving away from the middle line for 10%, lets say a=1 m every L=10 m (i.e. 1 m to the left after 10 m, 0 m after 20 m, 1 m to the right after 30 m, etc)?

Lets denote a pecent of off-center deviation as f=a/L. Then, instead of every straight L meters you are doing L1=SQRT(L^2+a^2)=SQRT(L^2+f^2L^2)=L*SQRT(1+f^2) meters. Your path is longer for k=L1/L=SQRT(1+f^2) (in percents). So, for f=10% off-center deviation your path is longer for k=0.499 %. In 1 straight km you are doing 1.00499 km or 4.99 meters more.

The point of this is to show that inaccuracy of the calculator due to the oscilating path of the wheel is negligible (0.5 % for off-center deviations as high as 10%). Manual instructions for calculator set-up are typically innacurate up to +3%.