Introduction

The mechanical advantage of a bicycle is defined as, \[MA =\frac{F_{\mathrm{out}}}{F_{\mathrm{in}}}\]

Here, \(F_{\mathrm{in}}\) is the force applied by cyclist on the pedal and \(F_{\mathrm{out}}\) is the forward (frictional) force applied by ground on the bike.
The force on pedal leads to the front gear experiencing a torque given by: \[\tau=F_{\mathrm{in}}a\] Here, a is the length of the crank arm. Now, this torque leads to a tension on the upper chain(lower chain is loose). So, we get the force applied on the upper chain is \[F_{\mathrm{chain}}=\frac{\tau}{r_f}=\frac{F_{\mathrm{in}}a}{r_f}\] Here, \(r_f\) is the radius of the front gear. If the radius of the rear gear is \(r_b\) then the torque applied at the rear wheel, \(\tau_b\) is given by: \[\tau_b= F_{\mathrm{chain}}\,r_b=\frac{F_{\mathrm{in}}a\,r_b}{r_f}\] The force at the circumference of the rear wheel caused by this torque is given by: \[F_{\mathrm{out}}=\frac{\tau_b}{R}=\frac{F_{\mathrm{in}}a\,r_b}{r_f\,R}\] This gives us our required result, \[\boxed{ MA=\frac{a\,r_b}{r_f\,R} }\] Since R is much larger than the other quantities in the expression, for a bicycle \(MA < 1\). Since, the work done by rider and static friction will be the same (by conservation of energy), we get an alternate expression for MA of a bicycle, \[MA= \frac{F_{\mathrm{out}}}{F_{\mathrm{in}}}=\frac{d_{\mathrm{in}}}{d_{\mathrm{out}}}\] Here \(d_{\mathrm{in}}\) is the distance moved by the pedal and \(d_{\mathrm{out}}\) is the distance moved by the bicycle.
At low gear(\(r_b\) small), MA is small so for a given \(F_{\mathrm{in}}\), the forward push would be low but the distance travelled by the bike would be high. Use this mode on steep roads.
At high gear(\(r_b\) larger), MA will be comparatively larger so for a given \(F_{\mathrm{in}}\), the forward push would be high but the distance travelled by bike would be low. Use this mode on flat, slightly elevated or sloping roads.