transmission of road vibration through bike tires
Consider half the rider + bike to be a point mass, suspended from the road by a spring, the bike tire. This is a classic spring-mass system . Spring-mass systems naturally resonate at an angular velocity ω₀ = sqrt[κ/M], where κ is the elastic constant of the spring (the ratio of force to displacement), and M is the total mass of the load (the bike + rider in this case). The frequency response z as a function of angular velocity ω is: z(ω) = 1 / [1 - (ω / ω₀)²] To go from angular velocity (radians per second) to frequency (oscillations per second, or Hz) divide by 2π. So well below the resonance, the frequency response is one: when riding over gradual rollers, the tire deflection barely changes. On the other hand, well above resonance, the transmission decreases proportional to the square of the frequency. In actuality, no spring is perfect: there is some energy loss with each oscillation. When this effect is included, the system becomes a "damped" spring-mass system. Whe...