Saturday, January 2, 2010

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. When a bike goes over a bump, the bike doesn't continue to oscillate at ω₀. Indeed, it quickly comes to rest again.

damped spring-mass system, from Wikipedia

When damping is included, three things happen. First, the tendency to oscillate at ω₀ is reduced, or for critical damping or greater, essentially eliminated Second, ω₀ increases somewhat. And third, the strongest (proportional to frequency squared) reduction in high-frequency vibrations occurs at frequencies increasingly above ω₀ (see Wikipedia's article).

Okay, so what's this mean? Basically if I calculate ω₀, without knowing the damping, for the bike-tire system I do a fairly good job at determining below which frequency road vibration is transmitted to the bike.

In this analysis I'll take a linear approximation to the tire, which is the "small signal" estimation. This applies to minor road roughness. Obviously a deep pot hole changes thing: the shape of the tire will substantially distort, and the rim may even come into contact with the road. Or if the bike falls off a cliff, this become even more nonlinear... So I'll calculate an effective spring constant for the tire (κ), combine that with the mass load on the tire (half the system mass M) and calculate ω₀.

According to Wilson's Bicycle Science 3rd edition, Whitt derived the following empirical formula based on measurements.

Lc / rw = [ Fv / (0.95 p rw²)]3/8

Lc is the half-length of the contact patch
rw is the outer radius of the tire
p is the tire pressure
Fv is the load force

The problem can also be treated analytically which yields a somewhat different result, but the analytic approach tends to over-simplify the tire (for example treating it as an idealized toroidal balloon).

One can then calculate the maximum vertical deflection of the tire (at the center of the contact patch):

yws =
   ½(Lc² / rw) =
   ½(Fv / 0.95 p)¾ (1 / rw)½

I can differentiate this with respect to Fv to get the inverse of the effective spring constant, then multiply by Fv/g = mass load on that tire, and I get τ², where τ is the resonant time constant, and g is the gravitational acceleration.


τ² = (3 / 8g) (Fv / 0.95 p)¾ (1 / rw)½

The resonant frequency is then

ω₀ = 1 / τ

which is radians/second. To get Hz:

f₀ = 1 / 2πτ

Assuming tires pumped to 140 psi, 2.1 meter rolling circumference, 65 kg total mass half of which is on a given tire, I get around f₀ ≈ 12.3 Hz. This is sort of an extreme case: high pressure and low mass load. But it compares nicely with the 10 Hz which I used the last time (indeed the result would have been around 10.8 Hz were I to have used a more typical tire pressure, say 100 psi).

Now it needs to be mentioned this is the resonant frequency of the spring-mass system, so shouldn't necessarily be equated to the frequency used in the Gaussian spectral function I used to model bike noise. The Gaussian was applied to the noise density measured on the bike (for example in a pedal spindle). This is the ratio of the tranmitted vibration to the source of vibration, considering just the tires. There's attenuation of vibrations within the bike itself, as well as a shape to the spectral distribution of road vibration. So it would be a mistake to assume the vibration spectrum were the same function as the transmission function. But certainly one might conclude the vibration function will be well attenuated where the transmission function is falling off rapidly. In other words I don't need to be apologetic for using a Gaussian function there versus spring-mass transmission function.

Next time I'll consider some experimental data, despite my reluctance to let experiments get in the way of a good theory. I suspect they will. One thing to watch out for is this analysis is for vibration directly transmitted from the road. If stuff oscillates on the bike itself, all bets are off. I'll avoid worrying too much about this just yet.


Alison Chaiken said...

Damping also broadens resonances. I suspect that the damping broadens the oscillation through the tires so much that the power spectrum of the road surface dominates the noise. A way to test for the dominance of road spectral noise is to ride the same patch of rough road over and over again at 5, 10 and 15 mph and see what noise power distribution you measure. In principle, if you believe the marketers, results will differ for Al bikes (low damping) and Ti/steel bikes (allegedly higher damping). Maybe with solid rubber tires, the intrinsic harmonic oscillator response will dominate.

djconnel said...

Now that I've actually looked up some references on the subject, I'll post something to the blog...