§12
Physics Foundations
§12 · Scaling and dimensionless numbers

Key examples — scaling & dimensionless numbers

Where the chapter’s machinery shows up across the bookshelf.

Example 1: cochlear perilymph is Stokes flow

For perilymph (water, ν=10−6 m2/s\nu = 10^{-6}\,\text{m}^2/\text{s}) at typical cochlear flow speeds (10−4 m/s10^{-4}\,\text{m/s}) over the basilar-membrane geometry (10−3 m10^{-3}\,\text{m}), the Reynolds number is Re∼10−1\mathrm{Re} \sim 10^{-1} — deep in the Stokes-flow regime. This is what lets the cochlear traveling-wave analysis use the linear Stokes equation rather than the full Navier-Stokes. See Hearing Ch 4.3.

Example 2: shock formation in nonlinear acoustics

A finite-amplitude sound wave steepens because its local sound speed depends on the local amplitude: clocal≈c0+βv′c_\text{local} \approx c_0 + \beta v'. After a distance Lshock∼c0/(βωv′)=1/(βωMaa)L_\text{shock} \sim c_0 / (\beta \omega v') = 1/(\beta \omega \mathrm{Ma}_a) the wave forms a shock. For a 1-Pa, 1-kHz audio signal (Maa∼10−5\mathrm{Ma}_a \sim 10^{-5}), Lshock∼10 kmL_\text{shock} \sim 10\,\text{km} — far beyond any acoustic propagation in air. For a sonic-boom-strength wave (Maa∼0.1\mathrm{Ma}_a \sim 0.1), Lshock∼1 mL_\text{shock} \sim 1\,\text{m} — essentially immediate. See Sound Ch 10.4.

Example 3: small vs large radiators (Helmholtz number)

A speaker of radius aa radiating at frequency f=ω/(2π)f = \omega/(2\pi) has Helmholtz number ka=ωa/cka = \omega a/c. When ka≪1ka \ll 1 (small compared to wavelength), it radiates as a point monopole with very inefficient coupling to free air. When ka≳1ka \gtrsim 1, it radiates as a piston, with peak directivity straight ahead and efficient coupling. For a 6-inch (15 cm) woofer at 100 Hz, ka≈0.14ka \approx 0.14 — small. At 5 kHz, ka≈7ka \approx 7 — large and beaming. This is why subwoofers need to be physically big and tweeters can be tiny. See Sound Ch 6.

Example 4: drag-crisis Reynolds number on a sphere

A smooth sphere has drag coefficient CD≈0.5C_D \approx 0.5 for 103<Re<3×10510^3 < \mathrm{Re} < 3\times 10^5, then drops abruptly to CD≈0.1C_D \approx 0.1 above Re=3×105\mathrm{Re} = 3\times 10^5 — the drag crisis. The transition is set by the laminar→turbulent transition of the boundary layer: a turbulent BL stays attached further around the back of the sphere, narrowing the wake. Golf balls are dimpled to trip the BL into turbulence at lower Re, getting more distance from the same swing.

Example 5: alveolar surface tension and the Bond number

A typical alveolus has L∼100 μmL \sim 100\,\mu\text{m} and water surface tension σ∼72 mN/m\sigma \sim 72\,\text{mN/m}. The Bond number is Bo=ρgL2/σ∼10−3\mathrm{Bo} = \rho g L^2/\sigma \sim 10^{-3} — capillary effects dominate over gravity. This is why surfactant matters: gravity is irrelevant, and surface tension alone sets the pressure needed to inflate the alveolus, which is the muscle workload of breathing.