1.3 History of Ocean Exploration

Step 1: Magnetic polarity reversals as a geological clock

Earth's magnetic field periodically reverses polarity. As new oceanic crust forms at a mid-ocean ridge, the cooling basalt records the ambient magnetic field direction. The result is a pattern of normal and reversed polarity stripes symmetric about the ridge axis. The timing of these reversals is known from radiometric dating of volcanic rocks on land, giving us a geomagnetic polarity timescale (GPTS).

Step 2: Relating distance to age

If a magnetic anomaly stripe at distance $x$ from the ridge axis corresponds to a reversal that occurred at time $t$ in the past, and crust has been produced symmetrically, then $x$ is the total distance the crust has moved in time $t$. Assuming a constant half-spreading rate $v$:

$$x = v \cdot t$$

Step 3: Solving for the half-spreading rate

Rearranging to solve for the half-spreading rate $v$ (the rate at which one plate moves away from the ridge):

$$v = \frac{x}{t}$$

Step 4: Full spreading rate

Since crust moves away on both sides of the ridge, the full spreading rate is twice the half-spreading rate:

$$v_{\text{full}} = 2v = \frac{2x}{t}$$

Step 5: Using multiple anomalies for regression

In practice, one identifies several anomaly stripes at distances $x_1, x_2, \ldots, x_n$ corresponding to known reversal ages $t_1, t_2, \ldots, t_n$. A linear regression of $x$ versus $t$ yields the half-spreading rate as the slope:

$$v = \frac{\sum_{i=1}^{n}(t_i - \bar{t})(x_i - \bar{x})}{\sum_{i=1}^{n}(t_i - \bar{t})^2}$$

Step 6: Age of the seafloor at any point

Once the spreading rate is known, the age of the seafloor at any distance $x$ from the ridge is:

$$\text{Age}(x) = \frac{x}{v_{\text{spread}}}$$

For example, the South Atlantic has a half-spreading rate of about 1.5 cm/yr. Crust 150 km from the ridge axis would have an age of $\frac{150 \times 10^3\,\text{m}}{0.015\,\text{m/yr}} = 10\,\text{Myr}$. Typical half-rates range from ~1 cm/yr (slow, e.g. Mid-Atlantic Ridge) to ~8 cm/yr (fast, e.g. East Pacific Rise).

Step 1: Force balance on a fluid element

Consider a thin horizontal slab of seawater at depth $z$ with thickness $dz$ and cross-sectional area $A$. In static equilibrium, the net upward pressure force must balance the weight of the slab:

$$p(z + dz)\,A - p(z)\,A = \rho\,g\,A\,dz$$

Step 2: Differential form

Dividing both sides by $A\,dz$ and taking the limit as $dz \to 0$, we obtain the hydrostatic equation:

$$\frac{dp}{dz} = \rho\,g$$

where $z$ is depth (positive downward), $\rho$ is water density, and $g$ is gravitational acceleration.

Step 3: Integration for constant density

For an ocean of approximately constant density $\rho_0$, integrate from the surface ($z = 0$, $p = p_{\text{atm}}$) to depth $h$:

$$\int_{p_{\text{atm}}}^{p} dp' = \int_0^{h} \rho_0\,g\,dz'$$

Step 4: Result

Evaluating both integrals:

$$p = p_{\text{atm}} + \rho_0\,g\,h$$

The gauge pressure (pressure above atmospheric) is simply $\rho_0 g h$. Since $p_{\text{atm}} \approx 1.013 \times 10^5$ Pa corresponds to one atmosphere, each additional 10.06 m of seawater ($\rho_0 = 1025$ kg/m³) adds roughly one atmosphere of pressure.

Step 5: Application to the Challenger Deep

At the Challenger Deep ($h = 10{,}916$ m), using $\rho_0 = 1025$ kg/m³ and $g = 9.81$ m/s²:

$$p = 1025 \times 9.81 \times 10916 \approx 1.098 \times 10^8 \;\text{Pa} \approx 1{,}083\;\text{atm}$$

In reality the pressure is slightly higher (~1,100 atm) because seawater density increases with depth due to compressibility. A more accurate calculation integrates $\rho(z)$ using the equation of state for seawater (TEOS-10).

Step 1: Spherical law of cosines

Two points on a sphere of radius $R$ are specified by their latitudes and longitudes: $(\phi_1, \lambda_1)$ and $(\phi_2, \lambda_2)$. The central angle $\sigma$ between them follows from the spherical law of cosines:

$$\cos\sigma = \sin\phi_1\sin\phi_2 + \cos\phi_1\cos\phi_2\cos(\Delta\lambda)$$

where $\Delta\lambda = \lambda_2 - \lambda_1$. The great circle distance is $d = R\sigma$. However, this formula suffers from numerical rounding errors for small angles (short distances).

Step 2: Introduce the haversine function

The haversine function is defined as:

$$\operatorname{hav}(\theta) = \sin^2\!\left(\frac{\theta}{2}\right) = \frac{1 - \cos\theta}{2}$$

This function is always positive and varies smoothly near zero, avoiding the cancellation errors that plague the cosine formula for small angles.

Step 3: Rewrite the cosine formula using haversines

Apply the identity $\cos\theta = 1 - 2\operatorname{hav}(\theta)$ to both sides of the spherical law of cosines. Substituting for $\cos\sigma$ and $\cos(\Delta\lambda)$:

$$1 - 2\operatorname{hav}(\sigma) = \sin\phi_1\sin\phi_2 + \cos\phi_1\cos\phi_2\bigl(1 - 2\operatorname{hav}(\Delta\lambda)\bigr)$$

Step 4: Simplify using the cosine addition formula

Expand the right side and use $\cos(\phi_1 - \phi_2) = \sin\phi_1\sin\phi_2 + \cos\phi_1\cos\phi_2 = 1 - 2\operatorname{hav}(\Delta\phi)$:

$$\operatorname{hav}(\sigma) = \operatorname{hav}(\Delta\phi) + \cos\phi_1\cos\phi_2\;\operatorname{hav}(\Delta\lambda)$$

Step 5: Substitute back the sine form

Replacing $\operatorname{hav}$ with $\sin^2(\cdot/2)$ and defining the intermediate quantity $a$:

$$a = \sin^2\!\left(\frac{\Delta\phi}{2}\right) + \cos\phi_1\cos\phi_2\sin^2\!\left(\frac{\Delta\lambda}{2}\right)$$

Step 6: Solve for the central angle and distance

Since $a = \sin^2(\sigma/2)$, we have $\sigma = 2\arcsin(\sqrt{a})$. Multiplying by the Earth's radius gives the Haversine formula:

$$d = 2R\arcsin\!\left(\sqrt{\sin^2\!\left(\frac{\Delta\phi}{2}\right) + \cos\phi_1\cos\phi_2\sin^2\!\left(\frac{\Delta\lambda}{2}\right)}\right)$$

This is numerically stable for all distances. For Earth, $R = 6{,}371$ km. The formula gives the shortest path on a sphere (great circle arc), which is the route Polynesian navigators intuitively approximated and which modern ships and aircraft follow to minimize travel distance.