By the end of this lecture…

• [LO-6.1] Sketch a refraction survey geometry and identify direct, reflected, and head wave ray paths
• [LO-6.2] Derive travel-time equations for direct and head waves in a two-layer model
• [LO-6.3] Distinguish the critical distance $x_\text{crit}$ from the crossover distance $x_\text{cross}$
• [LO-6.4] Calculate $t_i$, $x_\text{crit}$, and $x_\text{cross}$; invert the T(x) plot for $V_1$, $V_2$, and $H$
• [LO-6.5] Identify assumptions of the two-layer model and when they fail

The discovery: Mohorovičić, 1909

A Croatian geophysicist examines seismograms from an earthquake near Zagreb.

• Nearby stations: P-wave slope $\approx 1/5.6$ km/s — crustal rock
• Distant stations: a faster P-wave appears — slope $\approx 1/8.1$ km/s
• Beyond ~200 km, the fast wave arrives before the direct wave

From slopes and intercept → crust-mantle boundary ≈ 54 km depth

This is the Mohorovičić discontinuity — discovered by seismic refraction.

The refraction survey

Three types of arrivals

The refraction method exploits first arrivals — the earliest energy at each geophone.

Equipment

Source: Sledgehammer on plate (~30 m) · weight drop (~100 m) · explosives (km-scale)

Receivers: 12–48 geophones at 2–5 m spacing, or DAS fiber-optic cable

Recording: Multichannel seismograph — digitizes all channels simultaneously

At crustal scale: explosions or earthquakes as sources; 100s of stations over 100s of km profiles.

Direct wave travel time

The direct wave travels horizontally through layer 1 at velocity $V_1$:

$$T_\text{direct}(x) = \frac{x}{V_1}$$

A straight line through the origin with slope $= 1/V_1$.

The near-offset branch of the T(x) plot gives $V_1$ directly from its slope.

Head wave: three-segment ray path

1. Down through layer 1 at $\theta_c$: horizontal reach $= H\tan\theta_c$
2. Along the interface: distance $= x - 2H\tan\theta_c$, speed $V_2$
3. Up through layer 1 at $\theta_c$: symmetric to segment 1

$$T_\text{head} = \frac{2H}{V_1\cos\theta_c} + \frac{x - 2H\tan\theta_c}{V_2}$$

Apply $\sin\theta_c = V_1/V_2$ and $1 - \sin^2\theta_c = \cos^2\theta_c$:

$$\boxed{T_\text{head}(x) = \frac{x}{V_2} + \underbrace{\frac{2H\cos\theta_c}{V_1}}_{t_i}}$$

$\cos\theta_c$ is in the numerator.

Intercept time and layer depth

$$t_i = \frac{2H\cos\theta_c}{V_1} = \frac{2H\sqrt{V_2^2 - V_1^2}}{V_1 V_2}$$

Solving for $H$:

$$\boxed{H = \frac{t_i\,V_1\,V_2}{2\sqrt{V_2^2 - V_1^2}}}$$

Once slopes give $V_1$ and $V_2$, the intercept time $t_i$ completely determines the layer thickness $H$.

Critical distance and crossover distance

$x_\text{crit}$ vs. $x_\text{cross}$

Always: $x_\text{crit} < x_\text{cross}$

Between these two offsets the head wave exists but arrives after the direct wave.

Survey design: the geophone array must reach at least $x_\text{cross}$ or the inversion fails.

How the T(x) diagram is built

The plasma color = interface lag time. Same scale in both panels.

The T(x) plot: all three branches

The slope-intercept inversion

From first arrivals to Earth model:

1. Pick first arrivals on the shot gather
2. Plot $T$ vs. $x$ and identify the two linear branches
3. Near-offset slope $\Rightarrow$ $V_1$; far-offset slope $\Rightarrow$ $V_2$
4. Read $t_i$ from the T-axis intercept of the head-wave line
5. Compute $H = t_i V_1 V_2 / (2\sqrt{V_2^2 - V_1^2})$

Synthetic shot gather and inversion

Field inversion: water table at 5 m

Near-offset slope $\Rightarrow$ $V_1 = 350$ m/s (dry sand)

Far-offset slope $\Rightarrow$ $V_2 = 1500$ m/s (saturated sand)

Intercept: $t_i = 27.8$ ms, $\theta_c = \arcsin(350/1500) = 13.5°$

$$H = \frac{0.0278 \times 350 \times 1500}{2\sqrt{1500^2 - 350^2}} = 5.0 \text{ m}$$

The velocity jump arises because water's bulk modulus dominates the pore space — the $V_P$ physics from Lecture 4 applied at field scale.

Worked example: UW campus bedrock

$V_1 = 800$ m/s (glacial till) · $V_2 = 3200$ m/s (bedrock) · $H = 12$ m

$$\theta_c = 14.5° \qquad t_i = 29.0 \text{ ms} \qquad x_\text{crit} = 6.2 \text{ m} \qquad x_\text{cross} = 31.0 \text{ m}$$

The crossover at 31 m falls between these two receivers — consistent with the prediction.

Assumptions and failure modes

Hidden layer: a thin low-velocity unit between two faster layers generates no first-arrival head wave and is entirely invisible to refraction. Addressed in Lecture 7.

Societal relevance: PNW applications

Depth to bedrock: Sound Transit used refraction to map bedrock along Seattle light rail alignments — directly informing cut-and-cover vs. bored-tunnel decisions.

Water table: The 350 → 1500 m/s jump is one of the strongest refraction signals in near-surface geophysics; used routinely for contamination plume monitoring.

Crustal thickness: Moho depth beneath the Pacific Northwest — ~10 km under the ocean, ~40 km under the Cascades — is mapped from earthquake refraction arrivals recorded by the PNSN.

AI as a reasoning partner

Prompt to evaluate:

"Derive the head-wave travel time for a two-layer model. Show the three path segments, simplify using $\sin\theta_c = V_1/V_2$, and derive $x_\text{crit}$."

Criteria for a correct response:

• $\cos\theta_c$ appears in the numerator of $t_i$
• $x_\text{crit} = 2H\tan\theta_c$ is derived geometrically, not confused with $x_\text{cross}$
• The step $1 - \sin^2\theta_c = \cos^2\theta_c$ is shown explicitly

Concept Check

1. $V_1 = 2000$ m/s, $V_2 = 5500$ m/s, $H = 50$ m. Calculate $\theta_c$, $t_i$, $x_\text{crit}$, $x_\text{cross}$.

2. In three sentences: why does the head wave — which travels a longer total path — arrive first at distant receivers?

3. Slopes 5.0 ms/m and 1.25 ms/m, $t_i = 30$ ms. Determine $V_1$, $V_2$, $H$.

4. A survey records no head-wave arrivals. List three physically distinct explanations.

Next time

Lecture 7 — Seismic Refraction II

What happens when the interface is dipping? When there are multiple layers? When a layer is hidden?

Forward and reverse shooting · dipping-layer geometry · the plus-minus method · hidden layers and velocity inversions

Lab 2 (Friday): Python II — implementing the travel-time equations and forward ray tracing