Chapter 71 Seismic Travel Times and Subsurface Models

Year 4 · Geophysics · G442-L2

Calculate simple seismic travel times and depths while examining velocity assumptions, resolution and interpretation uncertainty.

Wavelength, amplitude and travel-time relationships in a simple wave model. Original Geology Hub teaching diagram.

This independent lesson follows the topic framework of the supplied University of Peshawar BS Geology curriculum, session 2004–05 onward. It is newly written study material, not an official university course or qualification.

Learning objectives

Convert a simple two-way travel time to depth.
Read a time-distance graph under stated assumptions.
Explain resolution nonuniqueness and common corrections.

A reflection section begins in time

Reflection data record returning energy as a function of time and survey position. A horizontal reflector beneath a constant-velocity layer gives a simple depth relation: depth equals velocity multiplied by two-way time divided by two. Real paths are usually more complicated because velocity changes, interfaces dip and source-receiver offsets vary. Corrections and imaging methods account for parts of this geometry. A time section and a depth section therefore answer different questions and should be labeled accordingly.

Reflections arise from contrasts in acoustic properties, commonly represented through acoustic impedance involving density and velocity. One geological boundary may be weak if the contrast is small. Closely spaced interfaces can interfere and appear as one event; seismic resolution is finite. Fault interpretation benefits from consistent offsets, terminations and regional geometry rather than one broken-looking reflection. Processing artifacts or poor signal can also interrupt continuity.

Refraction and the assumptions behind straight lines

Refraction interpretation often starts with first-arrival times plotted against source-receiver distance. In a simple horizontal-layer model with increasing velocity downward, the slope of a straight travel-time segment is related to reciprocal velocity. A later branch can represent critically refracted energy returning from a faster layer. Intercept-time or more advanced methods estimate layer properties under their model assumptions. The line on the graph is a fit to measurements, not the actual subsurface surface.

Dipping layers, lateral changes, faults and gradual velocity gradients complicate the pattern. A low-velocity layer beneath a faster one can be poorly resolved by standard first-arrival methods. Reflection and refraction offer complementary constraints, and both need geological context. Elevation, near-surface weathering and geometry can require corrections. Report the adopted velocity model and investigate how changing it alters depth rather than hiding this uncertainty behind a precise-looking image.

A reflection depends on contrasts and path geometry

For a simple normal-incidence acoustic comparison, impedance is density multiplied by P-wave velocity. The reflection coefficient can be written as (Z2 − Z1)/(Z2 + Z1). In an invented example with impedances 4 and 6 in consistent units, the coefficient is 2/10 = 0.2. Reversing the order reverses the sign. This idealization explains why a geological contact with little impedance contrast may be weak and why the same lithological label does not guarantee a fixed reflection strength.

The depth relation d = vt/2 assumes the relevant down-and-back path can be represented by the chosen velocity and geometry. In layered material, travel time is accumulated through different velocities, so one arbitrary rock velocity may be inappropriate. A depth error can be systematic even when a reflection is picked consistently to a small time increment. Present picking uncertainty separately from uncertainty in the velocity model and structural geometry.

Reading first arrivals in a simple two-layer case

For a direct wave in a uniform upper layer, the time-distance slope is 1/v1. If velocity increases downward and the interface geometry permits critical refraction, a more distant first-arrival branch can have slope 1/v2. The two line segments describe different paths, not a sudden change in instrument speed. Their crossover and intercept contain information under the layered model assumptions. A dipping interface or lateral velocity change can produce a more complex pattern.

The critical-angle relation for this simple faster-underlying-layer case is sin ic = v1/v2, with angles measured from the interface normal. If v1 is 2,000 and v2 is 4,000 metres per second, ic is 30 degrees. This is an idealized relation, not a complete depth calculation. A hidden slower layer, gradual velocity gradient or three-dimensional structure may violate the assumptions used to fit simple straight segments.

Why corrections cannot be treated as decoration

Receiver elevation and the near-surface low-velocity zone can change arrival times independently of the deeper target. Geometry-related processing aims to account for such effects and combine compatible observations, but every correction uses assumptions. Record the purpose and input model rather than describing the final image as corrected and therefore exact. A misplaced near-surface correction can create apparent structure where no corresponding deep deformation exists.

Compare two depth models that honor the same reflection times but use different justified velocities. Identify the borehole, refraction or other observation that would distinguish them. Then inspect whether interpreted faults or terminations persist in both versions. Features robust across reasonable models deserve more confidence than those dependent on one uncertain conversion. This connects seismic processing to geological testing rather than treating time sections as transparent photographs of the subsurface.

Worked activity

A fictional near-vertical reflection arrives at 0.80 seconds two-way time. With an assumed constant velocity of 2,500 metres per second, depth is 1,000 metres. Using 2,200 or 2,800 metres per second changes the estimate to 880 or 1,120 metres. This 240-metre spread comes from velocity assumptions alone. For a second exercise, a direct-wave time-distance line increases by 0.10 seconds over 200 metres, implying 2,000 metres per second in the simplified model. Show units and state why these results cannot establish a unique lithology.

Practice and interpretation

Annotate an interpreted section with observed reflection events, proposed horizons and uncertain correlations using different line styles. Write one alternative velocity model and calculate its effect on a selected horizon. Identify which borehole, refraction measurement or independent geological constraint would most reduce the ambiguity. This separates picking precision from geological accuracy: repeatedly selecting the same event does not eliminate a systematic error in the assumed subsurface model.

Self-check

Why divide two-way time by two?
What does a travel-time slope represent in a simple uniform model?
Can a clear seismic event give an exact depth without a velocity model?

Check your answers

The reflected wave travels down to the interface and back in the simplified geometry.
The reciprocal of the relevant propagation velocity.
No. Conversion from time to depth requires velocity and geometrical assumptions.

Sources and further study

The explanations and teaching examples are original. Use these institutional resources to check terminology and explore the subject further.

Chapter 71 multiple choice questions

Choose one best answer for each question. Open each answer panel after making your choice.

A near-vertical reflection has 0.80-second two-way time in a constant 2,500-metres-per-second medium. What depth follows?
1. 1,000 metres
2. 2,000 metres
3. 500 metres
4. 3,125 metres
Answer and explanation
A. 1,000 metres Depth is velocity × two-way time / 2. The result is 2,500 × 0.80 / 2 = 1,000 metres.
For the same 0.80-second reflection, which depths follow from velocities of 2,200 and 2,800 metres per second?
1. 1,760 and 2,240 metres
2. 880 and 1,120 metres
3. 440 and 560 metres
4. 2,750 and 3,500 metres
Answer and explanation
B. 880 and 1,120 metres Applying depth = velocity × 0.80 / 2 gives 880 and 1,120 metres. The spread illustrates sensitivity to assumed velocity.
A simple direct-wave branch increases by 0.10 second over 200 metres. What velocity does this imply?
1. 20 metres per second
2. 200 metres per second
3. 2,000 metres per second
4. 20,000 metres per second
Answer and explanation
C. 2,000 metres per second Velocity is distance divided by time: 200/0.10 = 2,000 metres per second. The slope itself is reciprocal velocity in this simple model.
Why can two nearby interfaces appear as a single seismic event?
1. Every interface must have exactly the same geological age
2. The receiver spacing directly determines mineral composition
3. Reflection data cannot contain any information about layering
4. Their responses may interfere because seismic resolution is finite
Answer and explanation
D. Their responses may interfere because seismic resolution is finite Closely spaced interfaces may not be resolved separately. Interference and finite resolution limit the detail visible in a seismic image.
A low-velocity layer lies beneath a faster layer. What caution applies to standard first-arrival refraction methods?
1. The slower layer may be poorly resolved
2. The layer must generate the earliest arrival at every distance
3. Its exact thickness follows without a velocity model
4. Refraction must show every geological layer individually
Answer and explanation
A. The slower layer may be poorly resolved The lesson highlights limitations where velocity does not increase simply with depth. First-arrival interpretation may miss or poorly resolve slower underlying layers.

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