Seismic Refraction

Seismic Refraction Method

Applications:

depth of weathering zone (used for statics correction to seismic reflection data)
depth bedrock
depth of groundwater table
depth of basement
depth of Moho
depth of any faster unit

For our purposes, assume flat (not necessarily horizontal), homogeneous layers. In order to get a head wave, V₂>V₁!

The critical angle is the incident angle where the head wave begins:

Snell's law for multiple horizontal layers

Refracted angle into one layer becomes incident angle into next layer:

Travel-time curve for single horizontal layer on a half-space:

so velocities gotten from reciprocal of the slopes of the direct and refracted segments, and depth gotten from reflected time intercept (or cross-over distance). However, often only first arrivals are recorded:

Single horizontal layer on a half-space, V₂>V₁:

Alternatively, in terms of T_i2, the intercept time from the second travel-time segment,

Two horizontal layers on a half-space, V₃>V₂>V₁

where the depth to the lower interface is the sum of z₁ and z₂, where z₁ is computed by the single-layer formula above.

Single dipping layer on a half-space, V₂>V₁:

Example of ambiguity problem: Shooting up-dip gives apparent velocity that is too fast; vice-versa. VA, up-dip velocity, is too fast (shallow slope), VB, down-dip velocity is too slow (steeper dip). Note that, without reversing the profile, could not distinguish from horizontal case. Note also the total travel-time from end to end is same in either direction: reciprocity theorem.

Low Velocity Layers

never get a refracted head-wave from a slow layer underlying a fast layer
eventually get head-wave when faster layer (V>V₁) encountered: e.g., V₂<V₁<V₃

The travel time curve will look like this (another example of ambiguity):

interpreter assumes a layer-over--half-space model (can't "see" layer 2)
t₀ (delay in getting down to layer 3 and back) is large because V₂ is slow
causes layer to be interpreted thicker than it really is