L18_SeismicRefractionI-page9

L18_SeismicRefractionI-page9 - Dipping layer traveltime...

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9 Applied Geophysics – Refraction I Dipping layer traveltime ) sin( 1 φ + = c d i V V Given ) sin( 1 = c u i V V We can solve for: + = = u d c u d V V V V i V V V V 1 1 1 1 2 1 1 1 1 1 2 1 sin sin sin sin Finally, the intercept times can be used to determine the perpendicular distance to the reflector: 2 1 sin V V i c = V 2 then obtained from: 1 cos 2 V i h T c d id = 1 cos 2 V i h T c u iu = Applied Geophysics – Refraction I Dipping layer Example Direct arrivals Velocities from slopes: 1780 m/s and 2250 m/s
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