Question

A stacked seismic section shows a single dipping event with a slope of \(0.5\ \text{s/km}\). Stolt migration with a constant velocity of \(2\ \text{km/s}\) is applied to the data. The dip of the event in the migrated section in degrees is __________ (rounded off to one decimal place).

Hint:

Remember \(p=\partial t/\partial x\) (s/km) and \(\sin\theta=\dfrac{v}{2}p\) for zero-offset events. This lets you read true dip directly from time slopes once a migration velocity is chosen.

Answer 1

Step 1: Use the time-slope/angle relation.
For a planar reflector, the zero-offset time slope is \[ \frac{\partial t}{\partial x}= \frac{2\sin\theta}{v}, \] where \(\theta\) is the reflector dip (equal to the incidence angle at the reflection point) and \(v\) is the migration velocity. Step 2: Solve for \(\theta\).
Given \(\partial t/\partial x=0.5\ \text{s/km}\) and \(v=2\ \text{km/s}\), \[ \sin\theta=\frac{v}{2}\,\frac{\partial t}{\partial x} =\frac{2}{2}\times 0.5=0.5. \] Hence, \[ \theta=\sin^{-1}(0.5)=30^\circ. \] \[ \boxed{30.0^\circ} \]