Normal moveout

Remarks

The formula follows from the Pythagorean theorem: the reflected path consists of two segments of length $\sqrt{z^2 + (x/2)^2}$. So $t(x)^2 - x^2/v^2 = t_0^2$, and the reflection is a branch of a hyperbola in the $(x, t)$ plane. Example: $z = 1000$ m and $v = 2000$ m/s give $t_0 = 1$ s; at $x = 2000$ m, $t = \sqrt{1 + 1} \approx 1.414$ s, so the moveout is about $0.414$ s. Non-example: for a reflector dipping at angle $\theta$ in the plane of the profile, the time curve is $t(x)^2 = t_0^2 + x^2 \cos^2\theta / v^2$. So the formula with the true speed $v$ fails; the curve is still a hyperbola, but with moveout velocity $v / \cos\theta$. For horizontally layered media the formula holds approximately at small offsets, with $v$ replaced by an NMO velocity, close to the root-mean-square speed of the layers.