The NMO velocity

Figure shows a dipping reflector and the position of the source, geophone and CMP. The constant-offset traveltime from the source at point A to the receiver at point B is represented by the segments AR+RB. The zero-offset traveltime from the CMP to the reflector and back is given by the segment FG. The dipping reflector serves as the axis of symmetry for the figure.

 

dmofig1
Figure 3 A dipping reflector in a constant velocity medium. The dipping angle is $\theta$. The raypath from the source to the receiver is sketched by the segments AR and RB. The zero-offset raypath is equal to the segment FG.

For the geometry in Figure we have

$${\bf AC}={\bf AR}+{\bf RB}= {v t_h}$$

where v is the velocity of the medium and t_h is the shot-receiver traveltime. The segment FG (which has the length of the zero-offset raypath) is equal to the segment AE. From Figure

$${\bf FG}= {({\bf AD}+ {\bf BC}) / 2}={{\bf BC}+{\bf ED}}= {\bf AE}.$$

From the triangle $\bigtriangleup {\bf AEC}$ we have  

$${\bf FG}^2={\bf AE}^2={\bf AC}^2-{\bf EC}^2={(v t_h)^2-(2h \cos \theta)^2}$$ (1)

where $\theta$ is the angle of the dipping reflector and 2h is the distance between source and receiver. Finally, by dividing the segments with the velocity we have the zero-offset traveltime from the CMP to the reflector:  

$$t_0^2={t_h^2-\left( {{2h \cos \theta} \over v} \right)^2}= {t_h^2-\left({2h \over v_{nmo}}\right)^2}.$$ (2)

From equation (2) we easily see that the expression for t_h is a hyperbola, with the top at t₀. The NMO velocity necessary to flatten the hyperbola for a dipping reflector is  

$$v_{nmo}={ v \over {\cos \theta}}.$$ (3)