# Raypath for velocity linear with depth

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Series Geophysical References Series Digital Imaging and Deconvolution: The ABCs of Seismic Exploration and Processing Enders A. Robinson and Sven Treitel 2 http://dx.doi.org/10.1190/1.9781560801610 9781560801481 SEG Online Store

The differentials of horizontal distance and vertical distance are, respectively,

 {\begin{aligned}&dx{\ =\ sin\ }\theta \ ds{\ =}\ \rho {\rm {\ sin\ }}\theta \ d\theta {\ ,\ }dy{\ =\ cos\ }\theta \ ds{\ =}\ \rho {\rm {\ cos\ }}\theta \ d\theta .\end{aligned}} (42)

Let a particle of energy start at the origin $\left(x{\rm {=0,\ }}y{\rm {=0}}\right)$ , where the path makes at an initial angle ${\theta }_{0}$ with the vertical. The particle travels along a circular arc to point $\left(x{\rm {,\ }}y\right)$ , where the path makes an angle $\theta$ with the vertical. We wish to find the coordinates $\left(x{\rm {,\ }}y\right)$ . The horizontal distance is given by

 {\begin{aligned}x&{\ =\ }\rho \int _{{\theta }_{\rm {o}}}^{\theta }{\rm {\ sin\ }}\theta \ d\theta {\ =}-\rho {\rm {\ cos\ }}\theta {\rm {+}}\rho {\rm {\ cos\ }}{\theta }_{0}\end{aligned}} (43)

and the vertical distance is given by

 {\begin{aligned}y&{\rm {=}}\rho \int _{{\theta }_{0}}^{\theta }{\rm {\ cos\ }}\theta \ d\theta {\ =\ }\rho {\rm {\ sin\ }}\theta -\rho {\rm {\ sin\ }}{\theta }_{0}.\end{aligned}} (44)

Equations 43 and 44 are the parametric equations of a circle (Figure 11). The center of this circle is point C, with coordinates Figure 11.  The arcs of the circles through the origin represent raypaths with different initial angles. The circles with centers on the depth axis represent wavefronts with different traveltimes.

 {\begin{aligned}C&{\ =\ }\left(x_{C}{\rm {,\ }}y_{C}\right){\rm {=}}\left(\rho {\rm {\ cos\ }}{\theta }_{0}{\rm {,\ }}-\rho {\rm {\ sin\ }}{\theta }_{0}\right){\rm {=}}\left({\frac {v_{0}}{a{\rm {\ sin\ }}{\theta }_{0}}}{\rm {\ cos\ }}{\theta }_{0}{\rm {\ ,\ }}-{\rm {\ }}{\frac {v_{0}}{a{\rm {\ sin\ }}{\theta }_{0}}}{\rm {\ sin\ }}{\theta }_{0}\right)\\&{\rm {=}}\left({\frac {v_{0}}{a{\rm {\ tan\ }}{\theta }_{0}}}{\rm {,\ }}-{\frac {v_{0}}{a}}\right).\end{aligned}} (45)

In terms of Snell’s parameter p, the center is

 {\begin{aligned}C&{\rm {=}}\left(x_{C},y_{C}\right){\rm {=}}\left({\frac {\rm {1}}{ap}}{\rm {\ cos\ }}{\theta }_{0}{\rm {,\ }}-{\frac {v_{0}}{a}}\right){\rm {=}}\left({\frac {\rm {1}}{ap}}{\left({\rm {1}}-p^{\rm {2}}v_{0}^{\rm {2}}\right)}^{\rm {1/2}}{\rm {,\ }}-{\frac {v_{0}}{a}}\right).\end{aligned}} (46)

As observed by Slotnick (1959), the raypath is given by the equation of a circle of radius 1/ap, whose center is at the point with coordinates

 {\begin{aligned}x_{c}{\rm {=}}{\left({\rm {1}}-p^{\rm {2}}v_{0}^{\rm {2}}\right)}^{\rm {1/2}}/{ap},\ y_{C}{\rm {=}}-v_{0}\ {\textit {la}}.\end{aligned}} (47)

Note that the y-coordinate of the center is independent of the parameter p.

We therefore can write the parametric equations of the circular raypath as

{\begin{aligned}x&{\rm {=}}-\left({\frac {v_{0}}{a{\rm {\ sin\ }}{\theta }_{\rm {o}}}}\right){\rm {\ cos\ }}\theta {\rm {+}}{\frac {v_{0}}{a{\rm {\ }}{\rm {tan\ }}{\theta }_{0}}}\end{aligned}} {\begin{aligned}&y{\rm {=}}\left({\frac {v_{0}}{a{\rm {\ sin\ }}{\theta }_{0}}}\right){\rm {\ sin\ }}\theta -{\frac {v_{\rm {o}}}{a}}.\end{aligned}} (48)

Note that the vertical component $-v_{0}/a$ does not depend on the angle ${\theta }_{0}$ . As a result, the family of rays with the given velocity function is made up of the origin and that have their centers on the line $y{\rm {=}}-v_{0}/a.$ . The circular raypath starts at the origin (which is the shot point). The tangent to the circle at the shot point makes an angle ${\theta }_{0}$ with the positive y-axis.

Often, we will know only that the ray passes through a certain point, say, $B{\rm {=}}\left(x{\rm {,\ }}y\right)$ , and we must determine the equation of the ray. This problem can be solved as described here and shown in Figure 12. Draw line OB. The midpoint of this line is $A{\rm {=}}\ {\rm {(}}x/{\rm {2}},{\textit {y/2}}{\rm {)}}$ . The center of the raypath circle must lie on the perpendicular bisector of the line OB. The center of the raypath circle also must lie on the horizontal line through the point $\left({\rm {0,\ }}-v_{0}/a\right)$ . The intersection of these two lines of determination gives the center E $E{\rm {=}}\left(x_{E}{\rm {,\ }}-v_{0}/a\right)$ of the raypath circle. From the geometry, we see that Figure 12.  Determination of the equation of a ray that passes through a point B. Use the fact that the origin O is also a point on the ray. Connect these two known points with line OB. Use the fact that the ray is a circle. Hence the center lies on the perpendicular bisector AE of line OB. Use the fact that this center also lies on line JE on which the velocity would be zero. Thus the center is at the intersection E of these two lines. The radius is given by line EB. The raypath is the circle with this center and radius.

 {\begin{aligned}x_{E}{\rm {=}}JE{\rm {=}}JG{\rm {+}}GE{\rm {=}}OF{\rm {+}}GE{\rm {=}}OF{\rm {+}}AG{\rm {\ tan\ }}\alpha \\{\rm {=}}{\frac {x}{\rm {2}}}{\rm {+}}\left({\frac {y}{\rm {2}}}{\rm {+}}{\frac {v_{0}}{a}}\right){\frac {y}{x}}{\rm {=}}{\frac {x^{\rm {2}}{\rm {+}}y^{\rm {2}}}{{\rm {2}}x}}{\rm {+}}{\frac {v_{0}}{a}}{\frac {y}{x}}.\end{aligned}} (49)

Thus, we have determined the center E. The radius is

 {\begin{aligned}&\rho {\rm {=}}OE{\rm {=}}{\sqrt {OJ^{\rm {2}}{\rm {+}}JE^{\rm {2}}}}{\rm {=}}{\sqrt {{\left({\frac {v_{0}}{a}}\right)}^{\rm {2}}{\rm {+}}{\left({\frac {x^{\rm {2}}{\rm {+}}y^{\rm {2}}}{{\rm {2}}x}}{\rm {+}}{\frac {v_{0}}{a}}{\frac {y}{x}}\right)}^{\rm {2}}}}.\end{aligned}} (50)

The initial angle of the raypath is

 {\begin{aligned}{\theta }_{0}&{\rm {=\ arcsin\ }}{\frac {v_{0}}{a\rho }}\end{aligned}} (51)