Two point raytracing for reflection off a 3D plane

Offset-vector map demigration

Another application of the coordinate neutral approach for 3D reflection point calculation that arose at SEP recently is offset-vector map demigration. For this, the aim is to model where a point, $P$ , on a planar subsurface reflector will appear in a constant-offset, constant-azimuth survey.

For this calculation, there is one fixed coordinate, the depth axis, with the sources and receivers on the surface, described by an arbitrary point $Q_0$ with (downward) normal ${\bf z}$ . We are further given the source-to-receiver offset vector $2$ $h$ ${\bf x}$ and the reflector inward normal ${\bf n}$ from the point $P$ on the reflector.

We know the ellipsoid of specular reflection has its major axis through the source and receiver, and that the inward normal bisects the reflection angle between the source and receiver. Therefore the normal line through the reflection point intersects the source-receiver axis somewhere between the source and receiver. Let $Q$ be the point on the surface where the normal ray would reach. Then we may write

$$Q = P +$$   $$\mbox{\unboldmath$\gamma$}$$$${\bf n}$$

for some scalar $\gamma$ . As before we calculate

$$\begin{array}{ccc} {\bf z} \cdot ( Q - Q_0) & = & {\bf z} \c... ...{\bf z} \cdot ( Q_0 - P )}{{\bf z} \cdot {\bf n}} } \end{array}$$

and the horizontal distance of $Q$ from the vertical plane through P as

$${\bf x} \cdot [ (P - \{{\bf z} \cdot (P - Q_0)\} \, {\bf z} ) - (P +$$   $$\mbox{\unboldmath$\gamma$}$$$${\bf n}) ] =$$   $$\mbox{ \unboldmath$\gamma$\ }$$$${\bf x} \cdot {\bf n}$$

thereby fixing the source-receiver axis and the relative location of $Q$ . What still remains is to ascertain the source-receiver midpoint relative to $P$ . This we can determine by means of tedious algebra, the way I did it, or by a succinct bit of trigonometry provide by Daniel Kane (pers. comm.) of the Stanford Department of Mathematics.

KaneProof Figure 2. Diagram used to obtaining a quadratic relation for calculating $x$ from $z$ , $h$ , and $\theta _0$ .

Due to symmetry, we may rotate the reflection point around the source-receiver axis until it is directly below that axis. This does not change the unknown distance to the source-receiver midpoint, but does reduce the computation to one on a planar ellipse. Let $x_0$ and $z_0$ denote the respective horizontal and vertical distances from the source-receiver midpoint to the reflection point. The dip angle $\theta _0$ is implicitly determined by $\sin \theta_0 = -{\bf n \cdot {\bf x}}$ and $\cos \theta_0 = \sqrt{1 - \sin^2 \theta_0}$ . Using this dip angle, $z_0$ may be written as $\gamma \cos \theta_0$ . Referring to Fig. 2, Fermat's principle of extremal traveltime tells us that reflecting a focus of the ellipse around the tangent produces an image point on the straight line connecting the reflection point and the other focus. Hence we know that $AB'C$ forms a triangle. Denoting the three angles $\alpha$ , $\beta$ , and $\theta _0$ as illustrated in the figure, we have

$$\alpha + \beta + 2 \theta_0 = \pi$$

whence

$$\tan 2\theta_0 = - \frac{\tan \alpha + \tan \beta }% { 1 - \tan \alpha \tan \beta }$$    .

But

$$\begin{array}{rcl} \displaystyle \tan \alpha & = & \displays... ...playstyle \frac{z_0}{ x_0 - \mbox{\unboldmath$h$} } \end{array}$$

hence

$$\tan 2 \theta_0 = - \frac{ z_0 (x_0 - \mbox{\unboldmath$h$}) + z_... ...$} - z_0^2 } = \frac{ - 2 x_0 z_0 }{ x_0^2 - z_0^2 - \mbox{\unboldmath$h^2$} }$$

and so we have the quadratic relation

$$x_0^2 + 2x_0 z_0 \cot 2 \theta_0 - ( z_0^2 +$$   $$\mbox{\unboldmath$h^2$}$$$$) = 0$$    .

Solving the quadratic equation we get

$$x_0 = - z_0 \cot 2 \theta_0 + \sqrt{z_0^2 \cot^2 2\theta_0 + z_0^2 +h^2}$$

$$= - z_0 \cot 2 \theta_0 + \sqrt{z_0^2 \csc^2 2\theta_0 +h^2}$$

$$= - \frac{z_0^2 ( \csc^2 2\theta_0 - \cot^2 2\theta_0 ) + h^2}% {-z_0 \cot 2\theta_0 - \sqrt{z_0^2 \csc^2 2 \theta_0 + h^2}}$$

$$= \frac{\sin 2 \theta_0 ( z_0^2 + h^2 )}{z_0 \cos 2\theta_0 + \sqrt{z_0^2 + h^2 \sin^2 2 \theta_0}}$$

in a form that does not exhibit a numerical singularity at $\theta_0 = 0$ .

The relation $\alpha + \beta + 2 \theta_0 = \pi$ is actually a special case of the more general proposition attributed to Boškovic (Boscovich) (1754):

From any point $H$ outside an ellipse with foci $F$ and $f$ , with $F$ being no farther from $H$ than $f$ , draw two tangents, touching the ellipse at $P$ and $p$ respectively. Then the interior angle $PHp$ is half the difference of the interior angles $PFp$ and $Pfp$ .

A translation of his original Latin demonstration appears in Appendix D.

So, in summary, only the dot products ${\bf z} \cdot {\bf n}$ and ${\bf x} \cdot {\bf n}$ are needed to find the demigration location of point $P$ .

Two point raytracing for reflection off a 3D plane