Two point raytracing for reflection off a 3D plane |

For this calculation, there is one fixed coordinate, the depth axis, with the sources and receivers on the surface, described by an arbitrary point with (downward) normal . We are further given the source-to-receiver offset vector and the reflector inward normal from the point 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 be the point on the surface where the normal ray would reach. Then we may write

for some scalar . As before we calculate

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

thereby fixing the source-receiver axis and the relative location of . What still remains is to ascertain the source-receiver midpoint relative to . 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
Diagram used to obtaining a quadratic relation for calculating
from
,
, and
.
Figure 2. | |
---|---|

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 and denote the respective horizontal and vertical distances from the source-receiver midpoint to the reflection point. The dip angle is implicitly determined by and . Using this dip angle, may be written as . 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 forms a triangle. Denoting the three angles , , and as illustrated in the figure, we have

whence

.

But

hence

and so we have the quadratic relation

.

Solving the quadratic equation we get

in a form that does not exhibit a numerical singularity at .

The relation is actually a special case of the more general proposition attributed to Boškovic (Boscovich) (1754):

*From any point
outside an ellipse with foci
and
, with
being no farther from
than
,
draw two tangents, touching the ellipse at
and
respectively.
Then the interior angle
is half the difference of the
interior angles
and
.*

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

So, in summary, only the dot products
and
are
needed to find the demigration location of point
.

Two point raytracing for reflection off a 3D plane |

2012-10-29