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![]() | Two point raytracing for reflection off a 3D plane | ![]() |
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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
and the horizontal distance of
thereby fixing the source-receiver axis and the relative location of
KaneProof
Figure 2. Diagram used to obtaining a quadratic relation for calculating ![]() ![]() ![]() ![]() |
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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
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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
.
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![]() | Two point raytracing for reflection off a 3D plane | ![]() |
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