Journal of the Optical Society of America A
Vol. 15,
Issue 11,
pp. 2903-2912
(1998)
•https://doi.org/10.1364/JOSAA.15.002903

Phenomenological description of bidirectional surface reflection

Jan J. Koenderink and Andrea J. van Doorn

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Jan J. Koenderink and Andrea J. van Doorn, "Phenomenological description of bidirectional surface reflection," J. Opt. Soc. Am. A 15, 2903-2912 (1998)

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Abstract

General surface scattering is characterized through the bidirectional reflection distribution function (BRDF). The BRDF is a function of the directions of the incident and remitted beams and thus depends on four parameters. Under very general assumptions one shows that the BRDF is invariant under interchange of the incident and remitted beams, the so-called Helmholtz reciprocity. For isotropic surfaces the BRDF depends only on the absolute value of the difference between the azimuths of the incident and remitted beams. Since these exhaust the symmetries, the BRDF is a very complicated function. For many applications it would be advantageous to be able to summarize empirical data or to smooth and/or interpolate (often even extrapolate) BRDF data. We present a principled way to do this, exactly respecting the symmetry properties.

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Equations (46)

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$I_{00}^{0} (θ_{i}, θ_{r}, Δ ϕ_{ir})$ a = $\frac{1}{2 π}$
$I_{11}^{1} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{2 \sqrt{2}}{π} sin \frac{θ_{i}}{2} sin \frac{θ_{r}}{2} cos Δ ϕ_{ir}$
$I_{20}^{0} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{1}{π} \sqrt{\frac{3}{2}} (2 {sin}^{2} \frac{θ_{i}}{2} + 2 {sin}^{2} \frac{θ_{r}}{2} - 1)$
$I_{22}^{0} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{3}{2 π} (4 {sin}^{2} \frac{θ_{i}}{2} - 1) (4 {sin}^{2} \frac{θ_{r}}{2} - 1)$
$I_{22}^{2} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{6 \sqrt{2}}{π} {sin}^{2} \frac{θ_{i}}{2} {sin}^{2} \frac{θ_{r}}{2} cos 2 Δ ϕ_{ir}$
$I_{31}^{1} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{4 \sqrt{2}}{π} sin \frac{θ_{i}}{2} sin \frac{θ_{r}}{2} (3 {sin}^{2} \frac{θ_{i}}{2} + 3 {sin}^{2} \frac{θ_{r}}{2} - 2) cos Δ ϕ_{ir}$
$I_{33}^{1} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{16 \sqrt{2}}{π} sin \frac{θ_{i}}{2} sin \frac{θ_{r}}{2} (3 {sin}^{2} \frac{θ_{i}}{2} - 1) (3 {sin}^{2} \frac{θ_{r}}{2} - 1) cos Δ ϕ_{ir}$
$I_{33}^{3} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{16 \sqrt{2}}{π} {sin}^{3} \frac{θ_{i}}{2} {sin}^{3} \frac{θ_{r}}{2} cos 3 Δ ϕ_{ir}$
$I_{40}^{0} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{1}{π} \sqrt{\frac{5}{2}} (12 {sin}^{4} \frac{θ_{i}}{2} + 12 {sin}^{4} \frac{θ_{r}}{2} - 6 {sin}^{2} \frac{θ_{i}}{2} - 6 {sin}^{2} \frac{θ_{r}}{2} + 1)$
$I_{42}^{0} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{1}{π} \sqrt{\frac{15}{2}} (2 {sin}^{2} \frac{θ_{i}}{2} + 2 {sin}^{2} \frac{θ_{r}}{2} - 1) (24 {sin}^{2} \frac{θ_{i}}{2} {sin}^{2} \frac{θ_{r}}{2} - 6 {sin}^{2} \frac{θ_{i}}{2} - 6 {sin}^{2} \frac{θ_{r}}{2} + 1)$
$I_{42}^{2} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{4 \sqrt{15}}{π} {sin}^{2} \frac{θ_{i}}{2} {sin}^{2} \frac{θ_{r}}{2} (4 {sin}^{2} \frac{θ_{i}}{2} + 4 {sin}^{2} \frac{θ_{r}}{2} - 3) cos 2 Δ ϕ_{ir}$
$I_{44}^{0} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{5}{2 π} (24 {sin}^{4} \frac{θ_{i}}{2} - 12 {sin}^{2} \frac{θ_{i}}{2} + 1) (24 {sin}^{4} \frac{θ_{r}}{2} - 12 {sin}^{2} \frac{θ_{r}}{2} + 1)$
$I_{44}^{2} (θ_{i}, θ_{r,} Δ ϕ_{ir}) = \frac{10 \sqrt{2}}{π} {sin}^{2} \frac{θ_{i}}{2} {sin}^{2} \frac{θ_{r}}{2} (8 {sin}^{2} \frac{θ_{i}}{2} - 3) (8 {sin}^{2} \frac{θ_{r}}{2} - 3) cos 2 Δ ϕ_{ir}$
$I_{44}^{4} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{40 \sqrt{2}}{π} {sin}^{4} \frac{θ_{i}}{2} {sin}^{4} \frac{θ_{r}}{2} cos 4 Δ ϕ_{ir}$

SpectralComponent
n	1	222	333	444444
m	1	022	133	022444
l	1	002	113	002024

$I_{00}^{0} (θ_{i}, θ_{r}, Δ ϕ_{ir})$ a = $\frac{1}{2 π}$
$I_{11}^{1} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{2 \sqrt{2}}{π} sin \frac{θ_{i}}{2} sin \frac{θ_{r}}{2} cos Δ ϕ_{ir}$
$I_{20}^{0} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{1}{π} \sqrt{\frac{3}{2}} (2 {sin}^{2} \frac{θ_{i}}{2} + 2 {sin}^{2} \frac{θ_{r}}{2} - 1)$
$I_{22}^{0} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{3}{2 π} (4 {sin}^{2} \frac{θ_{i}}{2} - 1) (4 {sin}^{2} \frac{θ_{r}}{2} - 1)$
$I_{22}^{2} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{6 \sqrt{2}}{π} {sin}^{2} \frac{θ_{i}}{2} {sin}^{2} \frac{θ_{r}}{2} cos 2 Δ ϕ_{ir}$
$I_{31}^{1} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{4 \sqrt{2}}{π} sin \frac{θ_{i}}{2} sin \frac{θ_{r}}{2} (3 {sin}^{2} \frac{θ_{i}}{2} + 3 {sin}^{2} \frac{θ_{r}}{2} - 2) cos Δ ϕ_{ir}$
$I_{33}^{1} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{16 \sqrt{2}}{π} sin \frac{θ_{i}}{2} sin \frac{θ_{r}}{2} (3 {sin}^{2} \frac{θ_{i}}{2} - 1) (3 {sin}^{2} \frac{θ_{r}}{2} - 1) cos Δ ϕ_{ir}$
$I_{33}^{3} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{16 \sqrt{2}}{π} {sin}^{3} \frac{θ_{i}}{2} {sin}^{3} \frac{θ_{r}}{2} cos 3 Δ ϕ_{ir}$
$I_{40}^{0} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{1}{π} \sqrt{\frac{5}{2}} (12 {sin}^{4} \frac{θ_{i}}{2} + 12 {sin}^{4} \frac{θ_{r}}{2} - 6 {sin}^{2} \frac{θ_{i}}{2} - 6 {sin}^{2} \frac{θ_{r}}{2} + 1)$
$I_{42}^{0} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{1}{π} \sqrt{\frac{15}{2}} (2 {sin}^{2} \frac{θ_{i}}{2} + 2 {sin}^{2} \frac{θ_{r}}{2} - 1) (24 {sin}^{2} \frac{θ_{i}}{2} {sin}^{2} \frac{θ_{r}}{2} - 6 {sin}^{2} \frac{θ_{i}}{2} - 6 {sin}^{2} \frac{θ_{r}}{2} + 1)$
$I_{42}^{2} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{4 \sqrt{15}}{π} {sin}^{2} \frac{θ_{i}}{2} {sin}^{2} \frac{θ_{r}}{2} (4 {sin}^{2} \frac{θ_{i}}{2} + 4 {sin}^{2} \frac{θ_{r}}{2} - 3) cos 2 Δ ϕ_{ir}$
$I_{44}^{0} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{5}{2 π} (24 {sin}^{4} \frac{θ_{i}}{2} - 12 {sin}^{2} \frac{θ_{i}}{2} + 1) (24 {sin}^{4} \frac{θ_{r}}{2} - 12 {sin}^{2} \frac{θ_{r}}{2} + 1)$
$I_{44}^{2} (θ_{i}, θ_{r,} Δ ϕ_{ir}) = \frac{10 \sqrt{2}}{π} {sin}^{2} \frac{θ_{i}}{2} {sin}^{2} \frac{θ_{r}}{2} (8 {sin}^{2} \frac{θ_{i}}{2} - 3) (8 {sin}^{2} \frac{θ_{r}}{2} - 3) cos 2 Δ ϕ_{ir}$
$I_{44}^{4} (θ_{i}, θ_{r}, Δ ϕ_{ir}) = \frac{40 \sqrt{2}}{π} {sin}^{4} \frac{θ_{i}}{2} {sin}^{4} \frac{θ_{r}}{2} cos 4 Δ ϕ_{ir}$

SpectralComponent
n	1	222	333	444444
m	1	022	133	022444
l	1	002	113	002024

Phenomenological description of bidirectional surface reflection

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Abstract

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Figures (6)

Tables (2)

Equations (46)

Journal of the Optical Society of America A