## Abstract

A Monte Carlo computer simulation has been used to determine the ratio of the scalar irradiance *E*_{0} to the downwelling irradiance *E*_{d}. These *E*_{0}/*E*_{d} ratios were calculated at depths corresponding to the 100, 10, and 1% downwelling irradiance levels. A range of volume reflectance 0 ≤ *R* ≤ 0.14 was considered, as were six conditions of incident radiation (collimated beams with incident angles 0 ≤ *θ*′ ≤ 89° plus a diffuse cardioidal distribution). Mathematical expressions were curve fitted to the Monte Carlo outputs to yield relationships between *E*_{0}/*E*_{d} and *R* for the depths and incident conditions considered. It was found that in many cases a single relationship would not accommodate the entire range of volume reflectances and that *R* = 0.055 provided an appropriate demarcation for mathematical curve fitting. Curves, tables, and equations are presented which indicate (a) for all *R* > ~0.02, the *E*_{0}/*E*_{d} ratio at the 1% downwelling irradiance depth is the same for *θ*′ = 0° as for diffuse cardioidal incidence, and (b) for *R* > ~0.08, the *E*_{0}/*E*_{d} ratio at the 10% downwelling irradiance depth for *θ*′ = 0° is nearly the same as the *E*_{0}/*E*_{d} ratio at the 1% downwelling irradiance depth for diffuse cardioidal incidence.

© 1988 Optical Society of America

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### Equations (20)

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(1)
$$\frac{{E}_{0}}{{E}_{d}}(R,0\xb0)=1+3.13R,$$
(2)
$$\frac{{E}_{0}}{{E}_{d}}(R)=\frac{1}{{\overline{\mu}}_{d}}+\frac{R}{{\overline{\mu}}_{u}},$$
(3)
$$\frac{{E}_{0}}{{E}_{d}}(R,{\theta}^{\prime})=\left(\frac{1.068}{{\mu}_{0}}-0.068\right)\frac{{E}_{0}}{{E}_{d}}(R,0\xb0),$$
(4)
$${\rho}_{\text{INT}}=\frac{0.249}{{\mu}_{0}}+0.271$$
(5)
$$\frac{{E}_{0}}{{E}_{d}}(R,{\theta}^{\prime})=\left(\frac{1.068}{{\mu}_{0}}-0.068\right)\hspace{0.17em}[1+3.13R].$$
(6)
$$\frac{{E}_{0}}{{E}_{d}}(R,{\theta}^{\prime})=\frac{1}{{\mu}_{0}}{[1+({C}_{1}+{C}_{2}R+{C}_{3}{R}^{2})R]}^{{C}_{4}},$$
(7)
$$\frac{{E}_{0}}{{E}_{d}}(R,0\xb0)={[1+(28.0-50.5R)R]}^{0.5}$$
(8)
$$\frac{{E}_{0}}{{E}_{d}}(R,89\xb0)=1.512[1+(7.39-149R+1376{R}^{2})R]$$
(9)
$$\frac{{E}_{0}}{{E}_{d}}(R,89\xb0)=1.60+3.43R$$
(10)
$$\frac{{E}_{0}}{{E}_{d}}(R,{\theta}^{\prime})=\frac{{E}_{0}}{{E}_{d}}(R,0\xb0)+\left(\frac{\text{cos}48.6\xb0}{1-\text{cos}48.6\xb0}\right)\hspace{0.17em}\left(\frac{1-{\mu}_{0}}{{\mu}_{0}}\right)\times \left[\frac{{E}_{0}}{{E}_{d}}(R,89\xb0)-\frac{{E}_{0}}{{E}_{d}}(R,0\xb0)\right],$$
(11)
$$\frac{{E}_{0}}{{E}_{d}}(R,0\xb0)={[1+(39.1-176R)R]}^{0.5},$$
(12)
$$\frac{{E}_{0}}{{E}_{d}}(R,89\xb0)=1.512{[1+(-2.10+117R-582{R}^{2})R]}^{0.8},$$
(13)
$$\frac{{E}_{0}}{{E}_{d}}(R,{\theta}^{\prime})=\frac{{E}_{0}}{{E}_{d}}(R,0\xb0)+\left(\frac{\text{cos}48.6\xb0}{1-\text{cos}48.6\xb0}\right)\hspace{0.17em}\left(\frac{1-{\mu}_{0}}{{\mu}_{0}}\right)\times \left[\frac{{E}_{0}}{{E}_{d}}(R,89\xb0)-\frac{{E}_{0}}{{E}_{d}}(R,0\xb0)\right].$$
(14)
$$\frac{{E}_{0}}{{E}_{d}}(R,{\theta}^{\prime})=1.37+4.93R$$
(15)
$$\frac{{E}_{0}}{{E}_{d}}(R)=1.177(1+3.13R),$$
(16)
$$\frac{{E}_{0}}{{E}_{d}}(R)=1.177{[1+(11.8+96.8R)R]}^{0.5}.$$
(17)
$$R(0\xb0)=0.319\frac{Bb}{a}$$
(18)
$$R(0\xb0)=0.267\frac{Bb}{a}+0.013$$
(19)
$$R({\theta}^{\prime})=\frac{R(0\xb0)}{{\mu}_{0}},$$
(20)
$${R}_{D}=1.165R(0\xb0),$$