*3.2. Combined Attenuation along Both Rays*

Let the flakes be opaque thin disks of radius *R* aligned horizontally or with small inclination the horizontal plane. Figure 2 shows the cross-section of paint layer by a horizontal plane. As it is seen there, a horizontal disk of radius *R* at depth *z* intersects a ray if its center is closer than *R* from the ray point at this depth, i.e., the center is within the circle of radius *R* and center in the ray point.

**Figure 2.** Cross-section of paint layer by a horizontal plane at depth z . The top half is top view, the bottom half being side view. Flakes are represented as nearly horizontal thin disks of radius R (green). We show the three cases when the flake at depth z intersects (shadows) only the incident ray (**a**), when it intersects both rays (**b**) and when it intersects only the reflected ray (**c**).

Figure 3 shows the domain where the flake center intersects with the incident or reflected ray. If the flake center is in the red domain, the flake intersects the incident ray (red arrow). If the flake center is within in the blue domain, the flake intersects the reflected ray (blue arrow). If the center is in the intersection of the domains, the flake shadows both. The area of the horizontal section of the intersected domains at depth *z* is denoted by S(*z* ). As it is seen from the top panel of Figure 3, the area S(*z* ) is twice circle's area *S* = *πR*<sup>2</sup> minus area of intersection of these circles. The intersection is twice the area of the sector *s*:

$$\mathfrak{G} = 2\pi R^2 - 2\mathfrak{s} \tag{6}$$

**Figure 3.** The center of a horizontal flake of radius R must belong to the blue domain if flake intersects the reflected ray (blue arrow) and red domain if flake intersects the incident ray (red arrow). The domain to intersect both rays is mixture (violet). The centers of the circles are the points of intersection with the reflected ray (left blue dot) and the incident ray (right red dot) at depth z .

The area of each of this sector spanning the angle *α* = 2arccos *<sup>l</sup>*/2 *<sup>R</sup>* is

$$s = \begin{cases} R^2 \left( \arccos\frac{l/2}{R} - \frac{l/2}{R} \sqrt{1 - \left(\frac{l/2}{R}\right)^2} \right), & l \le 2R \\ 0, & l > 2R \end{cases} \tag{7}$$

where *l*(*z* ) is the distance between the points of the incident and reflected rays at depth *z* and equals to

$$l(z') = c(z - z')$$

where *z* is the depth of the reflection point where both ray segments join; and *c* depends only on the directions of the incident and reflected rays [7].

To calculate transmission for the whole path (whose deepest point, i.e., the point of reflection is at depth *z*), we slice the paint's layer [0, *z*] into horizontal thin sub-layers of thickness *dz* .

Notice that if the flake has a non-zero transmittance *t* then if a ray intersects that flake, it still can pass with probability *t*. Thus, if both (incident and reflected) rays intersect a flake, the whole path still transmits with probability *t* 2. Then, from simple geometric considerations we can infer the following:


Since S = 2*S* − 2*s*, see (6), the total probability that the whole path is blocked by sublayer [*z* , *z* + *dz* ] is then

$$\Pr(z, z') = (1 - t)\Pr\_1 + \left(1 - t^2\right)\Pr\_2 = 2(1 - t)D(\mathcal{S} - (1 - t)\mathbf{s})dz'$$

In the case of horizontal flakes, the flakes at different depths are independent and so the shadowing events at different depths are independent. The probability that the path transmits the whole layer [0, *z*] is then the product of probabilities of transmission of each sublayer:

$$\begin{aligned} a(v', u', z) &= \prod\_{z'=0}^{z} (1 - \Pr(z, z')) = \prod\_{\substack{z'=0\\z'=0}}^{z} (1 - 2(1 - t)D(S - (1 - t)s)dz') \\ &= e^{-2(1 - t)\int\_{0}^{z}D(S - (1 - t)s)dz'} \end{aligned}$$

In case there is a mixture of flakes of different area *S* (in different concentrations *D*(*S*)), the path can be blocked by either of them, so the total probability of blocking is a sum over flake species, i.e.,

$$\begin{array}{rcl} \Pr(z, z') &= 2(1 - t) \sum\_{\text{spcics}} D(S)(S - (1 - t)s) dz' \\ &= 2(1 - t) (\int D(S)S dS - (1 - t) \int D(S)s(l, S) dS) dz' \\ &\equiv 2(1 - t) \left( \overline{DS} - (1 - t) \overline{Ds}(l) dS \right) dz' \end{array}$$

where *<sup>s</sup>* is given by (7) for *<sup>R</sup>* <sup>=</sup> <sup>√</sup>*S*/*<sup>π</sup>* and

$$\begin{array}{rcl} \overline{DS} & \equiv & \int D(S)S dS\\ \overline{DS}(l) & \equiv & \int D(S)s(l,S)dS \end{array}$$

So, attenuation becomes

$$\begin{array}{rcl} a(\upsilon', \mu', z) &=& e^{-2(1-t)\overline{D}\overline{S}z + 2(1-t^2)\int\_0^z \overline{D}\overline{s}(c(z-z'))dz'} \\ &=& e^{-2(1-t)\overline{D}\overline{S}z + 2c^{-1}(1-t)^2\int\_0^z \overline{D}\overline{s}(l)dl} \\ &=& e^{-4c^{-1}(1-t)\overline{D}\overline{S}Z + 2c^{-1}(1-t)^2a(Z)} \end{array} \tag{8}$$

where

$$\begin{array}{rcl}\mathfrak{a}(Z) & \equiv & \int\_0^{2Z} \overline{\mathrm{D}s}(l) dl = \int\_0^{2Z} (\int D(S)\mathfrak{s}(l,S) dS) dl\\ Z & \equiv & \frac{\overline{c\_2}}{2} \end{array} \tag{9}$$

Let us calculate the above *<sup>α</sup>*, which in view of (7) and recalling that *<sup>R</sup>* <sup>=</sup> <sup>√</sup>*S*/*<sup>π</sup>* can be written as

$$\mathfrak{a}(Z) \equiv \int\_0^{2Z} \overline{Ds}(l) dl = \int\_0^{2Z} \left( \int\_{\pi(l/2)^2}^{\infty} D(S) \frac{S}{\pi} \mathcal{S} \left( \frac{l}{2\sqrt{\frac{S}{\pi}}} \right) dS \right) dl$$

where

$$g(t) \equiv \arccos t - t\sqrt{1 - t^2} \tag{10}$$

Changing the order of integration

$$\int\_0^{2Z} \left( \int\_{\pi(l/2)^2}^{\infty} (\cdot \cdot \cdot) \, dS \right) dl = \int\_0^{\pi Z^2} \left( \int\_0^{2\sqrt{\frac{S}{\pi}}} (\cdot \cdot \cdot) dl \right) dS + \int\_{\pi Z^2}^{\infty} \left( \int\_0^{2Z} (\cdot \cdot \cdot) dl \right) dS$$

this becomes

$$\begin{split} a(Z) &= \quad \int\_{0}^{\pi Z^{2}} \Big( \int\_{0}^{2\sqrt{\frac{S}{\pi}}} D(S) \frac{S}{\pi} \mathcal{S} \left( \frac{l}{2\sqrt{\frac{S}{\pi}}} \right) dl \Big) dS + \int\_{\pi Z^{2}}^{\infty} \Big( \int\_{0}^{2Z} D(S) \frac{S}{\pi} \mathcal{S} \left( \frac{l}{2\sqrt{\frac{S}{\pi}}} \right) dl \Big) dS \\ &= \quad \frac{2}{\pi^{\frac{2}{3}}} \Big( \int\_{0}^{1} g(x) dx \Big) \int\_{0}^{\pi Z^{2}} D(S) S^{\frac{2}{3}} dS + \frac{2}{\pi^{\frac{2}{3}}} \int\_{\pi Z^{2}}^{\infty} D(S) S^{\frac{2}{3}} \left( \int\_{0}^{\frac{Z}{\sqrt{\pi}}} g(x) dx \right) dS \\ &= \quad \frac{2}{\pi^{\frac{2}{3}}} \Big( \frac{2}{3} \int\_{0}^{\pi Z^{2}} D(S) S^{\frac{2}{3}} dS + \int\_{\pi Z^{2}}^{\infty} D(S) S^{\frac{2}{3}} G \Big( \sqrt{\frac{\pi Z^{2}}{S}} \Big) dS \Big) \end{split} \tag{11}$$

where

$$G(\mathbf{x}) \equiv \int\_0^\mathbf{x} g(y) dy = \frac{2}{3} + \frac{\pi}{2} \mathbf{x} - \mathbf{x} \cdot \arcsin \mathbf{x} - \frac{2 + \mathbf{x}^2}{3} \sqrt{1 - \mathbf{x}^2} \tag{12}$$

Additionally, *<sup>s</sup>* ≡ *<sup>π</sup>Z*2.

Complexity of (12) prevents us from obtaining the analytic dependence of the integral <sup>∞</sup> *<sup>π</sup>Z*<sup>2</sup> *<sup>D</sup>*(*S*)*S*3/2*<sup>G</sup>* \$*<sup>π</sup>Z*<sup>2</sup> *S dS* on its parameter *Z* even for a simple distribution *D*(*S*). This can be overcome if we find a good polynomial approximation to *G*, then *Z* could be moved out of the integral of each its term. It is reasonable to require that this approximation preserves the following properties of the exact function *G*(*x*): it is 0 for *x* = 0, it is <sup>2</sup> <sup>3</sup> for *x* = 1, and it has zero derivative at *x* = 1. Then, the lowest degree polynomial is

$$\frac{1}{3}\left(\left(1-\varepsilon\right)\left(1-\left(1-\varkappa\right)^{3}\right)+\left(1+\varepsilon\right)\left(1-\left(1-\varkappa\right)^{2}\right)\right)^{2}$$

The value of *ε* can be found from the least square fit that gives *ε* = 0.091649 and thus

$$G(\mathbf{x}) \approx \frac{(5-\varepsilon)\mathbf{x} - (4-2\varepsilon)\mathbf{x}^2 + (1-\varepsilon)\mathbf{x}^3}{3} = 1.6361\mathbf{x} - 1.2722\mathbf{x}^2 + 0.30278\mathbf{x}^3$$

Substituting it in (11) we arrive at

$$\begin{array}{ll} \mathfrak{a}(Z) & \approx & \frac{2}{\pi^{2}} \Biggl( \frac{2}{3} \int\_{0}^{\pi Z^{2}} D(S) S^{\frac{3}{2}} dS + \int\_{\pi Z^{2}}^{\infty} D(S) S^{\frac{3}{2}} \Big( 1.6361 \sqrt{\frac{\pi Z^{2}}{S}} - 1.2722 \frac{\pi Z^{2}}{S} + 0.30278 \left( \frac{\pi Z^{2}}{S} \right)^{\frac{3}{2}} \Big) dS \Big) \\ & = & \frac{4}{3\pi^{2}} \int\_{0}^{\infty} D(S) S^{\frac{3}{2}} dS - \frac{4}{3\pi^{2}} \int\_{\pi Z^{2}}^{\infty} D(S) S^{\frac{3}{2}} dS \\ & + \frac{2}{\pi^{2}} \Big( 1.6361 \sqrt{\pi} Z \Big[ \int\_{\pi Z^{2}}^{\infty} D(S) dS - 1.2722 \pi Z^{2} \Big]\_{\pi Z^{2}}^{\infty} D(S) \sqrt{S} dS + 0.30278 \pi^{\frac{3}{2}} Z^{3} \Big[ \frac{\infty}{\pi Z^{2}} D(S) dS \Big] \\ & = & \frac{4}{3\pi^{2}} (\gamma\_{0}(0) - \gamma\_{0}(Z)) + \frac{2.16361}{\pi} Z \gamma\_{1}(Z) - \frac{2.1222}{\sqrt{\pi}} Z^{2} \gamma\_{2}(Z) + 2 \cdot 0.30278 Z^{2} \gamma\_{3}(Z) \Big] \end{array} \tag{13}$$

where

$$\gamma\_{\mathfrak{m}}(Z) \equiv \int\_{\pi Z^2}^{\infty} S^{\frac{3-\mathfrak{m}}{2}} D(S) dS$$

**Remark 1.** *In case the distribution of flake area is Gaussian Ce* <sup>−</sup> (*S*−*S*) 2 2*δ*2 *SS*2 *then*

$$\gamma\_m(Z) \;= \; \_C\overline{\mathbb{S}}^{\frac{5-m}{2}} \int\_{\frac{nZ^2}{S}}^{\infty} t^{\frac{3-m}{2}} e^{-\frac{(t-1)^2}{2\beta\_S^2}} \, dt$$
