## 2 Materials and Methods
The optical air mass, \(m\), is estimated from the equations hereunder:  
\[m=\frac{1}{\cos\psi}\left(\frac{\sqrt{r^{2}\cos^{2}\psi+2 rH+H^{2}}-r\cos\psi}{ H}\right)-\frac{2.8}{\left(90-\psi\right)^{2}}\text{ for }60^{\circ}<\psi<80^{\circ} \tag{4}\]  
and  
\[m=0.3885\psi^{2}-63.705\psi+2618\text{ for }80^{\circ}<\psi<90^{\circ} \tag{5}\]  
where \(r\) is the mean radius of the earth (6.390 \(\times\) 10\({}^{6}\) m); \(H\) is the mean atmospheric depth (7.991 \(\times\) 10\({}^{6}\) m). Equation (5) is the only case where \(\psi\) is not an independent variable of a sinusoidal function and expressed in degrees. Equation (4) is based on theory and corrected by empirical data from [PERSON] [17]. Although it can be used for angles greater than 80\({}^{\circ}\), the zenith angle should not be too close to 90\({}^{\circ}\) for obvious reasons. Equation (5) was obtained by curve fitting to experimental data of the optical air mass as a function of the zenith angle given by [PERSON] [18]. Quadratic regression analysis gave the best results with a correlation coefficient of 0.986. The equations are valid for an atmospheric pressure of one bar. Hence, for an atmospheric pressure, \(p\), the optical air mass given by Equations (4) and (5) is corrected by multiplying it by \(p\) (in bars) as performed by [PERSON] [17]. The solar declination, \(\delta\), is given by  
\[\delta=23.5\cos\left(\frac{2\pi}{365}[172-d]\right) \tag{6}\]  
The latter equation is applicable to the Northern Hemisphere. Note that on 21 June (summer solstice), i.e., Julian day 172, and 22 December (winter solstice), the solar declination angle is at its maximum and minimum, i.e., +23.5\({}^{\circ}\) and \(-23.5^{\circ}\), respectively. \(\delta\) is a sinusoidal (cosine) wave with an amplitude of 23.5. The expression for the solar declination in Equation (6) does not account for the small precession of the polar axis, which is considered inconsequential for the intents and purposes of the solar radiation model being discussed here. The hour angle, \(h_{s}\), is given by  
\[h_{s}=\frac{2\pi}{24}t \tag{7}\]  
where \(t\) is the time of the day from solar noon, i.e., positive values for time before noon and negative values for time after noon (e.g., 11:00 would be \(-1\), while 13:00 would be \(+1\)). The resulting \(h_{s}\) is in radians. The next step after calculating \(S^{c}_{dir}\) is to calculate the instantaneous diffuse solar irradiance at the earth's surface for a turbid atmosphere and cloudless sky (\(S^{c}_{diff}\)). The latter is expressed in terms of \(S^{c}_{dir}\), the zenith angle and the turbidity coefficient. The equation has been used previously by [PERSON] and [PERSON] [19] and is given by \[S^{c}_{diff}=\left(\frac{0.46}{\pi^{\text{sec}\,\varphi}}-0.5\right)S^{c}_{dir} \tag{8}\]  
Cloud cover can have a significant effect on the amount of solar (short-wave) radiation reaching the earth's surface [20]. Hence, \(S^{c}_{dir}\) and \(S^{c}_{diff}\) are adjusted to \(S_{dir}\) and \(S_{diff}\), respectively, in order to account for cloud cover \(C\), i.e., the fraction of sky covered by clouds (in oktas) as follows:  
\[S_{dir}=(1-0.75C)S^{c}_{dir} \tag{9}\]  
and  
\[S_{diff}=S^{c}_{diff}+C\mathcal{S}_{dir} \tag{10}\]  
The adjustment is intuitively obvious. Note that the reason for scaling down the cloud cover \(C\) by 25% (to 0.75C) is that clouds do not have zero transmissivity for short-wave radiation. In this case, a 25% transmissivity is assumed. This measured value is strictly speaking for the Australian continent and is variable [20]. Hence, the instantaneous total (direct plus diffuse) solar irradiance at the earth's surface in the presence of a turbid atmosphere and clouds (\(S_{total}\)) is obtained by summing Equations (9) and (10) to give  
\[\beta\approx 1+2\epsilon\cos\left(\frac{2\pi}{365}d\right) \tag{2}\]0  
It is worth digressing at this point and stating that one utility of \(S_{total}\) is in the determination of the atmospheric stability class.