The following is a theoretical estimation of solar radiation at different locations on the planet as well as different times of year. To investigate an investment in solar energy, we are interested in the amount of radiation received at each day of the year. The following such estimation is mainly for clear-sky conditions, we'll however also have a brief look at the effect of various cloud conditions. At the bottom of the page you can download an Excel tool that implements the calculations. It compares surface orientations with respect to received radiation, being horizontal plane, fixed slope, variable slope (1-axis) and tracker (2-axis). Calculated values seem to match measurements quite well, however values for the tracking solution are on the high side. If you find any mistakes, or have suggestions for improvement, feel free to let me know. Disclaimer: this is a work in progress.

Solar Constant

Before entering the earth's atmosphere, solar radiation can be considered to have equal energy at any measured surface. This power density is referred to as the solar constant S. However, multiple other factors need to be taken into account when calculating how much of this radiation arrives at different location on the planet. First, the distance to the sun varies throughout the course of a year, which is expressed in terms of orbital eccentricity E0. Next, how high the sun stands in the sky (zenith angle θ) depends on the latitude of the location in question, the declination of the earth at the day in question, and the time of that day. Both latitude and day of the year also determine how many hours of sunlight this location will get at any given day. Using these parameters, we can estimate how much radiation reaches the surface at a given location if there would be no atmosphere.

$S=\frac{1364W}{m^2}=\frac{1364J}{m^2.s}=\frac{4,910MJ}{m^2.hr}=\frac{117,8MJ}{m^2.day}$

Elliptical Orbit

As the earth has an elliptical orbit around the sun, with a distance between the two that varies over the course of a year, we'll need to account for the resulting variation in power density. The relative variation in distance is expressed as orbital eccentricity E0, defined as the square of the ratio of average distance to the distance at any given moment in time. Orbital eccentricity can be described as a function of day angle Γ, in which day number J is 1 at the 1st of january and 365 at the 31st of december. As you see, all we do with this formula is equate a full orbit of the earth around to sun to 2π radians, and then cut it in pieces of one day. The formula for orbital eccentricity can be expressed in terms of day angle Γ.

\begin{align} \Gamma & = 2.\pi.\left(\frac{1-J}{365}\right) \\ E_0 & =\left(\frac{r_0}{r}\right)^2=1,000110 + 0,034221.cos(\Gamma) + 0,001280.sin(\Gamma) + 0,000719.cos(2.\Gamma) + 0,000077.sin(2.\Gamma) \end{align}

Declination

Even though the earth travels in an elliptical orbit around the sun, with varying distance between the two, this isn't the reason for our seasons. We are closest to the sun in january and furthest from it in june, so this point should be obvious for those of us living in the northern hemisphere. Seasons, on the contrary, are caused by the tilt in earth's rotational axis, which has the north pole pointing towards the sun in june and the south pole in december. This impacts hours of sunlight a given location sees in a day, as well as their power density when hitting the curved surface of the earth. Declination δ, defined as the angle between the rays of the sun and the plane of the equator, captures this effect; its calculation uses previously specified concept of day angle Γ and returns a value in degrees.

$δ = 0,006918 - 0,399912.cos(\Gamma) + 0,070257.sin(\Gamma) - 0,006758.cos(2.\Gamma) + 0,000907.sin(2.\Gamma) - 0,002697.cos(3.\Gamma) + 0,00148.sin(3.\Gamma)$

Latitude

A very important determinant for how much solar radiation a place gets is it's latitude, Λ. Taking the plane of the earth's equator as a reference, latitude is measured as the angle at the center of the earth, between the equator and the location in question. This angle is said to be positive in the northern, and negative in the southern hemisphere. As you can see in below image, not only does the radiation need to travel though more atmosphere with increasing positive or negative latitude, because of the curvature of the earth the same amount of radiation also gets spread over a larger area.

Zenith Angle

At a given location, the zenith is the line that extends straight up from the observer, and the angle it makes with a celestial body (in our case the sun) is the zenith angle θ (in radians). At solar noon (moment, as seen by the observer, in which the sun reaches it's highest point of the day) it can simply be calculated as zenith angle θ = latitude Λ - declination δ (which is negative in above image), however the formula gets more complex if we want to trace the solar angle from sunrise to sunset. Below formula does just this, and it requires additional variables, which are the angular velocity of the earth's rotation ω (15°/hr or 0,2618 rad/hr) and the number of hours before (-) and after (+) solar noon t. The so-called hour angle ω.t gives us the angle before and after solar noon at that particular time of day, which needs to be between rising (TR) and setting (TS) of the sun. At both sun rise and sun set, the zenith angle is 90°, in between it reaches its lowest value at solar noon.

$θ = acos[sin(\Lambda).sin(\delta) + cos(\Lambda).cos(\delta).cos(\omega.t)], T_R ≤ t ≤ T_S$

Solar noon, Sunrise and Sunset

When we take time to be zero at solar noon (t = 0 hr), the moments of sunrise TR and sunset TS are represented as the number of hours before and after solar noon. At both moments the zenith angle θ of the sun equals 90°. Taking the cosine of both sides of above equation for zenith angle, and knowing that for the moments of interest cos(θ) = 0, moments TR and TS can be calculated as follows:

\begin{align} T_R & = -\frac{acos[-tan(\delta).tan(\Lambda)]}{\omega} \\ T_S & = +\frac{acos[-tan(\delta).tan(\Lambda)]}{\omega} \end{align}

For above formulae, however, we need to be conscious of the two polar cases. No sun rises at a location for which |Λ-δ| > 90°, and no sun sets if |Λ+δ| > 90°. We need to catch these conditions in our calculations.

Based on previous properties, we can now calculate solar radiation at a given location on the earth's surface, however not taking the effect of the earth's atmosphere into account yet. To acknowledge this, we're speaking of extraterrestrial solar radiation. In the latter's definition, the area of interest on the earth's surface is moved to a point at the top of the atmosphere, keeping it parallel with the original. The result is the same as considering the area on the surface of the earth, however not figuring for the atmosphere's influence.

Following is the formula for instantaneous extraterrestrial solar radiation kET. It takes into account previously calculated parameters: the solar constant S, orbital eccentricity E0, declination δ, latitude Λ, and time of day; the last 3 being necessary to calculate zenith angle θ.

$k_{ET} = S.E_0.cos[\theta(\Lambda,\delta,t)]$

In order to get extraterrestrial solar radiation KET as a daily total for each square meter of surface area, however, above formula needs to be integrated (summed) between times of sunrise TR and sunset TS.

\begin{align} K_{ET} & = S.E_0.\int\limits_{T_R}^{T_S}cos[\theta(\Lambda,\delta,t)].dt \\ & = 2.S.E_0.\int\limits_{0}^{T_S} [sin(\Lambda).sin(\delta)+cos(\Lambda).cos(\delta).cos(\omega.t)].dt \\ & = 2.S.E_0.\left[sin(\Lambda).sin(\delta).T_S+\frac{cos(\Lambda).cos(\delta).sin(\omega.T_S)}{\omega}\right] \end{align}

Clear-Sky Radiation on a Horizontal Surface

When solar radiation travels through the atmosphere, it is absorbed and reflected by both gasses and solid particles. How much gets absorbed and reflected depends on the distance it travels through the atmosphere, which is called optical air mass m and depends on declination δ, latitude Λ and time of day t. It can be calculated in relative terms, in which the shortest distance through the atmosphere, measured perpendicularly to the earth's surface at sea level, is considered to be 1. Being able to calculate the zenith angle θ and using basic trigonometry, one could figure out the distance through the atmosphere as follows:

$m \approx \frac{1}{cos(\theta)} = \frac{1}{sin(\Lambda).sin(\delta)+cos(\Lambda).cos(\delta).cos(\omega.t)}$

However, as can be seen from below images, this approximation gets more incorrect with increasing zenith angle θ.

The following formula (Yin 1997) gives more accurate results. First instantaneous optical air mass is given:

$m=\frac{1,021}{0.008307 + sin(\Lambda).sin(\delta)+cos(\Lambda).cos(\delta).cos(\omega.t)}-0.01259$

Next the above formula is integrated between sunrise TR and sunset TS to give the daily average optical air mass M.

\begin{align} \text{If } T_S&≠ 0 \text{ and } A \gt B, \\ M & = \left(\frac{1}{T_S}\right).\left\{\left(\frac{1}{\omega}\right).\left( \frac{1,021}{(A^2-B^2)^{1/2}}\right).acos\left[\frac{B+A.cos(\omega.T_S)}{A+B.cos(\omega.T_S)}\right]+C.T_S \right\} \\ \\ \text{If } T_S&≠ 0 \text{ and } A \lt B, \\ M & = \left(\frac{1}{T_S}\right).\left\{\left(\frac{1}{\omega}\right).\left( \frac{1,021}{(B^2-A^2)^{1/2}}\right).ln\left[\frac{\{[B+A].[1+cos(\omega.T_S)]\}^{1/2}+\{[B-A].[1-cos(\omega.T_S)]\}^{1/2}}{\{[B+A].[1+cos(\omega.T_S)]\}^{1/2}-\{[B-A].[1-cos(\omega.T_S)]\}^{1/2}}\right]+C.T_S \right\} \\ \\ \text{If } T_S&≠ 0 \text{ and } A = B, \\ M & = \left(\frac{1}{T_S}\right).\left\{\left(\frac{1}{\omega}\right).\left( \frac{1,021}{A}\right).tan\left(\frac{\omega.T_S}{2}\right)+C.T_S \right\} \\ \\ \text{In which:} \\ A & =0,008307+sin(\Lambda).sin(\delta) \\ B & =cos(\Lambda).cos(\delta) \\ C & =-0.01259 \end{align}

Previous values are calculated at sea level. If you would like to get values at elevation z, optical air mass can be adjusted using the following formula, in whihc z is the elevation in m.

$M(z)=M.exp\left(-\frac{z}{7000}\right)$

Based on the daily average optical air mass calculated above, the atmosphere's influence on solar radiation can now be quantified. We do this by use of so-called transmissivities, factors with a maximum of 1 figuring for the effect of the entire optical air mass on the amount of solar radiation that reaches the earth's surface. Following is the formula used to calculate how much of the extraterrestrial solar radiation reaches the earth's surface (on a horizontal plane) in terms of transmissivities τwa, τda, τws, τrs, τds due to absorption and scattering.

$K_{dirh}=K_{ET}.\tau_{wa}.\tau_{da}.\tau_{ws}.\tau_{rs}.\tau_{ds}$

Respective transmissivities are calculated as follows; αw being the fraction of solar radiation absorbed by water vapor, αd the fraction of solar radiation absorbed by dust and other solid aerosols, ρw the fraction of solar radiation scattered by water vapor, ρrs the fraction of solar radiation scattered by air molecules, and ρd the fraction of solar radiation scattered by dust and solid aerosols.

\begin{align} \tau_{wa}&=(1-\alpha_w)=1-0,077.(M.W)^{0,3} \\ \tau_{da}&=(1-\alpha_d)=0,965^{M} \\ \tau_{ws}&=(1-\rho_w)=1-0,0225.M.W \\ \tau_{rs}&=(1-\rho_{rs})=0,972-0,08262.M+0,00933.M^2-0,00095.M^3+0,0000437.M^4 \\ \tau_{ds}&=(1-\rho_d)=0,965^M \end{align}

In above formulae, precipitable water content W is expressed in cm; relative humidity RH in % and air temperature Ta in K.

$W=0,00493.\left(\frac{RH}{T_a}\right).exp\left(26,23-\frac{5416}{T_a}\right)$

$K_{dif}=0,5.K_{ET}.\tau_{wa}.\tau_{da}.(1-\tau_{ws}.\tau_{rs}.\tau_{ds})$

Of the radiation hitting the surface of the earth, a fraction is reflected back. This fraction, termed albedo a, relates to the reflectivity of the surface upon which the solar radiation falls. Being reflected away from the earth, it is again scattered on its way through the atmosphere and partly reflected back towards the earth.

$K_{BS}=a.(K_{dirh}+K_{dif}).[0,5.\tau_{wd}.\tau_{da}.(1-\tau_{ws}.\tau_{rs}.\tau_{ds})]$

Total clear-sky solar radiation KCSh hitting a horizontal surface under clear-sky conditions at a given location is thus equal to sum of direct, diffuse and backscattered radiation.

$K_{CSh}=K_{dirh}+K_{dif}+K_{BS}$

Clear-Sky Radiation on a Sloping Surface

Now what if the receiving surface is not horizontal, as it was taken to be in above calculations. You might want to calculate insolation for a hillside, the roof of a house, or a solar collector mounted to optimize the amount of solar radiation it receives. There are two angles to take into account when considering a non-horizontal surface, namely slope angle β and orientation angle α. Slope measures the angle a surface makes with the horizontal plane, orientation the angle it makes in the horizontal plane with the planet's north-south axis.

We'll be using the concepts of equivalent latitude and longitude to take slope and orientation of the surface into account. Given particular angles for slope and orientation, a new combination of latitude and longitude can be found at which the surface would be horizontal. In this case, the new location is said to have equivalent latitude Λeq and longitude Ωeq. In the following, difference in longitude ΔΩ is calculated, as there is no use in knowing the longitude itself.

\begin{align} \Lambda_{eq} & =asin\left[sin(\beta).cos(\alpha).cos(\Lambda)+cos(\beta).sin(\Lambda)\right] \\ \Delta\Omega & =atan\left[\frac{sin(\beta).sin(\alpha)}{cos(\Lambda).cos(\beta)-sin(\Lambda).sin(\beta).cos(\alpha)}\right] \end{align}

Slope and orientation angles don't influence optical air mass, solar radiation still needs to travel the same distance through the atmosphere. Times of solar noon TNS, sunrise TRS and sunset TSS however will be influenced by slope and orientation angles; these can be calculated using equivalent latitude and longitudinal difference.

\begin{align} T_{NS} & = -\Delta\Omega/\omega\\ T_{RS} & = -\frac{acos[-tan(\Lambda_{eq}).tan(\delta)]-\Delta\Omega}{\omega}\\ T_{SS} & = +\frac{acos[-tan(\Lambda_{eq}).tan(\delta)]-\Delta\Omega}{\omega} \end{align}

Adjusting slope and orientation can not increase length of day, their combination can cause a decrease. However, the amount of direct solar radiation can be improved, by letting the surface become more perpendicular to the sun's rays. Due to the combination of slope and orientation, solar noon can shift, meaning times of sunrise and sunset are no longer equal in absolute terms. As such, the below formula takes both times TRS and TSS into account.

$K_{ETS} = S.E_0.\left[sin(\Lambda_{eq}).sin(\delta).(T_{SS}-T_{RS})+cos(\Lambda_{eq}).cos(\delta).\frac{sin(\omega.T_{SS}+\Delta\Omega)-sin(\omega.T_{RS}+\Delta\Omega)}{\omega}\right]$

Only direct solar radiation is dependent on slope angle, diffuse and backscattered radiation are not. Applying transmissivities to extraterrestrial radiation KETs, we can calculate total clear-sky incident radiation on a slope KCSs:

$K_{CSS}=\tau_{wa}.\tau_{da}.\tau_{ws}.\tau_{rs}.\tau_{ds}.K_{ETs}+K_{dif}+K_{BS}$

Average Cloud Cover

It is possible to get average values for cloud cover from sites like weatherspark.com, however that information by itself isn't enough to determine solar radiation on the ground. Lacking the appropriate worldwide data at the moment, the excel tool available at the bottom of this page adds the possibility to adjust for different cloud conditions. The cloud conditions and values used can be found in David H. Miller, Energy at the Surface of the Earth: An Introduction to the Energetics of Ecosystems (1981, p. 77).

Condition of sky or cloud density Flux density (W m -2) Fraction
Cloudless 418 1,00
Scattered clouds, not covering sun 397 0,95
Sun's disk visible through cirrus 363 0,87
Sun's disk visible through stratus 286 0,68
Sun not visible, because of high clouds 307 0,73
Sun not visible, because of low clouds 203 0,49
Thick overcast 98 0,24

Resources

• An Excel sheet for calculating solar radiation using above method can be found here