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Hello, let’s talk about sunlight intensity, which is a topic that comes up frequently when comparing climates, but it’s seldom properly measured in order to compare the actual ‘strength’ of the sun in different places and during different times of the year. And I suspect there are some misjudgments about it, so let’s give it a try.
First of all, Solar Intensity on Earth is measured in power per unit of area (irradiance) and is usually expressed in W/m² (watts per square meter). As the Earth’s orbit is not entirely circular, but slightly elliptical, the power received varies throughout the year, from 1413 W/m² during the perihelion in January to 1321 W/m² during the aphelion in July. To simplify things, I’ll always use the mean value 1367 W/m², and as we will see, there will be further uncertainties that provide some errors in the calculations, so to the purposes of this thread, being very exact is not necessary.
Now straight to the point, and always considering clear skies:
What makes Solar Intensity vary across the region and time? Different inclination relative to the sunbeam, given by the latitude and the time of the year, which I guess it’s a very clear concept.
So, is the Solar Energy we get given by such inclination? Yes and no. In case of the Earth’s surface, yes, but it’s not necessarily what we experience, as the exposure of our body surface to Solar Radiation may change according to our position.
Let’s suppose we are lying on the ground sunbathing in the equator on March 21 at noon (I don’t recommend it). The sun is directly overhead, in the zenith, so that it hits us perpendicularly. Now we are in London, on June 21 at noon, sunbathing on an oblique platform that rises at 28º relative to the surface level. As London is at 51.5ºN and that the sun’s declination is 23.5ºN that day, the sun in the British capital will be 28º from the zenith, so that once again the sun is hitting us perpendicularly. So does it mean the Solar Energy our body receives is the same in both cases?
Intuition says NO, that the sun in London is anyway weaker than at the equator, and it’s true, but the cause can’t be the inclination… ok, indeed it is, but only secondarily. In case of London, as the sun is inclined 28º relative to the zenith, it will have to go through a greater volume of atmosphere until it reaches the surface than doing it directly from overhead as it happens in the equator. If we visualize it as a triangle, the adjacent leg is the thickness of the atmosphere and the hypotenuse is the sunbeam path. And as atmosphere attenuates the intensity of the sunbeam due to scattering and absorption, the bigger the layer to be crossed is, the bigger the loss of energy to the surface is.
From this concept we get our first formula, the air mass coefficient:
AM = 1 / (cos z)
In which ‘z’ is the angular distance to the sun from the zenith. This coefficient tells us how many ‘atmospheres’ the sunbeam must cross. It’s not an exact formula anyway, as it assumes a flat Earth and lets us some error in cases of inclination over 80º (let’s remember that as the Earth is round, the opposite leg is not straight, but it gradually bends toward inside, reducing the length of the hypotenuse). But we are not interested in such large inclination.
This formula itself doesn’t tell us anything about how much energy is lost, which is what we are ultimately looking for. For this it’s necessary a formula to calculate Solar Intensity depending on the atmosphere attenuation:
Ia = 1.1 x Io x [0.7^(AM^0.678)]
In which ‘Io’ is the mean irradiance at the top of the atmosphere, which I set in 1367 W/m². ‘0.7’ and ‘0.678’ are empirical figures and denote the proportion of radiation that gets to the surface, whereas the first number “1.1” expresses the approximate 10% extra diffuse radiation that is added to the direct radiation.
However, as long as we go at higher altitudes, the thickness of the atmosphere decreases, so this equation is no longer accurate, so we must add new variables, also empirical. The following is a formula to calculate Solar Intensity depending on the atmosphere attenuation with altitude considered:
Ia = 1.1 x Io x {(1- h/7.1) x [(0.7^(AM^0.678)] + h/7.1}
In which ‘h’ indicates altitude expressed in kilometers. I’d say this formula is useless for very high altitudes, let’s say over 5000 m. We are done with calculations based on the atmosphere attenuation, which are especially useful for photovoltaics. The solar panels themselves can be set up to track the sun and stay perpendicular relative to the sunbeams, maximizing its potential.
Now it’s time to deal with the most intuitive variable, which is the proper inclination of the Earth’s surface relative to the sun and that provides the power that a flat area of the Earth’s surface receives. If Earth didn’t have any atmosphere, this would be the formula to calculate the Solar Intensity on the surface by direct inclination:
I = (cos z) x Io
In which ‘z’ is the angular distance to the sun from the zenith again and ‘Io’ is the mean irradiance at the top of the atmosphere, which as I said I had established it in 1367 W/m². But the Earth does have an atmosphere that attenuates the radiation, so that ‘Io’ becomes ‘Ia’, giving us the formula to calculate the Solar Intensity on the surface by direct inclination and atmosphere attenuation:
I = (cos z) x {1.1 x Io x [(0.7^(AM^0.678)]}
We can obviously add the variables of altitude if necessary. Let’s see some examples:
Parallel 0º, March 21 at noon:
AM = 1 / (cos 0º) = 1.000
Ia = 1.1 x (1367 W/m²) x [(0.7^(1.000^0.678)] = 1053 W/m²
I = (cos 0º) x 1053 W/m² = 1053 W/m²
Buenos Aires, December 21 at noon (z = 34.5º - 23.5º = 11º):
AM = 1 / (cos 11º) = 1.019
Ia = 1.1 x (1367 W/m²) x [(0.7^(1,019^0,678)] = 1048 W/m²
I = (cos 11º) x 1048 W/m² = 1029 W/m²
Buenos Aires, June 21 at noon (z = 34.5º + 23.5º = 58º):
AM = 1 / (cos 58º) = 1.887
Ia = 1.1 x (1367 W/m²) x [(0.7^(1.887^0.678)] = 869 W/m²
I = (cos 58º) x 869 W/m² = 461 W/m²
London, June 21 at noon (z= 51.5º - 23.5º = 28º):
AM = 1 / (cos 28º) = 1.133
Ia = 1.1 x (1367 W/m²) x [(0.7^(1.133^0.678)] = 1020 W/m²
I = (cos 28º) x 1020 W/m² = 901 W/m²
London, December 21 at noon (z= 51.5º + 23.5º = 75º):
AM = 1 / (cos 75º) = 3.864
Ia = 1.1 x (1367 W/m²) x [(0.7^(3.864^0.678)] = 616 W/m²
I = (cos 75º) x 616 W/m² = 159 W/m²
North Pole, June 21 (z = 90º - 23.5º = 66.5º):
AM = 1 / (cos 66.5º) = 2.508
Ia = 1.1 x (1367 W/m²) x [(0.7^(2.508^0.678)] = 773 W/m²
I = (cos 66.5º) x 773 W/m² = 308 W/m²
South Pole, December 21 (z = 90º - 23.5º = 66.5º; h = 2.8):
AM = 1 / (cos 66.5º) = 2.508
Ia = 1.1 x (1367 W/m²) x {(1- 2.8/7.1) x [(0.7^(2.508^0.678)] + 2.8/7.1} = 1061 W/m²
I = (cos 66.5º) x 1061 W/m² = 423 W/m²
Well, it’s enough so far. Any comment, remark and correction are welcome. Still a lot of information is missing, namely the calculation of intensity at any time of any day of the year (not restricted to solstices and equinoxes and at noon), integration of the daily power, etc.
This is interesting stuff but I can't really understand it all properly. Re. altitude, the variable "h" wouldn't naturally correspond exactly to a man-made unit like a kilometre, surely?
One thing I have wondered is at what time of day (particularly referring to the evenings) is the sunlight intensity on 21 June here at 54N equivalent to what it is at midday on 21 December? I couldn't explain why but I guess it's more complicated than it just being all about the height of the sun in the sky (about 13 degrees)? Shame the elliptical orbit would make it more difficult to work out exactly.
Doesn't those calculating methods give negative values for very northern latitudes when the sun is shining in midwinter but the sun angle is too low to give proper irradiation? Like the Isles of Scilly have a larger annual solar irradiation than Helsinki, but during the actual growing season the difference is very small.
Doesn't those calculating methods give negative values for very northern latitudes when the sun is shining in midwinter but the sun angle is too low to give proper irradiation? Like the Isles of Scilly have a larger annual solar irradiation than Helsinki, but during the actual growing season the difference is very small.
I think they do, since the Sun is below the horizon, so the solar angle is negative.
This is interesting stuff but I can't really understand it all properly. Re. altitude, the variable "h" wouldn't naturally correspond exactly to a man-made unit like a kilometre, surely?
One thing I have wondered is at what time of day (particularly referring to the evenings) is the sunlight intensity on 21 June here at 54N equivalent to what it is at midday on 21 December? I couldn't explain why but I guess it's more complicated than it just being all about the height of the sun in the sky (about 13 degrees)? Shame the elliptical orbit would make it more difficult to work out exactly.
It more than likely is. Just so you know the intensity at solar noon on December 21 is equivalent to about an hour and a half before/after sunset/sunrise. So about 6 AM and 8 PM are the times it's at that sun angle? Would be interesting to see a comparison picture from those times on June 21 and solar noon on December 21.
One thing I have wondered is at what time of day (particularly referring to the evenings) is the sunlight intensity on 21 June here at 54N equivalent to what it is at midday on 21 December? I couldn't explain why but I guess it's more complicated than it just being all about the height of the sun in the sky (about 13 degrees)?
Yes, same angle means same intensity. There is another formula to determine the sun angle in a particular moment of the day:
cos(z) = sin(Φ) x sin(δ) + cos(Φ) x cos(δ) x cos(h)
In which z = angular distance to the zenith; Φ = latitude; δ = solar declination; h = hourly angle from -180º to 180º, where h = 0º is the noon, 15º = 1 hour, am gets negative figures and pm gets positive figures.
Let's use London again as an example. Zenith angle on December 21 is 75º. We need to know at what time of the day on June 21 (when the zenith angle reaches 28º) it gets to that angle. So we have:
cos(75) = sin(51.5) x sin(23.5) + cos(51.5) x cos(23.5) x cos(h)
h = cos^-1 {cos(75) - [sin(51.5) x sin(23.5)]} / [cos(51.5) x cos(23.5)] = 6,359 = 6 h 22 m
So the strength of the midday sun on Dec 21 equals the strength of the sun on Jun 21 when it's 6 h 22 m before or after the solar noon.
Quote:
Originally Posted by Ariete
Doesn't those calculating methods give negative values for very northern latitudes when the sun is shining in midwinter but the sun angle is too low to give proper irradiation? Like the Isles of Scilly have a larger annual solar irradiation than Helsinki, but during the actual growing season the difference is very small.
Not sure if I get what you mean, but if the sun is below the horizon, the power is obviously zero and the formula doesn't apply. However, if the sun is barely over the horizon, the result will be always positive.
For instance, a case where the sun is only 5º over the horizon:
AM = 1 / (cos 85º) = 11,468
Ia = 1.1 x (1367 W/m²) x [(0.7^(11,468^0,678)] = 233 W/m²
I = (cos 85º) x 233 W/m² = 20 W/m²
Yes, same angle means same intensity. There is another formula to determine the sun angle in a particular moment of the day:
cos(z) = sin(Φ) x sin(δ) + cos(Φ) x cos(δ) x cos(h)
In which z = angular distance to the zenith; Φ = latitude; δ = solar declination; h = hourly angle from -180º to 180º, where h = 0º is the noon, 15º = 1 hour, am gets negative figures and pm gets positive figures.
Let's use London again as an example. Zenith angle on December 21 is 75º. We need to know at what time of the day on June 21 (when the zenith angle reaches 28º) it gets to that angle. So we have:
cos(75) = sin(51.5) x sin(23.5) + cos(51.5) x cos(23.5) x cos(h)
h = cos^-1 {cos(75) - [sin(51.5) x sin(23.5)]} / [cos(51.5) x cos(23.5)] = 6,359 = 6 h 22 m
So the strength of the midday sun on Dec 21 equals the strength of the sun on Jun 21 when it's 6 h 22 m before or after the solar noon.
That makes sense, but it just doesn't match my observations. There are some gloomy overcast days around the winter solstice when I'd need a light on at midday to comfortably be able to read a book in my front room. According to what you said I would potentially need to put the light on to read some evenings at 8 pm in midsummer - admittedly we don't get that many June evenings with dense overcast like we get in December, but I can't imagine ever having the lights on that early at that time of year. Maybe it's just the oddness of how gloomy some winter days are that it sticks in my mind more.
Does same sun angle also = same sun power? I would be even more dubious about that. It's a sunny summer evening just after 8 pm now and the sun feels like more than just the lightbulb in the sky it feels like in midwinter.
Sun peaking in the mid 80's now. 86.1* today. Sunshine is like a fluorescent white glare. Of course, mid winter here is 40* so much like late March/ early April for UK posters.
That makes sense, but it just doesn't match my observations. There are some gloomy overcast days around the winter solstice when I'd need a light on at midday to comfortably be able to read a book in my front room. According to what you said I would potentially need to put the light on to read some evenings at 8 pm in midsummer - admittedly we don't get that many June evenings with dense overcast like we get in December, but I can't imagine ever having the lights on that early at that time of year. Maybe it's just the oddness of how gloomy some winter days are that it sticks in my mind more.
Does same sun angle also = same sun power? I would be even more dubious about that. It's a sunny summer evening just after 8 pm now and the sun feels like more than just the lightbulb in the sky it feels like in midwinter.
That's probably just your imagination. Though you probably get much more dense overcast in winter, maybe foggy/drizzly too.
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