Solar Energy ?
The second part of our solar installation is on it’s way. And it takes the form of several ‘tracking’ photo voltaic panels. For me this is tremendously exciting. Drawing energy from the sun without creating any carbon waste is awesome.
The radiation from the sun travels 93 million miles (149668992 km) to us here on Earth…
That radiation lands on our photovoltaic cells stimulating all those atoms producing electricity…
My problem, however, is where is the sun? To install the base frame correctly, and maximise the panels efficiency, I need it to be pointing to ‘true’ south, or ‘solar’ south, 180*, and here lies the problem.
Just as the Earth’s frozen Arctic and Antarctic have ‘true’ and ‘magnetic’ poles so does the Earth’s position to the Sun. What we have to do is calculate an ‘offset’ called ‘declination’.
Now ‘azimuth’ and ‘bearing’ are terms used to denote direction. The direction you are facing (your bearing) is normally measured in degrees from north (zero). For example if you are facing ‘true’ south your bearing or azimuth angle would be 180*. So, point your compass in the direction your panels are facing. Rotate the compass dial to align the orientating arrow and compass needle ensuring the red parts align then read the degree bearing nearest the travel arrow. Then adjust the panels to face south.
However this only gives you magnetic south. You then have to calculate the degree of ‘declination’.
Thankfully The University of Oregon has an on-line ‘solar position calculator’. Firstly you have to specify date and time followed by ‘Latitude’ (south negative) and ‘Longitude’ (west negative) and your time zone. Here we are Greenwich Mean Time + 1 hour.
They then require barometric pressure, temperature and aspect. i.e azimuth. Once this is all entered into their calculator the program returns solar zenith angle, declination, Julian day, equation of time, hour angle, instantaneous and daily extraterrestrial radiation values and sunrise and sunset times.
Solar position calculator results
(Using solar constant = 1367 W/m2)
Date: 1/1/2008 | Time: 12:00:00 | Zone: GMT + 1 |
Lat: 37.76° | Long: -2.07° | Aspect: 180° |
Pressure: 1013.0 mB | Temp: 10.0° C |
Declination | -23.0310° |
Solar zenith angle(no refraction) | 63.0794° |
Solar zenith angle(with refraction) | 63.0477° |
Julian day | 54466.9583 |
Equation of time | -3.2905 min |
Hour angle | -17.8928° |
Extraterrestrial global horizontal solar irradiance | 640.61 W/m2 |
Extraterrestrial direct normal solar irradiance | 1414.91 W/m2 |
Daily global ETR | 4226.4 W/m2 |
Daily direct normal ETR | 13351.6 W/m2 |
Earth radius factor | 1.0351 |
Sunrise | 08:28:28 |
Sunset | 17:54:39 |
And at this point I should thank my friend Nick for putting his brain power to work on programming this calculator with the right figures.
However, as far as I understand, the solar zenith angle is the vertical winter/summer angle of the sun in the sky and the declination is the solar angle of the sun ‘off’ the north south magnetic line your compass will give you.
So why does the angle of declination change from summer to winter?