Astronomy

Why doesn't the solar azimuth vary uniformly

Why doesn't the solar azimuth vary uniformly

For the past few dates, I've been observing the rotation of the earth around the sun and noticed that the azimuth of the sun doesn't vary uniformly throughout the day. The azimuth varies faster during some parts of the day, depending on the position link.

Why does this happen? I've been thinking about the cause, but can't come up with a proper explanation.


The Sun's hour angle, measured around the celestial equator, changes at a nearly constant rate. The gnomon of a sundial is aligned with the polar axis to take advantage of this. The celestial equator is oblique to the horizon except at the Earth's equator and poles.

The lines of azimuth converge at the observer's zenith and nadir, so the Sun's azimuth changes more rapidly near one of those points (e.g. summer midday, winter midnight) and more slowly near the horizon. The amplitude of this fluctuation depends on the Sun's declination and the observer's geographic latitude.

For an observer in the tropics, on a day when the Sun passes directly overhead, its azimuth swings abruptly from 90° to 270° at noon. For an observer near the north or south pole, the Sun remains near the same altitude all day, and its azimuth rate stays near the average 15°/hour.


Contents

The solstice may have been a special moment of the annual cycle for some cultures even during Neolithic times. Astronomical events were often used to guide activities, such as the mating of animals, the sowing of crops and the monitoring of winter reserves of food. Many cultural mythologies and traditions are derived from this.

This is attested by physical remains in the layouts of late Neolithic and Bronze Age archaeological sites, such as Stonehenge in England and Newgrange in Ireland. The primary axes of both of these monuments seem to have been carefully aligned on a sight-line pointing to the winter solstice sunrise (Newgrange) and the winter solstice sunset (Stonehenge). It is significant that at Stonehenge the Great Trilithon was oriented outwards from the middle of the monument, i.e. its smooth flat face was turned towards the midwinter Sun. [8]

The winter solstice was immensely important because the people were economically dependent on monitoring the progress of the seasons. Starvation was common during the first months of the winter, January to April (northern hemisphere) or July to October (southern hemisphere), also known as "the famine months". In temperate climates, the midwinter festival was the last feast celebration, before deep winter began. Most cattle were slaughtered so they would not have to be fed during the winter, so it was almost the only time of year when a plentiful supply of fresh meat was available. [9] The majority of wine and beer made during the year was finally fermented and ready for drinking at this time. The concentration of the observances were not always on the day commencing at midnight or at dawn, but at the beginning of the pagan day, which in many cultures fell on the previous eve. [ citation needed ]

Because the event was seen as the reversal of the Sun's ebbing presence in the sky, concepts of the birth or rebirth of sun gods have been common. [ citation needed ] In cultures which used cyclic calendars based on the winter solstice, the "year as reborn" was celebrated with reference to life-death-rebirth deities or "new beginnings" such as Hogmanay's redding, a New Year cleaning tradition. [ citation needed ] Also "reversal" is yet another frequent theme, as in Saturnalia's slave and master reversals.

Indian Edit

Makara Sankranti, also known as Makaraa Sankrānti (Sanskrit: मकर संक्रांति) or Maghi, is a festival day in the Hindu calendar, in reference to deity Surya (sun). It is observed each year in January. [10] It marks the first day of Sun's transit into Makara (Capricorn), marking the end of the month with the winter solstice and the start of longer days. [10] [11] In India, this occasion, known as Ayan Parivartan (Sanskrit: अयन परिवर्तन), is celebrated by religious Hindus as a holy day, with Hindus performing customs such as bathing in holy rivers, giving alms and donations, praying to deities and doing other holy deeds.

Iranian Edit

Iranian people celebrate the night of the Northern Hemisphere's winter solstice as, "Yalda night", which is known to be the "longest and darkest night of the year". Yalda night celebration, or as some call it "Shabe Chelleh" ("the 40th night"), is one of the oldest Iranian traditions that has been present in Persian culture from ancient times. In this night all the family gather together, usually at the house of the eldest, and celebrate it by eating, drinking and reciting poetry (esp. Hafez). Nuts, pomegranates and watermelons are particularly served during this festival.

Judaic Edit

An Aggadic legend found in tractate Avodah Zarah 8a puts forth the talmudic hypothesis that Adam first established the tradition of fasting before the winter solstice, and rejoicing afterward, which festival later developed into the Roman Saturnalia and Kalendae.

Germanic Edit

The pagan Scandinavian and Germanic people of northern Europe celebrated a winter holiday called Yule (also called Jul, Julblot, jólablót). The Heimskringla, written in the 13th century by the Icelander Snorri Sturluson, describes a Yule feast hosted by the Norwegian king Haakon the Good (c. 920–961). According to Snorri, the Christian Haakon had moved Yule from "midwinter" and aligned it with the Christian Christmas celebration. Historically, this has made some scholars believe that Yule originally was a sun festival on the winter solstice. Modern scholars generally do not believe this, as midwinter in medieval Iceland was a date about four weeks after the solstice. [12]

Roman cult of Sol Edit

Sol Invictus ("The Unconquered Sun/Invincible Sun") was originally a Syrian god who was later adopted as the chief god of the Roman Empire under Emperor Aurelian. [13] His holiday is traditionally celebrated on December 25, as are several gods associated with the winter solstice in many pagan traditions. [14] It has been speculated to be the reason behind Christmas' proximity to the solstice. [15]

East Asian Edit

In East Asia, the winter solstice has been celebrated as one of the Twenty-four Solar Terms, called Dongzhi in Chinese. In Japan, in order not to catch cold in the winter, there is a custom to soak oneself in a yuzu hot bath (Japanese: 柚子湯 = Yuzuyu). [16]

Although the instant of the solstice can be calculated, [17] direct observation of the solstice by amateurs is impossible because the Sun moves too slowly or appears to stand still (the meaning of "solstice"). However, by use of astronomical data tracking, the precise timing of its occurrence is now public knowledge. One cannot directly detect the precise instant of the solstice (by definition, one cannot observe that an object has stopped moving until one later observes that it has not moved further from the preceding spot, or that it has moved in the opposite direction). Furthermore, to be precise to a single day, one must be able to observe a change in azimuth or elevation less than or equal to about 1/60 of the angular diameter of the Sun. Observing that it occurred within a two-day period is easier, requiring an observation precision of only about 1/16 of the angular diameter of the Sun. Thus, many observations are of the day of the solstice rather than the instant. This is often done by observing sunrise and sunset or using an astronomically aligned instrument that allows a ray of light to be cast on a certain point around that time. The earliest sunset and latest sunrise dates differ from winter solstice, however, and these depend on latitude, due to the variation in the solar day throughout the year caused by the Earth's elliptical orbit (see earliest and latest sunrise and sunset).

Neolithic site of Goseck circle in Germany. The yellow lines indicate the directions in which sunrise and sunset are seen on the day of the winter solstice.

Sunrise at Stonehenge in southern England on the winter solstice

Other related festivals Edit

    (Ancient Rome): Celebrated shortly before winter solstice (Christian): Used to coincide with the winter solstice day : Takes place shortly after winter solstice, absorbed tradition from winter solstice celebration. Speculated to originate from solstice date, see Christmas#Solstice date and Dies Natalis Solis Invicti (Korea, Greater China): 105 days after winter solstice / Pongal (India): Harvest Festival – Marks the end of the cold months and start of the new Month with longer days.

The following tables contain information on the length of the day on December 22nd, close to the winter solstice of the Northern Hemisphere and the summer solstice of the Southern Hemisphere (i.e. December solstice). The data was collected from the website of the Finnish Meteorological Institute on 22 December 2015, as well as from certain other websites. [18] [19] [20] [21] [22] [23]

The data is arranged geographically and within the tables from the shortest day to the longest one.

The Nordic countries and the Baltic states
City Sunrise
22 Dec 2015
Sunset
22 Dec 2015
Length of the day
Murmansk 0 h
Bodø 11:36 12:25 0 h 49 min
Rovaniemi 11:08 13:22 2 h 14 min
Luleå 9:55 13:04 3 h 08 min
Reykjavík 11:22 15:29 4 h 07 min
Trondheim 10:01 14:31 4 h 30 min
Tórshavn 9:51 14:59 5 h 08 min
Helsinki 9:24 15:13 5 h 49 min
Oslo 9:18 15:12 5 h 54 min
Tallinn 9:17 15:20 6 h 02 min
Stockholm 8:43 14:48 6 h 04 min
Riga 9:00 15:43 6 h 43 min
Copenhagen 8:37 15:38 7 h 01 min
Vilnius 8:40 15:54 7 h 14 min
Europe
City Sunrise
22 Dec 2015
Sunset
22 Dec 2015
Length of the day
Edinburgh 8:42 15:40 6 h 57 min
Moscow 8:57 15:58 7 h 00 min
Berlin 8:15 15:54 7 h 39 min
Warsaw 7:43 15:25 7 h 42 min
London 8:04 15:53 7 h 49 min
Kyiv 7:56 15:56 8 h 00 min
Paris 8:41 16:56 8 h 14 min
Vienna 7:42 16:03 8 h 20 min
Budapest 7:28 15:55 8 h 26 min
Rome 7:34 16:42 9 h 07 min
Madrid 8:34 17:51 9 h 17 min
Lisbon 7:51 17:18 9 h 27 min
Athens 7:37 17:09 9 h 31 min
Africa
City Sunrise
22 Dec 2015
Sunset
22 Dec 2015
Length of the day
Cairo 6:47 16:59 10 h 12 min
Tenerife 7:53 18:13 10 h 19 min
Dakar 7:30 18:46 11 h 15 min
Addis Ababa 6:35 18:11 11 h 36 min
Nairobi 6:25 18:37 12 h 11 min
Kinshasa 5:45 18:08 12 h 22 min
Dar es Salaam 6:05 18:36 12 h 31 min
Luanda 5:46 18:24 12 h 38 min
Antananarivo 5:10 18:26 13 h 16 min
Windhoek 6:04 19:35 13 h 31 min
Johannesburg 5:12 18:59 13 h 47 min
Cape Town 5:32 19:57 14 h 25 min
Middle East
City Sunrise
22 Dec 2015
Sunset
22 Dec 2015
Length of the day
Tehran 7:10 16:55 9 h 44 min
Beirut 6:39 16:33 9 h 54 min
Baghdad 7:02 16:59 9 h 57 min
Jerusalem 6:35 16:39 10 h 04 min
Manama 6:21 16:51 10 h 30 min
Doha 6:15 16:49 10 h 34 min
Dubai 7:00 17:34 10 h 34 min
Riyadh 6:32 17:10 10 h 37 min
Muscat 6:43 17:23 10 h 41 min
Sana'a 6:25 17:38 11 h 13 min
Americas
City Sunrise
22 Dec 2015
Sunset
22 Dec 2015
Length of the day
Inuvik 0 h
Fairbanks 10:58 14:40 3 h 41 min
Nuuk 10:22 14:28 4 h 06 min
Anchorage 10:14 15:42 5 h 27 min
Edmonton 8:48 16:15 7 h 27 min
Vancouver 8:05 16:16 8 h 11 min
Seattle 7:55 16:20 8 h 25 min
Ottawa 7:39 16:22 8 h 42 min
Toronto 7:48 16:43 8 h 55 min
New York City 7:16 16:32 9 h 15 min
Washington, D.C. 7:23 16:49 9 h 26 min
Los Angeles 6:55 16:48 9 h 53 min
Dallas 7:25 17:25 9 h 59 min
Miami 7:03 17:35 10 h 31 min
Honolulu 7:04 17:55 10 h 50 min
Mexico City 7:06 18:03 10 h 57 min
Managua 6:01 17:26 11 h 24 min
Bogotá 5:59 17:50 11 h 51 min
Quito 6:08 18:16 12 h 08 min
Recife 5:00 17:35 12 h 35 min
Lima 5:41 18:31 12 h 50 min
La Paz 5:57 19:04 13 h 06 min
Rio de Janeiro 6:04 19:37 13 h 33 min
São Paulo 6:17 19:52 13 h 35 min
Porto Alegre 6:20 20:25 14 h 05 min
Santiago 6:29 20:52 14 h 22 min
Buenos Aires 5:37 20:06 14 h 28 min
Ushuaia 4:51 22:11 17 h 19 min
Asia and Oceania
City Sunrise
22 Dec 2015
Sunset
22 Dec 2015
Length of the day
Magadan 8:54 14:55 6 h 00 min
Petropavlovsk 9:36 17:10 7 h 33 min
Khabarovsk 8:48 17:07 8 h 18 min
Ulaanbaatar 8:39 17:02 8 h 22 min
Vladivostok 8:40 17:40 8 h 59 min
Beijing 7:32 16:52 9 h 20 min
Seoul 7:44 17:17 9 h 34 min
Tokyo 6:47 16:31 9 h 44 min
Shanghai 6:48 16:55 10 h 07 min
Lhasa 8:46 19:01 10 h 14 min
Delhi 7:09 17:28 10 h 19 min
Hong Kong 6:58 17:44 10 h 46 min
Manila 6:16 17:32 11 h 15 min
Bangkok 6:36 17:55 11 h 19 min
Singapore 7:01 19:04 12 h 03 min
Jakarta 5:36 18:05 12 h 28 min
Denpasar 5:58 18:36 12 h 37 min
Darwin 6:19 19:10 12 h 51 min
Papeete 5:21 18:32 13 h 10 min
Brisbane 4:49 18:42 13 h 52 min
Perth 5:07 19:22 14 h 14 min
Sydney 5:41 20:05 14 h 24 min
Auckland 5:58 20:39 14 h 41 min
Melbourne 5:54 20:42 14 h 47 min
Invercargill 5:50 21:39 15 h 48 min

Length of day increases from the equator towards the South Pole in the Southern Hemisphere in December (around the summer solstice there), but decreases towards the North Pole in the Northern Hemisphere at the time of the northern winter solstice.


Celestial equator

The celestial equator is the great circle of the imaginary celestial sphere on the same plane as the equator of Earth. This plane of reference bases the equatorial coordinate system. In other words, the celestial equator is an abstract projection of the terrestrial equator into outer space. [1] Due to Earth's axial tilt, the celestial equator is currently inclined by about 23.44° with respect to the ecliptic (the plane of Earth's orbit), but has varied from about 22.0° to 24.5° over the past 5 million years [2] due to perturbation from other planets.

An observer standing on Earth's equator visualizes the celestial equator as a semicircle passing through the zenith, the point directly overhead. As the observer moves north (or south), the celestial equator tilts towards the opposite horizon. The celestial equator is defined to be infinitely distant (since it is on the celestial sphere) thus, the ends of the semicircle always intersect the horizon due east and due west, regardless of the observer's position on Earth. At the poles, the celestial equator coincides with the astronomical horizon. At all latitudes, the celestial equator is a uniform arc or circle because the observer is only finitely far from the plane of the celestial equator, but infinitely far from the celestial equator itself. [3]

Astronomical objects near the celestial equator appear above the horizon from most places on earth, but they culminate (reach the meridian) highest near the equator. The celestial equator currently passes through these constellations: [4]

These are the most globally visible constellations.

Over thousands of years, the orientation of Earth's equator and thus the constellations the celestial equator passes through will change due to axial precession.

Celestial bodies other than Earth also have similarly defined celestial equators. [5] [6]


From the description (minor phrasing edits):

A short in-depth investigation and experimental method which looks at calculating the distance to the Sun with as few assumptions as possible. The method for calculating the distance to the Sun is described in detail in the video.

Basically June 21st the Summer Solstice is taken as a reference, and trigonometry is attempted in order to triangulate the position and distance of the Sun from the Earth. Unfortunately I failed to triangulate the Sun on this day, and the reasons will become obvious while watching the video.

The video raises some significant questions regarding the accuracy of our elevations angles, and distances to the known celestial bodies.
— conandrum74


The author followed up with a couple of additional videos in 2018:


Contents

Solar radiation closely matches a black body radiator at about 5,800 K. [1] As it passes through the atmosphere, sunlight is attenuated by scattering and absorption the more atmosphere through which it passes, the greater the attenuation.

As the sunlight travels through the atmosphere, chemicals interact with the sunlight and absorb certain wavelengths changing the amount of short-wavelength light reaching the Earth's surface. A more active component of this process is water vapor, which results in a wide variety of absorption bands at many wavelengths, while molecular nitrogen, oxygen and carbon dioxide add to this process. By the time it reaches the Earth's surface, the spectrum is strongly confined between the far infrared and near ultraviolet.

Atmospheric scattering plays a role in removing higher frequencies from direct sunlight and scattering it about the sky. [2] This is why the sky appears blue and the sun yellow — more of the higher-frequency blue light arrives at the observer via indirect scattered paths and less blue light follows the direct path, giving the sun a yellow tinge. [3] The greater the distance in the atmosphere through which the sunlight travels, the greater this effect, which is why the sun looks orange or red at dawn and sundown when the sunlight is travelling very obliquely through the atmosphere — progressively more of the blues and greens are removed from the direct rays, giving an orange or red appearance to the sun and the sky appears pink — because the blues and greens are scattered over such long paths that they are highly attenuated before arriving at the observer, resulting in characteristic pink skies at dawn and sunset.

The air mass number is thus dependent on the Sun's elevation path through the sky and therefore varies with time of day and with the passing seasons of the year, and with the latitude of the observer.

A first-order approximation for air mass is given by

The above approximation overlooks the atmosphere's finite height, and predicts an infinite air mass at the horizon. However, it is reasonably accurate for values of z up to around 75°. A number of refinements have been proposed to more accurately model the path thickness towards the horizon, such as that proposed by Kasten and Young (1989): [5]

A more comprehensive list of such models is provided in the main article Airmass, for various atmospheric models and experimental data sets. At sea level the air mass towards the horizon ( z = 90°) is approximately 38. [6]

Modelling the atmosphere as a simple spherical shell provides a reasonable approximation: [7]

These models are compared in the table below:

Estimates of airmass coefficient at sea level
z Flat Earth Kasten & Young Spherical shell
degree (A.1) (A.2) (A.3)
1.0 1.0 1.0
60° 2.0 2.0 2.0
70° 2.9 2.9 2.9
75° 3.9 3.8 3.8
80° 5.8 5.6 5.6
85° 11.5 10.3 10.6
88° 28.7 19.4 20.3
90° 37.9 37.6

This implies that for these purposes the atmosphere can be considered to be effectively concentrated into around the bottom 9 km, [8] i.e. essentially all the atmospheric effects are due to the atmospheric mass in the lower half of the Troposphere. This is a useful and simple model when considering the atmospheric effects on solar intensity.

The spectrum outside the atmosphere, approximated by the 5,800 K black body, is referred to as "AM0", meaning "zero atmospheres". Solar cells used for space power applications, like those on communications satellites are generally characterized using AM0.

The spectrum after travelling through the atmosphere to sea level with the sun directly overhead is referred to, by definition, as "AM1". This means "one atmosphere". AM1 ( z =0°) to AM1.1 ( z =25°) is a useful range for estimating performance of solar cells in equatorial and tropical regions.

Solar panels do not generally operate under exactly one atmosphere's thickness: if the sun is at an angle to the Earth's surface the effective thickness will be greater. Many of the world's major population centres, and hence solar installations and industry, across Europe, China, Japan, the United States of America and elsewhere (including northern India, southern Africa and Australia) lie in temperate latitudes. An AM number representing the spectrum at mid-latitudes is therefore much more common.

"AM1.5", 1.5 atmosphere thickness, corresponds to a solar zenith angle of z =48.2°. While the summertime AM number for mid-latitudes during the middle parts of the day is less than 1.5, higher figures apply in the morning and evening and at other times of the year. Therefore, AM1.5 is useful to represent the overall yearly average for mid-latitudes. The specific value of 1.5 has been selected in the 1970s for standardization purposes, based on an analysis of solar irradiance data in the conterminous United States. [9] Since then, the solar industry has been using AM1.5 for all standardized testing or rating of terrestrial solar cells or modules, including those used in concentrating systems. The latest AM1.5 standards pertaining to photovoltaic applications are the ASTM G-173 [10] [11] and IEC 60904, all derived from simulations obtained with the SMARTS code.

The illuminance for Daylight (this version) under A.M.1.5 is given as 109,870 lux (corresponding with the A.M. 1.5 spectrum to 1000.4 W/m 2 ).

AM38 is generally regarded as being the airmass in the horizontal direction ( z =90°) at sea level. [6] However, in practice there is a high degree of variability in the solar intensity received at angles close to the horizon as described in the next section Solar intensity.

The relative air mass is only a function of the sun's zenith angle, and therefore does not change with local elevation. Conversely, the absolute air mass, equal to the relative air mass multiplied by the local atmospheric pressure and divided by the standard (sea-level) pressure, decreases with elevation above sea level. For solar panels installed at high altitudes, e.g. in an Altiplano region, it is possible to use a lower absolute AM numbers than for the corresponding latitude at sea level: AM numbers less than 1 towards the equator, and correspondingly lower numbers than listed above for other latitudes. However, this approach is approximate and not recommended. It is best to simulate the actual spectrum based on the relative air mass (e.g., 1.5) and the actual atmospheric conditions for the specific elevation of the site under scrutiny.

Solar intensity at the collector reduces with increasing airmass coefficient, but due to the complex and variable atmospheric factors involved, not in a simple or linear fashion. For example, almost all high energy radiation is removed in the upper atmosphere (between AM0 and AM1) and so AM2 is not twice as bad as AM1. Furthermore, there is great variability in many of the factors contributing to atmospheric attenuation, [12] such as water vapor, aerosols, photochemical smog and the effects of temperature inversions. Depending on level of pollution in the air, overall attenuation can change by up to ±70% towards the horizon, greatly affecting performance particularly towards the horizon where effects of the lower layers of atmosphere are amplified manyfold.

One approximation model for solar intensity versus airmass is given by: [13] [14]

This formula fits comfortably within the mid-range of the expected pollution-based variability:

Solar intensity vs. zenith angle z and airmass coefficient AM
z AM range due to pollution [12] formula (I.1) ASTM G-173 [11]
degree W/m 2 W/m 2 W/m 2
- 0 1367 [15] 1353 1347.9 [16]
1 840 .. 1130 = 990 ± 15% 1040
23° 1.09 800 .. 1110 = 960 ± 16% [17] 1020
30° 1.15 780 .. 1100 = 940 ± 17% 1010
45° 1.41 710 .. 1060 = 880 ± 20% [17] 950
48.2° 1.5 680 .. 1050 = 870 ± 21% [17] 930 1000.4 [18]
60° 2 560 .. 970 = 770 ± 27% 840
70° 2.9 430 .. 880 = 650 ± 34% [17] 710
75° 3.8 330 .. 800 = 560 ± 41% [17] 620
80° 5.6 200 .. 660 = 430 ± 53% 470
85° 10 85 .. 480 = 280 ± 70% 270
90° 38 20

This illustrates that significant power is available at only a few degrees above the horizon. For example, when the sun is more than about 60° above the horizon ( z <30°) the solar intensity is about 1000 W/m 2 (from equation I.1 as shown in the above table), whereas when the sun is only 15° above the horizon ( z =75°) the solar intensity is still about 600 W/m 2 or 60% of its maximum level and at only 5° above the horizon still 27% of the maximum.

At higher altitudes Edit

One approximate model for intensity increase with altitude and accurate to a few kilometres above sea level is given by: [13] [19]

Alternatively, given the significant practical variabilities involved, the homogeneous spherical model could be applied to estimate AM, using:

where the normalized heights of the atmosphere and of the collector are respectively r = R E / y a t m >/y_ >> ≈ 708 (as above) and c = h / y a t m >> .

And then the above table or the appropriate equation (I.1 or I.3 or I.4 for average, polluted or clean air respectively) can be used to estimate intensity from AM in the normal way.

These approximations at I.2 and A.4 are suitable for use only to altitudes of a few kilometres above sea level, implying as they do reduction to AM0 performance levels at only around 6 and 9 km respectively. By contrast much of the attenuation of the high energy components occurs in the ozone layer - at higher altitudes around 30 km. [20] Hence these approximations are suitable only for estimating the performance of ground-based collectors.

Silicon solar cells are not very sensitive to the portions of the spectrum lost in the atmosphere. The resulting spectrum at the Earth's surface more closely matches the bandgap of silicon so silicon solar cells are more efficient at AM1 than AM0. This apparently counter-intuitive result arises simply because silicon cells can't make much use of the high energy radiation which the atmosphere filters out. As illustrated below, even though the efficiency is lower at AM0 the total output power (Pout) for a typical solar cell is still highest at AM0. Conversely, the shape of the spectrum does not significantly change with further increases in atmospheric thickness, and hence cell efficiency does not greatly change for AM numbers above 1.

Output power vs. airmass coefficient
AM Solar intensity Output power Efficiency
Pin W/m 2 Pout W/m 2 Pout / Pin
0 1350 160 12%
1 1000 150 15%
2 800 120 15%

This illustrates the more general point that given that solar energy is "free", and where available space is not a limitation, other factors such as total Pout and Pout are often more important considerations than efficiency (Pout/Pin).


Panels are usually oriented towards the south in the northern hemisphere because the sun mostly is in the southern part of the sky. The sun sometimes is in the northern part of the sky, e.g. during sunrise and sunset in spring and in summer. It only happens when the sun is relatively low so it doesn't have a huge influence on the total yield.

Here's a sun-path diagram for New Delhi (28.6°N, Northern India):

When solar panels are installed on buildings, they sometimes have to be integrated directly in the roof, so the orientation will be dictated by the architecture.

Depending on whether the electricity will be used on location, stored in batteries or sold to the grid, it might be interesting to produce less electricity per year but to produce it when it is most useful, e.g. during the afternoon for air conditioning. In that case, solar panels could be turned towards the west.

Finding the best tilt angle is a compromise :

  • too low and the panels won't be cleaned by rain.
  • too low and the panels won't produce much in winter.
  • too high and the panels won't produce much in summer. This can be desired for solar thermal collectors, because boiling water could damage the pumps.
  • too high and the rows will shadow each other.
  • too high and the panels and mount will have to withstand higher forces in windy conditions.

45° tilt seems to be too high in India for photovoltaic panels:

It could be about right for hot water production:

Finally, this angle might have been dictated by architectural choices.

Here's an average irradiance vs tilt diagram for New-Delhi (28.6°N):

In both case, the curves are pretty flat around the maximum, so the tilt angle could be chosen to be 20° or 25° in New-Delhi in order to avoid shadows. It shouldn't be much flatter than 10° in Kamuthi in order to avoid soiling.


5 Answers 5

First of all, the scheme of only single scattering seems to be an oversimplification: the light direction should be changed more than once. Can we prove that this is negligible by calculation, or is it not negligible?

This is an oversimplification, but for a clear sky at daytime it's not too wrong. See the following comparison of an atmosphere model computed with only single scattering and that including 4 orders of scattering (basically, 4 direction switches per light ray). The projection here is equirectangular, so you can see all the directions in one picture.

This becomes a much more problematic simplification when the sun is under the horizon, particularly noticeable under the belt of Venus, where the Earth's shadow is located:

Assuming the sun is in zenith, it follows from symmetry that the color of the sky in the directions having the same zenith angle must be the same, but closer to the horizon the way of the scattered light differs a lot from the rays coming near zenith — so is it possible to derive theoretically a formula which would predict the color of sky given the azimuth angle and the position of the Sun (at least in a simple geometrical setup when the Sun is in zenith)?

If we neglect non-uniformity of the atmosphere with latitude and longitude, this scenario will lead to the colors independent of azimuth. It's not quite clear what you mean by "position of the Sun" though, if you already put it into zenith. Also, if by "derive theoretically a formula" you mean some closed-form expression, then it's unlikely, given that the atmosphere is not a simple distribution of gases and aerosols. But it's possible to calculate the colors numerically, and the above pictures demonstrate this calculation done by my (work in progress) software, CalcMySky.

It's not clear why the color should not rapidly change from near blue at horizon to almost red near the Sun position: after all, the atmosphere is thicker along the lines going closer to horizon!

It shouldn't be bluer at the horizon than at the zenith. After all, you have relatively small thickness near zenith, which makes most of the light scattered to you not too extincted due to Beer-Lambert law, while near the horizon the thickness is much larger, and the light scattered into the observer, in addition to becoming bluer due to Rayleigh scattering depending on wavelength, becomes also redder due to extinction along this long path. The combination of this bluing and reddening effects gives a color closer to white (which you can see in the daytime simulation above), or reddish-orange (in the twilight).

Further, it follows from the usual explanation that blue light is partially reflected back into the space. Due to this, about half of all scattered light should be lost, so the total amount of red light coming from sun should be greater than the amount of blue light, which seems to contradict the observable reality.

Yes, the Earth indeed looks bluish from space, so the total radiation incoming from above should be redder at the ground level than at the top of atmosphere. But this is modified by the ozone layer, without which we'd have sandy color of twilight instead of blue. See for details the question Why is there a “blue hour” after the “golden hour”?

Short explanation is this. Red light comes directly from Sun almost un-scattered or scattered to small degree. And when blue light enters atmosphere it gets scattered by air molecules a lot in each direction, thus according to Huygens-Fresnel principle making each point in atmosphere as a secondary sources of blue light. These blue light sources adds-up along the direction of view, which in the end increases intensity of blue waves, compared to the red ones which reaches us only directly from Sun. So speaking by analogy, Earth atmosphere acts as a some sort of optical lens, focusing blue light towards direction of view. Schematics :

Of course this is a bit oversimplified, because blue light is scattered in ALL directions across the air. You can imagine a thousands of blue light bulbs turned-on in the sky. Maybe this would be better analogy, because each point in air acts as an ambient light source for blue waves.

Here are some answers, albeit back-of-the-envelope.

At a reasonably good site with a low amount of atmospheric aerosols and dust, the "extinction" is about 0.3 magnitudes per airmass at 400 nm, in astronomers units, compared to about 0.1 mag/airmass at 550 nm and about 0.04 mag/airmass at 700 nm.

What this means is that if light travels through the atmosphere at zenith, then a factor of $10^<-0.3/2.5>=0.758$ of blue light makes it to the ground, compared with a factor of 0.912 for green light and 0.963 for red light. Most of the remainder will be Rayleigh scattered (although there is some component from atmospheric absorption and scattering by aerosols in these numbers).

From this you can see that multiple scattering cannot be negligible for blue light, because at least a quarter of it is scattered by just travelling through the minuimum possible amount of air between space and the observer.

The next point: yes, it is possible to calculate the spectrum of the daylight sky given the appropriate atmospheric conditions (the run of density with height) and the aerosol content (the latter is important because the dependence of the scattering cross-section on wavelength is much more uniform than for Rayleigh scattering). Is there a simple formula - no. An example of where detailed calculations have been set out in great detail can be found here.

Then, why doesn't the sky become red near the Sun? Why would it? Red light is not effectively scattered, so red light that is emitted by the Sun does not get scattered towards the observer. On the other hand if you look directly towards the Sun (please do not do this) then blue light is preferentially scattered out of the direct sunlight, and indeed the Sun is "redder" than it would appear from space (plot below).

The only source of illumination from directions that are not towards the Sun are from scattered light. If we ignored multiple scattering and aerosols then that scattered light would have a spectrum that was proportional to the illuminating light multiplied by the Rayleigh scattering cross-section. The illuminating light does get progressively redder as the zenith angle increases (because the illuminating beam has to travel further and deeper through the atmosphere), so you would expect a whiter colour near the horizon, transitioning to a deeper blue higher above the horizon. However, this is not a very strong effect because only a quarter of blue light is scattered per airmass (and the eye has a pseudo-logarithmic response to spectral flux). Note though that in practice aerosols are not absent and that scattering from aerosols and particulates has some concentration in the forward scattering direction, which messes up this simple prediction, by making the sky whiter near to the Sun. Multiple scatterings also make the sky whiter near to the horizon because some of the blue light coming from that direction is then scattered out of the line of sight.

This is perfectly illustrated by a calculated sky image that shows the separate contribution of Rayleigh and aerosol (Mie) scattering (taken from this website, which does quantitative calculations, but which does not take account of multiple scattering). The sky is quite white near the horizon, then becomes a deeper blue at higher angles and is finally quite white again near to the Sun because of Mie scattering.


Disclaimer: The following material is being kept online for archival purposes.

A short but important section, deriving centripetal acceleration for motion at constant speed around a circle.

Part of a high school course on astronomy, Newtonian mechanics and spaceflight
by David P. Stern

This lesson plan supplements: "Motion in a Circle," section #19 http://www.phy6.org/stargaze/Scircul.htm

"From Stargazers to Starships" home page: . stargaze/Sintro.htm
Lesson plan home page and index: . stargaze/Lintro.htm


Goals : The student will learn

    About uniform circular motion, and the relation of its frequency of N revolutions/sec with the peripheral velocity v and with the rotation period T.

Terms: uniform circular motion, frequency, peripheral velocity, centripetal acceleration and force.

    (Illustrate the following by a drawing on the board, to which details are added as the discussion progresses.)
    [If a students says "because of the centrifugal (or centripetal) force," say "that is just a word, a technical term. What is actually happening?"]

The string does not allow it to do so, but pulls it back towards your hand, to keep it in its circle. We will show today that motion in a circle can be viewed as the combination of two motions taking place at the same time--like the motion of the airplane, flying in a cross-wind (Section #14).

One motion is the continuation of the existing velocity along a straight line (show on the drawing)--the way the weight would move by Newton's first law, if no outside force acted on it.

The other is a motion towards the center of the circle (draw it, too), returning the weight to its circular path. (Figuratively returning it in actuality, both motions are simultaneous and the weight never leaves the circle.) That second motion, it will be shown, is an accelerated one.

Now for the details. (Continue on the board with the derivations, while the students copy into their notebooks.)

Guiding questions and additional tidbits

(Suggested answers, brackets for comments by the teacher or "optional")

--Why is motion at a constant speed around a circle an accelerated motion?

    The speed, the magnitude of the velocity, is indeed constant. But the direction changes all the time. Velocity is a vector quantity, and any change of its direction also involves acceleration.

--When you whirl a stone at the end of a string and let go, how does the stone move?

-- Why doesn't the released stone move outwards, in the direction in which it pulled?

    Because from the moment when it is released, no forces exist any more in the direction of the string. The stone strains against the centripetal force only as long as it moves in a circle.

--Riding over a dirty road, the wheels of your car acquire a coating of mud, which soon flies off again. How does it fly off?

    Along a line tangent to the wheel. If, as is likely, the mud flies off soon after the wheel has picked it up from the road, it will fly backwards and upwards from the rim of the wheel--in the direction of the mudguards which truckers hang behind their wheels to intercept it.
    [Draw schematic on the blackboard].

--What is the acceleration of a stone rotating with speed v around a circle of radius r?

--In the derivation of the formula a = v 2 /r we neglected a small quantity x. Does that mean that the formula is only approximate, not exact?

--If a stone makes N circuits per second around the center, what is its rotation period, T?

    T = 1/N. The total time spent in those N circuits is found by multiplying the number of circuits (=N) by the length of each one (=T). But that time, by definition, is one second, so NT = 1 from which T = 1/N.

--If a stone makes N circuits per second around a circle of radius R, what is its centripetal acceleration?

    The distance covered in each revolution is 2 πR The distance covered in one second is (2 πR)N--which by definition is also its peripheral speed. Hence

Astronauts are subjected to large accelerations during launch and re-entry. The forces associated with such accelerations are often called "g forces" because they are measured in gravities, i.e. the acceleration is measured in units of g, the acceleration due to gravity for which we will use the approximate value 10 m/sec 2 .

In another lesson in "'Stargazers" it was noted that the V2 rocket of World War II started with an acceleration of 1 g and ended at "burn-out", with its mass (mostly fuel) greatly reduced, at about 7 g. The space shuttle (I believe) pulls 3g before burnout, which is quite uncomfortable, even for someone lying flat on the back on a contoured surface. Re-entry has comparable (negative) accelerations.

To get astronaut used to taking such forces, they are whirled around during their ground training in a centrifuge, inside a small cabin mounted at the end of a horizontally rotating arm. (Anyone seen such centrifuges on TV?) They are a bit like some amusement park rides, but can create greater stresses, and have TV cameras that monitor the rider.

    10 = (2πN) 2 R = (6.28) 2 N 2 (6) = 236.63 N 2
    N 2 = 0.04226
    N = 0.206 rev/sec (larger than N 2 , of course, since N --If you double the rotation speed, so that each revolution only takes 2.5 seconds, what will the g-force due to the rotation be?


I understand from the quote that those were the measurements at the time of how the poles were magnetically positioning themselves as configuration. Has this configuration remained as an equatorial placement of the N/S poles still lying E/W? Are there further documents indicating the precise current positioning as actual measurements taken that I can in fact reference in the interests of scientific accuracy?

Answered by Caty Pilachowski and Kevin Croxall

We were able to answer this question with the help of solar astronomers John Leibacher and Jack Harvey at the National Solar Observatory. The magnetic field of the sun is quite complex and variable. But the more distant one gets from the surface, the simpler it appears. Far away it resembles the field from a bar magnet. This simple approximation is used for comparison with solar wind measurements and other features in the distant heliosphere, where the ESA spacecraft Ulysses makes its measurements. Because the fields at the surface of the Sun are constantly changing, the strength and tilt angle of the hypothetical bar magnet used to approximate the real field also constantly change.

At the times of weak solar activity, such as now, the poles of the hypothetical bar are oriented N-S, coincident with the Sun's rotation axis, and the strength of this hypothetical bar magnet is also at a maximum. As solar activity increases, the patterns of real magnetic fields at the surface of the Sun become stronger and more complex so that the hypothetical bar magnet field weakens and no longer describes the solar magnetic field at large distances from the Sun. Eventually, the solar magnetic field reorganizes itself back into a bar magnet-like state at large distances from the Sun, but its polarity has flipped, north to south and south to north. Of course, this behavior should not be thought of as a giant magnet inside the Sun actually changing strength and flipping 180 degrees, but rather is a mathematical approximation to the changing patterns on the surface of the Sun.

To answer the question directly, the period of time when the bar magnet approximation was lying in the equator, as reported in the press release that prompted this question, was brief, and the overall magnetic field re-established itself quickly into the expected, strong NS orientation. Thus, the Sun is behaving just as the solar astronomers think it should be. [top ]


Watch the video: Sun-Earth angles. Declination,Altitude, Longitude,Amizuth Angle,Hour Angle,Zenith Angle REE GTU (September 2021).