Talk:Position of the Sun

Reason for the page

There seems to be a number of bits and pieces of "how to calculate the Sun's position" scattered around many different articles. This article is a home for them, so we can stop cluttering up every other astronomy article with this information. Tfr000 (talk) 17:19, 19 April 2012 (UTC)[reply]

Split a section from Declination to here

Text and/or other creative content from Declination was copied or moved into Position_of_the_Sun. The former page's history now serves to provide attribution for that content in the latter page, and it must not be deleted as long as the latter page exists.

It was more appropriate for this article. Tfr000 (talk) 14:53, 30 May 2012 (UTC)[reply]

Split a section from Horizontal coordinate system to here

Text and/or other creative content from Horizontal_coordinate_system was copied or moved into Position_of_the_Sun. The former page's history now serves to provide attribution for that content in the latter page, and it must not be deleted as long as the latter page exists.

I am putting this here for now. We already have a section very similar to this in the article. Tfr000 (talk) 16:20, 2 July 2012 (UTC)[reply]

The position of the Sun

There are several ways to compute the apparent position of the Sun in horizontal coordinates.

Complete and accurate algorithms to obtain precise values can be found in Jean Meeus's book Astronomical Algorithms.

Instead a simple approximate algorithm is the following:

Given:

the date of the year and the time of the day
the observer's latitude, longitude and time zone

You have to compute:

The Sun declination of the corresponding day of the year, which is given by the following formula which has less than 2 degrees of error:

$\delta =-23.45^{\circ }\cdot \cos \left({\frac {360^{\circ }}{365}}\cdot \left(N+10\right)\right)$

where $N$ is the number of days spent since January 1.

The true hour angle that is the angle which the earth should rotate to take the observer's location directly under the sun.
- Let hh:mm be the time the observer reads on the clock.
- Merge the hours and the minutes in one variable $T = hh + mm /60$ measured in hours.
- hh:mm is the official time of the time zone, but it is different from the true local time of the observer's location. $T$ has to be corrected adding the quantity $+ (Longitude/15 - Time Zone)$ , which is measured in hours and represents the difference of time between the true local time of the observer's location and the official time of the time zone.
- If it is summer and Daylight Saving Time is used, you have to subtract one hour in order to get Standard Time.
- The value of the Equation of Time in that day has to be added. Since $T$ is measured in hours, the Equation of Time must be divided by 60 before being added.
- The hour angle can be now computed. In fact the angle which the earth should rotate to take the observer's location directly under the sun is given by the following expression: $T$ . Since $T$ is measured in hours and the speed of rotation of the earth 15 degrees per hour, $H$ is measured in degrees. If you need $H$ measured in radians you just have to multiply by the factor 2π/360.

Use the Transformation of Coordinates to compute the apparent position of the Sun in horizontal coordinates.

The above formula for the Sun's decination is horribly inaccurate compared with others that are already in the article. DOwenWilliams (talk) 18:35, 2 July 2012 (UTC)[reply]

Wrong translations?

The article contains the following translations of its subject:

ca:Declinació solar

es:Declinación solar

Obviously, these mean "declination", not "position". I've read the Spanish (es) one. (The Catalan is very similar.) It definitely describes the Sun's declination.

DOwenWilliams (talk) 16:20, 24 December 2012 (UTC)[reply]

At which latitude the (centre of the) sun reaches zenith - Sun Declination

The article descibes where on the sky the sun can be found. But sun declination is a main part of this, which gives the excact latitude, at which the centre of the sun (somewhere at this latitude) currently stands in zenith position. This is always at one (and one only) signular point of the surface of the Earth. A singular point is none-dimentional and not an area. But a point that is located at a certain latitude between the tropics. Sun declination borders the equator at the equinoxes (which are excact moments, not time-periods), and tropic of cancer at northern hemisphere summer solstice (which also is a moment) , and tropic of capricorn at northern hemishpere winter solstice. The sun declination follows a sinewave (and a sinewave is a displaced cosinevawe, and reversed) whith an amplitude that eaquals the angle between the ecliptic and the equator - approx. 23.5 degrees. I belive this even is the definition of eqinoxes and solstices (?). The orbit of the Earth around the sun isn't circular, hence local noon differs between months. But - if I'm right - the sun declination isn't affected by the annual elliptic path of the Earth !? In any case this part of "position of the sun", the sun declination ought to have an article of its own, I think. It tells us at which latitude the (centre of the) sun reaches zenith. Now "sun declination" points to this page, and the issue isn't delt with. (I do not suggest that this page is wrong, just that sun declination itself is importaint enough for an article of its own) Boeing720 (talk) 20:34, 29 August 2013 (UTC)[reply]

The solar declination does not follow a sine wave. Imagine a planet that moves in a circular orbit, but has an axial tilt of 90 degrees, so its rotation axis is in its orbital plane. At some point on its orbit, the Sun will be overhead at its north pole. From that point, the solar declination will decrease linearly with time, from +90° to -90° when the Sun is overhead at the south pole. Then it will increase linearly back to +90°, and so on. In short, the solar declination will follow a sawtooth wave. If the axial tilt decreases, the shape of the wave becomes less like a sawtooth and more like a sine wave, but even when the tilt equals that of the real Earth, the wave will not be exactly a sine wave. The maxima and minima will be more pointed than those of a sine wave. Also, since the Earth moves in an elliptical orbit, faster near perihelion, in January, than near aphelion, in July, the wave will be distorted in another way, being more stretched out in July than January. So the solar declination graph is a lot more complicated than a simple sine wave. DOwenWilliams (talk) 22:18, 28 January 2016 (UTC)[reply]

Take a look at this:

http://green-life-innovators.org/tiki-index.php?page=The+Latitude+and+Longitude+of+the+Sun+by+David+Williams

It's a piece I wrote years ago to explain this stuff.

I think we should delete the degrees signs you put in. They're just confusing. But maybe we should add a line to say that, if a calculator is used to evaluate the expressions, it should be set to treat angles in degrees.

DOwenWilliams (talk) 01:43, 30 January 2016 (UTC)[reply]

Wrong formules

\delta _{\odot }=\arcsin \left[\sin \left(-23.44^{\circ }\right)\cdot \cos \left({\frac {360^{\circ }}{365.24}}\left(N+10\right)+{\frac {360^{\circ }}{\pi }}\cdot 0.0167\sin \left({\frac {360^{\circ }}{365.24}}\left(N-2\right)\right)\right)\right]

which can be simplified by evaluating constants to:

\delta _{\odot }=-\arcsin \left[0.39779\cos \left(0.98565\left(N+10\right)+1.914\sin \left(0.98565\left(N-2\right)\right)\right)\right]

is wrong, first of all, one is -arcsin and the other is +arcsin? and in adittion I have tried to use both and I am getting numbers like 0.0003º when was expected around -17º.

--79.109.245.126 (talk) 18:11, 12 November 2014 (UTC)[reply]

The constants in the cosine expression of the second formula evaluate to degrees. I added the added degree symbols in the article to clarify. --Jbergquist (talk) 18:56, 13 January 2016 (UTC)[reply]

I checked the formulas against the Sun's positions using the MICA , the US Naval Observatory's computer almanac, and they appear to work ok. This might help clarify things a little. The formulas are simpler if the angles are in radians.

{\begin{aligned}&{\text{Jan 1, 2000 12:00 UT}}\\&{{\lambda }_{E}}=100.466{\text{ }}\!\!{}^{\circ }\!\!{\text{ }}\quad \quad {{\lambda }_{S}}={{\lambda }_{E}}+180{\text{ }}\!\!{}^{\circ }\!\!{\text{ }}=280.466{\text{ }}\!\!{}^{\circ }\!\!{\text{ }}\\&\\&{\text{Jan 1, 2000 00:00 UT}}\\&{{{\lambda }'}_{S}}={{\lambda }_{S}}-0.5\cdot n=279.974{\text{ }}\!\!{}^{\circ }\!\!{\text{ }}=4.88646\ radians\\&{\text{n = mean daily motion of the Sun along the ecliptic}}\\&\\&\lambda ={{{\lambda }'}_{S}}+n\cdot \Delta t+2\cdot e\cdot \sin \left[n\cdot \left(\Delta t-2.010\right)\right]\\&\\&\sin \left(\delta \right)=\sin \left(\iota \right)\sin \left(\lambda \right)\\&\quad \quad \ \ =\sin \left(\iota \right)\sin \left({{{\lambda }'}_{S}}+n\cdot \Delta t+2\cdot e\cdot \sin \left[n\cdot \left(\Delta t-2.010\right)\right]\right)\\&\quad \quad \ \ =-\sin \left(\iota \right)\cos \left({{{\lambda }'}_{S}}-{\frac {3\pi }{2}}+n\cdot \Delta t+2\cdot e\cdot \sin \left[n\cdot \left(\Delta t-2.010\right)\right]\right)\\&\quad \quad \ \ \approx -\sin \left(\iota \right)\cos \left\{n\cdot \left(\Delta t-10\right)+2\cdot e\cdot \sin \left[n\cdot \left(\Delta t-2\right)\right]\right\}\\\end{aligned}}

The J2000.0 value for the ecliptic longitude of the Earth λ_E is from the Explanatory Supplement to the Astronomical Almanac. Δt is the elapsed time from the start of the year 2000. The maximum error found for the longitude, λ, was about 1.5 minutes of arc. --Jbergquist (talk) 21:21, 28 January 2016 (UTC)[reply]

The mean motion, n, is strictly speaking an angular velocity so one would expect it to be expressed in ° per unit time or radians per unit time but the product nΔt is just an angle and the article expresses product as the mean daily angular change and the number of days and fractions thereof if necessary. The correction for aberration is about 20 arc seconds or less and has a sinusoidal dependence on the longitude and the light time from the Sun also has an annual variation. The formulas for longitude and declination don't appear to be corrected for changes in the position of the vernal equinox. These corrections aren't necessary if one is working at the level of 1 minute of arc accuracy. --Jbergquist (talk) 17:44, 29 January 2016 (UTC)[reply]