Talk:Declination

Sun's declination

In the Southern Hemisphere, because our seasons are switched is the formula for the declenation of the sun now:

d = +23.45 * cos(360/365 * (N +10)) ?

Is the Sun Dec now -23°27' at the Southern hemisphere summer solstice and +23°27' at the Southern hemisphere winter solstice?\ - 08:37, 4 March 2006 219.88.199.135

no, it's just a matter between the sun and the earth, dec is + when the sun is overhead in the northern hemisphere (northern summer, southern winter) and dec is - when the sun is overhead in the southern hemispere (southern summer, northern winter). hope that helps...

- FourBlades 19:48, 26 July 2006 (UTC)[reply]

Hello, The "more precise formula" requires one to know the fractional year. For this quantity, N is required. Is this really the same as N in the previous paragraph? If so, then this formula appears to be much less accurate then the linked table. Can anybody add a comment on how accurate that formula is? Thank you,

Roger Jeurissen. 19:09, 30 May 2007 86.82.146.207

The Spencer formula is more accurate than the top formula, but indeed is somewhat less accurate than the linked table. I wouldn't say "much less accurate" though. For example, on Oct 15 of a 365 day year, the first formula gives a declination of -9°30'; the Spencer formula gives a declination of -8°11'; and the table gives a declination of -8°18'. I did these calculations in Excel 2003, but I did notice that the the first formula gives a value opposite in sign to that given by the Spencer formula or the linked table. --Rpresser 20:38, 31 October 2008 (UTC)[reply]

The solar declination does not follow a sine wave. It can't be calculated by plugging the time of year into a sine (or cosine) function.

Imagine an extreme case. Suppose that the earth's orbit is exactly circular, so the planet goes around the sun at a uniform speed, and that the tilt of its axis ("obliquity") is 90 degrees. At some time of the year the sun would be exactly overhead at the South Pole, so its declination would be -90 degrees. Six months later, the sun would be overhead at the North Pole, with a declination of +90 degrees. During the six months, the sub-solar point would travel northward at a uniform speed, so the solar declination would move *linearly* from -90 to +90 degrees. A graph of the solar declination against time, for that period, would be a straight line. After passing over the North Pole, the sun would go south, also in a linear fashion. So, instead of a sine wave, the overall graph of the solar declination would be a "saw-tooth".

If the axial tilt becomes less than 90 degrees, the graph becomes smoother, approaching a sine wave, but never quite reaching it. The maxima and minima are always "pointier" than a true sine wave.

Further complexity is added because the real earth is closer to the sun, and moving faster around its orbit, in early January than at other times of the year. It moves slowest in early July. The result is to stretch out the variation of solar declination with time in July, and compress it in January. The declination graph is even pointier when the declination is near its southernmost value, and less pointy in the northern summer.

Using a simple sine wave to calculate the solar declination can lead to errors of a degree or two. That's several times greater than the angular diameter of the sun.

Of course, the computer routine I've posted below is a whole lot better. (On Oct 15, it calculates the solar declination to be -8.22 degrees, or -8°13′.)

DOwenWilliams (talk) 02:49, 31 August 2010 (UTC) David Williams[reply]

Dubious

The following statement is dubious:

The Moon also has an annual cycle, with maximum declination at northern hemisphere midwinter and minimum at midsummer.

This statement is only accurate for the full moon. At other times of the lunar month, the Moon reaches its highest declination at different times of the year.
The statement should be reworded in a more globally-neutral manner. The full moon would appear highest in the sky closest to the winter solstice and at its lowest closest to the summer solstice - and this statement is true regardless of hemisphere, or which winter or summer solstice is being considered here. From an equatorial perspective, the most northerly full moon is the one closest to the southern solstice, and vice versa. -- B.D.Mills (T, C) 04:19, 19 June 2008 (UTC)[reply]

When the moon runs the highest and lowest:

Around Mid-winter, the moon runs the highest when in its waxing gibbous phase, and the lowest when in waning crescent.

Around the spring equinox, the moon runs the highest when its around first quarter (waxing half-moon), and the lowest when it's around last quarter (waning half moon).

Around mid-spring, the moon runs the highest when in its waxing crescent phase, and the lowest when in waning gibbous.

Around the summer solstice, the moon runs the highest when it's around the new moon, and the lowest when around the full moon.

Around mid-summer, the moon runs the highest when in its waning crescent phase, and the lowest when in waxing gibbous.

Around the fall equinox, the moon runs the highest when its around last quarter, and the lowest when it's around first quarter.

Around mid-fall, the moon runs the highest when in its waning gibbous phase, and the lowest when in waxing crescent.

Around the winter solstice, the moon runs the highest when it's around the full moon, and the lowest when around the new moon.

There is a bit of variation, regarding the exact points that the moon runs the highest and lowest each time the moon orbits Earth. This is because the moon's orbit deviates slightly from the ecliptic. However, since that deviation is less than 1/4 of Earth's axis tilt, Earth's tilt is strongly the dominating factor. Therefore, the above generalizations are always at least close to actuality.

I'm not sure if there's a word that is used to identify the moon's highest and lowest declinations for each time it orbits Earth. However, I think that "lunastice" would be a good word, because it replaces the "sol" (Latin for sun) in "solstice" with "luna" (Latin for moon). I also think that it would be better to use the words "high" and "low", rather than "summer" or "winter", with "lunastice", because the moon's declination, of course, doesn't dictate the seasons like the sun's does.

DKStuntz (talk) 00:45, 7 June 2009 (UTC)[reply]

Several sources (the Oxford English Dictionary and Ruggles, Astronomy in Prehistoric Britain and Ireland, p. 36) give lunistice, rather than lunastice, for the monthly lunar extremes. SteveMcCluskey (talk) 15:03, 27 July 2013 (UTC)[reply]

Moon's declination

"Declination of the Moon is computed by adding Sun's declination (which is called Declination of Place while computing declination of other planets and Moon) to Moon's latitude."

It most certainly is not, for 2 reasons. 1) Latitude is measured perpendicular to the ecliptic, while the declination is measured perpendicular to the equator. Thus they cannot simply be added. 2) The Moon may be far from the Sun and thus its declination may be much farther from the Sun's that the latitude difference. Moreover, as I write this 20090807 04:25:57 (UT), the Sun's declination is 16°23 N (of the equator), the Moon's Latitude is 2°24 N (of the ecliptic), and the Moon's declination is 10°05 S (of the equator). There is no simple formula which relates these 3 quantities. You need to know the longitude of the Moon, the obliquity of the ecliptic, and the Moon's latitude in order to determine the Moon's declination.

The above is more properly a method for approximating the MAXIMUM declination, or the declination of the Moon's plane rather than the Moon itself. At the moment that the Sun reaches the tropic of Cancer, the equatorial and ecliptic meridian are the same, and thus latitude of the Moon can be added the declination of the Sun. MistySpock (talk) 04:38, 7 August 2009 (UTC)[reply]

Removed entire section on the Moon, added link to Lunar_standstill MistySpock (talk) 05:25, 23 August 2009 (UTC)[reply]

Declination as latitude

I'm hoping an astronomer can clarify something I found confusing on this page. According to the text of the article, declination corresponds to terrestrial latitude; but the accompanying diagram indicates that lines of declination are not parallel, but rather meet at the celestial poles, which would make them longitudinal and not latitudinal. Is the text wrong, is the diagram wrong, or is there something about the way the earth is projected onto the sky that somehow makes longitude and latitude "switch?"

98.208.141.126 (talk) 06:52, 6 December 2009 (UTC)[reply]

IANAA. The diagram is showing declination measured along a line through the celestial pole, as terrestrial latitude is angular displacement from the equator measured perpendicular to circles of latitude. The orange text pertains to the orange arrow. The gray arc of celestial longitude is for reference. -Ac44ck (talk) 08:23, 6 December 2009 (UTC)[reply]

What does φ (phi) have to do with anything

I don't understand what φ (phi) is referring to in this article. What is phi? —Preceding unsigned comment added by 70.68.138.99 (talk) 05:44, 13 January 2010 (UTC)[reply]

Computer code to calculate Equation of Time and Solar Declination

The following subroutine, here written in QBasic but easily translatable to other languages, calculates the Equation of Time and the Solar Declination on any day of the year. It is quite accurate. Its Root-Mean-Square error for the Equation of Time is only 3.7 seconds. This is a small fraction of the time, about 120 seconds, that the sun takes to cross its own diameter as it moves across the sky. The RMS error for the Declination is 4.7 arc-minutes, which is a small fraction of the angular diameter of the sun as seen from the earth, which is about 30 arc-minutes.

I originally posted this on the main pages entitled "Equation of time" and "Declination", here on Wikipedia. However, it was removed by editors because I could not provide a citation to a previous publication, other than to ones I had written myself. I could, and did, refer to a computer program including this routine, which demonstrates its accuracy. But apparently direct observation does not satisfy The Rules.

Anyone who wants further information should follow the following link: [Link] The program that includes the routine is ETIMSDEC. There are instructions how to run it. The article titled "The Latitude and Longitude of the Sun", which I wrote several years ago, describes the astronomical logic behind the routine.

DOwenWilliams (talk) 21:29, 29 August 2010 (UTC) David Williams[reply]

FUNCTION ET.Dec (D, F%) STATIC 
  ' Calculates equation of time, in minutes, or solar declination, 
  ' in degrees, on day number D of year. (D = 0 on January 1.) 
  ' F% selects function: True (non-zero) for Equation of Time, 
  ' False (zero) for Declination. 
  ' STATIC means variables are preserved between calls of function 
 
  IF PI = 0 THEN ' first call, initialize constants 
 
    PI = 4 * ATN(1) 
    W = 2 * PI / 365 ' earth's mean orbital angular speed in radians/day 
    DR = 180 / PI ' degree/radian factor 
    C = -23.45 / DR ' reverse angle of earth's axial tilt in radians 
    ST = SIN(C) ' sine of reverse tilt 
    CT = COS(C) ' cosine of reverse tilt 
    E2 = 2 * .0167 ' twice earth's orbital eccentricity 
    SP = 12 * W ' 12 days from December solstice to perihelion 
    D1 = -1 ' holds last D. Saves time if D repeated for both functions 
 
  END IF 
 
  IF D <> D1 THEN ' new value of D 
    A = W * (D + 10) ' Solstice 10 days before Jan 1 
    B = A + E2 * SIN(A - SP) 
    D1 = D 
  END IF 
 
  IF F% THEN ' equation of time calculation 
    C = (A - ATN(TAN(B) / CT)) / PI 
    ET.Dec = 720 * (C - INT(C + .5)) ' this is value of equation of time
    ' in 720 minutes, earth rotates PI radians relative to sun 
 
  ELSE ' declination calculation 
    C = ST * COS(B) 
    ET.Dec = ATN(C / SQR(1 - C * C)) * DR ' this is value of declination
    ' arcsine of C in degrees. ASN not directly available in QBasic 
 
  END IF 
 
END FUNCTION

The consensus formed at Wikipedia:Reliable sources/Noticeboard/Archive 74#Source for QBasic program indicated that there is no source for this program that meets the requirements of WP:IRS. Doubt was also expressed about whether any computer source code should be added.

I believe the need for computer source has greatly diminished since QBasic was in its heyday, because many websites will provide astronomical data. A more plausible requirement for source code would be to build into a larger program, perhaps a program that aims solar panels. But before the source code could be used in such a way, it would have to be translated to a more modern programming language. Since the calculation is rather straightforward, I doubt the need to provide source code at all; the mathematical description in the article or in reliable sources such as the Astronomical Almanac could easily be implemented in the programming language of choice, without the need to put on a dust mask and search for the QBasic programming manual. Jc3s5h (talk) 00:40, 30 August 2010 (UTC)[reply]

Hmmm... The link I gave above leads to a program that uses this code and can be used to point anything at the sun. It is available in Basic, Perl, and C. These last two are still enjoying their 15 minutes of fame. Actually, the differences beween the Basic and C versions (I haven't looked at the Perl one) are quite minor. Translating from one to the other is very simple.

Incidentally, programmable "controllers" - little computers that are designed to control machinery such as solar trackers - are still quite often programmed in Basic. For this kind of purpose, the language is far from obsolete.

The philosophical question as to whether *any* program source code should be in Wikipedia is one which, perhaps, should be put to a vote of Wikipedians.

DOwenWilliams (talk) 14:03, 30 August 2010 (UTC) David Williams[reply]

The latitude and longitude of the sun

I wrote the following article at about the same time as the above computer code. Basically, the article describes how the code was derived, and how it works.

Note that the calculations of the Equation of Time and the Solar Declination are done from first principles. The only data required are the earth's orbital parameters, the orbital period, eccentricity and the date of perihelion, and the earth's rotational parameters, the rotational period, the tilt of its axis ("obliquity") and the date of the December solstice. No use is made of derivative quantities such as the solar "anomaly".

Despite the unconventional approach and the simplicity of the calculations, they are quite accurate.

Of course, Wikipedia generally hates unconventionality...

DOwenWilliams (talk) 23:02, 31 August 2010 (UTC) David Williams[reply]

For solar-energy purposes, it is often important to know, and to be able to predict, where the sun is in the sky at any time. As seen from most latitudes (except in the arctic and antarctic), it rises every day in the eastern part of the sky, and sets in the western part. But, of course, it does some more complicated motions with the passing seasons.

Everyone knows that the sun is further north in the sky in June than it is in December. This variation of the solar latitude, or the sun's "declination" as astronomers like to call it, is responsible for the seasons. Generally, the weather is warmer when the noon-day sun is higher in the sky than when it is lower.

It is less widely known that the sun's longitude, relative to its mean position, also varies with the time of year. The effects of this variation are not as obvious as the effects of the variation of solar declination, but they are significant in some circumstances. For example, there is a difference between "solar time", as shown by a sundial, and "mean time", as shown by a clock. We generally regulate our lives by clocks, so sundials appear to run too fast or too slowly at different times of the year. In early November, a sundial is more than a quarter of an hour "fast", compared with a clock that shows local mean time. By mid-February, it is nearly a quarter of an hour "slow". Not only do sundials lose time during this period, other sun-related events such as sunrise and sunset show the same effect. This is why, for those of us in north-temperate latitudes. the date of the earliest sunset of the winter is several weeks before the date of the latest sunrise. The evenings are already beginning to become brighter when the mornings are still becoming more gloomy.

There are two reasons for this. One is that the earth's orbit around the sun is not exactly circular. Around the time of perihelion, which happens on or about January 3 each year, the earth is closest to the sun and moves slightly faster around its orbit than it does at other times of the year. At aphelion, in early July, the movement is slowest. The speed at which the sun appears to move across the sky each day is essentially the difference between the earth's angular speed of rotation and the angular speed of its orbital motion. When the orbital speed is fastest, at perihelion, the apparent speed of the sun is slowest, so sundials lose time.

The other reason is related to the tilt of the earth's axis. If we imagine that lines of latitude and longitude are marked on the earth's surface, and that its rotation is then stopped, relative to the fixed stars, the path of the sub-solar point, where the sun is vertically overhead, moves around the tilted circle of the ecliptic as the earth moves around its orbit. At the equinoxes, the sub-solar point is at the equator, where the lines of longitude are spaced furthest apart, and is moving at an angle to these lines. At the solstices, the sub-solar point is moving perpendicularly across the lines of longitude, which are closer together than they are at the equator. So the longitude of the sub-solar point changes most rapidly at the solstices and most slowly at the equinoxes. The result when this effect is superimposed on the rotation of the real earth is to make sundials lose time, relative to clocks, near the solstices, and gain time near the equinoxes.

The two reasons act in the same direction during December and January, which are close to both a solstice and perihelion. This is why sundials lose time most rapidly during these months. In June and July, the two effects act in opposite directions, so sundials do not lose time as rapidly then.

The difference between solar time and local mean time is called the Equation of Time. Unfortunately, there is no universally accepted convention as to its sign. The usage I will follow is to regard the Equation of Time as the correction that must be subtracted from the reading of a sundial to get mean time, so it is positive when the sundial is ahead of the clock and vice versa. This is the normal convention. Some writers, however, use the opposite signs. Of course, it does not matter which sign is used, so long as the meaning is clear. In fact, some authors avoid the ambiguity by not using signs at all, but by writing words such as "sundial fast" and "sundial slow" instead.

Understanding the causes of the Equation of Time in a qualitative way is simple. Calculating it quantitatively is a bit more complex.

In this article, I will outline a reasonably accurate way to calculate the sun's longitude and hence the Equation of Time and also the declination (latitude) of the sun on any date in a year. For simplicity, I will ignore all the slow motions of the earth: precessions, nutations, and so on. Over a long period of time, these simplifying approximations will certainly have significant effects. But for this present year, or any year fairly close to it, the results of the calculation should be (and are) quite accurate.

In order to understand the logic behind the calculation, first imagine a planet in a perfectly circular orbit, with its axis perpendicular to the orbital plane, but which is not rotating with respect to the fixed stars. The sub-solar point would move around the equator at a constant speed, taking one year to make a circuit. We can imagine some sort of vehicle that moves so as to remain at the sub-solar point at all times. This vehicle would change its longitude at a uniform speed, so its longitude is directly proportional to time. Let us call this vehicle the "clock car", and imagine that it continues to move around the equator at the same uniform speed even when we make the planet's motions become like those of the earth.

Now let us imagine that the planet's orbit becomes slightly elliptical, but the orbital period remains the same. The angular orbital velocity is no longer quite constant, so the sub-solar point sometimes advances ahead of the clock car, and sometimes falls behind it. The position of the clock car shows "mean time", and the position of the sub-solar point shows "solar time". If the eccentricity of the orbit is small, as is true in the case of the earth which has an orbital eccentricity of only 1.67 percent, the oscillation of the sub-solar point with respect to the clock car is very close to sinusoidal. The amplitude of the variation of the planet-sun distance equals the eccentricity times the mean distance, but the amplitude of the variation of the angular velocity equals twice the eccentricity times the mean velocity. The factor of two comes from the fact that, to conserve angular momentum, the orbital angular velocity is inversely proportional to the square of the sun-planet distance. So we can write equations like:

A = W * T

B = A + 2 * E * SIN(A - SP)

where T shows time, W is the mean orbital angular velocity (so A, which equals W * T, is the longitude of the clock car relative to its longitude when T is zero), E is the orbital eccentricity, B is the longitude of the sub-solar point, and SP is a constant which is zero if we make the value of T zero at perihelion. Otherwise, SP is the angle the planet (the earth) travels around its orbit between the time of perihelion and the time that we choose to use as zero. In this calculation, relating to the earth, I use the December solstice as the zero, which is 12 days before perihelion, and solar days as the units of time, which gives SP the value W * 12.

The longitude of the sub-solar point on our non-rotating planet shows the position of the sun in the planet's sky or, conversely, the position of the planet as seen from the sun. Therefore, the equations above can be used to calculate the angle between the planet-sun vectors at any two times. The difference between the two values of B, at the two times, is this angle.

Let us define a set of Cartesian axes, with the centre of the planet as the origin. The Z axis is perpendicular to the plane of the orbit, the X axis passes through the clock car, and the Y axis is perpendicular to the other two. Let us first calculate the coordinates of the sun in this frame of reference. If we call the distance between the planet and the sun 1 unit, and if the angle between the X axis and the planet-sun vector is B, then the sun's Z is zero, since the Z axis is perpendicular to the orbital plane. The sun's X coordinate is COS(B), and its Y coordinate is SIN(B).

Now let us imagine the X and Z axes and the set of lines of latitude and longitude on the non-rotating planet to be turned by a moderate angle about the Y axis. We have now given our planet an axial tilt. The clock car's position is also rotated, so it continues to move around the equator, which is now tilted with respect to the orbital plane. This rotation of the frame of reference changes the X and Z coordinates of the sun. (The Y coordinate is not changed, since the rotation is around the Y axis.) The X coordinate was COS(B) and the Z coordinate was zero. After the rotation, through the angle of axial tilt which I will call L (we are already using T for time), the X coordinate becomes COS(B).COS(L) and the Z coordinate becomes COS(B).SIN(L). Actually, since we are talking about the northern-winter solstice, so the angle of tilt is away from the sun, the sun's Z coordinate becomes COS(B).SIN(-L), where L is the positive angle (23.45 degrees) of tilt of the earth's axis.

If we write ST for SIN(-L), we are now in a position to write an expression for the declination of the sun:

SIN(Declination) = COS(B).ST

This is because the sun-planet distance is still 1 unit, and the Z coordinate of the sun is COS(B).ST. Nothing could be simpler! The sun's declination is therefore given by:

Declination = Arcsine(COS(B).ST)

Unfortunately, programming this in a computer language such as QBasic is hindered because the arcsine function is not directly supported. Instead, we have to use a roundabout method using the arctangent function, ATN:

C = COS(B).ST

Declination = ATN(C / SQR(1 - C * C))

We can now also derive an expression for the sun's longitude:

TAN(Longitude) = Y / X = SIN(B) / (COS(B).COS(L)) = TAN(B) / COS(L)

Writing CT for COS(L) (or COS(-L), if we want to use the reverse angle), this becomes:

TAN(Longitude) = TAN(B) / CT

Taking the arctangent of the right hand side of this equation essentially gives an expression for the solar longitude.

We are now very close to an expression for the solar longitude relative to its mean position, which is basically the difference between the longitudes of the sub-solar point and of the clock car. The clock car's longitude is A, so if we calculate the difference between A and the expression for the solar longitude, we get, in BASIC:

Q = A - ATN(TAN(B) / CT)

(I have done the subtraction in the order that gives the conventional sign to the Equation of Time. However, it also makes the positive direction for the longitude westward.)

Now we encounter a little problem. Arctangent is a many-valued function. For any value of the tangent, there are many (an infinite number) of angles, differing from each other by multiples of "half turns", 180 degrees or PI radians. The function ATN produces one of these, but not necessarily the one we want. We may have to add or subtract a multiple of half turns to get the correct angle. Fortunately, in the case of the earth, which has an orbit that is nearly circular, the sun's longitude relative to its mean value is always small. The difference is always less than 0.03 of a half turn. So, if we first divide Q, in the last equation, by PI, to get its value in half turns, and then subtract the nearest integer, we will get an expression for the sun's longitude relative to its mean position with no ambiguity:

C = (A - ATN(TAN(B) / CT)) / PI

Relative_Longitude = C - INT(C + 0.5)

It turns out that the integer that is subtracted is 0, 1, or 2, depending on the time of year.

This expression for relative longitude is in half turns. To get it back in radians, we just have to multiply by PI.

SL = PI * (C - INT(C + 0.5))

Alternatively, we can get the Equation of Time, in minutes, by multiplying by 720 instead of PI, since the earth takes 720 minutes (12 hours) to rotate a half turn relative to the sun.

ET = 720 * (C - INT(C + 0.5))

As the saying goes: The proof of the pudding is in the eating. Theory must always be tested against reality. I wrote a little program that, on 36 days of the year (the 1st, 11th, and 21st days of every month), compares the values of the solar declination and the Equation of Time, calculated as shown above, with values that I got from a published table. (See Reference 1) The agreement between the calculated and published values is striking. The Root Mean Square difference between the calculated and published values of the Equation of Time comes to only 3.7 seconds of time! The greatest difference is 6.0 seconds, on July 11. Compare these with the approximately 130 seconds that the sun takes to move a distance equal to its own diameter in the sky. In the case of the Solar Declination, the R.M.S. difference is 4.7 arc-minutes (60ths of a degree), and the largest difference, on April 11, is 8.8 arc-minutes. At that time of year, the solar declination changes at a rate of about 24 arc-minutes per day. Also, the angular diameter of the sun as seen from the earth is about 32 arc-minutes. The discrepancies between the calculated and published values of the solar declination are only small fractions of either of these.

In my program "SunAlign", which calculates the position of the sun in the sky and the alignment of a heliostat mirror, I used the expressions derived above for the declination and longitude of the sun. These are somewhat better than the approximations that are used in some other programs. For example, some assume that the solar declination varies sinusoidally with time. Others approximate the Equation of Time as the sum of two sine waves, one with a period of a year and the other with a period of half a year. Both of these approximations produce considerable errors.

However, I should not pretend that the expressions derived above are precisely accurate. In addition to the slow variations of the earth's motions mentioned earlier, they do not take account of the non-spherical shape of the earth, the perturbations of its orbit by the moon and other planets, or the effect of parallax when the sun is observed from different points on the earth. Also, the correction for the eccentricity (elliptical shape) of the earth's orbit is only a first-order one. The expressions I have derived are therefore not exact. However, for the purposes of solar energy, they should be amply good enough.

Reference 1

There are many published tables of the Equation of Time and solar declination. The ones I used are from the book:

Sundials, Their Theory and Construction, by Albert E. Waugh, Dover Publications, New York, 1973, ISBN 0-486-22947-5

Counting days in Varying declination

The value of 'N' is described as "the number of days elapsed since January 1."

This is ambiguous. Is N=1 on January 1 or on January 2? To me, "day number in the year" is different from the number of days elapsed.

The formula for "fractional year in radians" suggests N=1 on January 1. Otherwise its value is negative for only one day in the year when N=0 on January 1.

Is N=1 on January 1 for all formulas in this section?

If so, it seems likely that a number of readers will make an off-by-one error using the description as "the number of days elapsed since January 1." - Ac44ck (talk) 18:01, 1 April 2011 (UTC)[reply]

If all formulas are using the same basis for 'N', a reference to the ordinal date, neglecting leap years, may be useful. - Ac44ck (talk) 18:19, 1 April 2011 (UTC)[reply]

Ordinal date seems right for all formulas in the section.

Tables of ordinal dates:

http://disc.gsfc.nasa.gov/julian_calendar.shtml

The "fractional year in radians" needs N=1 at the beginning of January 1.

The maximum value of the simpler cosine formula occurs at ((360°/365)*(172.5+10)) = 180°. This is consistent with an approximation that the summer solstice happens at noon on June 21 (N=172.5). - Ac44ck (talk) 16:47, 2 April 2011 (UTC)[reply]

Subscript

The subscript in "δ_☉" renders as a square in my installations of both Opera and Internet Explorer. The subscript is used in the section on varying declination of the sun. The square is apparently intended to be some symbol representing the sun, but it is not effective in my browsers. If this subscript is necessary, please use an alternate encoding for this symbol. If it is not necessary, please delete it.

In a search that perhaps was too brief, I did not find a document using a solar symbol as a subscript for declination. Is this symbology found in general publications? In the context of this brief Wikipedia article, a cryptic symbol (a custom glyph as a subscript to a lower case Greek letter) is probably more jargon than the reader needs. I suggest deleting the illegible and arcane subscript in the section which is obviously discussing only solar declination. -Ac44ck (talk) 13:55, 3 April 2011 (UTC)[reply]

This is standard orthography. I doubt it's a matter of your browser, but whether you have adequate fonts installed.

The other symbol (δ_ʘ) was the letter for a labial click, which besides probably not displaying for other readers, is factually wrong. The symbol for the Sun should be used for the Sun. I don't know why it wouldn't display for you, since AFAIK just about every word processor has a font which supports this. — kwami (talk) 06:14, 5 June 2011 (UTC)[reply]

Comparison of decimal and degrees-minutes values for solar declination

Both forms are accurate to the precision stated:

The degrees-minutes value, 23°26', is rounded to the nearest minute.
The constant in the cosine formula, 23.44, is rounded to the nearest 0.01°.

The axial tilt article cites:

http://hpiers.obspm.fr/eop-pc/models/constants.html

Where the value is given as:

23°26'21".4119

This is 23+(26+(21.4119/60))/60 = 23.4392810833°.

The rounded degrees-minutes value, 23°26' = 23.4333°, is slightly less accurate than the decimal value in the cosine formula.

The source cited for the cosine formula uses 23.45. - Ac44ck (talk) 16:54, 10 April 2011 (UTC)[reply]

Solar declination is off-topic

While an interesting subject, and useful if you use solar energy, it is only indirectly related to the concept of declination, which should be a definition of part of a celestial coordinate system. The Sun's declination should be in a separate article, or perhaps in one of: Solar_elevation_angle, solar_energy, solar_tracker, heliostat. Opinions? I will move it to one of the above within a few weeks if there are no objections. Tfr000 (talk) 14:35, 18 April 2012 (UTC)[reply]

I disagree. While there are other aspects of the concept of declination that are certainly interesting and relevant, the declination of the sun, and its seasonal variation, is probably the most interesting of all. It is relevant not only to solar energy, but also to navigation, meteorology, and a host of other topics. If people who are reading about these things follow a link to Declination, and find that the solar declination is not mentioned, they will be frustrated and annoyed.

Conceivably, solar declination should be covered in a whole separate article, with a disambiguation page to guide people to it, or to a more abstract article. It should definitely not be subsumed into any of the pages you mentioned.

DOwenWilliams (talk) 15:26, 18 April 2012 (UTC)[reply]

We would, of course, fix any links broken by moving it. I agree that it probably needs its own article, or maybe there should be a general "position of the Sun" article covering how to figure the alt/az for any place on Earth and any time - I didn't find anything like that with a brief search. I would like to get the present article cleaned up, improved, and on-topic of celestial coordinates, along with right ascension. Tfr000 (talk) 16:20, 18 April 2012 (UTC)[reply]

In the Equation of time article, there's a calculation of the EoT on any date, which also has, as an addendum, one extra line that calculates the solar declination. The two quantities come out from the same calculation. If you want the sun's alt/az at any time, as seen from any place, there's a calculation here: http://green-life-innovators.org/tiki-index.php?page=Sunalign which does the job, and also calculates the correct alignment of a heliostat mirror. I wrote it, years ago, so I can't post it here because of the prohibition of original research, but if you want to post it, you're welcome to do so. There's also an explanation of the logic of the calculation, line by line. DOwenWilliams (talk) 21:11, 18 April 2012 (UTC)[reply]

I have created the article Position_of_the_Sun. Looking around some, many Wiki articles have little bits of "how to compute the Sun's position", so now we have a place to put that. So if there are no further objections, I will be moving the Solar declination material to Position_of_the_Sun within a few weeks, fixing any links broken on the way. Tfr000 (talk) 22:45, 2 May 2012 (UTC)[reply]

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.

Done. I also moved some of the External links, and all of the References, since they belonged with that text. This article now has no references! Tfr000 (talk) 14:58, 30 May 2012 (UTC)[reply]

Declination and latitude

Unfortunately some people seem to like this sentence, which is confused and misleading at best:

"The zenith at a particular location is perpendicular to the ellipsoid, and this [what?] varies by as much as 11.5 minutes of arc (0°.192) from geocentric latitude, which does match declination."

Maybe the writer meant to say "If an observer is at geocentric latitude 29.00 degrees, a moon/planet/star directly above him must have a declination of 29.00 degrees, assuming no vertical deflection." -- which isn't true. Whatever the writer meant, we can't blame the reader for thinking that's what the sentence means. Want to try to reword it so it's telling the truth? (Best plan is not to mention geocentric latitude, which is irrelevant to this article and a waste of the reader's time.) Tim Zukas (talk) 20:45, 5 August 2012 (UTC)[reply]

Unfortunately, there are three meanings of zenith, and many meanings of declination. The Astronomical Almanac Online explains at the zenith entry:

in general, the point directly overhead on the celestial sphere.

zenith, astronomical:

the extension to infinity of a plumb line from an observer’s location.

zenith, geocentric:

The point projected onto the celestial sphere by a line that passes through the geocenter and an observer.

zenith, geodetic:

the point projected onto the celestial sphere by the line normal to the Earth’s geodetic ellipsoid at an observer’s location.

The declination is analogous to latitude, but there are dozens, or perhaps hundreds, of ways to precisely define it. Basically you can select many combinations of the following (at least):

Center	Name or date
Sun	B1875.0
Solar System barycenter	B1950.0
Earth's center	International Celestial Reference System
observer on the surface of the Earth	J2000.0
	true equator and equinox of date

So it is probably too complex to describe which variety of declination is most comparable to geocentric zenith, and how that differs from geodetic zenith. Probably the whole section should be removed.

However, the change made by Tim Zukas is wrong for many of these combinations. The context rules out the Sun or Solar System barycenter as the center for the declination measurement frame, so the only factor that would change declination appreciably because of the proximity of the object being measured would be shifting from a measurement frame centered on the Earth's center to a measurement frame centered on the observer on the Earth's surface.

For a discussion of this see

U.S. Naval Observatory Nautical Almanac Office (1992). P. Kenneth Seidelmann (ed.). Explanatory Supplement to the Astronomical Almanac. University Science Books, Mill Valley, CA. pp. 126–7, 200–7. ISBN 0-935702-68-7.

Jc3s5h (talk) 23:32, 5 August 2012 (UTC)[reply]

No one said declination changed as the object gets closer. It's the difference between declination and geodetic latitude that increases as the object gets closer-- you should be able to figure that out. For the moon, that difference can reach 0.003 degrees; for the planets and the stars the difference is much less.

"The declination is analogous to latitude, but there are dozens, or perhaps hundreds, of ways to precisely define it."

In this article we'll define declination as the angle the line from the earth's center to the center of the object makes with the earth's equatorial plane. If you want to tell folks there are thousands of other definitions feel free, but you ought to mention which definition the almanacs are using when they give "declination".

"there are three meanings of zenith"

Think it's hard to pick which one to use for this article?128.32.104.164 (talk) 21:19, 6 August 2012 (UTC)[reply]

"In this article we'll define declination as the angle the line from the earth's center to the center of the object makes with the earth's equatorial plane." Well, it wouldn't to to use only one of three acceptable definition for the article, but we can use it for purposes of this discussion. With that definition, the declination is indeed independent of the distance of the celestial object.

"It's the difference between declination and geodetic latitude that increases as the object gets closer-- you should be able to figure that out." Geodetic latitude is determined by the observer's location and the choice of reference spheroid, such as WGS84. Celestial objects don't come into it at all. So the difference between declination and geodetic latitude is independent of the distance to the celestial object. What is your source for your statement? What is your source for 0.003 degrees? Jc3s5h (talk) 22:48, 6 August 2012 (UTC)[reply]

Say the object's declination is 29.00000 degrees and it's 10000 light-years away. If it's directly overhead (assuming no vertical deflection) the observer must be at geodetic latitude 29.00000 degrees-- right?

Now say the object's declination is 29.00000 degrees and it's 2 meters above the Earth's surface. If it's directly overhead what must the observer's geodetic latitude be, assuming no vertical deflection? (Answer: 29.16 deg)

"Well, it wouldn't to to use only one of three acceptable definition for the article"

Use as many definitions as you like-- just let the reader know which definition the almanacs use. (Won't be hard to choose.)

"With that definition, the declination is indeed independent of the distance of the celestial object." No one said different.Tim Zukas (talk) 19:26, 7 August 2012 (UTC)[reply]

{od|5} "Say the object's declination is 29.00000 degrees and it's 10000 light-years away. If it's directly overhead (assuming no vertical deflection) the observer must be at geodetic latitude 29.00000 degrees-- right?" Right.

"Now say the object's declination is 29.00000 degrees and it's 2 meters above the Earth's surface. If it's directly overhead what must the observer's geodetic latitude be, assuming no vertical deflection? (Answer: 29.16 deg)"

Yes. I had intended to fix the 1 degree citation I had put in, realizing that the effect is far less when the observer is nearly on a line between the object and the center of the earth, but Wikipedia was acting up and I wasn't able to edit. Jc3s5h (talk) 13:39, 8 August 2012 (UTC)[reply]

Checking the claim in the article, "For the moon this discrepancy can reach 0.003 degree...", I figured the greater the declination of the moon, the greater the discrepancy. The new moon with the greatest declination in 2012 is on Nov. 28 at 14:46 UT1. The moon's apparent geocentric RA and declination are 4h 20m 28s and 20° 28' 14" respectively according to the Multiyear Computer Interactive Almanac. By trial and error one can find the point with less than 1 arcsecond zenith distance at that time is at location E135°42'48.0", N20°28'22.0". The difference between that geodetic longitude and the moon's geocentric declination is 8 arcseconds or .0022°, which is comparable to the value in the article. Still, it would be good to have a citation for the article's value. Jc3s5h (talk) 14:55, 8 August 2012 (UTC)[reply]

I am more or less in favor of removing the section completely. Most of this discussion is about topocentric positions, that is, the differences in position of celestial objects as seen from different locations on the Earth, which is only indirectly related to declination. Probably more confusing than anything, here in the declination article. The section is a relic of the original declination article, which was sort of a patchwork of bits and pieces of astronomy information. There is no current "topocentric positions" article... maybe someone would like to write one. Horizontal coordinate system or parallax are closest things we have now. Tfr000 (talk) 18:21, 2 December 2012 (UTC)[reply]

The article definitely needs to mention (in passing at least) that when the object's declination equals the observer's latitude the object will pass just about directly overhead. That's by far the simplest way to define declination. And if it mentions (as it should) that the declination of an object directly overhead isn't exactly equal to the observer's latitude, then it should give some idea what the difference can amount to. Tim Zukas (talk) 20:59, 3 December 2012 (UTC)[reply]

Link in diagram caption?

The caption for the diagram near the top of the article links "vernal equinox" to the article about the equinox event. Would it be better to link it to http://en.wikipedia.org/wiki/Equinox_(celestial_coordinates) article? 209.131.230.182 (talk) 19:53, 7 May 2013 (UTC)[reply]

"Hour angle" used incorrectly

Around 2013, someone changed this sentence in the introduction:

In astronomy, declination (abbreviated dec; symbol δ) is one of the two direction coordinates of a point on the celestial sphere in the equatorial coordinate system, the other being either right ascension or hour angle.

to:

In astronomy, declination (abbreviated dec; symbol δ) is one of the two direction coordinates of a point on the celestial sphere in the equatorial coordinate system, the other being hour angle.

In fact, hour angle is not used in the equatorial coordinate system, see our own article. This should be changed to right ascension. Tfr000 (talk) 14:04, 11 November 2017 (UTC)[reply]