MODELLING THE SUN’S ALTITUDE FROM SUNRISE TO NOON
A note by Eamon Henry, 5 August 2009
Foreword:This note follows the Henry note of 27 July 2009 “How warm is the summer sun!”, now available on
http://www.asits.blogspot.com/. The trigonometric model which gives results presented below uses two formulae appearing in a 14-page document “Spherical Trigonometry” available on the web at
www.krysstal.com/sphertrig.html. This interesting article even gives 2 pages of (in effect) a “do it you yourself kit” for building your own sundial (just what we all need!!).The Henry Excel 97-2003 worksheet “Sunrise to Noon Sun Altitudes_rev1”(3 July 2009) gives model results for midsummer day and midwinter day, at 15-minute intervals, computed by him. At present the ASITS blogspot is not able to carry an Excel worksheet as such. Table 1 following gives summary results for 30-minute (0.5- hour) intervals.
Results:
Table 1: Model estimates of Sun’s altitude at Dublin from Sunrise to Noon, Midsummer and Midwinter
Time, local, hours
Midsummer altitude, degrees
Midwinter altitude, degrees
12.00
60.0
13.0
11.30
59.8
12.7
11.00
59.3
11.8
10.30
58.5
10.2
10.00
57.3
8.0
9.30
55.7
5.1
9.00
53.6
1.5
8.48
0.0
8.30
51.1
8.00
48.0
7.30
44.3
7.00
40.0
6.30
35.1
6.00
29.7
5.30
24.0
5.00
18.3
4.30
12.7
4.00
7.4
3.30
2.6
3.12
0.0
“Local time” means calibration so that noon (Sun due south) is 12.00 hours, and thus for Dublin some 25 minutes later than Greenwich Mean Time (GMT) noon, because Dublin is about 6.25 degrees West (longitude); Sun due east will be at 6.00 hours local time, for Earth’s rotation 15 degrees per hour. Thus, in “Azimuth” terms, direction East is 90 degrees away from South, and direction West is 90 degrees away in the opposite direction, as for “Azimuth” angles to be used for model applications described below. “AZIMUTH is the angular distance from the north or south point of the horizon to the intersection with the horizon of a vertical circle passing through a given celestial body” (page 97 of the Oxford Encyclopedic English Dictionary, 1991 edition).
How the Model differs from the Real-World Situation:The model gives 21 June Sunrise as occurring at 3h 12m a.m. The actual Dublin sunrise time is about 3h31m local time, some 19 minutes later. For 21 December, the model gives Sunrise as at 8h48m local time, compared with actual 8h15m local time, some 33 minutes earlier. On page 11 of “Spherical Trigonometry” a calculation is made for Sunrise times at London. The azimuth angles are 39.62 degrees north of east (3h21.2m am local time) and 39.62 degrees south of east (8h38.5m am local time). My Diary 2009 gives London 20 June sunrise as 3h43m, about 22 minutes later, and London 19 December sunrise as 8h03m a.m., some 35 minutes earlier. So the model is wrong in a consistent way, it seems.
One reason for these differences is that the model implicitly assumes that the Earth’s orbit is a circle about the Sun, and thus completely regular. But in fact the orbit is an ellipse with the Sun at one focus, and our distance from the Sun ranges from 147 million kilometers in December to 152 million in June (page 9 of Patrick Moore: “Exploring the Night Sky with Binoculars”). However, the model gives no discrepancy for the Spring and Autumn equinoxes, 21 March and 23 September, with sunrise at 6.00 hours local time everywhere. Altitudes are not calculated for one of these, in the attached Excel sheet, but can be readily derived by changing a few parameters of the model, as will be indicated below.
Note that the model gives exactly 12 hours as the sum of the times from sunrise to noon, 8h48m midsummer plus 3h12m midwinter. The actual sum is about 12h15m, meaning that the Northern Hemisphere does in fact get more Sun-time than the Southern Hemisphere.
Because of these model discrepancies, I am re-calculating (in the Postscript section below) the relative per-hour intensities of the Sunshine for midsummer versus midwinter, as given in my previous note of 27 July 2009. For the summer sunrise situation, one might estimate actual Sun’s altitudes (angles of elevation) by going to the model altitude at least 15 minutes earlier: take the 5.45 hours altitude 26.9 degrees as estimating actual altitude at 6.00 hours, and take model 8.45 hours altitude 52.4 degrees as estimating actual altitude at 9.00 hours. My own estimates (for these times) used in my note of 27 July were 23 degrees and 47 degrees, respectively.Of course, both model and actual altitudes coincide at local noon, so they must gradually come closer as the morning progresses.
Technical description of the Model:The main equation I use is on page 8 of “Spherical Trigonometry” as follows:
sin(D) = sin(L)*cos(Z) – cos(L)*sin(Z)*cos(A)………………………………….. (1)
The sign * means “multiplied by”.I use latin letters where the text uses greek letters. D is the angle of declination (latitude on the Celestial Sphere) of some star, in our case the Sun, at 0 degrees for the equinoxes, 23.5 for summer solstice and -23.5 degrees for winter solstice. L is the Latitude of the point of observation, for Dublin about 53.5 degrees North, for Oxford about 51.6 degrees North. Z is the “Zenith distance”, given as 90 degrees less the Altitude(angle of elevation) of a star (in our case the Sun). A is the Azimuth, measured in degrees from direction South (e.g. direction East has azimuth value “minus 90 degrees”). Here a comment on the Celestial Sphere is in order. At Greenwich at noon on 21st March (spring equinox), the Sun is at location (0,0) on the Celestial Sphere, whose equator coincides with the imaginary circle where the plane of Earth’s equator cuts this visual sphere of stars etc. around the Earth. Longitude is called “Right Ascension” and latitude is called “Declination”. The north pole of the Celestial Sphere is directly above Earth’s north pole, its axis of rotation coinciding with that of Earth.
Let us get the Sunrise equation from equation (1) above. For altitude zero degrees, Z has value 90 degrees, Cos (Z) is zero and sin (Z) has value 1. Thus equation (1) reduces to:
Sin(D) = -cos(L)* cos(A), which can be written cos(A) = - sin(D) / cos(L),………..(2)
And we can write it as cos (A)= sin(D)/cos (L), if we measure A from North direction clockwise, i.e. East is 90 degrees. So, for D and L given we have A, measured from North direction. A gives the Sunrise azimuth angle, as each 15 degrees equal to one hour of time, measured from midnight onwards. For example, find the summer sunrise direction and time for Dublin:
cos(A) =sin 23.5/cos 53.5 (in degrees) = 0.398749/0.594823 =0.670366.A has value 47.90 degrees X 4 = 191.62 minutes of time after midnight (North direction) =3h11.6m a.m.
To use equation (1), see the column headings and function expressions on the Excel worksheet. D has value 23.5 for midsummer (Dublin). All trig calculations must have radian measure of angles, hence the columns of radian values. L has value 53.5 degrees. Angle A (azimuth) is progressed backwards in degrees negative from noon of value 0 degrees. Find Z from this equation for each specified value of azimuth A. In column I you put in estimated altitude values from noon known value 60 degrees back to 0 at time of sunrise. (At equinoxes, Sun’s noon altitude is given by 90-latitude 53.5=36.5; add 23.5 more for Sun coming up by 23.5 degrees (declination) so as to have noon altitude 60 degrees at midsummer).
Now calculate value of right-hand-side (RHS) of (1), and subtract it from left-hand-side to give error. Calculate value of Z-differential of RHS, which is –sin(L)*sin(z) – cos (L)*cos(Z)*cos (A), and divide the error by the latter, to give a Z correction, added on to first Z estimate. (This approach is called Newton’s Method of solving equations.) Re-calculate the RHS of (1) and again find error. These resulting errors are so small (see column R) that a second iteration (to find a further Z increment) is not warranted. So the amended Z is our solution in each row; subtract it (in degrees) from 90, and this is the required altitude (angle of elevation) of the Sun given by the model for that time of morning (as A expressed in time before noon) and shown in column X.
Obviously, Sun’s altitude calculations could be made for other times of year, for any specified Sun declination value D. The Sun’s declination throughout the year is shown on the “Ecliptic” curve on star maps. Just now (early August) it has an approximate value of 12.5 degrees, well below the 23.5 degrees of 21 June.
Postscript (21 August 2009)The right-hand columns Z to AI of the Henry Excel worksheet give detailed calculations of the average per-hour Sun intensity, midsummer and midwinter. In the note “How warm is the summer sun!” rough estimates of per-hour average square- metre areas were given as 3.39 for midsummer and 11.35 for midwinter. Revised figures are 3.419 and 8.587, respectively, as shown on the Excel worksheet. The midwinter area is considerably reduced, due to the Sun’s apparent curve (as given by the model) being well above the previous assumed straight-line movement from sunrise to noon, thus giving larger sine values with matching smaller 1/sine values and thus smaller areas.
These new results are calibrated to match 15-minute intervals from sunrise to noon, but again omitting first 1 degree altitude (elevation) of the Sun after sunrise. The time-factor multiplier is taken to be (for each 15-minute interval) the average time per 1 extra degree of the Sun’s altitude, based on the model results. The relative per-hour midsummer/midwinter intensity is now 2.512, given by 8.587/3.419.
