ANSWER.TXT Questions, Answers and Comments about Sunlight 2006/05/30 1. Question: Why does the longest day of the year occur near the summer solstice at the equator and even just south of the equator? Answer: Near the equator (including into the southern hemisphere) the seasonal variations of the longest of the day are dominated by the Earth's position in its orbit and not by the tilt of its axis. Seen from the North Star, the Earth rotates around its axis and revolves around the Sun in a counterclockwise motion. Near perihelion (currently January 3), the Earth's speed in its orbit is fastest and the angular frequency (angular velocity) of the Sun (as seen from a nonrotating Earth) is fastest. Thus the Sun's daily traverse from east to west is fastest, or the length of daytime is shortest. Near aphelion (currently July 5), the reverse is true; the length of daytime is longest. Aphelion is currently near the summer solstice (about June 21), but it is the occurrence of aphelion that dominates the longest day of the year near the equator. 2. Comment from Dr. Robert A. Nelson: The calculation of daylight on this web site is a useful source of tabular information. However, the web designer should be able to better model the length of the tropical year without assuming a constant value (the table takes it to be the average length of the year in the Gregorian calendar). Referring to the Explanatory Supplement of the Astronomical Almanac [1961], page 99, the tropical year is given by 365.24219879 - 0.00000614*T days, where T is measured in Julian centuries of 36525 days from the epoch 1900. This formula follows from Newcomb's formula for the mean longitude of the Sun. The tropical year thus decreases by 0.00000614 days per century (this is actually an extremely long periodic variation). An alternative formula is given on page 576 of the current edition of the Explanatory Supplement [1992] edited by P.K. Seidelmann. In addition, one needs to consider the variable rate of rotation of the Earth to assign a time of day in the distant past. Due to tidal friction, the length of the day has been increasing over the very long term at the rate of about 1.7 ms per day per century. The difference between uniform time and UT1 (the true measure of the Earth's angle of rotation) is called Delta T. An approximation to Delta T is given by F.R. Stephenson in Historical Eclipses and Earth's Rotation [Cambridge, 1997] as Delta T = 0.0031*(Y - 1820)^2 - 20 seconds, where Y is the year. Thus the length of the day was exactly 86400 SI seconds in about 1820. (Today the length of day exceeds 86400 seconds by about 2.5 ms, which is the reason why leap seconds are inserted into Coordinated Universal Time with a frequency on the order of once per year.) The correction to time for a date 2000 years ago is about 3 hours. This web site's table would be more useful for historical research involving astronomical phenomena if the calendar dates were expressed in terms of the Julian calendar for dates prior to October 15, 1582. It would be interesting to observe the regression of the date of the vernal equinox through the Julian calendar, in which the length of the calendar is exactly 365.25 days. This regression was arrested by the adoption of the Gregorian reform in 1582. The average calendar year in the Gregorian calendar is 365.2425 days, which is a better approximation to the tropical year. If the web designer were able to make these modifications, the table would be greatly improved. 3. Question: If the Earth's axis were not tilted, would there be 12 hours of daylight and 12 hours of nighttime everywhere? Answer: Yes, there would be approximately 12 hours of daylight and nighttime each day, but there are three reasons that this would only be approximate. a. The Sun is not a point but is a sphere with a radius. At any given time exactly half the points on the Earth's surface are blocked by the Earth itself from forming a line to the center of the Sun. Less than half the points, however, are blocked by seeing none of the Sun's large disk. Thus, on average, each point on the Earth's surface will be able to see a fraction (or all) of the Sun for more than 12 hours. b. Sunlight is refracted by the Earth's surface causing twilight. Near sunrise and sunset, although all of the Sun's disk is blocked by the Earth in a geometric line, the Sun is still visible because rays of sunlight are curved around the Earth. This increases the amount of daytime and decreases the amount of nighttime. c. Because the Earth is in an elliptical orbit with varying speed, the angular movement of the Sun, as seen from a non-rotating Earth, is greatest near perihelion and least near aphelion. The Earth, however, is rotating and at a fixed rate even faster than its revolution, but in the same direction. It is the difference between these two angular movements that are inversely proportional to the amount of daytime or nighttime. Near perihelion, when the difference of the angular movements is least, the length of daytime and nightime is several seconds longer than it is near aphelion. When the daytime - nighttime oscillation is chopped into fixed twenty four hour days and a local time zone is used, daytime always gets its full share and nighttime gets what is left over. 4. Question: How can I calculate insolation at a particular location on Earth for a particular time of day? Answer: This is how the Model calculates insolation: a. Knowing the YEAR, call subroutine ORBPAR which returns the three orbital parmaeters: ECCEN, OBLIQ and OMEGVP. OBLIQD (in degrees) is used in some places instead of OBLIQ (in radians). Sometimes fixed values for the year 2000 AD are used. b. Knowing ECCEN, OBLIQ, OMEGVP, and DAY (days measured since 2000 Janary 1, hour 0; may be + or -), call subroutine ORBIT which returns SIND, COSD and SUNDIS. The Model assumes that at 12 noon GMT, the Sun is directly over the Greenwich meridian which is not quite correct. This small inaccuracy is returned as an additional output variable called EQTIME. c. Call subroutine COSZ0 which stores the Model's SINJ and COSJ (sine and cosine of latitude) and SINI and COSI (sine and cosine of longitude measured east from the International Date Line). Here is where a user can replace the Model's 90 x 60 grid cells with his own latitude and longitude. Remember that the COSZ subroutines do not use normal longitude, but measure longitude from the IDL. Southern latitudes are negative. d. Call subroutine COSZT which requires as input: ROT1 and ROT2 (the initial and final time in radians, Greenwich Mean Time multiplied by 2*pi divided by 24 hours), SIND and COSD (the declination angle of the Sun), SINJ and COSJ (the latitude), and SINI and COSI (the longitude). COSZT returns COSZ (local cosine of the zenith angle) weighted by time. (COSZS returns COSZ weighted by sunlight.) e. Knowing the solar constant, SOLARC (global annual insolation), calculate the present insolation at top-of-atmosphere (W/m^2) as: SOLARC * COSZ / SUNDIS^2. The Model assumes SOLARC is 1367 (W/m^2) which may be a slight overestimation.