APPARENT POSITION OF THE SUN IN THE SKY [Dave Rusin, math.niu.edu. Last modified 2/24/95] History: Got tired of seeing this circle of questions raise in sci.math, rec.gardens, and misc.consumers.house (!) in 1993. Of course, all this material is already known, but I don't know by whom so I have no references. This paper describes the calculation of the apparent sun position in local coordinates for an earth observer. The first section sets up notation; section 2 sets up a solution; section 3 is to remove the effect of one variable and simplify the solution. Section 4 provides qualitative descriptions of solar motion. In section 5, I address the conversion of these variables to local time, in section 6 the length of a day, and in section 7 the amount of sunlight falling (i.e. energy gain). There is a short discussion of errors in section 8. An accompanying BASIC program 'suncalc.ub' carries out these calculations, although I have not altered older routines which are not consistent in form with this essay. (As far as I know, though, the conclusions are correct.) We assume the earth is a sphere; one may choose units so the radius is 1. We also assume the earth orbits the sun in a circle (this approximation is less good, as the earth-sun distance varies about 3% over the year). We let L be the distance from earth to sun (roughly 25000 earth radii). We will assume the observer is not at the north or south poles (so the the direction "north" will make sense). =========== 1. Expressing true sun position======================== We note that in our problem it is equivalent to assume the sun travels in a circular orbit around the earth, and so we set up spherical coordinates, using the (north) axis of the earth's rotation as the (positive) z-axis. We take the x-axis to be the intersection of the earth's equatorial plane and the plane of the sun's transit. (These planes meet at an angle \alpha of about 23 degrees, familiar from all desk globes.) A person on the earth's surface at a latitude of \theta will always have this fixed spherical coordinate \theta. However, the other spherical coordinate \phi will run through the interval [0, 2\pi] once per sidereal day (the sidereal day is the time it takes a given star to return to a given position in the sky; due to the earth's revolution about the sun, this is not quite the same as the time between successive noons). Using standard formulas for spherical coordinates, we then see that the cartesian coordinates for the observer's position P are x = cos(\theta) cos(\phi) y = cos(\theta) sin(\phi) z = sin(\theta) where \theta is the fixed latitude and \phi changes linearly in time. The position of the sun is always in the plane defined by z = y . tan(\alpha) and always a fixed distance from the earth (x^2+y^2+z^2=L^2); we deduce that there is an angle \psi such that the sun's position is x = L cos(\psi) y = L sin(\psi) cos(\alpha) z = L sin(\psi) sin(\alpha) Since we are assuming a circular orbit, it is true that \psi will change linearly in time, ranging from 0 to 2\pi in 1 year; the day when \psi is zero is the day the earth's axis is perpendicular to the line to the sun, i.e. the March equinox. =========== 2. Determining apparent sun position=================== Now let us relate these cartesian coordinates to the local coordinates used by an observer. To a (small) observer on the surface of the earth, the earth looks flat; that is, the observer does not distinguish (his portion of) earth from the tangent plane at his location. Positions on the tangent plane are usually measured against two axes labelled "east" and "north". "East" is the direction the observer is travelling as the earth rotates, that is, it is in the direction of the partial derivative \partial P / \partial t; since we know \theta is constant and \phi changes linearly, this is the same direction as \partial P / \partial \phi Take derivatives to see this is cos(\theta) times ( -sin(\phi), cos(\phi), 0 ) (we divide by that factor cos(\theta) to get a unit vector). "North" is the direction travelled when increasing latitude, that is, the direction of the partial derivative \partial P / \partial \theta = (cos(\theta)cos(\phpi), cos(\theta)sin(\phi), sin(\theta)) Finally, the observer uses a direction "Up", which is the same as the vector from the earth's center to his position, and so is the same as the vector P : ( cos(\theta) cos(\phi), cos(\theta) sin(\phi), sin(\theta) ) These three vectors are orthonormal, and so any other direction can be expressed as linear combinations of them, using the formula v = (v*"North")."North" + (v*"East")."East" + (v*"Up")."Up" (where "*" is the inner product or dot product.) In particular, we can do this for the vector joining the earth's center to the sun (since L is so large, this is essentially the vector from the observer to the sun). Since we are interested only in the sun's apparent position, the length of this vector is irrelevant; dividing by L gives us a unit vector, whose coordinates in the "North"-"East"-"Up" coordinate system are the dot products as above: N= - sin(\theta) cos(\psi) cos(\phi) - cos(\alpha) sin(\theta) sin(\psi) sin(\phi) + sin(\alpha) cos(\theta) sin(\psi) E= - cos(\psi) sin(\phi) + cos(\alpha) sin(\psi) cos(\phi) [E is theta-independent!] U= + cos(\theta) cos(\psi) cos(\phi) + cos(\alpha) cos(\theta) sin(\psi) sin(\phi) + sin(\alpha) sin(\theta) sin(\psi) (Remember, \alpha is a global constant; \theta is fixed by the observer's latitude; \psi changes across a year; \phi changes across a day). ============= 3. Using time to determine \phi================== In order to use these formulas, one needs to know the current value of \phi. Roughly this means "what time of day is it", but \phi=0 does not correspond to any fixed time of day. Instead, we *define* NOON (and midnight) to be moments when E=0 (i.e. when the sun is in the "north"-"up" plane). We usually describe this by saying the sun is due south (or north). This is not quite the same as saying the sun is at its maximal height in the sky: that would require d U / d t = 0, i.e. (d U / d \psi) (d \psi / d t ) + (d U / d \phi) (d \phi / d t ) = 0. As it happens, d U / d \phi = cos(\theta). E, so at the time we called NOON, the second term drops out. The first term is not zero, but is small since (d \psi / d t) is only 1/365 as big as (d \phi/ d t ). So it is not too bad of an estimate to think of NOON as being the moment the sun is highest. We observe that using E=0 as our definition of NOON gives the condition on \phi necessary to declare NOON: tan(\phi) = tan(\psi).cos(\alpha)..........................(*) (and when \psi is +-\pi/2, \phi is also +-\pi/2.) If we let \phi0 be the value of \phi which satisfies this condition, then we can write any other \phi in the form \phi = \phi0+\phi1 for some value \phi1; \phi1 measures the change in the position of the observer since NOON. Using the angle-addition formulas, the defining relation sin(\phi0) = tan(\psi).cos(\alpha).cos(\phi0), and massive trigonometric cancellations, we obtain simplified expressions for the apparent location of the sun. Again using (N,E,U) coordinates, its position is N= - sin(\theta) cos(\psi) sec(\phi0) cos(\phi1) + sin(\alpha) cos(\theta) sin(\psi) E= - cos(\psi) sec(\phi0) sin(\phi1) U= + cos(\theta) cos(\psi) sec(\phi0) cos(\phi1) + sin(\alpha) sin(\theta) sin(\psi) If you don't like Cartesian coordinates, you can express the sun's apparent position with two angles. Its angle of elevation off the (south) horizon is arctan(U/sqrt(N^2+E^2)); the angle it makes with the line to the east is arctan(N/E). Remember, \alpha is a global constant; \theta is fixed by the observer's latitude; \psi changes across a year; \phi1 is essentially the time since noon, and sec(\phi0) is sqrt( 1 + [ tan(\psi).cos(\alpha) ] ^ 2 ). Actually the above formula is ambiguous: there are two square roots. We choose the one with the same sign as cos(\theta) cos(\psi), so that when \phi1=0 (i.e., when it is NOON) we get the larger possible value of U -- the sun is higher in the sky at NOON than at midnight. ================ 4. Descriptive motions ================================ Having never been outside the north temperate zone, I can only comment on the extent to which the above formulas agree with my experience. But even in extreme cases it is easy to compute what the behaviour of the sun will appear to be as it appears to trace out motion in a unit sphere above the observer's head. From the formulas in section 3 we see that cos(\theta) N + sin(\theta)U = sin(\alpha) sin(\psi). This says the sun appears to travel in a certain plane in space, a plane that appears to cross the flat earth in an east-west line, making an angle equal to our latitude (=\theta) with the flat earth, and crossing the horizon (i.e. the plane U=0 of the flat earth) at a point with N coordinate sin(\alpha) sin(\psi) sec(\theta). (Strictly speaking, the plane moves north or south a bit during the day, since \psi changes.) You may have seen this plane if you have seen time-lapse pictures of the sun's transit in a day. The formulas predict what may not be immediately obvious: that the inclination of this plane never changes (at one point on earth), and that it is highest at its southernmost point. Observe that the sun can rise north of east even in the northern hemishpere (point of sunrise varies with sin(\psi) across the year to a maximum N coordinate of sin(\alpha) sec(\theta).) The motion of the sun in this plane is also easily given. We have that its position in (N,E,U) coordinates is of the form v1 + v2 cos(\phi1) + v3 sin(\phi1) for three vectors v1 v2 v3. Again discounting the change of \psi across a day, this motion is a circle, centered at the point sin(\alpha) sin(\psi) ( cos(\theta), 0, sin(\theta) ). This center point is always on a fixed line pointing due north and up, making an angle with the horizontal equal to the observer's latitude. As the year progresses, the center point moves along the line, either above or below the earth's surface according to season. At the equinoxes, the sun's circular path is centered on the earth's surface, and the sun rises due east, precisely at 6am local time. At the solistices, the center is a distance sin(\alpha) away from the observer. (This is the same displacement for all observers, although of course the direction of the displacement is different for observers at different latitudes.) At the equator (\theta=0) the plane is vertical, and the sun sweeps out a circular path in this vertical circle. Only at the equinoxes is the sun ever directly overhead (i.e. at coordinates (0,0,1) ). As the observer moves away from the origin, the plane of the sun's motion begins to tilt, and the magnitude of the seasonal variations becomes greater: the location of the sun at solistice dawn moves progressively farther from east, and the length of the day increases. At the arctic (or anarctic) circle (+-\theta = 90 - \alpha), the sun follows the semicircular arc at the equinox, but by summer solistice, the sun rises due north (or south) just after midnight, travel all the way around the compass, and set (briefly) the next midnight (hence the 'land of the midnight sun'). The circle describing the sun's motion is tangent to the plane of the earth. By winter solistice, the circle of the sun's motion has dropped to a parallel circle again tangent to the plane of the earth, but now under the observer. The sun that day will not rise until noon, it will do so due south (or north), and immediately set. In the polar regions, the sun's plane of motion changes during the year along an axis which is now even more vertical. At equinox, the sun still rises east at 6am, although it will not rise high by noon. At some point in the year, the sun's path is tangent to the earth's surface, and we have behaviour similar to that describe at the arctic circle at solistice (although the noon elevation is smaller). The condition which describes the time of year (i.e. \psi) is that cot(\psi) = sin(\alpha) sqrt( tan^2(\theta) - tan^2(90-\alpha)), which makes it clear that this can only happen past \theta=90-\alpha. Thereafter, the sun's path moves to a smaller parallel circle higher in the sky, so that although the sun moves from due south at noon to due north at midnight, its midnight location is still above the horizon. Naturally, in the sinter months the situation is reverse: the days grow shorter, with sunrise progressivly more southerly, until the day the sun fails to rise and prolonged darkness begins. At the north pole, (\theta=90), the sun's motion is in a circle parallel to the earth's surface. At equinox, the sun is indeed due east at the horizon at 6 am, but the rest of that day it travels around the compass points without leaving the horizon. As summer progresses, the sun's path is along a circle progressively more elevated (and of smaller radius) until by solistice it is of apparently elevation sin(\alpha) and radius cos(\alpha). Then it grows larger and lower until autumn equinox and then 6 months darkness. (I would appreciate corrections, corroborations, or questions, since these are rather specific predictions based only on the mathematics.) =============== 5. Dealing with clock time======================== Typically we want to find the sun knowing only the time on the clock, rather than the time since what we have called NOON. Clock time can be converted to Greenwich Mean Time by adding a conversion factor based (mostly) on longitude (and adjusting for summer time changes). If the observer's longitude is \beta degrees West, then the observer is _roughly_ (\beta)/15 time zones ahead of Greenwich, and so (local political decisions excepted) the clock time in Greenwich, when the observer records the time as T, should be T + round( \beta / 15). On the other hand, local NOON will occur precisely \beta/15 hours after NOON occurs in Greenwich, that is, NOON is at 12:00 + \beta/15 GMT Comparing these formulas we see that NOON occurs for an observer when the observer's clock reads 12:00 + {x-round(x)} where x=\beta/15 Thus, barring irregularities due to political considerations, NOON should occur roughly at 12:00, the difference directly calculable from longitude. Users ought to be able to add or subtract a whole number of hours to get correct correspondences (India, Saudi Arabia, and other areas excluded). =============== 6. Calculating time and place of dawn and dusk============= As was the case with NOON, it is helpful to treat \psi as a constant in these cases too (i.e. to ignore the progress of the earth around the sun within a single day). Thus, with a fixed observer (hence \theta) and on a fixed day (hence \psi) we may write U in the form (see section 3) U = A cos(\phi1) + C. "Dawn" and "Dusk" are the moments when U = 0; we solve to find \phi1. Clearly these are the values of \phi1 making cos(\phi1) = -C/A, i.e. \phi1 = arccos(-C/A). = arccos(-sin(\alpha) tan(\theta) tan(\psi) / sec(\phi0) ) where sec(\phi0) is the square root described in section 3: sqrt( 1 + [ tan(\psi).cos(\alpha) ] ^ 2 ), chosen to have the same sign as cos(\theta) cos(\psi). The two smallest solutions to this (in absolute value) are the values of \phi1 which describe the time before or after NOON to get dawn and dusk respectively. Thus the length of the (sunlit) day is the difference between these two, i.e., 2 arccos( -C/A ). We will write \phiD for the value of \phi1 corresponding to dusk, so that a day is 2\phiD (in hours, 2(\phiD * 24/(2\pi)) .) We also recall that the position of the sun at dusk is (N,0,E) where E = -sqrt(1-N^2) (we use the negative square root since the sun sets in the West.) The value of N was obtained at the end of section 3: N = sin(\alpha) sin(\psi) sec(\theta). It is traditional to report a position such as that of the setting sun by giving its bearing north (or south) of east (or west). The angle in question is arctan(N/|E|) north of west, or simply arcsin(N). This is not directly east/west, but clearly depends on the tilt of the earth (\alpha), the oberver's latitude (\theta), and the time of year (\psi). ================= 7. Solar Power=========================== The sunlight which falls can be measured as an amount of energy. This energy is useful in heating a house in winter, and an obstruction to cooling in the summer. The amount of sunlight translates rather directly to energy gain by plants in a garden, although gardeners will have to discover for themselves how effectively plants can use this energy for growth and flowering. Those interested in solar energy collection may wish to adjust the positions of their solar collectors based on the following information. The solar radiation reaching earth's orbit is quite constant over time; it varies most because the earth is at some times a bit further from the sun. The variations we contend with however are a function of energy loss between the light's arrival at earth orbit and our reception of it at the earth's surface. Radiation of photons from a hot source is the classic black-body problem of quantum mechanics. The total energy radiated is proportional to the mass of the object and to the fourth power of its temperature. However, those radiating photons have many different wavelengths; the distribution of how many photons are emitted of each wavelength is a well-known but complex function of temperature and wavelength which I will not reproduce. The intensity of the radiation diminishes by a factor proportional to the square of the distance from the sun. At the earth's orbit, the total energy from the sun, across all wavelengths, is 1.40 kw/m^2 (that is, 1.40 kwh per square meter of collector per hour of exposure). What happens to this energy? For the most part, the photons travel through the air and onto the surface of the earth, where they are either reflected or absorbed (the latter causing a gain in energy of a molecule, that is, the substance gets warmer). Not all wavelengths are absorbed or reflected equally (hence "opacity" and "color" of some objects). Among the substances which absorb the solar radiation are clouds and obstacles (trees, buildings) -- local variations for which I cannot account in this discussion. The one remaining variable is the angle of inclination of the surface to the line of the solar rays. If the surface is perpendicular to the incoming rays, the light collection is maximal; if tilted, one only gets as much energy as was in the photons of light whose path is intercepted. Thus we can measure the _percentage_ of the maximal amount of energy which a tilted surface will receive. This is just the area of the projection (shadow) of the collector onto the plane perpendicular to the solar rays, divided by the total are of the collector. Mathematically, this is the dot product of: the unit vector in the direction of the sun, and the unit vector perpendicular to the collector. The former is the vector (N,E,U) described at the end of section 3. For the latter, I will mention several special cases: (1) A solar collector trained to point always to the sun has unit normal vector also equal to (N,E,U). In this case, the dot product is 1, i.e., this solar collector is collecting energy at 100% of the maximal rate. (2) Flat, level ground has a unit normal (0,0,1). Thus, the energy collected in such a garden is (0,0,1)*(N,E,U) = U percent of maximal. (U is given at the end of section 3.) (3) A vertical wall facing south has unit normal (-1,0,0); other vertical walls have normals (a,b,0), where the relative values of a and b measure the orientation of the direction faced, north and east. The amount of sunlight falling on this surface is (aN+bE) of maximal. In all cases, if the dot product is negative, this indicates the sun is behind the collector. Possibly a solar collector can be 2-sided to take account of this, but a garden or a house wall will collect no energy when the sun is below the horizon or when the wall is in shadow, respectively. (Ignoring, of course, the collection of reflected light, moonlight, and so on.) Given the amount of solar flux at all times t in a day, one can integrate across all times to get the total solar energy collected daily. For example, a solar collector which may move to follow the sun collects an amount proportional to the length of the sunlit day (see section 6). For a fixed-direction, one-sided collector with normal vector (a,b,c), the integral must be done over the time during which the sun is on the appropriate side of the collector. Recalling our previous formulas: N= - sin(\theta) cos(\psi) sec(\phi0) cos(\phi1) + sin(\alpha) cos(\theta) sin(\psi) E= - cos(\psi) sec(\phi0) sin(\phi1) U= + cos(\theta) cos(\psi) sec(\phi0) cos(\phi1) + sin(\alpha) sin(\theta) sin(\psi) we will write these as N = A cos(\phi1) + B E = C sin(\phi1) U = D cos(\phi1) + F Then the condition that the sun be in front of a collector with normal (a,b,c) is that [Aa+Dc] cos(\phi1) + [Cb] sin(\phi1) + [Ba+Fc] >= 0. The easiest way to solve such an inequality K cos(\phi1) + M sin(\phi1) + P >= 0 is to divide by sqrt(M^2+N^2) and then find an angle \gamma with sin(\gamma) = K/sqrt(M^2+K^2) cos(\gamma) = M/sqrt(M^2+K^2) (which implies \gamma is an arctangent of K/M) and then use a trig identity to rewrite the condition as sin(\phi1 + \gamma) + P/sqrt(M^2+K^2) >=0. For P sufficiently large, the solution set is empty; if sufficiently negative, the solution set is the set of all \phi1. But in general, this will hold for all values of \phi1 in an interval, say the interval (\phi2, \phi3). Since in applications we get no sunlight when the sun is below the horizon (even if we point our collector in the "right" direction!) we will intersect this interval with (-\phiD, \phiD) so as to assume the light really is in front of the collector for times from \phi1= \phi2 to \phi3. The integral of the solar flux is then [Aa+Dc][sin(\phi3)-sin(\phi2)] - [Cb][cos(\phi3)-cos(\phi2)] + [Ba+Fc][\phi3-\phi2] An important special case is an upward facing collector, that is, the flat ground. In this case, \phi2 and \phi3 are the moments of dawn and dusk respectively, and we get simply 2D sin(\phiD) + 2F \phiD, that is, 2 cos(\theta) cos(\psi) sec(\phi0) sin(\phiD) +2 sin(\alpha) sin(\theta) sin(\psi) \phiD I will not address the issue of total energy gain across a year, which if nothing else can be found by adding 365 daily values. ================= 8. Estimation of Errors=========================== Throughout this analysis we have made assumptions which will introduce errors into our predictions. Probably the worst error here is ignoring the 1/4 of 1% change in the earth's position (\psi) within the course of a day. While the earth's orbit is very nearly circular, the ellipticity most noticeably means that the earth's orbit is not centered at the sun (small change in ratio of semi-axes can imply a rather larger change in separation of foci). While distance to the sun is more or less irrelevant here, Kepler's laws imply the earth's revolution around the sun is rather slower when the earth is at its furthest point. (I believe d\psi/dt is about 6% slower in June.) Presumably ignoring this effect is of less consequence than ignoring d\psi/dt altogether. Replacing the observer-sun vector with the earthcenter-sun vector changes the apparent position of the sun by about 1 minute of arc. (The mean earth-sun distance is about 23491 earth radii. This is about 3.4% greater on July 4 than January 4, however.) I am not sure what effect the non-sphericity of the earth has on the sun position, but the difference in observer-earthcenter distances is about 1/3 of 1%. I am always happy to hear of modifications I should make to this file. ============================================================================== [ jbartlo@ouchem.chem.oakland.edu (Joseph Bartlo) sent me some extremely ] [ helpful corrections to the documents and programs in this directory. We ] [ both agreed I ought to roll his corrections into my documents but I haven't ] [ had time to do so yet; I thought I ought to set them out to avoid the ] [ propogation of errors, though. -- djr ] ============================================================================== Date: Wed, 2 Aug 1995 05:37:26 -0400 To: rusin@math.niu.edu From: jbartlo@ouchem.chem.oakland.edu (Joseph Bartlo) Subject: SUNCALC Hi Dave, I find your SUNCALC explanation quite interesting. Your solution to the problem of finding solar position is well thought-out; and as you say, mathematically sound. However, in some ways I think you go about solving the problem 'the hard way'. Perhaps more importantly, a few things are neglected which significantly affect solar position; which I point out below. 2 references which are extremely useful for solar position and other astronomical calculations are: Explanatory Supplement to the Ephemeris (to the Astronomical Almanac) Astronomical Algorithms The top one is published by the United States Naval Observatory, and the one below was written by Jean Meuss. The Explanatory Supplement includes detailed diagrams, geometry, and equations. Jean Meuss' book does somewhat, and is more concerned with practical aspects of getting useful answers efficiently. I have FORTRAN programs which calculate solar position and sunrise/ sunset times based on those references; and ones to calculate solar energy flux considering scattering and absorption by our atmosphere and earth. Those are both for broadband (photon frequency not considered) and spectral (photon frequency considered) solar energy. I can send you any or all of those programs, if you wish. The 2 *main* problems with your method of calculation are: In calculation of solar position, earth's ellipsoidal orbit has a greater effect than you realize. True, earth's variation in orbital speed is small over a day; but over a sequence of days, change in \phi accumulates; to the point that *time of solar noon (NOON) varies by about 30 min in local time over the course of a year*. When expressing solar energy collection, atmospheric influenece is much too great to neglect without large error. Many atmospheric constituents both absorb solar energy and scatter it in all directions, a significant amount back to outer space. Thus, solar energy flux over our entire globe at locations under cloudless skies is only about 2/3 of what it would be if no atmosphere were present. Clouds increase that efffect to the point that slightly less than half reaches ground. Because of scattering and absorption of solar energy, solar incidence angle does not have as great of an effect on its collection than it otherwise would. Below I include comments to your discussion APPARENT POSITION OF THE SUN IN THE SKY and SUNCALC program, since you asked for them. I appreciate your effort at an original solution to this problem:) Joseph Bartlo Introduction: You say "we" when you mean "I". Unfortunately, 'we' are taught to do that:) Section 1: You mention "about 23 degrees, familiar from all desk globes". I state this because I grew up thinking that obliquity of the ecliptic was 23 1/2 degrees *exactly*, because it was written as such (or 23.5 deg) 'everywhere'; while it is 23.44 deg and gradually decreasing. It is intersting that the intersection of your 2 planes with our earth's orbit *define* equinoxes. Contrary to popular usage, equinoxes *are those point of intersection*, and the days we call equinoxes are those days our earth passes by them. You state in one place "earth's revolution about the sun" when you obviously mean 'about its axis'. It is not clear to me what \alpha is, since it is not clearly defined; but I think it is obliquity of the ecliptic. Section 3: Are you sure that NOON is not the time at which our sun is highest in the sky in some places? A physical analysis states that dU/dt would be exactly 0 at noon if solar declination did not change, which is related to /phi. Still, even though solar declination changes; increasing during our winter and spring and decreasing during our summer and fall; its incremental change just before or after solar noon may be less than that in arc of the ecliptic. That is stated in my terminology, if you can understand it. At the poles, where our sun continuously rises or sets, solar noon is obviously not the time at which solar elevation angle (angle of our sun's center above the horizon) is greatest; but at lower latitudes, I think it may be. Interesting question. Section 4: Neat explanation, but I have not picked at details regarding it. I learned a few things from it:) Section 5: Your discussion regarding clock time is true except for effects due to variations in d\phi/dt. As stated above, that causes a very significant change in local clock time of NOON over the course of a year. For example, if NOON occurs at 11:58 am at a location on 6 NOV, it will occur at 12:28 pm on 9 FEB. Mean solar time is defined as you have done, such that each day is exactly 24 hours; rather than being slightly more or less, depending on season. An 'equation of time' is used to account for that. It is difference between solar longitude and right ascension at a specific time; with angles expressed as time (15 deg per hour), as you do. Equations to calcuate it are usually expressed as Fourier series based on theory and observation, in which only a few terms are significant. That is the largest source of error in your calculation of solar position. Section 6: By "Dawn and Dusk", you obviously mean 'sunrise and sunset'. This is mainly a question of terminology, but I address it because dawn and dusk are often thought of as twilight times before sunrise and after sunset. You may consider the following definitions expressed in the Astronomical Almanac (beginning and ending times): Sunrise/sunset - solar zenith angle of 90 deg 50 min Civil twilight - solar zenith angle of 96 deg Nautical twilight - solar zenith angle of 102 deg Astronomical twilight - solar zenith angle of 108 deg One can note that all occur when solar zenith angle (angle between Up-axis and a ray to the apparent position of our sun) is > 90 deg. Sunrise/sunset occur at such an angle due to atmospheric refraction; bending of light by our atmosphere. Thus, when our sun's center is on the horizon, it would be about 50 arc min below the horizon if no atmosphere were present. Sunrise and sunset times are typically defined according to the position of our sun's center, to be consistent with other stellar observations. However, our sun's disc occupies about 32 arc min in our sky. I define sunrise and sunset times to be those at which our sun comes in or goes out of view. At low and middle latitudes, atmospheric refarction increases day length by 5-10 min, but by much more at higher latitudes, at which location our sun's apparent position remains very close to the horizon for greater time periods. At arctic locations, even calculations based on our sun's center and upper limb differ by tens of minutes in some circumstances; but only are 2-3 min different at low and middle latitudes. Section 7: In discussing solar energy, you state kWh in a place where you obviously mean kW. You may consider the following definition: solar constant - solar energy flux perpendicular to the solar beam at our earth's mean distance to our sun, in abscence of atmospheric influence The 'solar constant' is not constant, but slightly varies due to sunspots, solar flares, other solar acivity; and to a very small extent, interplanetary matter. However, it is about 1370 W/m^2; and typically varies by only a few W/m^2. Of course, the amount reaching our earth's atmosphere (extra-terrestrial) varies as you stated; from about 1330 W/m^2 at perihelion to 1415 W/m^2 at aphelion. Solar energy is greatly scattered and absorbed, even by a 'clear' atmosphere. Gas molecules scatter it approximately proprtional to frecuency^4. Thus, colors toward the blue end of the visible spectrum are preferentially scattered, and our sky appears blue. Aerosols scatter proportional solar energy to frequency. They also scatter more toward the blue end of the visible spectrum, but also significant amounts in other colors. Thus, on a day with many aerosol particles (dust, smoke, pollen, salt, etc.) in our atmosphere, a milky yellow-white appearance may is seen around the solar disc. Clouds scatter solar energy fairly equally in all frequencies, and thus appear whitish. Thier bases often appear dark beacuse very little solar energy is reflected up toward them and scttered back toward ground. You may consider typical summertime magnitude of solar energy flux (in W/m^2) near NOON in Illinois: Extra-terrestrial solar energy flux: 1335 cos(40-22) = 1270 CLOUDLESS SKY Global (all) solar energy flux incident to horizontal: 1000 Direct solar energy flux (solar beam) incident to horizontal: 850 Diffuse solar energy flux (scattered) incident to horizontal: 150 CLOUDY SKIES (averaged over a sufficient time period) About 20% cloud-covered : Global 900 Direct 720 Diffuse 180 About 50% cloud-covered : Global 700 Direct 450 Diffuse 250 About 80% cloud-covered : Global 500 Direct 280 Diffuse 220 Overcast : Global 200 Direct 2 Diffuse 198 Cloudy sky amounts are those typical, and depend very much on cloud types. Section 8: Parallax (difference between observations from earth center and surface) makes a difference in solar elevation angle of up to a little more than 8 arc min. That is small compared to some other inaccuracies. SUNCALC: I notice that you mistakenly put "June equinox" and "December equinox" in your output. It seems that "energy collection per day" includes nighttime periods as well as daytime, since that is much less than percentages at specific times. ============================================================================== Date: Fri, 11 Aug 1995 20:14:50 -0400 To: rusin@math.niu.edu (Dave Rusin) From: jbartlo@ouchem.chem.oakland.edu (Joseph Bartlo) Subject: Re: SUNCALC Hi Dave, Thank you for your kind comments. I think it would be best if you did rewrite my comments, so that they fit in better with the rest of your text (not seeming akward because of grammatical differences). It is fine with me if you do so, so long as you mention that I made a few suggestions. I have 3 corrections to make on what I wrote to you: I refered to earth's orbit as ellipsoidal, whereas now I recall that an ellpsoid is a 3-D trace of an ellipse about one of its axes. Quasi-elliptical is a better term for it. I got aphelion and perihelion reeversed, as I do too often. Jean Meeus' name is as spelled here. His is an excellent book to reference! I am happy that my comments are helpful to you. Please do not hesitate to send a messgae to me if you have further questions, comments, or ideas you wish to discuss. Joseph