INTRODUCTION TO CLIMATOLOGY Preface EAS 488, Introduction to Climatology, consists of three sections, 1. Physical Principles of Climate. The mathematical framework for understanding climate. *COMPUTER PROGRAMS FOR THE EQUATIONS WILL BE PUT ON THE WEB.* 2. Climates of the World. A descriptive section that includes the impact on and interaction of climate with vegetation, animal life, landforms and mineral content of soils. For this section you will memorize and draw the world climate map, as generations of students before you have. 3. Climate Change. The synthesis of earlier material including descriptions of past climates, evidence for climate change, and theories and models for why climates change. There is a test at the end of each section. There are also homework assignments due a week after being handed out. The assignments include calculations. There will be two field trips to the Bronx Botanical Gardens and the American Museum of Natural History. There will also be three PowerPoint presentations, given in class with some choice of topics, namely 1. 2. 3. 4. 5. 6. An anomalous Region (e. g., Venezuela coast) or Period (e. g., El Nio) of Climate. A Geological Climate Indicator. Satellite (MODIS) interpretation of a selected region of Earth A Climate of the Past. A society or civilization terminated by or with the help of a climate change. A simple climate model.

Much of the course material is spread over many popular books, textbooks and journals. This is one reason for these notes and for E31_CLIM.PPT. Some books on climate and/or weather are: 1. Physical Climatology, W. D. Sellers, 1965. 2. Earths Climate: Past and Future, W. Ruddiman, 2000 3. Introducing Physical Geography, A. Strahler and A. Strahler, 2000. 4. The Science and Wonders of the Atmosphere, S. D. Gedzelman, 1980. 5. Earth System History, S. M. Stanley, 2004. 6. Climate System Modeling, K. Trenberth, Ed, 1992. 7. Climatology: An Atmospheric Science, J. Hidore and J. Oliver, 2002 8. Weather and Climate, Aguado and Burt, 2001 9. The Climate Near the Ground, R. Geiger and R. Aron, 2003 10. Global Physical Climatology, D. Hartmann, 1994 11. Abrupt Climate Change, Nat. Res. Council, 2002 12. World Survey of Climatology, 15 vol 13. Climate, History, and the Modern World, H. Lamb, 1995 14. Climatology and the Worlds Climates, G. Rumney, 1968 Now we begin with the

PHYSICAL PRINCIPLES OF CLIMATE Introduction Climatology begins with the sun. The Sun shines down on the Earth, providing more heat where it is higher in the sky and where the day is longer. The Suns heat is (1) reflected or absorbed in the atmosphere and at the Earth's surface. It is then (2) reradiated as infrared waves, (3) conducted directly to the depths of the Earth or to the atmosphere by contact, (4) carried aloft by convection, or (5) used to evaporate water. Temperature differences arise from the unequal distribution of sunlight over the Earth and the variations in the Earth's surface. The temperature differences produce the winds, storms and ocean currents, which in turn stir the atmosphere or oceans and act to reduce the temperature differences. Climate, the synthesis of weather, is the end product of these interacting processes and phenomena. The first questions to answer involve the simple thermal response of the atmosphere or ocean to the various heating and cooling processes. Thus, we want to know how much the temperature will rise or how much water will be evaporated by a given amount of sunshine, or how much a motionless body of air or water will cool during the long polar night winter. These questions appear in a variety of forms. It is possible to ask A: how many kilograms of snow can be melted by 1000 joules of sunshine or B: how many centimeters of snow pack will melt after an hour of sunshine. The first question is more straightforward. Simply use the first law of thermodynamics. The second involves the density of snow and the time as well. Thus we must begin with the units, dimensions, and observed or measured values of many quantities. Units We use the Systeme Internationale (SI). This involves Quantity Abbreviation Basic Value Dimension Length L meters m Time t seconds s Mass m kilograms kg Temperature T Absolute Area A, dA m2 Volume V m3 Velocity u, v, w (Speed) m s-1 Acceleration a m s-2 Density kg m-3 Force F newtons kg m s-2 Pressure p pascals (Pa) kg m-1 s-2 Energy (Heat) E dQ joules kg m2 s-2 Power P watts (W) kg m2 s-3 Irradiance I (Flux density) kg s-3 Humidity w Mixing Ratio parts per thousand ()

By knowing the dimensions of the quantities in the left hand column it is possible to work out their interrelationships. Thus, force = mass acceleration, or F = ma The acceleration of gravity, g, is g = 9.8 m s-2 Pressure = force divided by area, p = F/A Mean sea level atmospheric pressure is p(MSL) = 101325 Pa Also, density = mass volume, = m/V The densities of several common substances are given below Substance Fresh Water Sea Water Pure Ice Sea Ice Granite Iron Limestone Density (kg m-3) 1000 1025 920 880-930 (depending on air bubble content) 2650 7700 2150-2860

The Atmospheric Column: Many quantities such as pressure (F/A) and solar irradiance (Power/A) involve the cross sectional area. It is useful when talking about the atmosphere to treat a column of air or water. An example is the depth of rain or snow, dz. The volume of an air column is the product of the cross-sectional area, A, and the height, dz, dV = Adz Time is crucial. Thus, Power = Energy divided by time. P = dE/dt

The solar irradiance is energy (or heat) divided by time and area, or I = dE/(Adt), or dE = dQ = IAdt Warning! You will be sorry when you forget these! Example: If solar irradiance is 1000 W m-2, how much energy can a pool with an area of 200 m2 absorb in an hour? dE = dQ = IAdt = 10002003600 = 7.2(10)8 joules For air, the density can be related to the temperature, pressure and the vapor content or mixing ratio, w, by a special form of the Ideal Gas Equation for Moist Air p = RT(1+.61w) = 287T (1+.61w) Example: What is the density of air with p = 105 Pa, T = 273 K, w = .001 = p/[287T(1+.61w)] = 105/[287273 (1+.00061)] = 1.2755 kg m-3 Since density is variable, the following is not straightforward: How much heat is needed to raise the temperature of the lowest kilometer of the atmosphere. This problem can be simplified with the Hydrostatic Equation dp = -gdz Example: A 1000 m thick layer of atmosphere has an average density of = 1.25 kg m-3. What is the pressure difference between the top and bottom of the layer? (Here we can neglect the minus sign since only the magnitude of the difference is needed.) dp = gdz = 1.259.81000 = 1.225(10)4 Pa The Solar Constant Satellites and earlier measurements have shown that the Suns output is almost constant but varies slightly over time, for example with the sunspot cycle (top figure next page). The bottom figure suggests that solar irradiance is now slightly larger than during the depths of the Little Ice Age. The Suns output is expressed as an irradiance or flux density. When the Sun is overhead at an average distance from Earth each square meter at the top of the atmosphere receives 1367 W.

I(SO) = 1367 W m-2 Source: http://en.wikipedia.org/wiki/Solar_variation

Thus, if a window had an area, A = 1 m2 and the Sun came directly through, the inside would be as bright as if almost 14 different 100 watt light bulbs were turned on. Since sunlight can either raise temperature or evaporate water, we want to be able to answer such questions as, how much sunshine does it take to raise the temperature of the atmosphere by 1 or to evaporate 1 m of water? The first law of thermodynamics, a form of the law of conservation of energy, provides the answer to these questions. The first law of thermodynamics usually includes only thermal energies. Potential and kinetic energies are generally excluded because they tend to be surprisingly small in comparison with the thermal energies. The First Law of Thermodynamics (without work due to compression) dQ = mcdT + Ldm where dQ is the heat input to the system, c is the specific heat capacity (E/(m*deg)), L is the latent heat (E/m), dT is the temperature change and dm is the amount of mass that has changed phase (by melting or boiling). Before giving any examples it is essential to have a table of values of specific heats and latent heats of various substances. Substance Ice Water Vapor Air Iron Quartz Specific heat Capacity 2100 j kg-1/K 4186 1847 1004 (at constant pressure) 760 (at constant volume) 210 840 Latent heat 3.34*105 j kg-1 2.50*106 2.25*106

Melting Vaporizing (0 C) Vaporizing (100 C)

Example: The Sun provides 106 joules to heat 104 kg of air. Find the air temperature increase. dQ = mcdT + Ldm = 106 = 1041004dT + 0 Solving for dT yields dT = 106/(1041004) = 0.0996 or almost 0.1 K. Example: The Sun provides dQ = 106 joules to evaporate water. How much water will evaporate if none of the heat is used to raise the water's temperature? dQ = 106 = mc0 + Ldm = Ldm = 2.5106dm

Solving for dm (the mass that changes phase by evaporation) yields dm = 106/(2.5106) = 0.4 kg Often it is more useful to know the amount of evaporation in meters rather than kilograms. Thus, an often asked question is, how many meters will a lake level decline as a result of evaporation. It is also often important to know the rate of evaporation or the rate of temperature change as a result of heating. Thus, how long will it take to heat the atmosphere by 1? To answer these questions, begin by dividing the first law of thermodynamics by the area, A, of a column of atmosphere or water. It then becomes dQ/A = mcdT/A + Ldm/A If the rate of heating or evaporation is wanted, divide by the time interval, dt. Thus, dQ/(Adt) = mcdT/(Adt) + Ldm/(Adt) Next, use the relationship between mass, density and volume, m = V = Adz The first law of thermodynamics then becomes dQ/(Adt) = dzcdT/dt + Ldz/dt To express in terms of pressure thickness of a layer of air, use the hydrostatic equation dz = dp/g and the definition of the mixing ratio w = vapor/air to get dQ/(Adt) = dpcdT/(gdt) + Ldwdp/(gdt) Since the equations divided by Adt have the same units as irradiance they are very useful for computing temperature changes and evaporation rates due to sunlight. Example: Bright sunlight (I = 1000 W m-2) heats the lowest 100 mb (dp = 104 Pa) of the atmosphere for an hour. How much does the temperature rise? The best form for the first law of thermodynamics for this example is,

dQ/(Adt) = dpcdT/(gdt) + Ldwdp/(gdt) = dpcdT/(gdt) Solving for dT gives dT = Igdt/(cdp) = 10009.83600/(1041004) = 3.51 K Example: The Sun shines down on the ocean for five hours with an irradiance of 700 w m-2. If the water is 10 meters deep, what is the increase of temperature? dQ/(A*dt) = 700 = cdzdT/dt = 1000104186dT/(53600) Solving for dT gives dT = 70053600/(1000104186) = 3.01*10-2 Considering that the ocean is a lot deeper than 10 meters it is little wonder that it takes so long to heat it up or cool it off. Now lets melt some snow. Example: New York gets a snowfall of 0.5 meters (fat chance this will happen again in our lifetimes). If the snow has 1/10 the density of water (or snow= 100 kg m-3), a typical value and the albedo, A, of the snow is 80% (the snow reflects 80% of the sunlight that strikes it) then how many hours of bright winter Sun at I = 300 W m-2 will it take to melt the snow? dQ/(Adt) = Lsnowdz/dt and the absorbed sunlight is what is not reflected, or dQ/(Adt) = I(1-albedo) = 300(1.0-0.8) = 60 W m-2 Combining these yields dQ/(Adt) = 60 = Ldz/dt = 3.34(10)51000.5/dt Solving for dt yields, dt = 2.78(10)5 s or about 77.2 hours. Considering how short the winter day is you can see that the snow will stick around quite a while unless there is a heat wave. Solar Radiation

The atmosphere derives more than 99.9% of its energy from the Sun. The remainder comes mostly from the bowels of the Earth. If the Sun turned off suddenly temperatures on Earth would gradually decrease to near absolute zero and the oceans and atmosphere would freeze solid. There are three basic laws regarding the amount of sunlight received at any place on Earth. Two of these laws are geometrical. They are the inverse square law and the cosine law. The third law describes (mathematically of course) how radiation weakens as it passes through the atmosphere. The inverse square law is almost intuitive. Light from any source spreads out in all directions. Just imagine a flash of light spreading out like an expanding spherical shell. The total energy remains the same but as the light spreads over an increasing area, its concentration or irradiance weakens. The area times the irradiance remains constant. Since the area is proportional to the radius (or distance) squared, we get The Inverse Square Law of Irradiance I(x) = I(y)[y/x]2 One of the best examples is finding the irradiance of direct sunlight on another planet. On Earth, the solar constant is 1367 W m-2 and the average distance between the Earth and Sun is 149.5106 km. Example: Find the solar constant on Mars at its average distance of 228106 km from the Sun. I(Mars) = 1367 [149.5/228]2 = 588 W m-2 The inverse square law can also be used to solve for distance. Example: You arrive on a planet and want to know how far you are from the Sun. You measure the direct solar irradiance as 125 W m-2. What is the distance to the Sun is I(planet) = 125 = 1367 [149.5/d]2 Solving for d yields d = 149.5*[1367/125]0.5 = 494106 km The distance from the Earth to the Sun varies over the year because the orbit is an ellipse. The Earth approaches as close as 147 106 km to the Sun around Jan 3 (perihelion) and gets as far as 152106 km from the Sun around July 3 (aphelion). A simple and reasonably accurate formula can be used to tell the distance of the Earth from the Sun on any day of the year. In its simplest form, give every month 30 days so that the year has 360 days. Then in millions of kilometers, we have

Earths elliptical orbit is shown in the diagram below with the eccentricity of the ellipse greatly exaggerated. Note that Earth moves faster around the Sun when it is closest to the Sun. Since the Earth is closest to the Sun on around January 3, this makes Fall and Winter shorter seasons than Spring and Summer. This can be seen from the table below of the timing of the Solstices and Equinoxes during 2006 and 2007. Many of these features change slowly over thousands of years. The Sun now lies 2.5 million km from the center of the orbit in one of the foci of the ellipse but gets as close as 0.7 million km and as far as 9 million km over complex cycles of 100,000 and 413,000 years. The tilt of Earths daily axis of rotation (now 23.45) varies between 22.2 and 24.5 over a cycle of 41,000 years. The dates of perihelion and aphelion advance with respect to the solstices and equinoxes by roughly 1 day every 60 years so that it advances an entire year (365.25 days) every 60365.25 22,000 years. This is the orbital cycle called precession.

Event Equinox Spring Solstice Summer Equinox Fall Solstice Winter

Time Duration Mar 20 18Z 92d 18h Jun 21 12Z 93d 16h Sep 23 04Z 89d 20h Dec 22 00Z 89d 00h

All the slow cycles of Earths orbit have produced strong climate cycles. For example, for the past 800,000 years, the Earth has experienced major Ice Ages every 100,000 years. Newtons 2nd Law provides the basis for finding Earths velocity of revolution about the Sun. The force of gravitation produces centripetal acceleration and leads to, GM S m E d SE2

= mE

vE d SE

2

Earths (or any planets) velocity is, vE = GM S d SE

Example: Compare Earths velocity at aphelion and perihelion. G = 6.67(10)-11 MS = 1.99(10)30 kg vE(aphelion) = [6.67(10)-11(1.99)(10)30/1.52(10)11]0.5 = 29.55 km s-1 vE(perihelion) = [6.67(10)-11(1.99)(10)30/1.47(10)11]0.5 = 30.05 km s-1

Earth Sun Distance Formula de = 149.5 + 2.5cos{number of days from July 3} Example: What is the distance of the Earth to the Sun on September 13. This is 70 days after July 3. Then we find de = 149.5 + 2.5cos(70) = 150.36 million km The above formula contains small errors that arise because Earth moves in an elliptical orbit that obeys Keplers law, moving fastest when it is closest to the Sun and moving slowest when it is furthest. At present, this makes fall and winter a few days shorter than spring and summer. Very slow and relatively small changes take place in Earths orbit over time. These changes are almost periodic and lead to climate cycles. We return to the issue of Earths orbital changes in Section III. The second geometrical law of light concerns the relation between the irradiance of light on a surface and the angle between the light and the surface. The law is known as the cosine law. It too makes sense although it is a bit harder to derive because it involves trigonometry. When an object faces a light directly (at right angles) it receives the maximum possible irradiance. As the object turns obliquely, it intercepts less of the light and is not illuminated as intensely. When tilted at angle, Z, it intercepts a fraction, cos(Z), of the maximum possible light and its irradiance is therefore only 1/cos(Z) of the maximum. This relates directly to sunshine striking the Earth. When the Sun is overhead it is at the zenith. The zenith angle, Z, measures how far the Sun is from overhead. Now it is time to write the Cosine Law of Irradiance IZ = I0cos(Z) Example: When the Suns zenith angle, Z = 60 (30 above the horizon) find the irradiance of sunlight striking level ground.

I(60) = I0cos(Z) = 13670.5 = 683.5 W m-2 It is easy to find the Suns zenith angle at noon. On any day of the year the Sun is overhead at a latitude, equal to the declination), . The declination gets as large as 23.45 on June 21st and gets as small as -23.45 deg on December 21. We often approximate this by 23.5. The noon zenith angle is then given by Noon Zenith Angle Z= Rule: Latitudes are negative in the Southern Hemisphere. Of course there is still a minor problem - you must know the declination on any given day of the year. The declination varies almost sinusoidally and is given approximately by the Declination Formula = 23.5cos[number of days from June 21]30 Solar Declination 20 10 0 -10 -20 -30 0 1 2 J J M A M 3 4 5 6S 7 8 9D 10 11 12 A O N J F M Month

Again, each month is given 30 days so that for the purpose of this approximate formula the year has 360 days. Example: Find the declination on October 21. October 21 is 120 days from June 21. Therefore, = 23.5cos(120) = -11.75 Thus the Sun is overhead at 11.75 S.

It is also easy to tell the Suns azimuth (compass) angle, , at noon. The azimuth angle for due south is 0. It increases clockwise so that west is 90, north 180, and east 270. At noon, the Sun is either due south or due north. How do you tell which? If you are north of the latitude where the Sun is overhead then you must look back toward the south to see the Sun. Thus, for example, if you are at latitude 10 N then on March 21 the noon Sun will be due south (Z = 10) while on June 21 (when Z = 10 23.5 = -13.5) the Sun will be due north.

It is far more difficult to find the zenith and azimuth angles at times other than noon. These are given by Sun path diagrams or equations. The 3-D Sun path diagram gives a rather quick and easy feeling for these angles. There are also 2-D Sun path diagrams for each latitude that can be used to find these angles once you learn how to read the diagrams. But you must also know the formulas that give these angles. Sun Path Diagrams Knowledge of the Sun path provides us with the means to calculate solar irradiance. It also has a number of other useful applications. From ancient times, people have realized that the Sun is also a remarkably accurate timepiece, and have constructed sundials. Informed architects consider the Suns position. For example, in most of the middle latitudes it is better to put windows on the south side of a building than the north side. South-facing windows allow the low winter Sun to enter. In summer, when the Sun is in the south it is so high that very little sunlight will come in through a south-facing window, especially if there is an overhanging ledge above the window. North-facing windows get no light in fall and winter, allowing much valuable heat to escape, but in spring and summer they allow a surprisingly large amount of sunlight in - just when you don't want it.Sun paths for 30 North (top) and 30 South latitude (bottom) on July 21 showing the Sun at sunset. In July it is summer in the North Hemisphere with long days and a high Sun but winter in the South Hemisphere with short days and a low Sun. In the North Hemisphere outside the tropics the Sun moves across the sky from left to right. In the South Hemisphere outside the tropics the Sun moves across the sky from right to left.

To draw the Sun path, enlarge the local view of the celestial hemisphere in the figure of the Earth on the previous page and make the ground horizontal. Then the observer stands at the center of his world, directly under the zenith. The perimeter of the ellipse represents the horizon or the apparent edge of the Earth. The infernal hemisphere can also be included to show where the Sun would be seen below ground if the Earth were transparent. Instructions For Drawing the Sun Path

1. Choose a latitude, . In the diagram below, = 30 N. 2. Begin with the equinox because on the equinoxes the Sun rises due EAST and sets due WEST. Put large dots at the eastern and western horizons to represent sunrise and sunset. 3. Put another large dot at zenith angle, Z = to represent the noon position of the Sun. Connect the three dots with a curve. This curve represents the Sun path for the equinox. This is done below for. 4. At all other times of the year the NOON Sun is displaced to the north or south by an angle equal to the declination. Thus, on July 21, the NOON Sun is 20.3 to the north of the equinox noon position. Thus it has a zenith angle equal to Z = 30 - (20.3) = 9.7 5. Finally, ALL SUN PATHS AT A GIVEN LATITUDE ARE PARALLEL. Draw a curve parallel to the equinox path that passes through the noon point, Z = 9.7

Now that we have the picture of the Sun path it is time to present the formulas for solar zenith angle and azimuth angle. After that I will present the 2-D Sun path diagrams First and most important is the formula for the cosine of the zenith angle. Zenith Angle Formula cos(Z) = sin()sin() + cos()cos()cos(h) where h is 2/24 radians hr-1 = 15 hr-1 from noon. Example: Find the zenith angle of the Sun on September 11 at 3 PM at latitude 41 N. First, note that at 3:00 PM, h = 45. Second, note that since September 11 is 80 days from June 21, = 23.5cos(80) = 4.08

Then, cos(Z) is given by, cos(Z) = sin(4.08)sin(41) + cos(4.08)cos(41)cos(45) = (.071)(.656) + (.997)(.755)(.707) = 0.579 Then find Z by taking the inverse cosine Z = cos-1(.579) = 54.6 The formula for the azimuth angle () is longer but is also just a plug in. Azimuth Angle Formula tan() = {cos()sin(h)/[cos()sin()cos(h) - sin()cos()]}

Example: Consider Sept 11 at 41 N at 3 PM once again. Then tan() =(. 997)(.707)/[(.997)(.656)(.707) + (-.071)(.755)] = 1.724 Taking the inverse tangent yields = tan-1{1.724} = 59.9 Now the 2-D Sun path diagrams will make sense. These are circular graphs that show both zenith angle (the center is the zenith and the outer perimeter is the horizon) and the azimuth angle. 2-D Sun Paths On any given day the Sun `burns' a path across the 2D Sun path diagram, as in the diagram to the right for July 21 at 30 N. Sun path diagrams for various latitudes and for any day of the year appear on the next two pages. Now, using the formula for the zenith angle it is a rather simple matter to solve for the time of sunrise or sunset. At sunrise or sunset the Sun is at the horizon, 90 from the zenith. Thus Z = 90 and cos(Z) = cos(90) = 0. Sunset hour is denoted as H. Then solving for H, we get,

Sunrise and Sunset Hour H (hr) = cos-1[-tan()tan()]/15 Example: Find sunset time at 41 N on September 11. H(hours) = cos-1[-tan(4.08)tan(41)]/15 = 6.24 hr = 6:14 PM Example: Find the Suns azimuth angle, , at sunset on January 21 at 45 N latitude. This is more difficult because we must first find the declination, on Jan 21 (210 days after June 21). = 23.5cos(210) = -20.35 and then find the hour of sunset, H = cos-1{-tan()tan()}/15 = cos-1(.371) = 68.2 Now, plug into the long formula for to get = 60.34 Thus the Sun sets in the WSW and agrees with the 2-D Sun path diagram.

Penetrating the Atmosphere The atmosphere scatters or absorbs some of the sunlight and hence allows only a certain fraction to reach the ground. The fraction of direct sunlight that penetrates to sea level when the Sun is directly overhead is called the transparency, a. A typical value for atmospheric transparency is a = 0.7, meaning that 70% of direct sunlight reaches sea level when the Sun is directly overhead (at the zenith) and 30% is either absorbed or scattered around the sky providing skylight. When the air is pure and clean (i. e, no aerosols) about half the light scattered in the atmosphere continues downward. When aerosol particles are present, more than half the scattered light reaches the ground because most of the light scattered by aerosol particles is deflected by small angles. When the Sun is not at the zenith it must pass through the atmosphere obliquely. Then its effective path through the atmosphere becomes longer so that less light penetrates to sea level than when the Sun is overhead. The lower the Sun gets in the sky the longer the effective optical path or the more atmosphere a sunbeam must penetrate. Consequently, when the Sun is very low in the sky it is often weak enough to look at directly without burning your eyes out instantly, particularly if the air is hazy and polluted. The diagram to the left shows that the path

length is proportional to sec(Z) so long as the Earth's curvature can be neglected. This is called the plane parallel atmosphere and it is a good approximation so long as the Sun is more than about 5 above the horizon. Earths curvature shortens the path noticeably when the Sun is near the horizon. A simple outgrowth of Lamberts Law describes the way radiation penetrates the atmosphere when the Sun is not overhead. (Note: To be accurate, Lamberts Law must be applied separately for every wavelength we do not do this here.) Lamberts Law works like calculating probabilities or odds in repeated betting. For example, when Z = 60 the path of a sunbeam is sec(60) = 2.0 or twice as long as when the Sun is overhead. It is as if the sunlight must penetrate two atmospheres before reaching sea level. When atmospheric transparency is a = 0.4, the solar irradiance is only 40% of its initial value after penetrating the first atmosphere. After passing through the second atmosphere the irradiance would drop to 40% of the 40%, or 16% of the original value. Thus, Lambert's Law or Beer's Law of Light Transmission I(Z,a) = I(Z)asec(Z) When combined with the cosine law, Lambert's Law becomes I(Z,a) = I(0)cos(Z)asec(Z) Example: Find the irradiance of sunlight on level ground at the bottom of the atmosphere when Z = 60 and the transparency, a = 0.7. I(60,.7) = 1367cos(60)0.7sec(60) = 341.75 W m-2 Now, we solve for solar irradiance at any instant at any latitude and on any day of the year. To do this we must take into account all the factors that affect sunlight irradiance, namely the inverse square law, the cosine law and the transparency law. The factors must be multiplied and thus the grand irradiance formula is, Sunlight Irradiance Formula I = 1367[149.5/d]2cos(Z)asec(Z) where formulas for and Z have already been given. Example: What is the solar irradiance on level ground at 3PM on Sept 11 at 41 N when atmospheric transparency, a = 0.7? Look above to find that most of the calculations have already been performed. Thus we have I = 1367[149.5/150.36]2(.579)0.71/.579 = 422.6 W m-2 Not on the Level Solar irradiance is altered when the ground is not level. The slope of the land thus has important consequences in climatology and architecture. For example, glaciers and snowfields tend to form or

last longer on northern slopes of mountains. It is also important to compute sunlight incident on vertical walls of buildings. Ironically, north facing windows of buildings in themid latitudes actually get more Sun during summer than south facing windows. Thus, it is important to present the equation for

Solar Zenith Angle on Sloped Surfaces (Z) cos(Z') = cos(Z)cos(i) + sin(Z)sin(i)cos(') where i = slope angle, = azimuth angle of Sun from south ' = azimuth angle pointing down slope from south. Note that and increase clockwise from S. Example: Find the irradiance on a west facing wall at 3 PM on Sept 11 at 41 N. Recall Z = 54.62, = 59.89. For the case of a vertical wall facing west, i = 90 and az' = 90. cos(Z') = sin(54.62)sin(90)cos(59.89 - 90) = .705 and Z' = cos-1{.705} = 45.15 To get the irradiance, simply replace Z with Z' in the total irradiance formula (but keep Z in the transparency term) to get I = 1367(149.5/150.36)2(.705)0.71/.579 = 514.6 W m-2 This is more than on level ground because the Sun is already heading towards the western horizon and will hit a west facing vertical wall more directly than level ground. Daily Solar Irradiance Next in the endless chain of calculations (will the Sun never set on these?) we calculate the daily average solar irradiance on level ground. To do this, integrate the irradiance equation from noon to sunset (H is given in radians) and then divide by the time from noon to midnight ( radians). Integration is straightforward on top of the atmosphere (when transparency, a = 1.0) and leads to, Average Daily Solar Irradiance Iavg = 1367[149.5/d]2{Hsin()sin() + cos()cos()sin(H)}/ Example: Find the average irradiance at the top of the atmosphere at latitude 41 N on Sept 11? For this case we found H = 1.633 radians. Substituting into the daily insolation formula yields Iavg = 343.8 W m-2

The average irradiance can also be found by numerical or graphical 1200 I(Sun) integration and must be done this way 1000 when the atmospheres transparency is included (a < 1). An integral is the 800 area under a curve. Here a graph is 600 constructed with time plotted on the x axis and irradiance, I is plotted on the 400 y axis. The area under the curve can be approximated by making 200 rectangles and then multiplying the 0 height of each rectangle by its width. 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Add the average irradiance for each full hour the Sun is up. For partial hour hours of sunlight multiply by the fraction of the hour that the Sun is up. Say the Sun sets at 6:14, as it does September 11 at 41 N. Then the last hour of the day has only 14 minutes 14/60th = 0.233 hours. Finally, divide the sum by 24 hours to get the average. The numerical integration for September 11 at 41 N is shown in the graph above. In such examples there are many lengthy calculations and the computer can be one of your best friends. Now that all the formulas have been presented it is time to see the resulting Distribution of Sunlight on Earth When the daily insolation striking level ground is computed for every day of the year and every latitude, the graphs below result. The insolation has been calculated for two cases, namely, a = 1.0 (no atmosphere) and a = 0.7 (a typical transparency) on the next page. In these complex diagrams the declination (suns latitude) is shown by the yellow circles. To read the diagram, select a day and a latitude. For example, on April 21 at 45 N, average irradiance is approximately 420 W m-2 if transparency, a = 1 and only 250 W m-2 if a = 0.7. Note that when a = 1, maximum solar irradiance for the year occurs at the Poles on the solstices. For a < 1 direct sunlight does not include scattered light and underestimates total solar heating. Somewhat more than half the scattered light reaches the ground as skylight. These graphs contain a load of information. This makes it simpler to look at how sunlight varies either with 1. Latitude for a particular day, 2. Over the year at a particular latitude Choice 1. The first diagram (p23) shows solar irradiance for 3 different values of transparency on the June solstice. When a = 1 the graph has two maxima - one around 30 N and one at the North Pole. The minor maximum is due mainly to the high Sun although it is modulated by the length of the day. The major maximum is due to the fact that the North Pole gets 24 hours of sunlight on theSolar Irradiance (W m -2)

600 Irradiance (W m-2) 500 400 300 200 100 0 -90 -60 -30 0 latitude ( )o

a = 1.0 a = 0.7 a = 0.4

solstice. When atmospheric transparency is included, the North Pole maximum disappears due to severe attenuation of the solar beam as it passes obliquely (i. e., at large Z) through the atmosphere. The diagram to the left shows how sunlight varies over the year at several different latitudes. The equator has the largest average value but the smallest annual variations. The result is steady warmth. The Pole has the smallest average irradiance, but the largest variation and the largest single value. The Poles are very cold on average and have a large annual temperature range. They remain cold in summer but only because they are ice covered or surrounded by water. Observed Distributions of Insolation The observed values of solar irradiance shown below are affected by clouds, atmospheric transparency, and reflection of light from the surface or albedo, R, shown on the next page. Absorbed insolation at the surface is largest over the subtropical oceans because cloudiness, humidity, haze and surface albedo are low. Note the high values for the sand-covered Sahara and other deserts and for the Greenland and Antarctic ice caps (gray means R > 50%.) Low values of absorbed insolation occur in the high latitudes. Solar irradiance reaching the surface is shown for the United States on the next page. Minimum values occur over the Pacific Northwest coast and the Great Lakes areas while maximum values occur over the clear air of the deserts of

30

60

90

600Mean Irradiance (Wm-2)

500 400 300 200 100 0 J J A S O N D J F M A MMonth

= 90 = 60 = 30 =0

the Southwest.

(http://modis-atmos.gsfc.nasa.gov/ALBEDO/browse.html) An excellent web source for all figures of global temperature and energetics is

http://geography.uoregon.edu/envchange/clim_animations/

The annual progression of daily insolation can be graphed at individual stations. Comparing graphs at Binghamton, NY and Yuma AZ show the influence of greater cloudiness at Binghamton. web sources (http://projectsol.aps.com/solar/data_insolation.asp), (http://climvis.ncdc.noaa.gov/cgi-bin/fsod_state)

Problems 1. 1000 W m-2 of sunlight shines for one hour on a solar panel that is able to convert all its energy to work. How high will it lift 70 kg in that hour? (Remember, PE = mgdz.) 2. What is the mass of air and the mass of water vapor in a volume, 2 m3 with T = 280 K, p = 105 Pa, w = .005. 3. If the water vapor in the above problem condensed how deep would it be? Assume the crosssectional area is 1 m2. 4. In an hour solar panels with an area 5 m2 convert 106 j to useful energy. What is their power? What is their effective irradiance? 5. What is the mass of a 1 m2 column of atmosphere with surface pressure, p = 101325 Pa. What is the mass of water vapor of this atmosphere if the average mixing ratio, w = .001? What is the precipitable water of this atmosphere? 6. Given the mixing ratio at two different values of pressure (e. g., w = .008 at p = 9(10)4 P and w = .006 at p = 7.5(10)4 Pa), calculate the average mixing ratio and then the amount of precipitable water in the layer. What would be a general formula for converting average mixing ratios in an atmospheric layer to values of precipitable water? 7. How much will the temperature of the atmosphere in problem 6 rise in an hour if the direct solar irradiance, I = 1367 w m-2 is all absorbed? What depth of liquid water would evaporate if all this energy had gone to evaporating water? 8. How long would it take sunlight of irradiance I = 500 W m-2 to raise the temperature of a column of pure ice 2 m thick from -10 C to 0 C and then melt the ice? Do this problem in two separate steps. 9. If heat is withdrawn from the air to raise its mixing ratio from 0 to .001 by evaporation of a water body, how much will the air temperature decrease? 10. What is the direct solar irradiance on the surface of the sun? The sun's radius is 6.95(10)8 m. 11. What is the solar constant on Pluto given its mean distance from the sun is d = 5.9(10)12 m? 12. How much solar energy does the earth intercept in an hour? The earth's radius is 6.37(10)6 m. 13. What is the distance of the earth to the sun on Feb 21? 14. What is the percentage reduction of direct solar irradiance from perihelion to aphelion? 15. On the planet, Jupiter find I when d = 7.78(10)11 m and the zenith angle, Z = 53?

16. What is Z at noon at latitude 50 N on Nov 21? What is I on level ground if atmospheric transparency, a = 0.7? 17. Draw the 3-D sun paths for the equinoxes and solstices at 3 different latitudes. 18. Describe the procedure for drawing the 3-D sun path for any day of the year at a given latitude. Give an example. 19. What is Z at 2 PM at latitude, 50 N on Nov 21. What is I if a = 1, a = 0.7? 20. What is sunset time in NYC on June 21. How long is the day? 21. Write a computer program that solves for cos(Z) and then for Z with input, latitude, day of the year and hour of the day. Since BASIC does not have the arccos function define your own using, acs(x) = -atn(x/sqr(-x*x+1))+pi/2 22. Extend the computer program of the last problem to compute I on level ground with an atmosphere of transparency, a, with input, latitude, day of the year and hour of the day. 23. Modify the computer program of the last problem by removing the time of the day as input and then creating a loop in which I is calculated every hour of the day until sunset time. 24. What is the azimuth angle of the sun, az, at 2 PM at latitude 50 N on Nov 21? (See prob. 19.) 25. Find I on south and west facing vertical walls in the last problem, given the a = 0.7. 26. What is the total Q/A on April 1 at 27 S latitude with a = 1.0? 27. Numerically integrate the function y = x2 from x = 0 to x = 6. In this problem you already know the answer, namely, 63/3 = 72. Thus, the procedure is paramount. Procedure: (Read Carefully!) First choose an interval, dx, that is some fraction of the total interval, 6 (i. e., 6 = ndx, where n is some integer.) Next, find the value of y at the central x value of the interval (i. e., y[(k+1/2)dx]). Then take the sum of the product, y[(k+1/2)dx] dx. Note: The smaller the width of an interval, dx, the more accurate the result. 28. Now that you can numerically integrate, find Q/A for problem 27 when a = 0.7? Warning! Do not integrate beyond sunset or you may get negative values! 29. Using your knowledge of concepts from calculus, find an expression for the subtropical latitude at which Q/A is a maximum on June 21. Evaluate that latitude and find its value of Q/A.