As preliminary background for the first few questions from Lab
Activity #2: The Seasons, you should become familiar with definitions of the terms listed below:
- Rate or rate of change
- Flux (of energy, mass, momentum, etc.)
- To calculate a flux, determine the rate at which the quantity of interest
encounters or passes through or emerges from or is absorbed or emitted or reflected by a surface, and divide that
rate by the area of the surface. The result is the rate at which
the quantity encounters/passes through/emerges or reflects from/is absorbed by each one unit area
surface (averaged over the surface). One way to think of it is as a measure of how concentrated or spatially "intense" the rate is on a surface.
- Energy flux
- Radiative intensity
- Question (1) in Lab Activity #2. In Figure
2-1, solar radiative energy strikes three different
at different angles. In each of the three panels of the diagram, two ambiguous
labels reading "1 unit" appear. In each panel,
the label along the Earth's surface represents units of surface area, whereas
surface (oriented perpendicular to the sun's rays) we'll interpret to represent
units of rate of energy transfer (that is, power) through the area indicated by the label.
In each panel, the solar radiative intensity on a horizontal surface (that is, insolation) at Earth's surface
is just the rate at which energy encounters the horizontal surface divided by
area that the energy encounters. (In this case,
the rate at which energy encounters the surface between evenly spaced adjacent arrows on the diagram is vague because we don't know the specific units of energy or time used: it's just one unit of energy per unit of time. Fortunately, the specific units don't matter for this purpose. The area that the solar radiation between adjacent arrows strikes is also somewhat vague, but at least we can compare the area in the three panel diagrams: 1 unit of area in the first panel, 1.4 units
in the second panel, and 2 units in the third panel). As a result, the solar
radiative intensity at the earth's surface in the first panel is (1 unit
of energy per unit time)/(1 unit of surface area) = (1 unit of energy/time)/area;
in the second panel it is (1 unit of energy per unit time)/(1.4 units of
area) ≈ (0.7 units of energy/time)/area; and in the third panel it is (1 unit of
energy per unit time)/(2 units of surface area) = (0.5 units of energy/time)/area.
Since the angle at which sunlight strikes the earth's surface in the three
panels is 90°, 45°, and 30°, respectively, we can, reasoning inductively (that is, from the particular to the general), conclude that there is a
general relationship: all else being equal, the greater
the sun angle (that is, the higher the sun is above a flat horizon), the
more intense the solar radiation is at the earth's surface. (It is possible to show using basic geometry and trigonometry that the relationship is a trigonometric one. If we define the symbol θsun to represent the sun angle and S(θsun) to represent the insolation expressed as a function of sun angle (θsun), then the relation is: S(θsun) = S(90°) × sin(θsun). As we can see by inspection of the diagram, this occurs because although in each panel solar energy between adjacent arrows might be coming in at the same rate, it is spread out over a larger surface area when the angle of the sun above the horizon is lower, which decreases the solar radiative intensity; or equivalently, the solar radiative energy is more concentrated on a smaller surface area when the sun is higher above the horizon, which increases the solar radiative intensity.
- Question (2) in Lab Activity #2. As solar radiation passes
through the atmosphere, it encounters gas molecules and larger particles
(cloud droplets or
ice crystals, solid particles of dust/ash/smoke/salt, etc.), which can
reflect or scatter it away, or absorb it, thereby reducing the amount of solar
reaches the earth's surface. In Figure 2-2, we can see that the smaller the
angle is between the incoming solar radiation and the earth's surface (that
is, the lower angle of the sun above the horizon), the greater the distance
that the radiation must travel through the atmosphere before reaching the
surface. As the distance that solar radiation travels through the
atmosphere increases, more of it will likely be scattered or reflected
away or absorbed, and hence less of it will reach the surface, which reduces
the solar radiative intensity at the surface (by reducing the
rate at which solar radiative energy strikes the surface).
Hence, we once again conclude—though for a different reason than above—that the
greater the sun angle (that is, the higher the sun is above a flat horizon),
the more intense the solar radiation is at the earth's surface.
Only one of the reasons would apply on the moon (which one, and why?), but
the relationship between sun angle solar and radiative intensity
at the surface would still apply —just
not quite as strongly.
Note that roughly speaking, outside the tropics, the angle of the sun above the horizon at solar
noon (the time of day when the sun is highest in the sky at any particular
location) decreases with increasing latitude.
Hence, outside the tropics (and at some times of the year within the tropics) we can say
that the intensity of solar radiation at solar noon decreases with increasing
latitude (a measure of distance from the equator).
- Question (3) in Lab Activity #2. The earth rotates around
it's axis from west to east. The "right hand rule" helps us remember this.
Hold out your right hand with the thumb aligned along the earth's axis of
rotation and pointing from the South Pole to the North Pole; your fingers
will curl naturally in the same direction in which the earth rotates around
If you pick a spot on the earth (other than the North or South Pole) and
follow its path through space as the earth rotates, the point will
move in a
That circle corresponds exactly to the latitude circle on which the spot lies.
Since it takes 24 hours for the earth to complete one rotation around the
axis of rotation, it follows that each moment (time) of the day corresponds
location on a latitude circle. We define solar noon to the be time
of day when the sun appears highest in the sky (that is, when the sun angle
at any particular moment there must be exactly one point on every latitude
circle corresponding to this time. On Figure 2-2 , the locations where it
is solar noon all lie along the left-hand edge of the earth, from the North
Pole to the South Pole.
Similarly, on each latitude circle it must be solar
midnight at exactly one point, and this point will be on the opposite side
of the latitude circle from where it is solar noon. These points lie along
the right-hand edge of the earth in Figure2-2, from the North Pole to the
South Pole. (Note that this makes the time on the North and South Poles ambiguous—it
appears to be both midnight and noon there at the place, and indeed all
other times as well! That is, the sun angle is the same at all times of the day! Some other way of specifying time of day other
than the angle of the sun above the horizon appears to be necessary at those
Six p.m. comes exactly half way between midnight and noon in time and hence
on each latitude circle as well. In the perspective shown in Figure 2-2,
the axis of rotation happens to split each latitude circle exactly in half,
so we know that it's 6 p.m. at points on each latitude circle where the latitude circle appears to intersect
the axis, but on which side of the earth—the one we can see in Figure
2-2 or on the other side? Using the right hand rule to determine that the
side of the earth we can see in Figure 2-2 moves from left to right (west
to east) as the earth rotates, we see that most locations on the earth at the
time of year shown pass from light to dark as they move from left to right.
The boundary between light and dark, called the terminator or the circle
of illumination, marks locations where it is either sunrise or sunset.
Places moving across the circle of illumination from the light side to the
dark side must be experiencing sunset, and since sunset occurs sometime
closer to 6 p.m. than to 6 a.m. everywhere on the earth that experiences sunrise
and sunset, we conclude that it must be 6 p.m. at locations in Figure 2-2 where
latitude circles appear to intersect the axis of rotation. We also note that
at the time of year shown, places in the Northern Hemisphere that experience
sunset at all (not all do) experience it earlier than 6 pm (they appear to
cross the circle of illumination, when the sun sets, before they cross the axis of rotation as viewed from this perspective, where
it's 6 p.m.), whereas places in the Southern
that experience sunset (not all do) experience it later than 6
p.m. (they cross the apparent intersection with the axis of rotation, where it is 6 p.m., before they cross the circle of illumination, where it is sunset).
- Question (4) in Lab Activity #2. Looking at the "side" view of the earth at the June solstice, the earth's axis of rotation splits every latitude circle exactly in half. Since it takes 24 hours for any particular point on the earth to move around the axis of rotation once, following the latitude circle on which it is located, it follows that it takes 12 hours for any spot to travel half way around it's latitude circle. Now notice that in the Northern Hemisphere, more than half of each latitude circle lies within the part of the earth lit by the sun, on the sun-lit side of the circle of illumination. (This happens when the Northern Hemisphere is tilted toward the sun, exposing more than half of each latitude circles in the N. Hem. to the sun.) Hence, places on these latitude circles must spend more than half the day—that is, more than 12 hours—in the sun-lit part of the earth, and so day length must be greater than 12 hours everywhere in the Northern Hemisphere at the June solstice.
Latitudes between the Arctic Circle and the North Pole lie entirely within the sun-lit part of the earth on the June solstice, so day length is 24 hours long on that day. (That is the only day of the year with 24 hours of daylight at the Arctic Circle, whereas on the North Pole there is 24 hours of daylight from the vernal equinox to the autumnal equinox—six months of daylight. Places at latitudes between the Arctic Circle and the North Pole experience intermediate numbers of days with 24 hours of daylight—the closer to the North Pole, the greater number of days each year with 24 hours of daylight.)
The equator is split exactly in half by the circle of illumination, so half of the equator lies in the sun-lit part of the earth and day length must be exactly 12 hours long. (In fact, this is true every day of the year at the equator—the only latitude on the earth where this is true.)
The Southern Hemisphere is tilted away from the sun on the June solstice, and as a result, less than half of each latitude circle in the S. Hem. is exposed to the sun. Hence, every place in the Southern Hemisphere must get less than 12 hours of daylight on the June solstice. Latitudes between the Antarctic Circle to the South Pole get no daylight at all on the June solstice. (That is the only day of the year with 24 hours of darkness at the Antarctic Circle, whereas on the South Pole there is 24 hours of darkness from the vernal equinox to the autumnal equinox—six months of darkness. Places at latitudes between the Antarctic Circle and the South Pole experience intermediate numbers of days with 24 hours of darkness—the closer to the South Pole, the greater number of days each year with 24 hours of darkness.)
Of course, from the autumnal equinox (Sept. 21 or 22) to the vernal equinox (March 21 or 22), the Northern and Southern Hemisphere each experiences what the other hemisphere experiences during the other half of the year.
- Question (5) in Lab Activity #2. Solar noon occurs at each latitude between the poles, around the edge of the earth that is (mostly) lit by the sun. You can determine where the sun strikes the earth directly (that is, where the sun is 90° above the horizon—that is, the sun angle is 90°) in several ways. One approach is to sketch the horizon line at each point around the sun-lit edge of the earth, examine the angle between incoming solar radiation and the horizon line, and look for the place(s) where the angle is 90°. (At any spot on the earth, horizon lines are tangent to the earth's surface. In the diagram. you can most easily draw horizon lines pointing to the north and south at places around the edge of the diagram, which includes all of the places where it is solar noon.) Another approach is first to identify all places where the sun is located at the zenith (the point directly overhead). To determine the direction of the zenith, simply draw a line from the center of the earth through the point on the earth's surface of interest to you. If sunlight is coming in from that direction, then the sun would appear directly overhead at that point. If sunlight is coming in at an angle to the zenith, then the sun won't appear directly overhead.
Either way, you should recognize that there is only one point on the earth where the sun is directly overhead at any particular moment (and one particular latitude where the sun appears directly overhead at solar noon). At the June solstice, that point lies on the Tropic of Cancer (23.5°N). At the December solstice, that point lies on the Tropic of Capricorn (23.5°S). At both equinoxes, that point lies on the equator. Between the vernal equinox and the autumnal equinox (our spring and summer), that point lies in the Northern Hemisphere tropics, between the equator and the Tropic of Cancer. Between the autumnal equinox and the vernal equinox (our fall and winter), that point lies in the Southern Hemisphere tropics, between the equator and the Tropic of Capricorn.
In San Francisco, at about 37.5°N, the sun is never directly overhead, and the same is true of all places outside the tropics.
To locate places where the sun is on the horizon at solar noon, first locate all places where the sun is on the horizon. These places would lie along the circle of illumination. Then locate all places where it is solar noon. Those places lie along the line of longitude between the North Pole and South Pole, along the edge of the diagram that is (mostly) lit by the sun.
Finally, look for the intersection of these two lines, which is where the sun will be on the horizon and it will be solar noon. At the June solstice, exactly one place meets the criteria: on the Antarctic Circle (where the sun touches horizon at solar noon and promptly goes back down again without ever getting above the horizon on that day). At the December solstice, the criteria are met only on the Arctic Circle. At the equinoxes, the sun is on the horizon at solar noon only at the North Pole and the South Pole, where the sun is on the horizon all day long (on those days).
- Question (6) in Lab Activity #2. Using straightforward geometry (and the definitions of sun angle, solar noon, and latitude), it is easy to show that in San Francisco, the sun is about 76° above the horizon at solar noon on the June solstice [90° – (37.5° – 23.5°)], whereas it is only 29° above the horizon at solar noon on the December solstice [90° – (37.5° + 23.5°)]. That's a difference of 2 × 23.5° = 47° from one solstice to the other. That difference in sun angle combines with the significant difference in the length of daylight that we experience from one solstice to the other to produce a significant difference between the solstices in the amount of solar radiation reaching the earth's surface in San Francisco each day, and helps explain why we have noticeable seasons here. (As you can see pretty easily by analyzing the diagrams in Figure 2-5, the length of daylight at any particular spot varies between solstices by an increasingly greater amount the closer that spot is the poles, which is one reason why the seasons are generally more dramatic the farther one gets from the equator. The effect of seasonal variations in sun angle takes more work to analyze from the diagrams in Figure 2-5, but outside the tropics that effect also tends to make seasonal variations more dramatic the farther one gets from the equator. Outside the tropics, the combined effects of length-of-daylight and sun angle variations make seasonal variations increasingly more pronounced the farther one gets from the equator.
Within the tropics, seasonal variations are often quite different from those at higher latitudes, because although the length of daylight varies in the same pattern as it does at higher latitudes (just less so), the sun angle at solar noon at these latitudes reaches a peak (of 90°) twice year, at times of year that differ from one latitude to another (and those times are not the solstices). The diagrams in Figure 2-5 make it possible to understand this point straightforwardly.
A plot of insolation at San Francisco vs. time of year would show a smooth curve with a peak on the June solstice, a minimum on the December solstice, and a slope that is greatest at the equinoxes, exactly between the solstices.
At the North Pole, it is possible to show using straightforward geometric relations, the definition of latitude, and the known tilt of the earth's axis or rotation (23.5°), that the highest the sun ever gets above the horizon is 23.5°, at the June solstice. For half the year (from the September equinox to the March equinox) the sun is at or below the horizon, so the range of sun angles that the North Pole experiences is 0 to 23.5°. That is, the difference between the maximum and minimum sung angles at solar noon is only 23.5°. This is only half the difference between the maximum and minimum sun angles that San Francisco (and every place between the Tropic of Cancer and the Arctic Circle) experiences.
With the help of Figure 2-5, the definition of latitude, the known angle of tilt of the earth's axis of rotation, and several basic concepts of geometry, it is possible to deduce that at the equator, the maximum sun angle occurs at solar noon twice a year,
namely at the two equinoxes. The minimum sun angle at solar noon occurs at the two solstices, at 66.5°N at the June solstice and at 66.5°S at the December solstice. Hence, although the sun angle at solar noon varies from 66.5°N to 66.5°S, a difference of 47°, the difference between the maximum and minimum sun angles at solar noon is only 23.5°. At latitudes between the equator on the one hand and the Tropics of Cancer and Capricorn on the other hand, the difference between maximum and minimum sun angles at solar noon increases with increasing latitude, from 23.5°N to twice that (47°). At those tropical latitudes, the sun appears directly overhead twice a year, though at different times of year at each latitude. (The Tropics of Cancer and Capricorn are the latitudes farthest north and south, respectively, where the sun ever appears directly overhead, at solar noon on the June and December solstices, respectively—that is, once per year at each of those two latitudes.)
- Question (7) in Lab Activity #2. The three plots on the graph below show north-south profiles of monthly-average incoming solar radiative energy flux (insolation) on a horizontal surface at the top of Earth's atmosphere in December, March, and June 1987, respectively. (These are the months in which the December solstice, the March equinox, and the June solstice, respectively, occur.) Insolation is expressed in Watts per square meter. (One Watt is one Joule per second, and a Joule is a unit of energy. Hence, insolation is expressed as an energy flux, as we expect.) In these plots, latitude (in degrees) is plotted on the horizontal axis: positive latitudes are in the Northern Hemisphere and negative latitudes are in the Southern Hemisphere.
- December plot. We note that the monthly-average insolation is zero within about 22° or so of the North Pole; it increases rapidly from there all the way into the lower latitudes of the Southern Hemisphere before almost flattening out; it has a slight dip at about 63°S, which is about 27° from the South Pole); and it peaks at the South Pole.
On the day of the December solstice, the sun angle is greatest at the Tropic of Capricorn (23.5°S; see Figure 2-5). On other days in December, the sun angle is greatest at latitudes just north of the Tropic of Capricorn but nearby. It follows that in December the sun at noon is the most intense near the Tropic of Capricorn, so the location of the peak of the curve (at the South Pole) might at first seem surprising. The key to understanding why the peak is not near the latitudes where the sun shines the most intensely at solar noon is to recognize that the plot is of monthly-average insolation, not instantaneous insolation (at noon or any other time on any particular day). The average is computed over the course of a month's worth of 24-hour days, including periods each day when the sun might be down. Hence, the monthly-average insolation depends not only on the maximum intensity during the day but also on the length of daylight (the number of hours that the sun is shining). At the South Pole, although the sun never gets more than 23.5° above the horizon, during the entire month of December the sun is above the horizon all day. The fact that the sun shines continuously during December at the South Pole must more than make up for the low intensity of the sun.
As one moves north (toward the equator) from the South Pole, the length of daylight begins to decrease while the maximum sun angle (occurring at solar noon) increases, and these two effects tend to compensate for each other all the way to near the Tropic of Capricorn, which is why the curve varies relatively little through those latitudes. Continuing north of the Tropic of Capricorn, the length of daylight continues to decrease and so does the maximum sun angle. Both effects cause monthly-average insolation to decrease, and so the monthly-average insolation decreases rapidly with latitude. Not far past the Arctic Circle, there is no insolation at all during December (see Figure 2-5), so the average becomes zero.
- June plot. For reasons that should be clear from Figure 2-5, the curve for the month of the June solstice is almost antisymmetric (reversed from left-to-right) relative to the curve in the December plot. The only departure from antisymmetry is the fact that the curve in June is slightly lower at every latitude than it is in December (the June peak at the North Pole is about 520 W/m2, while the December peak is about 550 W/m2, about a 5% difference). What accounts for this difference? A good bet is the fact that the Earth is closest to the sun in early January and farthest in early July, and insolation decreases with increasing distance from the sun. We'd therefore expect December average insolation to be a little greater than June insolation. (We could easily do some calculations based on the inverse square law to support (confirm) this possible explanation.)
- March plot. This curve peaks near the equator. We might expect this because on the March equinox (March 21 or 22), the length of daylight is the same everywhere (exactly 12 hours), whereas at solar noon the sun is most intense on the equator and the intensity decreases with distance from the equator. Hence, sun angle is the only factor affecting variations in average insolation on the day of the equinox, and the curve should peak where the sun is most intense, at the equator. Upon close inspection, however, we note that (1) the peak is actually slightly south of the equator, and (2) the curve isn't symmetric at the highest latitudes—the monthly-average insolation is greater at high latitudes in the Southern Hemisphere than it is at the same latitudes in the Northern Hemisphere. Why?
The key is to recognize again that the curve shows insolation averaged over the full month of March, and moreover, that the equinox doesn't occur in the middle of the month but rather later in the month. During the first 21 or 22 days of the month, the sun shines directly onto latitudes just south of the equator and the sun shines in the Southern Hemisphere for slightly longer than 12 hours per day, whereas the sun shines for slightly less than 12 hours/day in the Northern Hemisphere. During the last 9-10 days of the month (about half as many days), the opposite is true. Hence, in the March monthly average the Southern Hemisphere will get more insolation than the Northern Hemisphere. The asymmetry between the hemispheres is almost entirely a consequence of the period chosen for the averaging (the month of March). Months of the year don't happen to match the astronomically significant periods (demarked by the solstices and equinoxes) exactly, creating mathematical artifacts like this.
Notice that the peak monthly-average insolation in March (about 440 W/m2, near the equator) is lower than the peaks in either December or June (550 W/m2 and 520 W/m2 near the South Pole and North Pole, respectively), so that the high latitudes experience greater (monthly-average) insolation near the solstices than the equator experiences near the equinoxes, even though the equinoxes are the two times of the year when the sun shines directly on the equator. This again reflects the influence of the length of daylight on the monthly average insolation; the equator gets 12 hours of daylight every day of the year (see Figure 2-5), while the poles get 24 hours of daylight half the year, including the month of one of the solstices or the other.
Note also that in December and June, the insolation at the equator averages about 410 W/m2 and 385 W/m2, respectively. These values aren't much lower than the value near the equator during March (440 W/m2). That is, monthly-average insolation at the equator doesn't vary much from one time of the year to another (by only 55 W/m2), compared the variations experienced at the poles (from 0 W/m2 on one solstice to the low- to mid-500s W/m on the other solstice, a difference of over 500 W/m2). At midlatitudes the variations are somewhere in between those two extremes, which makes them much larger than at the equator. It shouldn't be any surprise that the high latitudes experience much more dramatic seasonal variations in temperature than the equator does! [Can you explain why the equator experiences greater insolation in March than it does in December of January? Why might it be greater in December than in June at the equator?]
- Question (8) in Lab Activity #2. The area-weighted global average insolation in June 1987 was about 340.0 W/m2, while in December it was about 361.8 W/m2, a difference of about 6%. Why? Once again, a good hypothesis is that the earth is closer to the sun in December than it is in June (by a small amount), and this accounts for the difference in insolation. We could test this hypothesis by calculating what the solar intensity facing the sun should be, using the inverse square law applied at the distances of the earth's closest and farthest distances from the sun and comparing these theoretical values to the observed ones. If the observed values are close to the theoretical ones, that lends support to the hypothesis.
[Note that the unweighted global averages of insolation, computed simply by adding up the insolation values at all of the 2.5° latitude x 2.5° longitude grid cells covering globe and dividing by the number of such cells [(360° longitude/2.5°) x (180° latitude/2.5°) = 10,368], gives values of 327 W/m2 and 307 W/m2 in December and June, respectively. Although again we'd conclude that the earth gets less solar radiative energy in June than it does in December, these values are both too low. The reason is because the grid cells are not all the same size. Lines of longitude converge on the poles, which makes the grid cells narrower and narrower the farther they are from the equator. When computing the global average, counting the insolation on narrow, high-latitude grid cells just as much as the insolation on the larger, low-latitude grid cells, over-represents (overweights) the insolation at high latitudes in the global average. Since insolation at high latitudes is less than it is at low latitudes (taking both hemispheres into account—try averaging the insolation at the North Pole and South Pole and compare that to the insolation at the equator on the plots for December or June above, for example), the unweighted average produces global-average values that are too low. Weighting the insolation in each grid cell by the area of each cell before computing the average compensates for this distortion and produces a much more accurate global average calculation.
There is another way to understand the role played by the size of the grid cells when calculating the monthly, global average insolation.
Another way to calculate the global average insolation is first to find the total (monthly average) rate at which solar radiative energy strikes the earth and divide that rate by the surface area of the earth. To determine the total rate at which solar radiative energy strikes the earth, we can calculate the rate at which it strikes each cell and simply add up the rates for all 72 × 144 = 10,368 cells to get the total. The rate at which solar radiative energy strikes each cell is just the insolation on each cell multiplied by the number of units of area in that cell (that is, the area of the cell). So, we get the total rate at which solar radiation strikes the earth = ∑Si×Ai, where Si is the insolation on the ith cell, Ai is the area of the ith cell, and i is an index that ranges in value from 1 to 10,368 (the total number of cells). The global average insolation would then be = (∑Si×Ai)/AEarth, where AEarth is the surface area of the earth. Since AEarth is the same for each member of the sum in the denominator, we can divide it into the area of each cell in the sum and rewrite it as ∑Si×(Ai/AEarth). Since Ai < AEarth, each ratio Ai/AEarth is much less than one, but it's easy to show that they add up to 1 (since ∑Ai = AEarth). The quantity Ai/AEarth multiplying the insolation on each cell, Si, is the weight we apply to each cell's insolation when we compute the global-average insolation. (We definitely wouldn't compute the global average insolation as (∑Si)/N, where N is the number of cells, because that would overweight the smaller cells and underweight the larger ones, as noted above.
- Question (9) in Lab Activity #2. Monthly-average insolation and surface temperature observations, averaged over the contiguous 48 U.S. states and over Argentina for each month of 1987, are summarized in the table below. (The contiguous 48 U.S. states and Argentina are at roughly similar latitudes, though in opposite hemispheres.)
|There are several points to note about these data:
- The greatest insolation and surface temperature observations in the U.S. and Argentina occur about six months apart. The same is true of the lowest insolation and surface temperature observations. This, of course, is why the Northern Hemisphere winter is from the December solstice to the March equinox and summer is from the June solstice to the September equinox, while in the Southern Hemisphere winter and summer occur at times opposite those in the Northern Hemisphere. (The reasons should be clear from Figure 2-5.)
- In the U.S., the month with the greatest average insolation is June while the month with the highest average temperature is a month later, in July. The month with the lowest average insolation is December while the month with the lowest average temperature is a month later, in January.
- In Argentina, the month with the highest average surface temperature (January) also lags behind the month with the highest average insolation (December), though the lowest average insolation and surface temperature occur in the same month (June).
If we'd thought initially that surface temperatures are highest when insolation is greatest (near a solstice) and lowest when insolation is the least (near the other solstice), these data cast some doubt on that simple idea, though it's probably not too far off. Evidently, additional factors help control temperature, and whatever factor or factors those are, they might tend to cause maximum and minimum average surface temperature to lag behind the maximum and minimum insolation, respectively, at least in some places sometimes. What might (an)other factor(s) be, and how might it (they) work to cause temperature to lag behind insolation? We'll have to find out....
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