Determine how many hours of daylight are received at 0°, 30°N, 60°N, and 90°N, at 30°S, 60°S, 90°S latitude

Question

http://spaceflight.nasa.gov/realdata/tracking/

I am supposed to use this site to answer the question below, but I can't figure out how to use the site. The vertical lines of longitude are separated by 60° or 2 hours. Determine how many hours of daylight are received at 0°, 30°N, 60°N, and 90°N and at 30°S, 60°S, and 90°S latitude.

Answer 1

The chart should have more info on it.  The very bottom horizontal line is the south pole or 90 deg.S and the top line of the chart is 90 deg.N or the north pole.  The other lines in between represent the equator in the center of the chart and 30 & 60 degrees north latitude in the top half of the chart.  The last two lines are 30 & 60 degrees south latitude in the bottom half of the chart.  The position of the sun is where it 12 noon right now.  The darker area of the chart where the black stippling is at represents the area of the world where it is currently nighttime.  And, conversely, the lighter area of the chart is where it is daylight right now.  

To find the number of hours at any given latitude, locate that latitude--let's use 30 deg S--and then start at one side of the chart and count the number of lines of longitude that are in the dark area.  Go all the way to the other side of the chart(that is 24 hours or a full day).  At 30 deg S, starting at the left hand side I count almost 7 full lines of dark area followed by five lines of the lighter area and then a small patch of darker area again just before hitting the right side of the chart.  Totaling the two patches of darkness gives us seven whole lines or as you said at 2 hours per line a total of 14 hours of darkness and, of course, 10 hours of daylight.

The other latitudes work the same way.  I will let you figure those out.

Note that the two poles are super easy to figure.

Answer 2

Not a simple answer and can be convoluted.

The calculations are referenced to the center point of the sun's disk and is the time that this center point is above the horizon.  The answer depends upon the observation latitude and the latitude of the sun's track across the surface of the earth.

The earth rotates once each 24 hrs, so at the equator (0deg) the center of the sun's disk is 12 hrs above the horizon and 12 hrs below the horizon.  We know that as the observation point moves further away from the equator, the period of daylight is less and is a cosine function of the latitude.

So, if the sun is at the equator and the observation point is also at the equator:

cos(0deg) * 12 = 1 * 12 = 12 hrs that the center of the sun's disk is above the horizon.

If the observation is taken at 60degN while the sun is at the equator, then:

cos(60) * 12 = .5 * 12 = 6 hrs that the center of the sun's disk is above the horizon.

Now let's keep the observation at 60degN but we'll move the sun northward to the Tropic of Cancer (TofC) which is 23.5degN.  Summer solstice.  We've effectively moved the sun's track 23.5 deg closer to the observation point.  So the observation point is 60degN - 23.5degN or 36.5deg.  

Therefore:  cos(36.5) = .8 * 12 = 9.6 hrs that the center of the sun's disk is above the horizon.

Again, let's keep the observation at 60degN but we'll move the sun to the Tropic of Capricorn (23.5 deg S).  Winter solstice.  We've effectively moved the sun's track to a point that is 83.5 deg away from the observation point.  

Therefore: cos(83.5) = .113 * 12 = 1.36 hrs that the center of the sun's disk is above the horizon.

So, to define what we did.  The cosine of the absolute value of the difference between the observation latitude and the sun's latitude * 12 equals the period of time that the center of the sun's disk is above the horizon.  This is true if the sum of the observation latitude and the sun's latitude is between 0 deg and 90 deg.  If the sum is >90 deg, you'll have 24 hrs daylight, less than 0 will be 24 hrs dark.  Also, remember that N latitudes are positive (+) numbers and S latitudes are negative (-) numbers.

Now, complicating factors:  

1.  This does not take into account the size of the sun's disk.  Sunrise is defined as the moment that the top of the sun's disk breaks the horizon in the east.  Sunset is the moment that the top of the sun's disk drops below the horizon in the west.  The time it takes for the sun's disk to traverse the horizon line adds time to the total daylight hrs observed.  

2.  Another factor is the bending of the sunlight's rays through the earth's atmosphere.  This makes the sun appear that it is above the horizon when it is not.  This adds time to the daylight hrs.  

3.  Also, the definition of daylight is a variable but usually means the ability to read a newspaper (or other printed material) without artificial light.  

4.  The earth is closer to the sun in Jan than in July.  The apparent size of the sun's disk is larger and takes longer to traverse a given point; adding to the amount of daylight.  Explanation:  The earth rotates at about 15deg per hour.  If the sun's disk occupies 15 deg field of view, it will take 1 hr to traverse the observation point.  If the sun's disk occupies 20 deg field of view because it is closer, it will take 1:20 minutes to traverse the field of view.