To calculate the noon sun's angle above the horizon at Placerita Canyon, identify today's date and find the corresponding solar declination using the analemma. Next, calculate the arc distance (AD) by subtracting this declination from 34°N, and finally subtract the AD from 90° to get the solar altitude (SA). This gives you the sun's angle above the horizon at noon.
To calculate the angle of the noon sun above the horizon at a location of approximately 34°N latitude, given that the solar altitude (SA) is 90º minus the arc distance (AD). Start with the latitude of 34°N. Identify today's date and find the corresponding declination of the sun using the analemma. Subtract this value from 34°N (or vice versa depending on which is greater) for the AD. Finally, subtract the AD from 90° to get the SA.
For example, if today's declination of the sun was 23°N: AD = 34°N - 23°N = 11°, then SA = 90º - 11º = 79º. So, at noon the sun would be 79 degrees above the horizon at Placerita Canyon if today's date corresponds to a declination of 23°N.
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Answer:
The latitude at which the noon sun is directly overhead is defined as the sun's declination. For example, if the sun appears directly overhead at 20°N latitude, the sun's declination is thus 20°N. Figure 7-3, known as analemma, shows the declination of the sun throughout the year. Thus, if you would like to know where the sun will be directly overhead on September 20, you can look on the analemma and find that it will be at 1°N. Given the declination, the noon sun angle at a particular latitude on any given day can be calculated.
Figure 7-3. The analemma, a graph illustrating the latitude of the overhead noon sun throughout the year.
Figure 7-4 shows that for each degree of latitude that the place is away from the latitude where the noon sun is overhead (declination), the angle of noon sun becomes one degree lower from being vertical (or 90°) above the horizon. Noon sun angle for any given latitude on any given day can be calculated using the following equation:
Noon Sun Angle = 90° - (Latitude +/- Declination)
In this equation, use " - " when the latitude and declination are on the same hemisphere and use " + " when they are on the opposite hemispheres. For example, on April 11, the declination is 8°N, the noon sun angle at Springfield, Illinois (40N) on April 11, therefore, should be 58° (90° - (40°N - 8°N).
Figure 7-4. Noon sun angle calculation. The example in this figure illustrates how to calculate the noon sun angle for a city 40° north of the equator (like Springfield, Illinois) on December 22.
Questions
Based on Figure 7-3, the sun shifts its position between which two important latitudes (find the terminologies for these two latitudes in Fig. 7-4). (2 points)
Using the analemma (Fig. 7-3), determine the declinations on each of the following dates. (6 points)
August 10 _________________
September 21 _______________
June 21 ______________
October 15 ____________
December 22 ______________
March 21 _________________
Calculate the noon sun angle in Grayslake, Illinois on each of the dates listed below. (6 points)
Noon Sun Angle in Grayslake on
August 10 _________________
September 21 _______________
June 21 ______________
October 15 ____________
December 22 ______________
March 21 _________________
Obtain the average temperature in Grayslake on each of the dates listed below using a weather station online. (6 points)
Average Temperature in Grayslake on:
August 10 ________________
September 21 _______________
June 21 _____________
October 15 ____________
December 22 ______________
March 21 _________________
Plot the noon sun angle and temperature on each of the dates you obtained in Questions 3-4 on the graph provided below and connect your temperature and noon sun angle plots using smooth lines. Discuss the relationship between noon sun angles and average temperature in Grayslake throughout the year based on your graph. Use the right vertical axis to show temperature variation from -30° F at bottom and +105° F on top (label temperature value at equal amount interval in between: 2 units for every 5°F). Use the left vertical axis to show noon sun angle, with 0° at the bottom and 90°F on top (label noon sun angle value at equal amount intervals in between: every 3 units per 10°). (4 points)
J F M A M J J A S O N D
Months
Figure 7-5. Variation of temperature and noon sun angle in Grayslake.
brief paragraph summarizing the yearly movement of the overhead noon sun and how the intensity of insolation aries in Grayslake throughout the year. (2 points)
Calculate the maximum, minimum, and average noon sun angles at the place in Grayslake. Show your calculations. (3 points)
Maximum noon sun angle:
Minimum noon sun angle:
Average noon sun angle:
When the noon sun is directly overhead a ship in the Pacific Ocean on October 16, the ship's latitude is _____________. Explain your answer. (2 points)
physical characteristics in Latin America?
Explanation:
as the ocean plate is pushed under the overriding continental plate the ocean sedimentary layers are heated compressed and melted the pressure of the converging plates that create the subduction zone forces the Continental play upwards creating mountain ranges such as Andes
Answer:
within a few hundred miles off the coast
Explanation:
Answer: Erosion can be harmful to a person's property, causing a negative effect since the person must have paid for the land and the risk of injury if it forms a manhole.
Explanation:
Hope this helps!
True
False
Answer:
Option (1)
Explanation:
Isohyets- This are the lines that connects the points of equal amount of rainfall that occurs at any given period of time.
Contour lines- This are the lines of equal elevation.
Isotherms- This are the lines that connects the points of equal temperature.
Isobars- This are the lines connecting the points of equal atmospheric pressure.
Isogrids- This are a type of half hollowed-out structure that is created mostly from a single metallic type of plate with hard triangular rib like structures that are commonly known as stringers. This are very light and hard in comparison to other substances, This are very costly to build, so they are basically used for the applications that are usually related to the spaceflight and also used in the field of aerospace.
Thus, the isogrids are not a common type of isoline.
Hence, the correct answer is option (1).
Answer and Explanation:
A 1.5 earthquake on the Richter scale is 15 times stronger than a level 0 earthquake. This is because the Richter scale is a logarithmic scale making a number on the scale symbolize an intensity 10 times greater than the previous number on the scale.
An earthquake measuring 1.5 on the Richter scale has tremors that are captured only by seismographs.