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Line 1= Noun (topic of poem)

Line 2 = Two adjectives that describe the noun in line 1

Line 3 = Three action verbs that end in -ing and relate to the noun in line 1

Line 4 = Four words: 2 words about the noun in line 1 and 2 words that relate to the noun in line 7

Line 5 = Three action verbs that in in -ing and relate to the noun in line 7

Line 6 = Two adjectives that describe the noun in line 7

Line 7 = Noun that is the opposite of noun in line 1

 

The Hermann grid illusion is an optical illusion, a visual perception that differs from
objective reality. The illusion is characterized by "ghostlike" grey blobs perceived at the intersections of a
white (or light-colored) grid on a black background. The grey blobs disappear when looking directly at an
intersection. The scintillating grid illusion, discovered by E. Lingelbach in 1994, is variation of the
Hermann grid illusion. It is constructed by superimposing white discs on the intersections of orthogonal gray
bars on a black background. Dark dots seem to appear and disappear rapidly at random intersections, hence
the label “scintillating”. When a person keeps his or her eyes directly on a single intersection, the dark dot
does not appear.
You can find information about the Hermann Grid illusion in my Lecture notes for Class 5 – Neurons and
Perception – Part 1.
Purpose: To familiarize yourself with the Hermann grid illusion, and those factors that affect it.
Task: This assignment will assess what properties of the stimulus affect both of these illusions. For this
assignment you will be using the Hermann Grid (curving) illusion from Michael Bach’s website:
http://www.michaelbach.de/ot/lum-herGridCurved/index.html
Optical illusions & visual phenomena and explanations (where possible):
Optical Illusions & Visual Phenomena (michaelbach.de)
Before you begin, disable the switching of the illusion from boxes to curved lines by removing the check on
the Auto-Run button.
Using your knowledge of how the retina is structured, along with the organization of receptive fields,
describe and explain your perceptions of the following questions:
Please write legibly!
1) How is your perception of the Hermann grid affected by changes in width of the grid lines (change the
street lines wheel to make bigger or smaller)? Why?

2) The illusion is the strongest when the grid lines are mid-sized. Why?

3) What happens to the illusion when you decrease the contrast between the lines and the boxes? (i.e. gray

lines, black boxes using the colors box to change the line and box colors

 

4) Are there any colour combinations that make the illusion weaker? Why?

 

5) Why do the blobs not appear at the point of fixation, and only in the periphery?

Question #1 - Example D: June, midnight and overcast.  Asphalt pavement at temperature 18oC and cloud with a base temperature of 2oC. (2.5 marks)

 

Solve the equations to determine an answer for EXAMPLE  D:        

Q         = Click or tap here to enter text.   W m-2

aQ       = Click or tap here to enter text.   W m-2

M out    = Click or tap here to enter text.   W m-2

M in      = Click or tap here to enter text.   W m-2

Rn        =  Q - aQ - Mout + Min = Click or tap here to enter text.   W m-2

 

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methods, so we use classes to organize our code. By the same token, we have a concept called package, and we use a package to group related classes.

As our applications grow, we're going to end up with hundreds or even thousands of classes, so we should properly organize these classes into packages. The base package for a Java project is the domain name of your company in reverse.

The code editor might look a little bit intimidating at first, but trust me it's really easy and you're going to learn about it throughout this course. For now, just type a base package for your project; it can be your name or whatever, it doesn't really matter.

Marsh has put together a comprehensive cheat sheet with summary notes that you can download below this video.

IntelliJ is building our application and we can see the result in this little terminal window. So, here's our "Hello World" message. That was our first Java program.

Next, I'm going to explain how Java code gets executed under the hood.

The energy budget, or the balance between energy receipts and losses (hence net radiation) at the surface can be expressed as:   Rn = Q - Q - Mout + Min    where Rn = the net radiation Q = the total incoming solar radiation or insolation Q = the solar radiation reflected from the surface, where   = albedo (reflectivity in %) Mout = the terrestrial radiation emitted from the surface Min = the terrestrial radiation incoming to the surface from clouds, water vapour and CO2 in the atmosphere, also called counter radiation.   As you can see from the formula, there are both positive terms (Q and Min) and negative terms (Q and Mout).  The positive terms add energy to the surface while the negative terms remove energy from the surface.  The net radiation is the balance between these positive and negative terms.  For example, the net radiation is usually positive during the day when the incoming solar radiation, or insolation (Q in the equation), is so large as to overwhelm the negative terms in the equation.  This positive value of Rn simply means the surface is warming up as there is more energy coming in than going out.  Conversely, at night when there is no sun, Q, and, of course, Q, both equal zero.  The negative term Mout dominates the equation resulting in a negative value of Rn, meaning the surface is cooling down as there is more energy going out than coming in.     The incoming solar radiation (Q) can be measured with a pyranometer (Figure 1), and we will do this outside.  While incoming solar radiation, or insolation, is measured with the device pointing upward, by flipping it over so it points downward, we can get a measure of the solar radiation reflected from the surface (Q).       Figure 1 – Pyranometer Diagram The terrestrial radiation (represented in the equation by the terms Min and Mout) can be calculated via the Stefan-Boltzmann Law, given as:   M = T4  W m-2   where  = Stefan-Boltzmann constant = 5.67 x 10-8 W m-2 K-4 T = temperature of the emitting surface in K  (K = °C + 273)   As can be seen from the formula, all we need is a surface temperature.  This is easily obtained using a thermal infrared thermometer.  For terrestrial radiation emitted by the surface (Mout), the thermometer is simply pointed directly downwards toward the ground surface.  The terrestrial radiation incoming to the surface, also called counter radiation (Min), originating from the base of clouds, as well as water vapour and carbon dioxide (CO2) in the atmosphere, can also be measured with the thermal infrared thermometer.  By simply pointing it straight upward, we can get a measure of the equivalent temperature at which the sky is radiating.  Once we have the temperatures of the emitting surfaces, either the ground surface or the sky itself, the terrestrial radiation (either Min or Mout) can be computed using the Stefan-Boltzmann Law.     EXAMPLES OF NET RADIATION COMPUTATIONS   Example A: Month, time and weather: June, midday, no cloud Solar radiation measured at surface (Q): 850.0 W m-2 Type of Surface: Brock’s asphalt parking lots Surface Temperature: 35 °C Sky radiation from CO2 and water vapour radiated at an equivalent temperature of: 10 °C   1. Solar radiation (Q): given as 850.0 W m-2   2. Solar radiation reflected (Q): From Table 1.1 (see pg. 7), the albedo of asphalt pavement is 3%, or in decimal form, 0.03.  Solar radiation (Q) is as given in Step 1 (Q = 850.0 W m-2).  Thus: Q= 0.03 x 850.0 W m-2 = 25.5 W m-2   3. Outward terrestrial radiation (Mout): The intensity of radiation from a surface is given by the Stefan-Boltzmann Law: M = T4  W m-2 where  = Stefan-Boltzmann constant = 5.67 x 10-8 W m-2 K-4 T = temperature of the emitting surface in  °K  (K = °C + 273) Thus: Mout = (5.67 x 10-8 W m-2 K-4) x ((35 °C + 273) K)4 =(5.67 x 10-8) x (308)4  W m-2 =510.3 W m-2   4. Inward terrestrial, or counter-radiation (Min): (again, using the Stefan-Boltzmann Law: M = T4  W m-2) Min = (5.67 x 10-8 W m-2 K-4) x ((10 °C + 273) K)4 =(5.67 x 10-8) x (283)4  W m-2 =363.7 W m-2   5. Combination of the various factors: We can now insert the above values into the original energy budget formula: Rn= Q - Q - Mout + Min  Thus: Rn = 850.0 - 25.5 - 510.3 + 363.7  = 677.9 W m-2 for Example A.             Now let us consider the same day and the same time, but a different surface.  As such, the only terms that change are the albedo (and the outward terrestrial radiation (Mout) as they are the only ones that are surface dependent.   Example B: Month, time and weather: June, midday, no cloud Solar radiation measured at surface (Q): 850.0 W m-2 Type of Surface: the water of Lake Moody, behind the Village Residence Surface Temperature: 12 °C Sky radiation from CO2 and water vapour radiated at an equivalent temperature of: 10 °C   1. Solar radiation (Q): given as 850.0 W m-2   2. Solar radiation reflected (Q): From Table 1.1 (see pg. 4), the albedo of water (sun overhead) is 2%, or in decimal form, 0.02.  Solar radiation (Q) is as given in Step 1 (Q = 850.0 W m-2).  Thus: Q= 0.02 x 850.0 W m-2 = 17.0 W m-2   3. Outward terrestrial radiation (Mout): The intensity of radiation from a surface is given by the Stefan-Boltzmann Law: M = T4  W m-2 where  = Stefan-Boltzmann constant = 5.67 x 10-8 W m-2 K-4 T = temperature of the emitting surface in K  (K = °C + 273) Thus: Mout = (5.67 x 10-8 W m-2 K-4) x ((12 °C + 273) K)4 =(5.67 x 10-8) x (285)4  W m-2 =374.1 W m-2   4. Inward terrestrial, or counter radiation (Min): (again, using the Stefan-Boltzmann Law: M = T4  W m-2) M in = (5.67 x 10-8 W m-2 K-4) x ((10 °C + 273) K)4 =(5.67 x 10-8) x (283)4  W m-2 =363.7 W m-2   5. Combination of the various factors: We can now insert the above values into the original energy budget formula: Rn = Q - Q - Mout + Min  Thus: Rn = 850.0 – 17.0 – 374.1 + 363.7 = 822.6 W m-2 for Example B.     Now let us consider the same day and the same surface, but a different time.  At midnight, of course the sun is not shining, so the solar radiation (Q) and albedo terms (Q) are reduced to zero.   Example C: Month, time and weather: June, midnight, no cloud Solar radiation measured at surface (Q): 0.0 W m-2 (it’s midnight!) Type of Surface: the water of Lake Moody, behind the Village Residence Surface Temperature: 12°C Sky radiation from CO2 and water vapour radiated at an equivalent temperature of: 10°C   1. Solar radiation (Q): given as 0.0 W m-2   2. Solar radiation reflected (Q): Since Q = 0.0 W m-2,  Q = 0.0 W m-2 as anything multiplied by zero is also zero. Thus: Q= 0.0 W m-2   3. Outward terrestrial radiation (Mout): The intensity of radiation from a surface is given by the Stefan-Boltzmann Law: M = T4  W m-2 where  = Stefan-Boltzmann constant = 5.67 x 10-8 W m-2 K-4 T = temperature of the emitting surface in K  (K = °C + 273) Thus: Mout = (5.67 x 10-8 W m-2 K-4) x ((12 °C + 273) K)4 =(5.67 x 10-8) x (285)4  W m-2 =374.1 W m-2   4. Inward terrestrial, or counter radiation (Min): (again, using the Stefan-Boltzmann Law: M = T4  W m-2) M in = (5.67 x 10-8 W m-2 K-4) x ((10°C + 273) K)4 =(5.67 x 10-8) x (283)4  W m-2 =363.7 W m-2   5. Combination of the various factors: We can now insert the above values into the original energy budget formula: Rn = Q - Q - Mout + Min  Thus: Rn = 0.0 – 0.0 – 374.1 + 363.7 = -10.4 W m-2 for Example C.       TABLE 1.1     Typical Albedos of Terrestrial Surfaces   Surface Albedo (%) Earth and Atmosphere (average value) 30   Sandy soil 25   Dense Forest 8   Field Crops 25   Grass 20   Water - Sun near horizon 40-75 Water – Sun overhead 2   Snow  - Fresh 90 Snow – Old 45   Concrete 25   Asphalt Pavement 3   Clouds – thick 60 – 90 Clouds – thin 30 - 50                     Figure 2a – c. Examples of Terrestrial Surfaces with varying albedos near Brock University     Example Calculation Summaries   EXAMPLE A: Q = 850.0 W m-2 (Given) Q = 0.03 x 850.0 W m-2 = 25.5 W m-2  M out = (5.67 x 10-8) x (35°C + 273)4 W m-2 = 510.3 W m-2 M in = (5.67 x 10-8) x (10°C + 273)4 W m-2 = 363.7 W m-2 Rn = Q - Q - Mout + Min = 850.0 – 25.5 – 510.3 + 363.7 = 677.9 W m-2    EXAMPLE  B: Q = 850.0 W m-2 (Given) Q = 0.02 x 850.0 W m-2 = 17.0 W m-2 M out = (5.67 x 10-8) x (12°C + 273)4 W m-2 = 374.1 W m-2 M in = (5.67 x 10-8) x (10°C + 273)4 W m-2 = 363.7 W m-2 Rn =  Q - Q - Mout + Min = 850.0 – 17.0 – 374.1 + 363.7 = 822.6 W m-2   EXAMPLE  C: Q = 0.0 W m-2 (Given) Q = 0.0 W m-2 (Given) M out = (5.67 x 10-8) x (12°C + 273)4 W m-2 = 374.1 W m-2 M in = (5.67 x 10-8) x (10°C + 273)4 W m-2 = 363.7 W m-2 Rn =  Q - Q - Mout + Min = 0.0 – 0.0 – 374.1 + 363.7 = -10.4 W m-2         Procedure and Questions   Assuming the conditions of Example A, B or C as appropriate, and unless otherwise specified, compute the net radiation for the following additional cases around the Brock University campus.     Note that from Brightspace you can download a file for your lab group that has the following information collected at each of the two sites outside: - Solar radiation (Q) in W m-2 - Solar radiation reflected (Q) in W m-2   You should also have values for Surface Temperature in oC and Sky Radiation from CO2 and water vapour expressed as an equivalent temperature in oC as measured during your field work.     For Questions #1 to #4, in the “fillable” areas beside each individual variable involved with the Net Radiation (Rn) equation, make sure to include all the values you are using to calculate your answers.  In other words, include as much information (i.e. values) within your calculations similar to that seen in the “Example Calculation Summaries” for EXAMPLES A, B, and C on the previous page.  The “fillable” areas will expand as you continue to type in your answers.     Question #1 - Example D: June, midnight and overcast.  Asphalt pavement at temperature 18oC and cloud with a base temperature of 2oC. (2.5 marks)   Solve the equations to determine an answer for EXAMPLE  D: Q = Click or tap here to enter text.   W m-2 Q = Click or tap here to enter text.   W m-2 M out = Click or tap here to enter text.   W m-2 M in = Click or tap here to enter text.   W m-2 Rn =  Q - Q - Mout + Min = Click or tap here to enter text.   W m-2         Question #2 - Example E: December, midday, cloudless and very dry so that downward long wave radiation (Min) is insignificant.  A newly snow covered surface at a temperature of -5oC. Assume a value of Q = 350 W m-2 (2 marks).   Solve the equations to determine an answer for EXAMPLE  E Q = Click or tap here to enter text.   W m-2 Q = Click or tap here to enter text.   W m-2 M out = Click or tap here to enter text.   W m-2 M in = 0  (given)   W m-2 Rn = Q - Q - Mout + Min = Click or tap here to enter text.   W m-2     Question #3 - Using the net radiation formula and the data collected outside, compute the net radiation over the grass covered surface (2.5 marks)    Calculations based on the Outside Data:   Over Grass: Q           =  Click or tap here to enter text.   W m-2 Q =  Click or tap here to enter text.   W m-2 M out =  Click or tap here to enter text.   W m-2 M in =  Click or tap here to enter text.   W m-2 Rn =  Q - Q - Mout + Min = Click or tap here to enter text.   W m-2           Question #4 - Using the net radiation formula and the data collected outside, compute the net radiation over the concrete sidewalk. (2.5 marks)     Over Concrete:  Q          =  Click or tap here to enter text.   W m-2 Q =  Click or tap here to enter text.   W m-2 M out =  Click or tap here to enter text.   W m-2 M in =  Click or tap here to enter text.   W m-2 Rn =  Q - Q - Mout + Min =  Click or tap here to enter text.   W m-2         With reference to the Net Radiation Equation, answer the following questions.    5. By examining the values for each of the four terms in the net radiation equation, explain why more energy is available at the surface over water (Example B) than at the surface over land (Example A). Hint, look at Q, Q, Mout and Min. Which are different between Example A and Example B? What makes them different? (2 marks)   Click or tap here to enter text.   6. Why, at midday in June, is the water temperature at Example B lower than that of the land at Example A? In answering, consider the demonstration using the sand and water filled beakers and the heat lamps performed during the lab, and Strahler_Text_Figure_4.18.pdf (found in the Assignment 1 folder on Brightspace). (2 marks)   Click or tap here to enter text.   7. In Example A, the net radiation is used to heat the asphalt pavement.  What is it used for in Example B? (Hint, only a small percentage is used to heat the water!) (2 marks)   Click or tap here to enter text.   8. Compare Examples C and D.  Which will be the warmer night?  Briefly explain, using your results from the net radiation equations. (2 marks)   Click or tap here to enter text.   9. Consider Example E.  Will the bright sunshine cause the snow to melt?  Briefly explain your answer using the results from the net radiation equations. (2 marks)   Click or tap here to enter text.     Global Perspective   Navigate to the following NASA website on Net Radiation:   Read the short description for the map series on the page.   Press the ‘play’ button on the left-hand side of the image and the movie will progress through data from 2006 until the most recent data in 2022. The area in orange indicates the region of positive net radiation, whereas the area in purple indicates the area of negative net radiation.  Watch the area covered by the orange and purple areas. Think about what is happening to create this pattern?   In the movie bar under the graphics that shows the progression of the images, you are able to slide the orange circle in the white bar (left or right) to select a specific month during the year. Slide the orange circle so you are looking at the data for January 2016.     10. Which location would have the highest net radiation loss in January? (1 mark)   a. North Pole (90oN) b. South Pole (90oS) c. Equator (0o)   Click or tap here to enter text.       11. Why would that location identified in the question above have the highest net radiation loss in January? (1 mark)   Click or tap here to enter text.     Now move the slider to July of 2016   12. Which location would have the highest net radiation gain in July? (1 mark)     a. Ushuaia, Argentina (54oS)   b. Miami Florida, USA (25oN)   c. Libreville, Gabon (0.4oN)   d. Reykjavik, Iceland (64oN)   Click or tap here to enter text.   13. Does the Niagara Region have a positive or negative net radiation gain in July? (1 mark)     a. Positive   b. Negative   Click or tap here to enter text.   14. When averaged over the entire year, the poles have which of the following? (1 mark) (click the corresponding checkbox for your answer)     a. Net energy deficit   b. Net energy surplus   Click or tap here to enter text.