Counting the number of cars in a dealerships lot
1.33 m. Assuming that the average car is 3 meters across, with 1 meter between cars, each car should have an allotted space of 4 meters across (giving half of the 1 m in-between space to each car, 3 0.5 0.5 4). To be able to count the cars, one must be able to distinguish between the cars, thus we divide the 4 m space into three parts two for the pixels that distinguish the spaces beside the car, and one to distinguish the space taken up by the car. 4 divided by 3 is 1 and 13, or 1.33.
Determining the extent of damage from pine beetle infestation in a forest
1-3 m. Since pine beetle infestation can turn entire groves a shade of red, there really is not much need for high resolutions. Although, it can be useful to see individual pines just in case there is one pine beetle attacked tree or so in the other unaffected parts of the forest, so it can be pinpointed and resolved before it spreads. The spatial resolution value of 1-3 meters is made from estimating the width of a small pine tree.
Measuring the amount of corn grown in Iowa
60-65 m. Corn production is usually measure in acres, with the number of bushels per acre usually already established. Assuming that the cornfields are regular (square or rectangular) shaped one can use the length of the side of a square-shaped acre as the spatial resolution. Since the units must be metric, one acre converts to around 4046 m2, the square root (length of the side of a square) of which is 63 m. For better error margins though, the spatial resolution adopted is 60 to 65 meters.
Identifying the make of car in a driveway
0.20 0.50 cm. The logoslabels on a car vary from a few centimeters to several inches. While a few have very unquestionable large insignias, like the Jaguar or the Chevrolet, some lesser known car manufacturers tend to have metal appliqu words on the back of the car. The estimated spatial resolution of 0.20 to 0.50 cm was chosen so that small labels for example those from the Asian-made company Nissan, can be read more or less clearly.
Counting the number of people in a crowd
10-15 cm. Assuming that the people have dark or vivid hair colors one should be able to count the number of people in a crowd, literally by the number of heads. Assuming that the count includes children, this should be the resolution as it is the smallest unit one needs to find. The average diameter of the top of the head of a child who can already walk in a crowd is estimated to be 10-15 cm. This is of course also assuming that the people are no less than 10 cm apart from each other.
Measuring the area covered by a city
1 m. Measuring area is different from counting, as one can simply count the dark pixels among the light pixels when for example, counting the number of black-haired people. A resolution of 1 m seems like an easy way to measure the area of a city with a margin of error of a few meters, or even parts of a meter if the city has a regular shape. One simply needs to count the number of pixels within the citys borders and that is a good estimate of the area in square meters.
3. The remote sensing process
What is the question I am trying to answer
The research question is What is the percent of damaged buildings within the Sri Lankan coast after the 2004 tsunami
What is my hypothesis
The hypothesis is that 95 of the damage was close to the coast, within a strip of 1 kilometer.
How will I test my hypothesis
The hypothesis will be tested by using data from the Quickbird sensor, with a spatial resolution of 0.61m. While the damaged structures will be identified through interpretation the total number of damaged buildings will be counted, as well the number of buildings damaged just along the coast. Solving for the percentage of damaged buildings just along the 1km strip relative to the total number of damaged buildings will answer whether it really was 95 of the damage.
What limitations do my methods have
A limitation is that, since the buildings surveyed will be damaged, the spatial resolution used, or the researchers interpretation may prove to be faulty.
How will I disseminate my results
By producing a map of the tsunami damage and giving copies of it to the Sri Lankan government as well the United Nations disaster management office.
4. Platform choice
For each of the following tasks, which platform would you chooseChoose your platform from the following list submarine, ship, truck, helicopter, airplane, kite, balloon, LEO satellite, GEO satellite.Indicate why you chose the platform you did above other competing platforms.In your answers, consider spatial resolution and coverage as well as any additional factors (such as survivability and cost) you think important.We havent covered the characteristics of sensors yet, so assume the same sensor is mounted on all of the platforms or that you are riding on the platforms.
a. Determining global weekly average cloud cover
GEO satellite. The GEO orbits the earth in a geosynchronous manner, and the spatial resolution need not be too high for cloud cover.
b.Tracking the growth and movement of hurricanes
LEO satellite. Although it would make sense to have w weather balloon track a hurricane, the fact that it is a hurricane can be a threat to the balloon itself and the equipment therein. The LEO satellite would be a good choice since the growth and movement will not only be tracked but its possible to trace its path given the scale afforded by the LEO, to predict where it will go.
c.Tracking icebergs in the Labrador Sea over the winter
Submarine. Icebergs might not be tracked properly from the air as only a small part of it is visible above water. Also, considering that it will be conducted in the winter, a submarine can move around much more as water has a property in which only the top of the water is solid ice will the underwater it is still liquid.
d.Counting prairie dogs over a 1-acre site
Kite. A 1-acre site is a relatively small area plus going among the prairie dogs in a truck will disrupt them and they will move around, thus making counting difficult.
e.Analyzing the health of crops on a weekly basis over a 160 acre farm
Truck. Analyzing the health of crops does not need an aerial view as one can see the crops from the ground and one can walk among them if need be. One really only needs to drive around the peripherals of a field in order to assess the health of the crops.
f.Monitoring a large forest fire in progress and assess the danger to homes
Helicopter. One can swoop around the burning area in a helicopter and when needed, swoop down to pick up the stranded camper of two. A helicopter would also be more useful like for this instance where smoke will be rising up its propellers will dispel some of the smoke while an airplane would be useless.
g.Continuous monitoring of all forests in Southern California for detecting fires
If its simply to detect fire one can use an airplane. A regular trip across the potentially afflicted every so often would be decently cheaper and relatively faster plus the agility of a helicopter will not be needed.
h.Monitoring traffic congestion on the freeways across a large urban area
Kite or balloon. A balloon can be set free to float over the area and provide the aerial view, although one needs to retrieve it and there is no guarantee that it will not be caught up in at the urban jungle. Its still a cheap and interesting method though, and if the balloon is released from a well-placed area, with all the bases in terms of getting it back, etc. are resolved, it seems like the best platform for this problem.
i.Monitoring the spread of an ash plume from an erupting volcano
LEO satellite. An ash plume from an erupting volcano can be seen from outer space with good enough resolution. One cannot be there in person as its too dangerous, and kites and the like would be ineffective for the kind of data needed.
j.Monitoring changes in the polar ice caps over a period of two decades
Ship. Considering a well-manned crew in a good ship is self-sufficient in theory with occasional trips to mainlands for supplies. Especially since the monitoring will span two decades.
5. EMR frequencies
10 m
Infra-red. Referring to the electromagnetic spectrum illustration in the notes, the 10 m is in the range of Infra-red rays.
0.01 m
First converting the wavelength to nanometers 0.01 m 0.01 1 m 0.01 1000 nm 10 nm. From the EM spectrum illustration again, 10 nm, or 10-8 m, falls within the range of Ultraviolet rays.
650 nm
Even without conversion, 650 nm is known to be within the (relatively) small visible light range. In fact, 650 nm is red.
1 10-10 m
1 10-10 m, or simply 10-10 m, falls within the range of X-rays.
10 km
10 km can be converted to meters 10 km 10 1 km 10 1000m 10, 000 m. 10, 000 m, or 104 m, falls within the range of AM transmissions, or simply, Long Waves.
6.EMR wavelengths
37 GHz
( c (. So, using this equation, the conversion of 37 GHz to 37000 Hz (also using 1s for consistency with units), and the standard value for speed of light c which is 3.0 108 ms we obtain the EMR wavelength
( (3.0 108 ms) (37 109 1s) 0.008108 m
which converts to 8.1 mm, or 8. 1 10-3 m.
This value falls within the range of Microwaves, although it is rather close to far Infra-red rays.
60 Hz
( c (. ( (3.0 108 ms) (60 1s) 5,000,000 m
5 106 m most definitely fall within the range of Long Waves.
97.3 MHz 97.3 106 1s
( c (. ( c (. ( (3.0 108 ms) (97.3 106 1s) 3.08324 m 3.08 m
3.08 m falls within the range of television (TV) and radio transmissions.
850 kHz 8.5 105 1s
( c (. ( c (. ( (3.0 108 ms) (8.5 105 1s) 352.94 m 350 m
350 meters falls within the range of shortwaves.
800 THz 800 1012 1s 8 1014 1s
( c (. ( c (. ( (3.0 108 ms) (8 1014 1s) 3.75 10-7 m 4 10-7 m or 400 nm.
The obtained EMR wavelength of 375 nm (rounded up 400 nm given the 1 significant figure), falls around the vicinity of the border between visible light and Ultraviolet rays.
7. Types of aerial photographs
4.01 Vertical
4.02 High-Oblique
4.03 Low-Oblique
4.04 High-Oblique
4.05 Vertical
8. Color combinations
a.white RGB high red, high green, high blue CYM low cyan, low yellow, low magenta b.black RGB low red, low green, low blue CYM high cyan, high yellow, high magentac.medium grey RGB medium red, medium green, medium blue CYM medium cyan, medium yellow, medium magentad.Blue RGB low red, low green, high blue CYM high cyan, low yellow, high magentae.green RGB low red, high green, low blue CYM high cyan, high yellow, low magenta
9. True-color and color-infrared film
a. The color would be green. True-color images are the same as what a normal human eye sees and since the human eye can only see what color an opaque surface reflects, it only will only see 500 nm, green in the visible spectrum.
Now, what color do you think this surface would appear in a color-infrared imageWhyIn a color-infrared image the surface would appear grey. Color-infrared images only see colors within the range 700 nm to about 900 nm the rest of the visible spectrum are seen as grey. And since the unknown surface is 500nm, not within the 700-900 range, it will be grey.
10. Airphoto interpretation5.01 It seems like an important government building complex. The courtyard and circular driveway are those similar to those of old grand mansion, but the parking lot on the left side of the picture instead of say, garages, shoots down the mansion theory. The sheer size of the complex also makes it unlikely to be a home. The architecture of the large main building is grand but not ornate. The lack of outposts and fences relays that though it may be a government complex, it is not a military-centric one. There are also other possibilities, such as a large university although the lack of a clear park makes it unlikely to be a place of neither leisure nor learning.
5.02 The buildings in the picture make up a coastal inn complex, or small resort. The swimming pool and the lounge chairs arranged in a row facing the sea are definite indications that this is a place of rest and leisure. One other possibility is that this is something along the lines of a retirement complex, due to the decently sized cottages, although retirement complexes are not known to have swimming pools. The inn or resort hypothesis is most likely due to the small, balconied rooms in both the seaside building and the larger roadside building.
5.03 The sheer number of cars behind the building, and their haphazard arrangement indicate that this is not a parking lot since the cars would incur major damage being parked that close to each other. This and the desolate state of the grounds around the building indicate that this might be a junkyard,
11. Interpretation of a color-IR image
Examine Figure 05.04 and answer the following questions
Are there any areas of crops being grown in the imageIf so, whereHow can you tell
Yes, there are active crop fields in the image. These are found in the upper left of the image, and some are even on the very bottom of the image, The regular rectangular-like shape of these areas, as well as the apparent texture of their surfaces indicate the presence of crops.
Are there any fallow fields (fields normally used for crops but now bare) in the image If so, whereHow can you tell
There are fallow fields in the image, and these are found on the mid- to lower left of the image. The barren texture of the evenly shaped areas marks them as such.
Are there any forest areas in the imageIf so, whereHow can you tell
The bottom left corner, and most of the right half of the image are occupied by forests. One can identify them as such from the darker hue of the red in those areas, and the irregular texture, these are undoubtedly foliage.
12. Calculating height using relief displacement
a.Using Figure 06.01, what is the height of the building in the lower left corner of the imageAssume that the aircraft taking the picture was flying at an altitude of 150 m and that the airphoto is vertical.
Drawing imaginary triangles from the 150 m altitude of the survey plane the difference between the distance of the center (reference) point to the top and the bottom of the building is 0.5 cm, or 5 mm. Using the estimated scale of 14,000 (calculation discussed below) 5 mm becomes 20, 000 mm or 20 m, the height of the building.
b.What is the scale of the imageExpress the scale as a representative fraction.How did you derive the scale
The roads in the image are about 3 mm across. In the real city, a 4-lane street would be roughly 12 m across. This gives us a ratio of 3 mm to 12 m. To be able to get the scale though, we need to have the unit for both values, so we convert 12 m to millimeters obtaining 12,000 mm. Now we have 3 mm is to 12,000 mm. Representing this as a fraction we get 312,000 then after simplifying the common denominator, 14,000.
13. Estimate area
Assuming a car has a length of 5 meters, and since the cars in the image are roughly 2 mm long the scale of the image is 12,500. Now estimating the diameter of the dome image to be 55 mm, thus giving it a radius of 27.5 mm using the 12,500 scale, the radius becomes 68.75 m. We can then calculate the area of it using the area of a circle, r2 where is given the value 3.14. This gives a rough area of 14,841.41 m2.
14. Tornado damage
First we estimate the scale of the image. Again, assuming that a regular car has a length of 5 meters, we look at the lengths of the cars in the image and estimate these to be 1 mm this gives us a scale of 15,000. Now inspection the image, we can see that the tornados damage can be clearly seen from the visible debris stretching from the lower left corner of the image to the upper right corner. We assume that the width of the path of destruction is the width of the tornado funnel. Picking out the widest part along this path of damage, the width of that part is measured to be around 20 mm. Converting this using the estimated scale of 15,000, we obtain a width of 100,000 mm or 100 m.
15. Crowd size estimate
For Figure 06.04, the image of the street is roughly 2 cm across and 16.5 cm long. This gives us an area of 33 cm2. Now we count the number of people in a random area of a particular size here weve chosen a 0.25 cm x 0.25 cm square since any larger than this is hard on the eyes to count. This is of course assuming the people are equally distributed along the street. We count 4 people per 0.0625 cm2 (or 116 of 1 cm2). So to get the total number of people we use (4 people per area) (16 area per cm2) (33 cm2). This gives us 2112 people.
We use the same method for obtaining the number of people in Figure 06.05. The street of this image is roughly around 2.5 cm across and 18 cm long, giving an area of 45 cm2. Again, assuming the people are evenly distributed, we count the people in a small area. Again, we use a 0.25 cm x 0.25 cm area, and again we count 4 people in that area. Calculating (4 people per area) (16 area per cm2) (45 cm2) gives us 2880 people.
16. Finding Landsat data HYPERLINK httplandsat.gsfc.nasa.govimages httplandsat.gsfc.nasa.govimages
1.33 m. Assuming that the average car is 3 meters across, with 1 meter between cars, each car should have an allotted space of 4 meters across (giving half of the 1 m in-between space to each car, 3 0.5 0.5 4). To be able to count the cars, one must be able to distinguish between the cars, thus we divide the 4 m space into three parts two for the pixels that distinguish the spaces beside the car, and one to distinguish the space taken up by the car. 4 divided by 3 is 1 and 13, or 1.33.
Determining the extent of damage from pine beetle infestation in a forest
1-3 m. Since pine beetle infestation can turn entire groves a shade of red, there really is not much need for high resolutions. Although, it can be useful to see individual pines just in case there is one pine beetle attacked tree or so in the other unaffected parts of the forest, so it can be pinpointed and resolved before it spreads. The spatial resolution value of 1-3 meters is made from estimating the width of a small pine tree.
Measuring the amount of corn grown in Iowa
60-65 m. Corn production is usually measure in acres, with the number of bushels per acre usually already established. Assuming that the cornfields are regular (square or rectangular) shaped one can use the length of the side of a square-shaped acre as the spatial resolution. Since the units must be metric, one acre converts to around 4046 m2, the square root (length of the side of a square) of which is 63 m. For better error margins though, the spatial resolution adopted is 60 to 65 meters.
Identifying the make of car in a driveway
0.20 0.50 cm. The logoslabels on a car vary from a few centimeters to several inches. While a few have very unquestionable large insignias, like the Jaguar or the Chevrolet, some lesser known car manufacturers tend to have metal appliqu words on the back of the car. The estimated spatial resolution of 0.20 to 0.50 cm was chosen so that small labels for example those from the Asian-made company Nissan, can be read more or less clearly.
Counting the number of people in a crowd
10-15 cm. Assuming that the people have dark or vivid hair colors one should be able to count the number of people in a crowd, literally by the number of heads. Assuming that the count includes children, this should be the resolution as it is the smallest unit one needs to find. The average diameter of the top of the head of a child who can already walk in a crowd is estimated to be 10-15 cm. This is of course also assuming that the people are no less than 10 cm apart from each other.
Measuring the area covered by a city
1 m. Measuring area is different from counting, as one can simply count the dark pixels among the light pixels when for example, counting the number of black-haired people. A resolution of 1 m seems like an easy way to measure the area of a city with a margin of error of a few meters, or even parts of a meter if the city has a regular shape. One simply needs to count the number of pixels within the citys borders and that is a good estimate of the area in square meters.
3. The remote sensing process
What is the question I am trying to answer
The research question is What is the percent of damaged buildings within the Sri Lankan coast after the 2004 tsunami
What is my hypothesis
The hypothesis is that 95 of the damage was close to the coast, within a strip of 1 kilometer.
How will I test my hypothesis
The hypothesis will be tested by using data from the Quickbird sensor, with a spatial resolution of 0.61m. While the damaged structures will be identified through interpretation the total number of damaged buildings will be counted, as well the number of buildings damaged just along the coast. Solving for the percentage of damaged buildings just along the 1km strip relative to the total number of damaged buildings will answer whether it really was 95 of the damage.
What limitations do my methods have
A limitation is that, since the buildings surveyed will be damaged, the spatial resolution used, or the researchers interpretation may prove to be faulty.
How will I disseminate my results
By producing a map of the tsunami damage and giving copies of it to the Sri Lankan government as well the United Nations disaster management office.
4. Platform choice
For each of the following tasks, which platform would you chooseChoose your platform from the following list submarine, ship, truck, helicopter, airplane, kite, balloon, LEO satellite, GEO satellite.Indicate why you chose the platform you did above other competing platforms.In your answers, consider spatial resolution and coverage as well as any additional factors (such as survivability and cost) you think important.We havent covered the characteristics of sensors yet, so assume the same sensor is mounted on all of the platforms or that you are riding on the platforms.
a. Determining global weekly average cloud cover
GEO satellite. The GEO orbits the earth in a geosynchronous manner, and the spatial resolution need not be too high for cloud cover.
b.Tracking the growth and movement of hurricanes
LEO satellite. Although it would make sense to have w weather balloon track a hurricane, the fact that it is a hurricane can be a threat to the balloon itself and the equipment therein. The LEO satellite would be a good choice since the growth and movement will not only be tracked but its possible to trace its path given the scale afforded by the LEO, to predict where it will go.
c.Tracking icebergs in the Labrador Sea over the winter
Submarine. Icebergs might not be tracked properly from the air as only a small part of it is visible above water. Also, considering that it will be conducted in the winter, a submarine can move around much more as water has a property in which only the top of the water is solid ice will the underwater it is still liquid.
d.Counting prairie dogs over a 1-acre site
Kite. A 1-acre site is a relatively small area plus going among the prairie dogs in a truck will disrupt them and they will move around, thus making counting difficult.
e.Analyzing the health of crops on a weekly basis over a 160 acre farm
Truck. Analyzing the health of crops does not need an aerial view as one can see the crops from the ground and one can walk among them if need be. One really only needs to drive around the peripherals of a field in order to assess the health of the crops.
f.Monitoring a large forest fire in progress and assess the danger to homes
Helicopter. One can swoop around the burning area in a helicopter and when needed, swoop down to pick up the stranded camper of two. A helicopter would also be more useful like for this instance where smoke will be rising up its propellers will dispel some of the smoke while an airplane would be useless.
g.Continuous monitoring of all forests in Southern California for detecting fires
If its simply to detect fire one can use an airplane. A regular trip across the potentially afflicted every so often would be decently cheaper and relatively faster plus the agility of a helicopter will not be needed.
h.Monitoring traffic congestion on the freeways across a large urban area
Kite or balloon. A balloon can be set free to float over the area and provide the aerial view, although one needs to retrieve it and there is no guarantee that it will not be caught up in at the urban jungle. Its still a cheap and interesting method though, and if the balloon is released from a well-placed area, with all the bases in terms of getting it back, etc. are resolved, it seems like the best platform for this problem.
i.Monitoring the spread of an ash plume from an erupting volcano
LEO satellite. An ash plume from an erupting volcano can be seen from outer space with good enough resolution. One cannot be there in person as its too dangerous, and kites and the like would be ineffective for the kind of data needed.
j.Monitoring changes in the polar ice caps over a period of two decades
Ship. Considering a well-manned crew in a good ship is self-sufficient in theory with occasional trips to mainlands for supplies. Especially since the monitoring will span two decades.
5. EMR frequencies
10 m
Infra-red. Referring to the electromagnetic spectrum illustration in the notes, the 10 m is in the range of Infra-red rays.
0.01 m
First converting the wavelength to nanometers 0.01 m 0.01 1 m 0.01 1000 nm 10 nm. From the EM spectrum illustration again, 10 nm, or 10-8 m, falls within the range of Ultraviolet rays.
650 nm
Even without conversion, 650 nm is known to be within the (relatively) small visible light range. In fact, 650 nm is red.
1 10-10 m
1 10-10 m, or simply 10-10 m, falls within the range of X-rays.
10 km
10 km can be converted to meters 10 km 10 1 km 10 1000m 10, 000 m. 10, 000 m, or 104 m, falls within the range of AM transmissions, or simply, Long Waves.
6.EMR wavelengths
37 GHz
( c (. So, using this equation, the conversion of 37 GHz to 37000 Hz (also using 1s for consistency with units), and the standard value for speed of light c which is 3.0 108 ms we obtain the EMR wavelength
( (3.0 108 ms) (37 109 1s) 0.008108 m
which converts to 8.1 mm, or 8. 1 10-3 m.
This value falls within the range of Microwaves, although it is rather close to far Infra-red rays.
60 Hz
( c (. ( (3.0 108 ms) (60 1s) 5,000,000 m
5 106 m most definitely fall within the range of Long Waves.
97.3 MHz 97.3 106 1s
( c (. ( c (. ( (3.0 108 ms) (97.3 106 1s) 3.08324 m 3.08 m
3.08 m falls within the range of television (TV) and radio transmissions.
850 kHz 8.5 105 1s
( c (. ( c (. ( (3.0 108 ms) (8.5 105 1s) 352.94 m 350 m
350 meters falls within the range of shortwaves.
800 THz 800 1012 1s 8 1014 1s
( c (. ( c (. ( (3.0 108 ms) (8 1014 1s) 3.75 10-7 m 4 10-7 m or 400 nm.
The obtained EMR wavelength of 375 nm (rounded up 400 nm given the 1 significant figure), falls around the vicinity of the border between visible light and Ultraviolet rays.
7. Types of aerial photographs
4.01 Vertical
4.02 High-Oblique
4.03 Low-Oblique
4.04 High-Oblique
4.05 Vertical
8. Color combinations
a.white RGB high red, high green, high blue CYM low cyan, low yellow, low magenta b.black RGB low red, low green, low blue CYM high cyan, high yellow, high magentac.medium grey RGB medium red, medium green, medium blue CYM medium cyan, medium yellow, medium magentad.Blue RGB low red, low green, high blue CYM high cyan, low yellow, high magentae.green RGB low red, high green, low blue CYM high cyan, high yellow, low magenta
9. True-color and color-infrared film
a. The color would be green. True-color images are the same as what a normal human eye sees and since the human eye can only see what color an opaque surface reflects, it only will only see 500 nm, green in the visible spectrum.
Now, what color do you think this surface would appear in a color-infrared imageWhyIn a color-infrared image the surface would appear grey. Color-infrared images only see colors within the range 700 nm to about 900 nm the rest of the visible spectrum are seen as grey. And since the unknown surface is 500nm, not within the 700-900 range, it will be grey.
10. Airphoto interpretation5.01 It seems like an important government building complex. The courtyard and circular driveway are those similar to those of old grand mansion, but the parking lot on the left side of the picture instead of say, garages, shoots down the mansion theory. The sheer size of the complex also makes it unlikely to be a home. The architecture of the large main building is grand but not ornate. The lack of outposts and fences relays that though it may be a government complex, it is not a military-centric one. There are also other possibilities, such as a large university although the lack of a clear park makes it unlikely to be a place of neither leisure nor learning.
5.02 The buildings in the picture make up a coastal inn complex, or small resort. The swimming pool and the lounge chairs arranged in a row facing the sea are definite indications that this is a place of rest and leisure. One other possibility is that this is something along the lines of a retirement complex, due to the decently sized cottages, although retirement complexes are not known to have swimming pools. The inn or resort hypothesis is most likely due to the small, balconied rooms in both the seaside building and the larger roadside building.
5.03 The sheer number of cars behind the building, and their haphazard arrangement indicate that this is not a parking lot since the cars would incur major damage being parked that close to each other. This and the desolate state of the grounds around the building indicate that this might be a junkyard,
11. Interpretation of a color-IR image
Examine Figure 05.04 and answer the following questions
Are there any areas of crops being grown in the imageIf so, whereHow can you tell
Yes, there are active crop fields in the image. These are found in the upper left of the image, and some are even on the very bottom of the image, The regular rectangular-like shape of these areas, as well as the apparent texture of their surfaces indicate the presence of crops.
Are there any fallow fields (fields normally used for crops but now bare) in the image If so, whereHow can you tell
There are fallow fields in the image, and these are found on the mid- to lower left of the image. The barren texture of the evenly shaped areas marks them as such.
Are there any forest areas in the imageIf so, whereHow can you tell
The bottom left corner, and most of the right half of the image are occupied by forests. One can identify them as such from the darker hue of the red in those areas, and the irregular texture, these are undoubtedly foliage.
12. Calculating height using relief displacement
a.Using Figure 06.01, what is the height of the building in the lower left corner of the imageAssume that the aircraft taking the picture was flying at an altitude of 150 m and that the airphoto is vertical.
Drawing imaginary triangles from the 150 m altitude of the survey plane the difference between the distance of the center (reference) point to the top and the bottom of the building is 0.5 cm, or 5 mm. Using the estimated scale of 14,000 (calculation discussed below) 5 mm becomes 20, 000 mm or 20 m, the height of the building.
b.What is the scale of the imageExpress the scale as a representative fraction.How did you derive the scale
The roads in the image are about 3 mm across. In the real city, a 4-lane street would be roughly 12 m across. This gives us a ratio of 3 mm to 12 m. To be able to get the scale though, we need to have the unit for both values, so we convert 12 m to millimeters obtaining 12,000 mm. Now we have 3 mm is to 12,000 mm. Representing this as a fraction we get 312,000 then after simplifying the common denominator, 14,000.
13. Estimate area
Assuming a car has a length of 5 meters, and since the cars in the image are roughly 2 mm long the scale of the image is 12,500. Now estimating the diameter of the dome image to be 55 mm, thus giving it a radius of 27.5 mm using the 12,500 scale, the radius becomes 68.75 m. We can then calculate the area of it using the area of a circle, r2 where is given the value 3.14. This gives a rough area of 14,841.41 m2.
14. Tornado damage
First we estimate the scale of the image. Again, assuming that a regular car has a length of 5 meters, we look at the lengths of the cars in the image and estimate these to be 1 mm this gives us a scale of 15,000. Now inspection the image, we can see that the tornados damage can be clearly seen from the visible debris stretching from the lower left corner of the image to the upper right corner. We assume that the width of the path of destruction is the width of the tornado funnel. Picking out the widest part along this path of damage, the width of that part is measured to be around 20 mm. Converting this using the estimated scale of 15,000, we obtain a width of 100,000 mm or 100 m.
15. Crowd size estimate
For Figure 06.04, the image of the street is roughly 2 cm across and 16.5 cm long. This gives us an area of 33 cm2. Now we count the number of people in a random area of a particular size here weve chosen a 0.25 cm x 0.25 cm square since any larger than this is hard on the eyes to count. This is of course assuming the people are equally distributed along the street. We count 4 people per 0.0625 cm2 (or 116 of 1 cm2). So to get the total number of people we use (4 people per area) (16 area per cm2) (33 cm2). This gives us 2112 people.
We use the same method for obtaining the number of people in Figure 06.05. The street of this image is roughly around 2.5 cm across and 18 cm long, giving an area of 45 cm2. Again, assuming the people are evenly distributed, we count the people in a small area. Again, we use a 0.25 cm x 0.25 cm area, and again we count 4 people in that area. Calculating (4 people per area) (16 area per cm2) (45 cm2) gives us 2880 people.
16. Finding Landsat data HYPERLINK httplandsat.gsfc.nasa.govimages httplandsat.gsfc.nasa.govimages
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