1. / File size is often an issue in storing and analyzing data. Why do you suppose the temperature data (e.g. MAXHOT) is stored in units of 0.1ºC?
When analyzing the various temperature data (e.g. MAXHOT, MINCOLD), a description of the data showed the value units stored in units of 0.1ºC. Consider figure 1:
Figure 1 – MAXHOT
file title : Maximum Monthly Mean TemperatureThe file size of this image file was 420,480 bytes, stored as a binary filetype and integer datatype. I believe that this data is stored in units of 0.1ºC in an attempt to conserve disk space. Had the file been converted to a real datatype, the size would have been 840,960 bytes, double in size. As a result, the temperature data is well compiled with supporting documentation, and file size kept to a relative minimum.
2. / Summer precipitation is highest in what parts of Africa?
The images from May to October and November to April comprised precipitation totals by season, with MAYOCTPR and NOVAPRPR respectively. I found summer precipitation to be highest in the most eastern region of Africa. Figure 2 shows the defined ar ea as an ASCII vector file.
Figure 2: TEMP.VEC
3. / What is the maximum value in MAYOCTPR?
The maximum value in MAYOCTPR is 5013mm of precipitation, total from May to October.
4. / Provide a list of the top three countries in terms of highest average summer precipitation?
The top three countries in terms of highest average summer precipitation are as follows:
|
IDR_ID |
ABBREVNAME |
AVPR (mm) |
|
11 |
Sierra Leone |
2324.439 |
|
12 |
Liberia |
1820.721 |
|
59 |
GuineaBissau |
1681.145 |
5. / Why do we have to add one to MAYOCTPR and NOVAPRPR (but especially NOVAPRPR)?
Scalar arithmetic was used on MAYOCTPR and NOVAPRPR by adding the pixels in the input image by a constant value of 1. The minimum value in both of the aforementioned images was 0. By adding 1 to each, zero values were eliminated. This is especia lly important to NOVAPRPR. As MAYOCTPR is to be divided by NOVAPRPR, adding 1 to NOVAPRPR is vital, as we cannot divide by zero.
6. / Range of Values in RATIO image
|
Hemisphere |
||
|
Rainfall Regime |
Northern |
Southern |
|
Summer |
z > 1.3 |
z < 1.3 |
|
Winter |
z < 0.7 |
z > 0.7 |
|
Uniform |
0.7 < z < 1.3 |
0.7 < z < 1.3 |
7.After multiplying CHANGER and RATIO, what happened to the values in the Northern Hemisphere? What about values in the Southern Hemisphere? Would this method be appropriate if RATIO had positive and negative values in both hemispheres? Explain.
After multiplying CHANGER and RATIO, all values in the Northern Hemisphere became 0 or positive, whereas all values in the Southern Hemisphere were either 0 or negative. I believe that this method would not have been appropriate if RATIO had posit ive and negative values in both hemispheres. If RATIO had negative values, they would have multiplied by –1 for the southern hemisphere, which would equate to a positive value, therefore outputting incorrect data, or a non-fatal error. In this case, all RATIO values were positive and all CHANGER values in the Southern Hemisphere were –1, thus this example worked well.
8.What are the minimum and maximum values of NSRATIO?
min. value : -5
max. value : 416
9.What can you say about the distribution of summer, winter, and uniform rainfall regimes in Africa?
The distribution shows that summer areas are located within a latitudinal strip just above the equator, winter rainfall areas are either south of the equator or north of the summer area, and unifrom rainfall occurs along the equatorial line.
10.What do you notice along the equator? What causes this? Can you think of a way to produce REGIMES without having this occur?
One notices that along the equator the uniform areas for rainfall. This is caused by the locational orientation of the areas along the equator. A way to produce REGIMES without having this occur is to use reclass use define areas for summer and w inter rainfall, ie two values.
11.If there are only 1's and 0's in the resulting image does that mean that the reclass was done
correctly? How might an error be made that wouldn't be caught by checking that there are only 1's and 0’s?
If there are only 1’s and 0’s, this may mean that the image was done correctly. An error may be made if there were values that were of a negative value multiplied by zero, which would equal zero. Thus boolean images with 0 or 1 values would be mo re suitable for better results.
12.Which OVERLAY operation should be used?
The overlay operation that should be used is first plus the second.
13.Which country has the largest area of land suitable for Eucalyptus? (HINT: Use EXTRACT).
Angola
14.Which statistic to you use for EXTRACT? Why?
I used the NVDI image as the image to be processed against the feature definition image of REGIME, using average as the summary type. I noticed that the standard deviation was high, and the data was skewed, so an averaging would produce better dat a. This extracted the NVDI statistics against the REGIME image which displayed the areas suitable for Eucalyptus.
In hindsight, I would use more rainfall regimes, to produce more ranges with less data skewing to the data, hence more precision.
15.Is there a difference in NDVI statistics for the rainfall regimes? Explain.
Yes.