Lab 2 Answer Sheet

Student Name: Elias Braunstein

Student Number:45667136

Date: October 13th, 2017

Instructions: Put your name, student number, and the date. You will lose marks if you do not. You are to print this document, along with your final map, and submit the report to Alex at the beginning of your lab during the week of Oct 9-13.

Total Possible Marks: This lab is worth 10% of your final course grade. There are 20 possible marks to earn in this lab.

Lab Questions

Q1. Can you suggest some reasons why contours are not typically used in GIS analyses but are often used to visualize topography? [1]

Contours are great for visualizing topography because they trace out lines of equal elevation and make it easy for the reader to visualize the area of interest. They aren’t good for GIS analyses because they will only capture data where elevation is changing in value. Areas that are flat won’t have many contours associated with them and thus data will be lacking in these areas.

Q2. Based on the feature attributes listed in your Geography Feature Catalogue query results, what type of lake is very unlikely to be found in this area? What types of lake are especially known from this area? Of the 13 possible feature classes, which ones are actually present in the water layer? [1]

Marshy and dry lakes are unlikely to be found in this area. This area is known for its intermittent lakes, which dry up during the summer months of the year. Of the 13 possible feature classes, definite lakes and intermittent lakes show up in the water layer.

Q3. What does your SQL statement look like? What would the SQL statements look like if you were working with a personal geodatabase, or a shapefile (DBF) rather than with a file geodatabase? What would the three statements look like if you were to use a wildcard that represented single characters rather than multiple characters? [2]

My SQL statement look like:

“FCODE” LIKE ‘GB15300%’

And

“FCODE” LIKE ‘GB15300###’ for single character wildcards.

If I was working with a shapefile or personal geodatabase it would look like:

“FCODE” LIKE ‘GB15300*’

Lab 2 Answer Sheet

And

“FCODE” LIKE ‘GB15300###’ for single character wildcards.

Q4. What is “XY Tolerance”? What could happen if you specified too large or too small a tolerance? (Use a figure in your answer to help clarify the issues.) [2]

XY tolerance is the minimum distance separating all feature coordinates. So if two features of the same class are within the XY tolerance distance away from one another, they will be joined together as one feature. The image below illustrates XY tolerance. Since the two points are within the XY tolerance from one another, they are joined as one point and the result is a connected line. A XY tolerance that is too small may not properly integrate features together. Conversely, a XY tolerance that is too high may have features collapsing on one another, resulting in an inaccurate visualization.

Q5. Create a layout showing the TIN you just created. You should include the lakes, rivers and roads (appropriately symbolized) on your map. Insert the map below. [3]

See next page

Lab 2 Answer Sheet

Lab 2 Answer Sheet

Q6. What effect would setting the Hedge to Somewhat have on the transformed values? [1] Defining a hedge changes the fuzzy membership values within the output fuzzy membership set. Setting the hedge to somewhat would increase the amount of values assigned high values (value of 1) within the output fuzzy membership set.

Q7. Why was 180° selected as the midpoint (rather than, say, 45°)? What effect does changing the spread from 1 to 0.1 to 0.001 have on the transformation of input values to output values? Including a figure in your answer would be helpful. [2]

180 degree was selected as the midpoint because the species of interest prefers southern facing surfaces, which correspond with having an aspect values of 180 degrees. Thus 180 is selected as the midpoint because it is the value that we want to be assigned to 1 upon normalization.

Decreasing the spread values makes the resulting fuzzy membership function spread larger. Thus, as you decrease the spread values, more aspect values close to 180 degrees are assigned high fuzzy membership values close to 1. This can be seen in the image below.

Final Map [4]:

See next page

Lab 2 Answer Sheet

Lab 2 Answer Sheet

Final Report [4]: The three main steps that were taken to produce the final map were to correct the inaccuracies in the data relating to Lake Osoyoos, create the DEM and preform two multiple criteria analyses (MCE) that were used to determine potential habitat for the plant species. In order to preform the MCEs, I had to normalize the elevation, slope and aspect values to values ranging from 0 to 1 in order to compare these three factors against each other in the MCEs. Values associated with 1 correspond to areas that are suitable for the plant species, such as southern facing slopes and elevations around 750.

The two MCEs were then created by applying weights of importance to each of the three factors. For the equal weights model, all three criteria were treated equally. In the weighted model, aspect was given the highest level of importance.

Hectares in each class (about 750ha for each MCE):

Weighted Model = 265.2 ha

Equal Weights Model = 255.1 ha

Both Models = 470.3 ha

Given these results, it is clear that the southern area of the map to the west of lake Osoyoos should be surveyed for potential habitat areas, as these areas are southern facing and not in close proximity to roads, like other suitable habitat areas area. Other suitable habitat areas that should be surveyed include the canyon about 8km to the west of the northern tip of Lake Osoyoos as well as the canyon about 2km east of the northern tip of Lake Osoyoos.

Lab 2 Answer Sheet

AHP Results: