Assignment # 1
Map Properties and Locating Objects

Part A: 1) The lag between the map’s geographic data and the map’s publication date is 3-4 years. This is important; as with in those 3-4 years there could be changes in the landscape of the mapped area that may not have been present at the time the geographic data of the area was taken or during the production stages of the map. 2) The NTS sheet number for the map is 30 M/5. Sheet numbers for the surrounding maps are: west, 40 P/8, north 30 M/12, and south 30 M/4. Part B: 1) The map projection used for this map is the Transverse Mercator Projection. The cartographer who invented this projection system is Gerardus Mercator.

2) The cartographer who invented this projection system is Gerardus Mercator. 3) The two coordinate systems used in this map are Universal Transverse Mercator system, using meters, and Longitude & Latitude, using degrees. 4) The distance of one side of a UTM grid cell is 1000m 5) UTM zone for this map is zone 17 6) The map is located east of the central meridian of its UTM zone 7) The datum of the NTS sheet is the North American Datum 1927 and the most recent North American Datum is 1983. Part C: 1) The scale of the NTS sheet is 150000 2) 5cm on the map is equivalent to 2.5km. 1cm: 50000cm = 1cm: 0.5km 5 cm: 2.5km 3) The surface are represented on the map is 1000 km2 4) Other scale used to publish NTS maps is the 1250000 scale.

6.2) The topographic profile for transects 1 tells us that it is a steep climb as the increase in elevation is rapid could be a mountainous region. The topographic profile for transect 2 shows a much less steepness but there is constant increase and decrease of elevation could be bumpy or uneven area. 6.3) The contour interval for this map is 10m. The importance of the interval is that it can be used to judge the hilliness or steepness of an area. 6.4) Euclidean Distancea = (xi-xj)2+(yi-yj)2 xi = 592000m.E yi = 4806000m.N xj = 595000m.E yj = 4806000m.N = (592000-595000)2+(4806000-4806000)2 = 30002 Euclidean distancea= 3000 m Euclidean Distanceb = (xi-xj)2+(yi-yj)2 xi = 597000m.E. yi = 4815000m.N xj = 599000m.E yj = 4813000m.N = (597000-599000)2+(4815000-4813000)2 Transect 1 | Transect 2 | Slopea = Change in Y / Change in X = Y2-Y1X2-X1 = 290-158595000-592000 X 100 Slopea = 4.4 | Slopeb = Change in Y / Change in X = Y2-Y1X2-X1 = 172-162599000-597000 X 100 Slopeb = -0.5 | = (-2000)2+(2000)2 = 800000 Euclidean distanceb= 2828.43 m 6.5)

The positive gradient for transects 1 shows a steep increase in elevation with respect to the horizontal distance as noted in the topographic profile. While the negative gradient, for transect 2 shows a very mild decrease in overall elevation with respect to horizontal distance but the decrease in elevation is very gradual.