GEOGRAPHY FORM 5 TOPOGRAPHICAL MAP INTERPRETATION-DETERMINE DIRECTIONS AND AREA
GEOGRAPHY FORM 5 TOPOGRAPHICAL MAP INTERPRETATION-DETERMINE DIRECTIONS AND AREA
UNAWEZA JIPATIA NOTES ZETU KWA KUCHANGIA KIASI KIDOGO KABISA:PIGA SIMU: 0787237719
ALSO READ;
TOPOGRAPHICAL MAP INTERPRETATION-DETERMINE DIRECTIONS AND AREA
DISTANCE AND SIZE AREA MEASUREMENTS
Topographical maps can be potentially used to asses the measurements of the geographical features observed with reflection to the respective actual areas represented. It is therefore important to learn how the measurements of the geographical features of represented areas can be established from the topographical maps.
The popular ground measurements which can be established from the topographical maps are of distance and size of different land structure.
DISTANCE DETERMINATION
Distance is defined as the length of an elongated object or space between the two points on the earth’s surface expressed in units of linear measurements like; meters, feet, kilometers and miles.
Distance measurements of any elongated object on the topographical map has to take into consideration of the following key issues.
- · Map scale; The ratio of distance between the map and the actual ground represented. This enables the change of convention distance (CD) into actual ground distance.
- · The appearance of the elongated object on the topographical map given whose distance is to be measured. This determines the technique to be applied to get the CD.
Distance determination
For straight elongated objects.
For straight elongated objects,one has to use a ruler more directly to get the conventional distance of the object on the map. i.e. a ruler has to be placed along the elongated object to be measured.
The distance of the road from above is determined as follows:-
· Map scale:- CD of the road = 8 cm
If 1cm represent 0.25 km
8cm represents ?
8cm x 0.25km = 2km
1cm
Thus, the distance of the roads is about 2km.
Distance measurements for the curved elongated objects.
It becomes more difficult to obtain the conventional distance for the curved elongated object with the use of a ruler directly from the topographical map. Up on this challenge and others, one has to resort the use of the following enhancing tools to get the conventional distance of what required to be established. The enhancing tools include the following:-
- A pair of divider
- A piece of paper
- A piece of thread
1. The use of a pair of divider
A pair of divider up on its nature is commonly used to measure short straight courses between points a long the elongated object. To make this, one has to do the following.
- Identify the two recommended points of the elongated object on the topographical map given. The points can be identified by considering the given grid references, place names. Sometimes, both grid references and place names and can be considered.
- Divide the elongated object on the topographical map into convenient short straight sections.
- Open your divider and measure the straight divisions (courses) along the recommended elongated object.
- Add up all the measures to get total convectional distance of the elongated object.
- The obtained convectional distance should be converted into the actual ground distance with respect to map scale provided.
The distance of the road above is obtained as follows:-
AB=1cm; BC = 1.5cm; CD = 0.5; DE = 2cm; EF = 2cm
· The total CD = 7 cm
· Map scale, 1 cm represents 0.5 km
Then;
If 1cm represent 0.5km
7 cm represent ?
7cms x 0.5km = 3.5 km
1 cm
Thus ; distance of the road is of about 3.5 km
2. The use of a piece of paper.
A piece of paper can be used to measure either a short or long distances.
Procedure;
- Take a piece of paper and fold it.
- Put the piece of paper on the map and measure each small distance along the route.
- Take the piece of paper to the linear scale on the map to have ground distance
Example;
The distance of the road below can be determined as follows:-
Distance determination with the use of a piece of paper is up on the following application:-
Thus; the distance of the road according to the application above is of about 4.5 km.
3. The use of a thread.
Procedure:-
- Identify the two recommended end points of the elongated object whose distance is to be measured on the map. The two ends can be identified by taking into consideration of grid references, place names or both.
- Take the thread and spread it along the elongated object from one end to another.
- Take the thread to a ruler to measure the conventional distance of the elongated object.
- Convert the conventional distance into ground distance with respect to map scale indicated on the map.
The length of the road from A to B is determined as follows.
- Map scale; 1 cm represents 0.5 km
- CD of the road = 11 cm
THEN;
If 1 cm = 0.5 km
11 cm = ?
11cm x 0.5 km = 5.5 km
I cm
Thus; the distance of the road is of about 5.5km
AREA SIZE DETERMINATION.
Area size refers to a bigness or extent of coverage of any part of the earth’s surface expressed in square unit of measurements. For instance; the bigness of a water body, plantation, forestland, country and others of the same consideration.
Area size determination of any thing from the topographical map should take into consideration of the following two keys issues.
- Map scale; The ratio of distance between the map and the actual ground represented. This enables to change of conventional measurements obtained on the topographical map into actual ground measurements.
- The shape of the feature on the topographical map given whose area size is to be measured whether regular or irregular. This determines the method to be used to get the size.
Area size determination for regular features.
Regular features are those whose appearance well defined. They include; triangle, rectangles, square, circle, trapeze, octogon and others of the same reflection. To calculate the size of these regular figures, some one has to apply a relevant mathematical application with respect to the shape of the features observed on the topographical map.
Consider the following:-
(a) Triangle.
Consider the following illustration.
The size of the area occupied by forest is calculated as follows:-
So long the forest land occupied the shape of a triangle, the relevant formula is as follows:-
Area size = ^{1}/_{2 }x base x height |
*Map scale; 1cm represents 0.5 km
*According to the measurements of the feature on the map.
· Base = 6 cms
· Heigh t = 14 cms
Then;
½ x (6 x 0.5 km) x (14 x 0.5 km)
½ x 3km x 7km
21 = 10.5
2
Thus; the area size of the forestland is of about 10.5 km^{2}
(b) Rectangle;-
Consider the following.
The size of the area occupied by forest from the given topographical map is determined by the following application.
So long, the forestland occupies the regular shape of a rectangle, the relevant formula is given as follows:-
Area size = width x length |
*Map scale; 1cm represents 0.5 km.
*According to measurements of the figure on the map,
*According to measurements of the figure on the map
- Width = 4 cm
- Length = 8cm
4 cm x 8 cm
(4 x 0.5 km) x (8 x 0.5 km)
2 km x 4 km = 8km^{2}
Thus; the area of the forestland is of about 8km^{2}
NOTE:
The two given above represent the other regular figures. It is therefore important for the students to be knowledgeable to relevant mathematical applications for calculating the size of the other regular figures up on observation on the topographical map.
Area size measurement for the irregular figures.
To determine size for the irregular features observed on the topographical maps, some onbe may resort the use of the following methods.
- Grid square method
- Stripping method
- Geometric method
The grid square method.
It is the most popular and widely used method of determining area for irregular figures observed on the topographical map with reflection to actual areas represented. It is mostly used, if the given map observed to have the vertical and horizontal grid lines on its face forming so perfect square as cross each other at right angles.
From above, the area size of the coffee plantation is calculated follows:-
* Map scale ……………… 1cm represents 0.5 km
*Method ………………… grid squares consideration
* Actual size of the ground by each grid square determination.
Area size = width x length |
According to the grid squares on the given map;
- Width = 2cm
- Length = 2cm
2 cm
2 cm
Then;
· 2cm x 2cm
(2 x 0.5 km) x (2 x 0.5km)
1km x 1km = 1 km^{2}
Hence; Grid sguare = 1 km^{2}
Complete squares = 3
3 x 1 km2 = 3 km^{2}
Incomplete squares = 11/2 = 5.5
* 5.5 x 1 km^{2} = 8.5 km^{2}
Thus; the area size of the recommended feature is of about 8.5 km^{2}.
Stripping method
By this method, part of the topographical map with reflection to the ground represented under consideration is divided into a number of convenient strips. It has to be followed by finding out the area size of each strip and have to be all added to get the entire area size of the feature.
Consider the following illustration:-
The area size of the lake is obtained as follows:-
*Map scale 1cm represents 0.5
- Strip A: 8cm x 1cm
(8 x 0.5 km) x (1 x 0.5 km)
4km x 0.5 km = 2km^{2}
- Strip B: 10 cm x 1 cm
(10 x 0.5 km) x (1 x 0.5km)
(10 x 0.5 km) x 0.5 km = 2.5 km^{2 }
5km x 0.5 km = 2.5 km^{2}
- Strip C: 8cm x 1cm
(8×0.5 km) x (1 x 0.5 = 2 km^{2}) = 2 km^{2}
- Strip D: 9cm x 1cm
(9×0.5 km) x 0.5km = 2.25km^{2 }
4.5km x 0.5km = 225km^{2}
- Strip E; 8cm x 1cm
(8 x 0.5 km) x (1 x 0.5 km)
4km x 0.5km = 2km^{2}
Then;
2km^{2} = 2.5km^{2} + 2km^{2} + 2km^{2} + 2.25km^{2} + 2 km^{2 }= 10.75km^{2}
Thus; the area size is of about 10.75km^{2 }
Thus; the area size of about 10.75km^{2}
Geometric method.
It is by dividing the boundary area of the feature into a convenient number of mathematical (geometric) figures of like rectangles, triangles or combination of these figures on\over the same. It should be followed by calculating the area size of the forced geometric figures with respect to the used scale. At last the distinctive size should be added to get the entire area size of the feature.
Consider the following illustration;-
*Map scale …………… 1 cm represents 0.5 km
Rectangle A
Area size = width x length |
8cm x 18 cm
(8 x 0.5 km) x (18 x 0.5 km)
4 km x 9 km = 36 km^{2}
Triangle B
Area size = ½ x base x height |
½ x 18cm x 10cm
½ x (18 x 0.5 km) x (10 x 0.5 km)
½ x 9km x 5km
½ x 45 km = 22 .5 km^{2}
Triangle C:-
Area size = ½ x base x height |
½ x 16cm x 6cm
½ x (16 x 0.5km) x (6X0.5km)
½ x 8km x km
½ x 24km = 12km^{2}
Then; 36km2 + 22.5 km2 + 12km^{2} = 70.5 km^{2}
Thus; the area size is of about 70.5km^{2}
EXERCISE
Qn. What is the length of the all weather road with loose surface from Grid 930352 to Grid 040339 in kilometers.
(NECTA 1999 —– Extract map Babati……….. Series Y 742)
Qn. Measure the lenghth of the major road from grd reference 886220 to grid reference 975192. Give your answer in kilometers. (NECTA 2004 —– Extract nmap of Mwanza sheer no. 33/2)
Qn. Calculate the area covered by forest in square kilometers.
(Necta 2009 —– extract map of MPWAPWA sheet no 163/4)
Qn. Calculate the area of kisangara estate in square kilometers
(EZEB —– 2009……. Extract map of Kisangara sheet no 73/1)
Qn. Calculate the area covered by plantation agriculture in square kilometers (NECTA 1999 —- Extract map Babati …… Series Y 742).
Qn. Find the actual distance of railway line from grid reference 860975 to 930039 in kilometers.
(NECTA 2006—– Extract map of Mpanda —– sheer no 153/3)
Qn; Calculate the area of Weru sisal estate. (NECTA 1997 – Extract map of Moshi —– series Y 742)
Qn. Measure the distance of the Moshi – mweka road all weather loose surface road (grid 154309 to 135400) (NECTA 1997 – Extract map of Moshi —– series Y 742)
Qn. What is the total area covered by the lake in square km?
Qn. Calculate the area of lake balangida gidaghangaat and give your answer in square kilometers. (NECTA 1998 —–Extra map of hanang — series Y 742)
Qn. Find the total area of the land of this Island found to the west of Easting 180 if the scale of the map to duced y half (DSM – Mock AN 2003—– Extract map Nansio Island sheet 22/1)
Qn. Calculate the area of seasonal swap, south of nothing 080 and west of easting 690. Give your answer is square km.. (JICA —- 2009) – Extract map of utete)
MAP REDUCTION AND ENLARGEMENT.
The cartographic processes of minimizing and expanding the size of any map understood as map reduction and enlargement respectively.
MAP REDUCTION
As it has been pre described, map reduction is a cartographic process of minimizing the size of a map to provide smaller representation of the area.
To reduce the map the following as rules should be followed:-
- All sides of the map i.e. the length and width should be reduced with the same number of times.
- All drawings on the map should be reduced with the same number of times.
- The scale has to be changed to become a comparatively smaller in order for it with respect to map to give the same ground measurement as the old map does.
- The map has to be redrawn by using the reduction measurements.
- Insert the details on the reduced map according to new reduction measurements. The transfer of details should take into consideration of the grid pattens, and if possible for making more accuracy, the diagonal lines can be drawn the grid squares.
Note:
The conventional symbols, signs and abbreviations should be reasonably small in order for them to fit onto the size of reduced map.
Impacts of map reduction.
- The map size becomes smaller and thus gives smaller representation of the area mapped.
- The map would become less detailed as it more selective due to the limitation of the map’s space.
- The cartographic symbols, signs and abbreviation appear smaller.
- The scale of the map becomes comparatively smaller.
Consider the given map and reduce it by ½
Procedure.
(i) Consideration of the given map by studying it clearly.
(ii) Determination of the amount of reduction reduction factor is by ½
(iii) reduction of the map length
- Length on the old map = 8cm
8 x 0.5 = 4cm
Hence;the new length = 4cm
(iv)reduction of the map width
- width on the old map = 8cm
8 x 0.5 = 4cm
Hence;the new length = 4cm
(v) Reduction of the grid squares
- Grid square on the old map = 2cm
2cm x 0.5 = 1 cm
Hence; the new grid square = 1 cm
(vi) Determination of the new scale
New scale = reduction factor x old scale |
½ x 1/50,000 = 1/100,000
Hence; the new scale = 1: 100,000
Alternatively the new scale can be determined by the following application
Map scale = Map distance Ground distance |
* 3cm = 1 cm to km
3 km
Or
4 cm = 1cm to 1 km
4km
Thus; the reduced map appears as follows:-
MAP ENLARGEMENT
It is a cartographic process of expanding the size of a map to provide larger representation of the mapped area.
To enlarge a map the following as rules should be followed:-
- All sides of the map, i.e. length and width should be enlarged with the same number of times.
- All drawings on the map should be enlarged with the same number of times.
- The scale has to be changed to become comparatively large in order for it with respect to the map to give the some ground measurements as the old map does.
- The map has to be redrawn by using the enlargement measurements.
- Insert the details. The transfer of details should take into consideration of the grid pattens, and if possible for making mre accuracy, the diagonal lines can be drawn across the grid squares.
Note
The conventional symbols signs and abbreviations or initial should be reasonably large in order for them to fit onto the enlarged map.
Impacts of enlargement.
- The map size become larger and thus, gives larger representation of the mapped area.
- The map is likely to be more detailed as it is less selective due to the larger space of the map.
- The cartographic symbols, signs and abbreviations may appear larger in size.
- The scale becomes comparatively larger.
Consider the given map below and it enlarge by 2.
1:50,000
Procedure
(i) Consideration of the given map studying it Clearly.
(ii) determine of the amount of enlargement. Enlargement is by 2.
(iii) Enlargement of the map length.
- Length on the old map = 4cm
4cm x 2 = 8cm
Hence ; the new length = 8cm
(iv)Enlargement of the map width
- width on the old map = 4cm
4cm x 2 = 8cm
Hence;the new width = 8cm
(v) Enlargement of the grid square.
- Grid square on the old map = 2cm
2cm x 2 = 4cm
Hence, the grid square = 4cm
(vi) Determination of the new scale
New scale = enlargement factor x old scale |
2 x 1/50,000 = 2/50,000 = 1/25,000
Hence; the new scale = 1:25,000
1:25,000
Alternatively the new scale can be determined by the following application.
Map scale = Map distance
Ground distance
8cm = 1cm to 0.25cm
2km
Qn. (a) Outlining your steps redraw provided (1:50,000) to a scale of 1:100,000 and indicate the following features:-
The forest area, Songoro and chimumbu hills, lake Duluti, Nkoaranga hospital and all weather road bound surface.
(b) Examine the two maps and comment on the impact the change of the map scale has on the map area and its contents.
(DSM – MOCK, 2002:Extract map of TENGERU: (sheet No./ 55/4Ed)
Qn. (a) Calculate the approximate area in square kilometers of the contoured map if its scale changed to 1:100,000 (NECTA 2001, Extract map of Kondoa – sheet # 104/4)
Qn. Redraw the map north of northing 120 and west of easting 650 by the scale of 1:25,000 and on it indicate woodland and seasonal swamp.
JICA ===2009) Extract map of ulete)
Qn. Draw a map of district found at the western part of the map by using the scale of 1:100,000 and on its show the following.
(i) Boundary of the district
(ii) Mwanza – Shinyanga road
(iii) usagara – Fela station road
(iv) Railway line
(v) Ngeleka hill, jijawenda hill and kagela hill
(vi) Out crop rock around jijawenda hill.
(vii) Bridges along Fela – Mwanza road (NECTA 2010)
No comments: