INTRODUCTION TO ENGINEERING SURVEYING
(CE 1305)
Sr Dr. Tan Liat Choon
Email: [email protected]
Mobile: 016-4975551
Principles Of
Surveying-2
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RECTANGULAR COORDINATES
Rectangular coordinates are the convenient method available for describing the horizontal position of survey points
With the application of computers, rectangular coordinates are used frequently in engineering projects
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SPHERICAL COORDINATES
• Longitude: Degrees East or West from the prime meridian
• Latitude: Degrees North or South from the Equator
LATITUDE AND DEPARTURE
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Geographic coordinates
Starts at the equator and the prime meridian
Units of measurement are degrees, minutes, and seconds– 60 minutes in a degree
– 60 seconds in a minute
LATITUDE AND DEPARTURE
Closure of traverse is initiated by computing the latitude and departure of each line
The latitude of course is its orthographic projection upon the north-south axis of the survey
The latitude of course is simply the N component of a line in the rectangular grid system
The departure of course is its orthographic projection upon the east-west axis of the survey
The departure of course is simply the E component of line in the rectangular grid system
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LATITUDE AND DEPARTURE
In traverse calculations, latitudes and departures can be either negative (-) or positive (+)
North latitudes and east departures are considered positive (+)
South latitudes and west departures are considered negative (-)
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LATITUDE AND DEPARTUREIn this example, the length of AB is 300 m and bearing is shown in figure below. Determine the coordinates of point 2
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DepartureP12 = (300) sin (42° 30’)= 202.677 m
LatitudeP12 = (300) cos (42° 30’)= 221.183 m
NP2 = 300 + 221.183 = 521.183 m
EP2 = 200 + 202.677 = 402.667 m
Departure = EastingLatitude = Northing
(Known)Coordinates from point A(E 200 , N 300)
N
E
P1
P2
(unknown = to be calculate)Coordinates point A(E 402.667 , N 521.183)
221.
183
m
202.677 m
- S
- W
LATITUDE AND DEPARTUREIn this example, it is assumed that the coordinates of points 1 and 2 are known and we want to calculate the latitude and departure for the line AB
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DepartureP12 = EP2 – EP1= 320 –(100)= 220 m
LatitudeP12 = NP2 – NP1= -100 –(300)= - 400 m
Departure = EastingLatitude = Northing
(known)Coordinates from point A(E 320 , S -100)
N
E
P1
P2
(known)Coordinates point A(E 100 , N 300)
- S
- W
-40
0 m
220 m
LATITUDE AND DEPARTUREConsider our previous example, determine the E and N coordinates of all the points
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N
E
P1
- S
- W
P2
P4
P3
P5
SiteBalanced
Departure Latitude
S1-S2 - 20.601 - 188.388
S2-S3 86.648 - 152.252
S3-S4 - 195.470 29.933
S4-S5 - 30.551 139.080
S5-S1 159.974 171.627
LATITUDE AND DEPARTUREConsider our previous example, determine the E and N coordinates of all the points
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E coordinatesP5 = 0 mP1 = P5 + 159.974 = 159.974 mP2 = P1 + (- 20.601) = 139.373 mP3 = P2 + 86.648 = 226.021 mP4 = P3 + (- 195.470) = 30.551 mP5 = P4 + (- 30.551) = 0 m
SiteBalanced
Departure Latitude
S1-S2 - 20.601
S2-S3 86.648
S3-S4 - 195.470
S4-S5 - 30.551
S5-S1 159.974
N
E
P1
- S
- W
P2
P4
P3
P5
LATITUDE AND DEPARTUREConsider our previous example, determine the E and N coordinates of all the points
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N
E
P1
- S
- W
P2
P4
P3
P5
SiteBalanced
Departure Latitude
S1-S2 - 188.388
S2-S3 - 152.252
S3-S4 29.933
S4-S5 139.080
S5-S1 171.627
N coordinatesP3 = 0 mP4 = P3 + 29.933 = 29.933 mP5 = P4 + 139.080 = 169.013 mP1 = P5 + 171.627 = 340.640 mP2 = P1 + (- 188.388) = 152.252 mP3 = P2 + (- 152.252) = 0 m
LATITUDE AND DEPARTURE
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N
E
P1 (E 159.974 , N 340.640)
- S
- W
P2 (E 139.373 , N 152.252)
(E 30.551 , N 29.933) P4
P3 (E 226.021 , N 0)
(E 0 , N 169.013) P5
IMPORTANCE OF DETERMINING AREA
To include acreage in a property deed
Determine the area of sections of interest
Determine the area to estimate required materials
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METHODS OF MEASURING AREA
Division of the area into simple figures (triangles, rectangles and trapezoids)
Offsets from a straight line
Double meridian distances
Rectangular coordinates
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SIMPLE FIGURES
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AREA BY TRIANGULATION
If you know the length of all three sides
Area = √ s(s - a) (s –b) (s – c)
Where:a, b and c are the sides of the triangle and
S = ½ (a + b + c)
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AREA BY TRIANGULATION
If you know the length of two sides and the angle in between the sides
Area = ½ ab sinC
Where:
C is the angle between sides a and b
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AREA BY TRIANGULATION
You will likely end up using some combination of the methods
With a traverse, you will know perimeter distances and interior angles
You can take additional angle measurements or additional distance measurements while in the field to simplify calculations
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AREA COMPUTED BY COORDINATES
The area of a traverse can be computed by taking each N coordinate multiplied by the difference in the two adjacent E coordinates (using a sign convention of + for next side and – for last side)
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AREA COMPUTED BY COORDINATES
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N
E
P1 (E 159.974 , N 340.640)
- S
- W
P2 (E 139.373 , N 152.252)
(E 30.551 , N 29.933) P4
P3 (E 226.021 , N 0)
(E 0 , N 169.013) P5
AREA COMPUTED BY COORDINATES
There are a simple variation of the coordinate method for area computation
Sum 1 = E1N2 + E2N3 + E3N4 + E4N5 + E5N1
Sum 2 = N1E2 + N2E3 + N3E4 + N4E5 + N5E1
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AREA COMPUTED BY COORDINATES
List E and N coordinates in two columns
Repeat coordinates of starting point at the end
Sum the products designated by arrows– Left to right (minus sign)– Right to left (plus sign)– Difference between the two sums = twice the area
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AREA COMPUTED BY COORDINATES
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P1 (E 159.974 , N 340.640)
P2 (E 139.373 , N 152.253)
P4 (E 30.551 , N 29.933)
P3 (E 226.021 , N 0)
P5 (E 0 , N 169.013)
AREA COMPUTED BY COORDINATES
Sum 1 = 159.974 (152.253) + 139.373 (0) + 226.021 (29.933) + 30.551 (169.013) + 0 (340.640)= 36285.364
Sum 2 = 340.640 (139.373) + 152.252 (226.021) + 0 (30.551) + 29.933 (0) + 169.013 (159.974)
= 108925.854
Subtract the smaller sum from the larger sum (since it doesn’t make any sense to have a negative area)So:2 (Area) = Sum 1 - Sum 2 or Sum 2 – Sum 1
= 108925.854 - 36285.364 = 72640.490
Remember that this is equal to twice the area, so divide this number by 272640.490 / 2 = 36320.245 m2
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T h a n k Yo u &
Q u e s t i o n A n d A n s w e r
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