introduction to engineering surveying (ce 1305) of surveying-2-2014.pdf · introduction to...

INTRODUCTION TO ENGINEERING SURVEYING

(CE 1305)

Sr Dr. Tan Liat Choon

Email: [email protected]

Mobile: 016-4975551

Principles Of

Surveying-2

1

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

2

SPHERICAL COORDINATES

• Longitude: Degrees East or West from the prime meridian

• Latitude: Degrees North or South from the Equator

LATITUDE AND DEPARTURE

4

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


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

5


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 (-)

6

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

7

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

8

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

9

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


10

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


11

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


12

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

13

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

14

SIMPLE FIGURES

15

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)

16


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

17


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

18

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)

19


20

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


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

21


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

22


23

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)


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

24

T h a n k Yo u &

Q u e s t i o n A n d A n s w e r

25

introduction to engineering surveying (ce 1305) of surveying-2-2014.pdf · introduction to...

Documents