BASIC SURVEY COMPUTATION

JKA, POLITEKNIK KUCHING SARAWAK

CG 103 SURVEY COMPUTATION 1

CG 103 SURVEY COMPUTATION 1 JKA, POLITEKNIK KUCHING SARAWAK

NOOR FAIZAH BINTI ZOHARDIN

Table of Contents

1.0BEARING AND ANGLE
Definition of bearing and angle conversion of bearing to angle conversion
and vice versa.

2.0 SURVEY TRIGONOMETRY
Problem solving of triangles, squares and trapeziums using trigonometric
formula.

3.0 TRAVERSE
Calculation of latitude and departure using Bowditch and Transit methods.

4.0 COORDINATE
Types of coordinate and calculation of surveyed area using coordinate
method.

PRAKATA

Alhamdullah, segala puji bagi Allah, tuhan pencipta sekalian alam.
Buku ini dihasilkan bagi memberi sedikit panduaan kepada pelajar-pelajar Diploma Ukur
Tanah Politeknik yang mengambil subjek CG103 Survey Computation 1.
Ia dihasilkan dengan ilustrasi dan jalan kerja yang mudah difahami untuk menarik minat
pelajar mempelajari Ilmu pengiraan dalam bidang Ukur dengan lebih mendalam. Ia juga
dihasilkan berdasarkan Kurikulum terbaharu Politeknik Malaysia, Kementerian Pengajiaan
Tinggi
Adalah diharap buku ini dapat menjadi panduaan bagi pelajar untuk memahami serta
mencintai ilmu ukur.

NOOR FAIZAH BINTI ZOHARDIN
PegawaiPendidikanPengajiaanTinggi
JabatanKejuruteraanAwam
Politeknik Kuching Sarawak
CETAKKAN PERTAMA :JUN 2013

CHAPTER 1 :
BEARING AND ANGLE

CG 103 SURVEY COMPUTATION 1 CHAPTER 1: BEARING AND ANGLE
JKA, POLITEKNIK KUCHING SARAWAK

CHAPTER 1
BEARING AND ANGLE
Bearing = angle from true north
Angle = different of 2 bearing

Figure : Bearing

BACK BEARING
Base on the above figure:-

a. Bearing from A to B is 58° 26’ 35”
Back bearing or bearing from B to A is 58° 26’ 35” + 180° = 238° 26’ 35”

b. Bearing from B to C is 109° 37’ 55”
Back bearing or bearing from C to B is 109° 37’ 55” + 180° = 289° 37’ 55”

c. Bearing from C to D is 232° 08’ 47”
Back bearing or bearing from D to C is 232° 08’ 47” - 180° = 52° 08’ 47”

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CG 103 SURVEY COMPUTATION 1 CHAPTER 1: BEARING AND ANGLE
JKA, POLITEKNIK KUCHING SARAWAK
Note:
CALCULATE ANGLE FROM 2 BEARING 1. Check value for each bearing
Base on following diagram, calculate angle y 2. + 180° to get right value

Angle y = 135° 34’ 54” - 39° 06’ 41”

= 96° 28’ 13”

Not right value, to get
right value - 180°

Angle y = 129° 53’ 12” - (237° 26’ 19” - 180°)
= 129° 53’ 12” - 57° 26’ 19”
= 72° 26’ 53”

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CG 103 SURVEY COMPUTATION 1 CHAPTER 1: BEARING AND ANGLE
JKA, POLITEKNIK KUCHING SARAWAK

Calculate angle θ1 = 360⁰ - 289⁰ 23’ 22”
= 70⁰ 36’ 38”

Calculate angle θ2 = 49⁰ 43’ 26”
Angle y = θ1 + θ2 = 70⁰ 36’ 38” + 49⁰ 43’ 26”

= 120⁰ 20’ 04”

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CG 103 SURVEY COMPUTATION 1 CHAPTER 1: BEARING AND ANGLE
JKA, POLITEKNIK KUCHING SARAWAK

Angle y = (127° 55’ 15” + 180°)- 223° 19’ 28” Angle y = (26° 03’ 24” + 180°) - 73° 01’ 42”
= 307° 55’ 15” - 223° 19’ 28” = 206° 03’ 24” - 73° 01’ 42”
= 84° 35’ 47” = 133° 01’ 42”

Angle y = (360°-326° 25’ 56”)+ (208° 13’ 25”-180°) Angle y = ( 35°36’43”+180°) - 18° 05’ 47”
= 33° 34’ 04” + 28° 13’ 25” = 215°36’ 43” - 18° 05’ 47”
= 61° 47’ 29” = 197° 30’ 56”

Angle y = (360° - 282° 38’ 16”) + 36° 59’ 50” Angle y = 213° 53’ 20” -138° 59’ 39”
= 77° 21’ 44” + 36° 59’ 50” = 74° 53’ 41”
= 114° 21’ 34”
POLIKU | DIPLOMA UKUR TANAH 4
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CG 103 SURVEY COMPUTATION 1 CHAPTER 1: BEARING AND ANGLE
JKA, POLITEKNIK KUCHING SARAWAK

CALCULATE BEARING
Base on following diagram, calculate bearing for line x to q

Note :-
To calculate bearing

 Check bearing
 if clockwise = bearing + angle
 anticlockwise = bearing - angle

Example 1

Clockwise, so to calculate
bearing = given bearing + angle

Bearing x to q = 196° 17’ 53” + 113° 52’ 26”
= 310° 10’ 19”

Example 2

Anticlockwise, so to calculate
bearing = given bearing - angle

Bearing x to q = 289° 59’ 48” - 44° 53’ 18”
= 245° 06’ 30”

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CG 103 SURVEY COMPUTATION 1 CHAPTER 1: BEARING AND ANGLE
JKA, POLITEKNIK KUCHING SARAWAK

Example 3

Calculate angle θ1 = 360⁰ - 299⁰ 43’ 22”
= 60⁰ 16’ 38”

Bearing xq = 105⁰ 30’ 10” - 60⁰ 16’ 38”

= 45⁰ 13’ 32”

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CG 103 SURVEY COMPUTATION 1 CHAPTER 1: BEARING AND ANGLE
JKA, POLITEKNIK KUCHING SARAWAK

EXAMPLE 4

Bearing x to q = 43° 42’ 08” + 78° 38’ 51” Bearing x to q = 74°02’ 07” – (360° - 305° 12’ 21” )
= 122° 20’ 59” = 19° 14’ 28”

Bearing x to q = (317° 54’ 05” -180° ) + 80° 31’ 20” Bearing x to q = 214° 43’ 25” – (360° -235° 16’ 35” )
= 137° 54’ 05” + 80° 31’ 20” = 90° 00’ 00”
= 218° 25’ 25”

Bearing x to q = 360° - ( 137° 23’ 33” - 99° 37’34”) Bearing x to q = (130°51’ 28” + 180° ) - 112° 59’ 15”
= 322° 14’ 01” = 310° 51’ 28” - 112° 59’ 15”
= 197° 52’ 13”

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CG 103 SURVEY COMPUTATION 1 CHAPTER 1: BEARING AND ANGLE
JKA, POLITEKNIK KUCHING SARAWAK

QUADRANT BEARING AND WHOLE CIRCLE BEARING

 Whole circle bearing- clockwise angle from 0˚ to 360˚

 Quadratic bearing

 angle lying between 0˚ to 90˚
 direction from north or south

E

quadrant bearing = N θ° E Quadrant bearing = S θ° E
Whole circle bearing = θ° Whole circle bearing = 180° - θ°

Quadrant bearing = S θ° W Quadrant bearing = N θ° W 8
Whole circle bearing = 180° + θ° Whole circle bearing = 360° - θ°

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CG 103 SURVEY COMPUTATION 1 CHAPTER 1: BEARING AND ANGLE
JKA, POLITEKNIK KUCHING SARAWAK DRAWING

CONVERT QUADRANT BEARING TO WHOLE CIRCLE BEARING

QUADRANT BEARING WHOLE CIRCLE BEARING
N 70° 35’ 40” E = 70° 35’ 40”

N 35° 45’ 23” W = 360° - 35° 45’ 23”
= 324° 14’ 37”

S 40° 37’ 30” E = 180° - 40° 37’ 30”
= 139° 22’ 30”

S 40° 37’ 30” W `= 180° + 40° 37’ 30”
= 220° 37’ 30”

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CG 103 SURVEY COMPUTATION 1 CHAPTER 1: BEARING AND ANGLE
JKA, POLITEKNIK KUCHING SARAWAK DRAWING

CONVERT WHOLE CIRCLE BEARING TO QUADRANT BEARING

WHOLE CIRCLE QUADRANT BEARING
BEARING = N 70° 35’ 40” E

70° 35’ 40”

135° 45’ 23” = 180° - 135° 45’ 23”
= S 44° 14’ 37” E

240° 37’ 30” = 240° 37’ 30” - 180°
= S 60° 37’ 30” W

340° 37’ 30” = 360° - 340° 37’ 30”
= N 19° 22’ 30” W

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CG 103 SURVEY COMPUTATION 1 CHAPTER 1: BEARING AND ANGLE
JKA, POLITEKNIK KUCHING SARAWAK

TUTORIAL

1. Base on the figure 1 and the data given, calculate all the internal angle for this traverse.

LINE BEARING
AB 100⁰ 08’ 01”
BC 140⁰ 28’ 18”
CD 237⁰ 53’ 12”
DE 270⁰ 44’24”
EF 357⁰ 12’ 14”
FA 28⁰ 17’ 02”

2. Base on the close traverse data given, sketch the traverse and calculate all the internal
angle for this traverse.

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CG 103 SURVEY COMPUTATION 1 CHAPTER 1: BEARING AND ANGLE
JKA, POLITEKNIK KUCHING SARAWAK

3. Base on figure, calculate bearing for BC and CD if bearing for line AB is 320⁰ 20’14”
and internal angle for ABC is 50⁰ 20’ 15” and internal angle for BCD is 114⁰ 20’29”

LINE QUADRANT BEARING
AB N 80⁰ 10’ 41” E
BC S 37⁰ 48’ 51” E
CD S 2⁰ 13’ 10’ E
DE S 82⁰ 12’ 49” W
EF N 24⁰ 48’ 46” W
FG N 84⁰ 05’ 33” W
GA N 17⁰ 05’ 45” E

4. Base on the data given:-
i. Convert the quadrant bearing to whole bearing
ii. Calculate the internal angle for whole traverse

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CHAPTER 2 :
TRIANGULATION

CG 103 SURVEY COMPUTATION 1 CHAPTER 2: TRIANGULATION
JKA, POLITEKNIK KUCHING SARAWAK
1. c² = a² + b² - 2ab cos C
CHAPTER 2
Triangulation 2. = 2+ 2− 2
2
Pythagoras theorem
3. area = 1 absin C

2

4. area = s(s − a)(s − b)(s − c)

s= a + b + c

2

5. sin A = sin B

ab

1. 2 = 2 + 2

2. sin = =
ℎ

3. cos = =
ℎ

4. tan =


= =
cot − cot + cot

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CG 103 SURVEY COMPUTATION 1 CHAPTER 2: TRIANGULATION
JKA, POLITEKNIK KUCHING SARAWAK

EXAMPLE 1: PYTHAGORAS THEOREM

30 Tan 83° 19’ 20” = 40

sin 60° 52’ 07” = AP

AB AP = 4.683

AB = 34.344

sin 70° 54’ 10” = AD sin 60°28'30" = BC
48.223
11.029
BC = 41.961
AD = 10.422
cos 70° 54’ 10” = XD cos θ = 12.352

11.029 60.231

XD = 3.608 θ = cos -1 12.352

PREPARED BY NFZ 60.231

θ = 78° 09’ 58”

= �(60.231)2 − (12.352)2
BC = 58.951

POLIKU | DIPLOMA UKUR TANAH 15

CG 103 SURVEY COMPUTATION 1 CHAPTER 2: TRIANGULATION
JKA, POLITEKNIK KUCHING SARAWAK

Example 2
Calculate distance for line AG and AB

FIND ANGLE A

Angle A = (23⁰17’17” + 180⁰)-56⁰10’32”

= 147⁰ 06’ 45”

FIND ANGLE B
Angle B=(56⁰ 10’ 32” +180⁰) – (38⁰ 20’ 18” + 180⁰)

= 17⁰ 50’ 14”

FIND ANGLE G
Angle G= 38⁰20’18’ - 23⁰17’17”

= 15⁰ 03’ 01”

FIND DISTANCE AG USING SIN METHOD

sin 147⁰06′45" sin 17⁰50′14"
84.178 =


AG = 47.487

FIND DISTANCE AB USING SIN METHOD

sin 147⁰06′45" sin 15⁰03′01"
84.178 =


AB = 40.255

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CG 103 SURVEY COMPUTATION 1 CHAPTER 2: TRIANGULATION
JKA, POLITEKNIK KUCHING SARAWAK

Example 3
Calculate distance AB and bearing AG

FIND ANGLE B POLIKU | DIPLOMA UKUR TANAH 17
= (23⁰17’17” + 180⁰)-(342⁰52’05”-180⁰)
= 40⁰ 25’ 12”
FIND ANGLE A USING SIN METHOD
sin 40⁰25′12" sin

36.759 = 56.234
A = 82⁰ 42’ 12”

FIND BEARING AG
BEARING AG= 23⁰17’17’ + 82⁰ 42’ 12”

= 105⁰ 59’ 29”
FIND ANGLE G
ANGLE G = 342⁰52’05” –(105⁰59’29” +180⁰)

= 56⁰ 52’ 36”
FIND DISTANCE AB USING SIN METHOD
sin 56⁰52′36" sin 40°25′12"

= 36.759
AB = 47.480

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CG 103 SURVEY COMPUTATION 1 CHAPTER 2: TRIANGULATION
JKA, POLITEKNIK KUCHING SARAWAK

Example 4
Calculate bearing and distance for line XD

Angle XED = (227⁰ 45’ 25” - 180° )- (200° 48’ 06” - 180° )
= 26° 57’ 19”

Distance XD = �(64.439)2 + (150.611)2 − 2(64.439)(150.611) cos 26° 57’ 19”
= 97.644

sin 26° 57’ 19” sin
97.644 = 64.439

Angle EXD = 17° 24’ 23”
Bearing XD = 200°48’ 06” - 17° 24’ 23”

= 183° 23 43”

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CG 103 SURVEY COMPUTATION 1 CHAPTER 2: TRIANGULATION
JKA, POLITEKNIK KUCHING SARAWAK

EXAMPLE 5

Given area for triangulation AXY is 1568.518m². Calculate distance AX and bearing and distance for line
XY

Calculate distance AX using area formula

Calculate angle A = 153° 46’ 44” - 88° 57’ 42”

= 64°49’02”

area = 1 × ( AX )(AY ) sin A
2

1568.518 = 1 × ( AX )(66.380) sin 64°49'02"
2

AX =52.222

Calculate bearing and distance for line XY

2 = 2 + 2 − 2( )( ) cos
2 = (66.380)2 + (52.222)2 − 2(66.380)(52.222) cos 64°49’02”
2 = 4183.403

= √4183.403

= 64.679

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CG 103 SURVEY COMPUTATION 1 CHAPTER 2: TRIANGULATION
JKA, POLITEKNIK KUCHING SARAWAK


sin 64°49’02”

64.679 = 52.222
sin = 0.731

= −10.731
= 46° 56′31"

Calculate bearing xy
Bearing YA = 153° 46’ 44” + 180°

= 333° 46’ 44”
Bearing XY = 46° 56′31" - (360° - 333° 46’ 44”)

= 20° 43’ 15”

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CG 103 SURVEY COMPUTATION 1 CHAPTER 2: TRIANGULATION
JKA, POLITEKNIK KUCHING SARAWAK

Example 6

Given vertical angle from station A to the top of building is 48⁰ 29’ 07” and vertical angle from station B
to the top of building is 46⁰29’28”. Calculate distance from station A to B. If height of building CD is
46.421m.

=
cot + cot


46.421 = cot 48° 29′07" + cot 46° 29′28"

H = 46.421 �tan 1 + tan 46°129′28"�
48° 29′07"

H = 46.421 (0.885 + 0.949)

H = 46.421( 1.834)

H = 85.136

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CG 103 SURVEY COMPUTATION 1 CHAPTER 2: TRIANGULATION
JKA, POLITEKNIK KUCHING SARAWAK

Example 7

Base on above figure, calculate height of building AB.

=
cot −

= 55.977 33°05′40"
cot 56°59′41"−cot

= 55.977 1
561°59′41"−tan 33°05′40"
tan

= 55.977

1 − 1
1.540 0.652

= 55.977
0.649 − 1.534

= 63.251

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CG 103 SURVEY COMPUTATION 1 CHAPTER 2: TRIANGULATION
JKA, POLITEKNIK KUCHING SARAWAK

EXAMPLE 8

Station A and B are used to observe the building height. Bearing from A to B is 84° 44’ 03” and distance
65.796 m. Bearing from station A to lowest tower is 49° 53’ 08” and bearing from station B to lowest
tower is 302° 41’24”.Vertical angle from station A to top of building is 35⁰ 18’ 20” and vertical angle
from station B to top of building is 37⁰ 19’ 02” . Calculate height of the building and distance from
station A and B to building.

Calculate distance AP and distance BP
Angle B = 302° 41’ 24” –( 84° 44’ 03” + 180° )

= 37° 57’ 21”

Angle P = (49° 53’ 08” + 180° ) –(302° 41’ 24” - 180°)
= 229° 53’ 08” - 122° 41’ 24”
= 107° 11’ 44”

Angle A = 84° 44’ 03’ - 49° 53’ 08”
= 34° 50’ 55”

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CG 103 SURVEY COMPUTATION 1 CHAPTER 2: TRIANGULATION
JKA, POLITEKNIK KUCHING SARAWAK

sin 37° 57’ 21” = sin 107° 11’ 44”
65.796

AP = 42.362

sin 34° 50’ 55” = sin 107° 11’ 44”
65.796

BP = 39.356

Height of building

Tan 35⁰ 18’ 20” =
42.362

PQ = 30.000

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CG 103 SURVEY COMPUTATION 1 CHAPTER 2: TRIANGULATION
JKA, POLITEKNIK KUCHING SARAWAK Calculate distance AD and bearing BC

TUTORIAL
Calculate distance BC ,AC and angle BAC

32⁰ 51’ 33”

Answer : BC: 25.116 , AC 45.578 Answer : AD: 29.834 bearing BC :127° 35’35”
angle BAC= 33° 26’22”

Given area for triangulatian ABD is 600.680m² and area for triangulation BDC is 502.849 m² . Calculate
distance for line DC.

Answer: DC : 26.919
Calculate distance EC ,AC and area CDE

Answer: EC : 22.060 , AC :33.372 Area CDE : 269.385m²

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CG 103 SURVEY COMPUTATION 1 CHAPTER 2: TRIANGULATION
JKA, POLITEKNIK KUCHING SARAWAK

1. Base on given situation, calculate:-
i. Height of tower BC
ii. Horizontal distance AD
iii. Slope distance AB

2. Base on diagram below, calculate height h2. Given data

Reduce lavel A and B = 30 m
Reduce lavel D = 135.5 m
Vertical angle α = 30⁰ 00 00
Vertical angle β = 36⁰ 20’ 00”
Instrument height ,hi = 1.5 m
Horizontal distance from station A to B = 128.5m

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CHAPTER 3:
TRAVERSE

CG 103 SURVEY COMPUTATION 1 CHAPTER 3: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK

CHAPTER 3

TRAVERSE

TRAVERSE is continues line with bearing and distance

OPEN TRAVERSE CLOSE TRAVERSE

 Start and close the traverse at same point
 Close the traverse at unknown

coordinate

 Start and close the traverse at known coordinate

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CG 103 SURVEY COMPUTATION 1 CHAPTER 3: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK
DEPARTURE
LATITUDE  Distance x sin (bearing)
 Different of E/ W coordinate
 Distance x cos (bearing)  Departure = distance x sin bearing
 Different of N/S coordinate
 Latitude = distance x cos bearing

Latitude = 62 - 100 Depart = 232 – 200
= - 38 = 32

ADJUSTMENT FOR LATITUDE AND DEPARTURE

BOWDITCH METHOD TRANSIT METHOD

LATITUDE

DEPARTURE

LINEAR MISCLOSURE
LINEAR MISCLOSURE =
(∆latit) 2 + (∆depart) 2

1:
total of dis tan ce

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CG 103 SURVEY COMPUTATION 1 CHAPTER 3: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK

Base on following diagram calculate latitude, departure and linear misclosure for this traverse

STN BEARING DISTANCE LATIT DEPART
E (+) W (-)
N (+) S (-)

1 10.658 7.510 5.750 7.562 10.061
2 45° 11’ 50” 14.468 11.887 13.276 10.952
3 113° 25’ 10” 11.888 21.451 11.319 0.172
4 179° 10’ 20” 15.144 28.961 21.013
5 221° 37’ 50” 24.085 28.956 21.010 0.003
1 332° 57’ 10” 0.005

Total 76.243
Different

Linear misclosure =1: (0.005) 2 + (0.003) 2 POLIKU | DIPLOMA UKUR TANAH 30

76.243

= 1 : 13 076
Latit = 10.658 x cos 45° 11’ 50” = 7.510
Depart = 10.658 x sin 45° 11’ 50” =7.562
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CG 103 SURVEY COMPUTATION 1 CHAPTER 3: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK

CALCULATE LATITUDE, DEPARTURE AND ADJUSTMENT USING BOWDITCH METHOD

STN BEARING DISTANCE LATIT DEPART
1 13.358 E (+) W (-)
16.510 N (+) S (-)
2 54° 13’ 30” 17.989
24.380 7.809 7.500 10.838 5.229
3 117° 01’ 00” 16.941 +0.000 -0.001 -0.001 +0.001
7.809 7.499 10.837 5.230
4 196° 53’ 50” 17.212 14.708 24.376
0.454 -0.001 -0.001 +0.001
5 271° 04’ 00” +0.000 17.211 14.707 24.377
0.454
1 13° 52’ 50” 16.446 24.712 4.064 29.605
+0.001 0.003 -0.001 0.005
Total 89.178 16.447 4.063
Different 24.709 29.610

Linear misclosure =1: (0.003) 2 + (0.005) 2 Note: 16.941 × 0.005 = 0.001
89.178
89.178 + for small value and
– for large value
= 1 : 15 294

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CG 103 SURVEY COMPUTATION 1 CHAPTER 3: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK

CALCULATE LATITUDE, DEPARTURE AND ADJUSTMENT USING TRANSIT METHOD

STN BEARING DISTANCE LATIT DEPART
1 13.358 E (+) W (-)
16.510 N (+) S (-)
2 54° 13’ 30” 17.989
24.380 7.809 7.500 10.838 5.229
3 117° 01’ 00” 16.941 +0.000 -0.000 -0.001 +0.000
7.809 7.500 10.837 5.229
4 196° 53’ 50” 17.212 14.708 24.376
0.454 -0.001 -0.001 +0.002
5 271° 04’ 00” +0.000 17.211 14.707 24.378
0.454
1 13° 52’ 50” 16.446 24.712 4.064 29.605
+0.001 0.003 -0.000 0.005
Total 89.178 16.447 49.421 4.064 59.215
Different 24.709 29.610

24.376 × 0.005 = 0.002
59.215

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CG 103 SURVEY COMPUTATION 1
JKA, POLITEKNIK KUCHING SARAWAK

CALCULATE AREA USING DOUBLE MERIDIAN METHOD

STN BEARING DISTANCE LATIT

N (+) S (-) E(
10.838
1 13.358 7.809 -0.001
2 54° 13’ 30” +0.000 10.837
7.809 14.708
3 117° 01’ 00” 16.510 7.500 -0.001
0.454 -0.001 14.707
4 196° 53’ 50” 17.989 +0.000 7.499
0.454 17.212 4.064
5 271° 04’ 00” 24.380 16.446 -0.001 -0.001
+0.001 17.211 4.063
1 13° 52’ 50” 16.941 16.447 29.610
24.709 24.712
Total 89.178 0.003
Different

Area = 1036.903 POLIKU | DIP
2

= 518.452

PREPARED BY NFZ

CHAPTER 3: TRAVERSE

DEPART 2 X LATIT 2 X DEPART (2XLATIT) X
(+) W (-) DEPART

8 7.809 10.837 84.626
1 8.119 36.381 119.406
7 -16.591 45.858 86.771
8 -33.348 16.251 812.924
1 -16.447 -4.063 -66.824
7 1036.903

5.229
+0.001
5.230
24.376
+0.001
24.377
4
1
3
0 29.605
0.005

16.251-24.377+4.063 -16.447 x 4.064
=-4.063 = - 66.824

IPLOMA UKUR TANAH 33

CG 103 SURVEY COMPUTATION 1 CHAPTER 3: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK

MISSING LINE

Distance = (latit) 2 + (depart) 2 Bearing = tan θ = depart

latit

IDENTIFY BEARING BASE ON LATITUDE AND DEPARTURE

LATIT AND DEPART DIAGRAM BEARING

Latitude (+) Bearing = θ
Departure (+)

Latitude (-) Bearing = 180 - θ
Departure (+)

Latitude (-) Bearing = 180 + θ
Departure (-)

Latitude (+) Bearing = 360 - θ
Departure (-) POLIKU | DIPLOMA UKUR TANAH 34
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CG 103 SURVEY COMPUTATION 1 CHAPTER 3: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK

Base on following diagram, calculate bearing and distance for station 4 to 5

STEP BY STEP :

1. 1st check all bearing
2. After that calculate latit and

depart for all line
3. Calculate total of latit and

depart
4. Using latit and depart to

calculate bearing and distance

1. Bearing from station 1 to 2 = 245° 45’ 10” - 180° = 65° 45’ 10”

STN BEARING DISTANCE LATIT DEPART

5 -7.627
24.152
1 347° 58’ 50” 35.625 35.822 21.625
10.878 -11.050
2 65° 45’ 10” 26.489 -19.235 -27.100
-19.235 0.000
3 131° 39’ 10” 28.942 -8.230

4 209° 52’ 30” 22.183 0.000
5

Total (cheking)

Distance = (8.230)2 + (27.100) 2

= 28.322

tan θ = 27.100
8.230

θ = 73° 06’ 25”
Bearing 4-5 = 180° + 73° 06’ 25”

= 253° 06’ 25”

PREPARED BY NFZ POLIKU | DIPLOMA UKUR TANAH 35

CG 103 SURVEY COMPUTATION 1 CHAPTER 3: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK

Base on following diagram, calculate bearing and distance for station 2 to 3

1. Bearing from station 3 to 4 = 285° 47’ 34” - 180° = 105° 47’ 34”

STN BEARING DISTANCE LATIT DEPART

3 -10.147 35.875
-37.204 14.665
4 105° 47’ 34” 37.282 -22.773 -41.741
38.106 -32.265
5 158° 29’ 10” 39.990 32.018 23.466
0.000 0.000
1 241° 23’ 00” 47.549

2 319° 44’ 44” 49.931

3

Total (cheking)

Distance = (32.018)2 + (23.466)2

= 39.696

tanθ = 23.466
32.018

θ = 36⁰ 14’ 16”
Bearing 2-3 = 36⁰ 14’ 16”

PREPARED BY NFZ POLIKU | DIPLOMA UKUR TANAH 36

CG 103 SURVEY COMPUTATION 1 CHAPTER 3: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK

TUTORIAL
1. Base on following diagram, calculate bearing and distance for station 4 to 5

Answer : 41.507 193⁰ 52’ 11”
2 . Base on following diagram, calculate bearing and distance for station 6 to 1

Answer : 35.556 146⁰ 39’ 58” POLIKU | DIPLOMA UKUR TANAH 37
PREPARED BY NFZ

CHAPTER 4:
COORDINATE

CG 103 SURVEY COMPUTATION 1 CHAPTER 4: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK

CHAPTER 4
Coordinate
North /south coordinate = know coordinate + value of latitude
East /west coordinate = know coordinate + value of departure

COORDINATE VALUE
N 100 E 200
S 200 W 200 Coordinate N = + 100
N 100 W 300 Coordinate E = + 200
S 250 E 250 Coordinate S = - 200
Coordinate W = - 200
Coordinate N = + 100
Coordinate W = - 300
Coordinate S = - 250
Coordinate E = + 250

Basic coordinate calculation

1. Check bearing , bearing must given from know coordinate to un know coordinate
2. After that, calculate latitude and departure
3. Finaly, calculate all the coordinate

POLIKU | DIPLOMA UKUR TANAH 39

CG 103 SURVEY COMPUTATION 1 CHAPTER 4: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK
CALCULATION
DIAGRAM 1. Calculate latit and depart for line 1 to 2
2. Latit = 90.972 cos 55⁰ 09’03”
= 51.983
3. depart = 90.972 sin 55⁰ 09’03”
= 74.657
4. Coordinate N = 200.230 + 51.983 =
252.213
5. Coordinate E = 320.210 + 74.657 = 394.867

So coordinate station 2 = N 252.213 E 394.867

1. Bearing 1 to 2 = 311⁰ 13’ 55” - 180⁰
= 131⁰ 13’ 55”

2. Latit = 55.723 cos 131⁰ 13’ 55”
= -36.728

3. Depart = 55.723 sin 131⁰ 13’ 55”
= 41.906

4. Coordinate N = 100 + (-36.728)
= 63.272

5. Coordinate W = - 250 + 41.906
= -208.094

Coordinate station 2 = N 63.272 W 208.094

POLIKU | DIPLOMA UKUR TANAH 40

CG 103 SURVEY COMPUTATION 1 CHAPTER 4: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK
1. Bearing 2 to 1 = 129⁰ 38’ 59” + 180⁰
= 309⁰ 38’ 59

2. Latit = 88.096 cos 129⁰ 38’ 59”
= -56.213

3. Depart = 88.096 sin 129⁰ 38’ 59”
= 67.830

4. Coordinate S = -100 + (-56.213)
= -156.213

5. Coordinate E = - 200 + 67.830
= 267.830

Coordinate station 2 = S 156.213 E 267.830
1. Bearing 2 to 1 = 53⁰ 12’ 20” + 180⁰
= 233⁰ 12’ 20”
2. Latit = 82.122 cos 233⁰ 12’ 20”
= -49.187
3. Depart = 82.122 cos 233⁰ 12’ 20”
= - 65.762
4. Coordinate S = - 150 + (-49.187)
= - 199.187
5. Coordinate E = 200 + (- 65.762)
= -134.238

Coordinate station 2 = S 199.187 E 134.238

POLIKU | DIPLOMA UKUR TANAH 41

CG 103 SURVEY COMPUTATION 1 CHAPTER 4: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK

Example 1
Calculate coordinate for all station. Given coordinate for station 1 is N 200 E 350.

Step by step

1. Check all bearing, bearing should be read from known station.
So bearing for line 3 to 4 is 229⁰ 18’ 10” - 180⁰ = 49⁰ 18’ 10”

2. Calculate latitude and departure for all line
3. Calculate coordinate using latitude, departure and given coordinate.

STN BEARING DISTANCE LATIT DEPART COORDINATE

1 46.823 29.047 36.724 N/S E/W
2 51⁰ 39’ 30” 54.661 -26.299 47.919 200 350
3 118⁰ 45’ 33” 42.740 27.869 32.404 229.047 386.724
4 49⁰ 18’ 10”
202.748 434.643

230.617 467.047

POLIKU | DIPLOMA UKUR TANAH 42

CG 103 SURVEY COMPUTATION 1 CHAPTER 4: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK

Example 2
Calculate coordinate for all station. Given coordinate for station 3 is S 200 E 200 .

Step by step

1. Check all bearing, bearing should be read from known station.
So bearing for line 2 to 1 is 55⁰ 37’ 33” + 180⁰ =235⁰ 37’ 33”

2. Calculate latitude and departure for all line
3. Calculate coordinate using latitude, departure and given coordinate.

STN BEARING DISTANCE LATIT DEPART COORDINATE

3 -37.136 N/S E/W
2 297⁰ 59’ 38” -36.939
1 235⁰ 37’ 33” - 200 200

42.057 19.741 -180.943 162.864
44.755 -25.268
-43.347 125.925

STN BEARING DISTANCE LATIT DEPART COORDINATE
29.038
3 N/S E/W
4 57⁰ 08’ 38”
- 200 200

34.567 18.754 -181.246 229.038

POLIKU | DIPLOMA UKUR TANAH 43

CG 103 SURVEY COMPUTATION 1 CHAPTER 4: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK

Example 3

Base on latitude and departure given, calculate coordinate for all station. Given coordinate for station
5 is N 320 W 200

STN LATIT DEPART

1 12.363 -60.192
2 53.103 -5.907
3 31.749 37.971
4 -3.371 45.003
5 -29.968 38.970
6 -63.876 -55.845
1

Answer

COORDINATE

STN LATIT DEPART N/S E/W

5 -29.968 38.970 320.000 -200.000
6 -63.876 -55.845
1 12.363 -60.192 290.032 -161.030
2 53.103 -5.907
3 31.749 37.971 226.156 -216.875
4 -3.371 45.003
5 238.519 -277.067

291.622 -282.974

323.371 -245.003

320.000 -200.000

POLIKU | DIPLOMA UKUR TANAH 44

CG 103 SURVEY COMPUTATION 1 CHAPTER 4: TRAVERSE
JKA, POLITEKNIK KUCHING SARAWAK

Example 4

Base on diagram given, calculate the coordinate for all station. Given coordinate for station 1 is S200 E
200

Answer COORDINATE

STN BEARING DISTANCE LATIT DEPART N/S E/W

1 -36.846 -200 200
2 318⁰ 44’ 01” 38.875
3 51⁰ 09’54” 55.864 41.990 40.703 -158.010 -163.154
4 118⁰ 27’49” 49.907 31.296 4.257
5 173⁰ 56’51” 46.300 -22.067 -46.989 -126.714 202.029
1 256⁰ 44’ 20” 40.370 -40.145
48.276 -11.074 -148.781 242.732

-188.926 246.989

-200.000 200

POLIKU | DIPLOMA UKUR TANAH 45