1. TRAVERSING

2. INTRODUCTION Literally means “passing across”
Definition( surveying) determine the lengths and directions of consecutive line
PURPOSE
To locate the features already existing in the area to be surveyed
To establish points or lines in accordance with predetermined measurements

3. INTRODUCTION
APPLICATION
Location plan
Setting out

4. CLASSIFICATION OF TRAVERSE
Closed traverse
returns back to the original position (starting point)
Application: Area determination, boundaries of lake
(widely use in engineering work, established control point)

5. CLASSIFICATION OF TRAVERSE
Open traverse
Does not return back to the original position (starting point)
Application: Path of highway,canal,pipeline
(Quality of traverse cannot be checked)

6. FREE OR LOOSE NEEDLE METHOD
METHOD OF TRAVERSING
CHAIN
MEASURING ANGLES
FAST NEEDLE METHOD
(tape & theodolite)
FREE OR LOOSE NEEDLE METHOD
(chain & compass)

7. - Represent points in a large area (point of Borehole in SI)
Coordinate System
Grid/Cartesan
P from O (Up,Tp)
Geographical
- Represent points in a large area (point of Borehole in SI)
Polar Coordinate
- P1 from O (d1,Ɵ1)

8. Calculation of Plan Distance
L = S sinz

9. Calculation of Plan Distance

10. LATITUDES AND DEPARTURES

11. LATITUDES AND DEPARTURES

12. LATITUDES AND DEPARTURES COMPUTATION
Latitudes = Length x cos θ Departures = Length x sin θ
Because a bearing angle never exceeds 90°, the Lat. and Dep. equations will always return positive values.

13. EXAMPLE 1
Compute the latitudes and departures for the following cases

14. EXAMPLE 1 (SOLUTION) CHECK: 4TH QUADRANT
Latitudes = cos (-25˚35’) = ft Departures = sin (-25˚35’)
= ft

15. EXAMPLE 1 (SOLUTION) CHECK: 3rd QUADRANT
Latitudes = cos (232˚50’) = ft Departures = sin (232˚50’)
= ft

16. SELF-EXERCISE
Fill up the table below by computing the latitude and departure

17. Type of Errors Systematic Errors Accidental Errors Bearing Distance
Angular error balancing bearing
Distance
Incorrect chain length
Temperature
Sloping ground
Tension (pull)
Accidental Errors
Error of closure coordinate adjustment

18. Closure
On a closed loop traverse, the sum of the Lats should equal zero as should the sum of the Deps.

19. Closure
But most traverses won’t close perfectly due to the presence of errors.
The Linear Closure, LC, /Closing error is the total misclosure distance. It is computed from the Lat. and Dep. errors:

20. Closure Reduced bearing of closing error, θ tan θ = 𝐷𝐸𝑃𝐴𝑅𝑇𝑈𝑅𝐸 𝐿𝐴𝑇𝐼𝑇𝑈𝐷𝐸
θ = 𝑡𝑎𝑛 −1 𝐷𝐸𝑃𝐴𝑅𝑇𝑈𝑅𝐸 𝐿𝐴𝑇𝐼𝑇𝑈𝐷𝐸

21. EXAMPLE 2
Determine the latitudes and departures for each side and compute the linear error of closure for the traverse.
Azimuth
Bearing
Distance (m)
48° 15 ′ 00"
𝑁 48° 15 ′ 00" 𝐸
423.58
79° 1 ′ 48"
𝑁 79° 1 ′ 48" 𝐸
366.20
178° 3′ 6"
𝑆 1° 56 ′ 54" 𝐸
343.28
241° 43 ′ 54"
𝑆 61° 43 ′ 54" 𝑊
502.03
329° 39 ′ 12"
𝑁 30° 20 ′ 48" 𝑊
330.10

22. EXAMPLE 2 (SOLUTION Azimuth Bearing Distance (m) Latitudes (m)
Departures (m)
48° 15 ′ 00"
𝑁 48° 15 ′ 00" 𝐸
423.58
79° 1 ′ 48"
𝑁 79° 1 ′ 48" 𝐸
366.20
69.686
178° 3′ 6"
𝑆 1° 56 ′ 54" 𝐸
343.28
11.671
241° 43 ′ 54"
𝑆 61° 43 ′ 54" 𝑊
502.03
329° 39 ′ 12"
𝑁 30° 20 ′ 48" 𝑊
330.10
𝑙𝑎𝑡=55.767
𝑑𝑒𝑝=78.259
𝑪𝒍𝒐𝒔𝒖𝒓𝒆 𝑬𝒓𝒓𝒐𝒓= 𝟓𝟓.𝟕𝟔𝟕 𝟐 + 𝟕𝟖.𝟐𝟓𝟗 𝟐
=96.096

24. Q: Calculate The Precision
EXAMPLE 3
TRAVERSE WITH BEARING
Q: Calculate The Precision

25. EXAMPLE 3

26. EXAMPLE 3
Precision = ′
Precision = 1 / 12000

27. Balancing Latitudes and Departures
Bowditch rule
Transit rule
Least square method

28. Balancing Latitudes and Departures

29. EXAMPLE 4
Calculate latitudes, departures and closing error for the following traverse . Adjust also the traverse using Bowditch’s rule.
Line
Length (m)
Azimuth AB
89.31
45˚ 10’ BC
219.76
72˚ 05’ CD
151.18
161˚ 52’ DE
159.10
228˚ 43’ EA
232.26
300˚ 42’

30. SOLUTION Line Length (m) Azimuth Bearings Quadrant AB 89.31 45˚ 10’
NE (I) BC
219.76
72˚ 05’ CD
151.18
161˚ 52’
18˚ 08’
SE (II) DE
159.10
228˚ 43’
48˚ 43’
SW (III) EA
232.26
300˚ 42’
59˚ 18’
NW (IV)
Line
Length (m)
Azimuth
Bearings
Latitudes
Departures AB
89.31
45˚ 10’
62.967
63.335 BC
219.76
72˚ 05’
67.605 CD
151.18
161˚ 52’
18˚ 08’
47.051 DE
159.10
228˚ 43’
48˚ 43’ EA
232.26
300˚ 42’
59˚ 18’
∑ (lat./dep.)
+0.508
+0.223

31. Closure Error tan θ = 𝑫 ∑𝑳 θ = 𝒕𝒂𝒏 − [ 𝑫 ∑𝑳 ] θ = 23˚ 42’ = 0.5547
( 𝟎.𝟓𝟎𝟖) 𝟐 + (𝟎.𝟐𝟐𝟑) 𝟐 =
tan θ = 𝑫 ∑𝑳
θ = 𝒕𝒂𝒏 − [ 𝑫 ∑𝑳 ]
θ = 23˚ 42’
The direction of the closing error is N 23˚ 42’. Since the error in latitude and departure are both positive, the corrections will be NEGATIVE.

32. Line
Length (m)
Latitudes
Departures
Correction L
Correction D AB
89.31
62.967
63.335 BC
219.76
67.605
-0.131 CD
151.18
47.051
-0.090 DE
159.10 EA
232.26 ∑
+0.508
+0.223
the corrections will be NEGATIVE so and should be adapted
Correction Latitudes
E.g: 𝟖𝟗.𝟑𝟏 𝟖𝟓𝟏.𝟔𝟏 ×−𝟎.𝟓𝟎𝟖 =
Correction Departures
E.g: 𝟖𝟗.𝟑𝟏 𝟖𝟓𝟏.𝟔𝟏 ×−𝟎.𝟐𝟐𝟑 =

33. SO HOW TO FIND CORRECTED LENGTH AND BEARING?
Line 1
Latitudes 2
Departures 3
C. Lat 4
C. Dep. 5
Adjusted Lat. [2+4]
Adjusted Dep. [3+5] AB
62.967
63.335 BC
67.605
-0.131
67.474 CD
47.051
-0.090 DE EA ∑
+0.508
+0.223
0.0004
0.0003
Correction is considered correct if the adjusted latitudes and departures finally approach zero value.
SO HOW TO FIND CORRECTED LENGTH AND BEARING?

34. Balance The Angle

35. Balance The Angle

36. Balance The Angle

37. Area Computed by Coordinates

38. *This note has been amended from the original note prepared by Puan Nur Fitriah Isa of JTKA.