1. Chapter 6
TRAVERSE

2. Horizontal Control
Horizontal control is required for initial survey work (detail surveys) and for setting out work.
The simplest form is a TRAVERSE - used to find out the co-ordinates of CONTROL or TRAVERSE STATIONS.

3. Grass N A C D E B

4. Horizontal Control
Co-ordinates of CONTROL or STATIONS.

5. Horizontal Control A F B E B C A C D E D F G
a) POLYGON or LOOP TRAVERSE
There are two types : -
b) LINK TRAVERSE A B C D E F B C D E F A G

6. A B C D E F G X Y Both types are closed. a) is obviously closed
b) must start and finish at points whose co-ordinates are known,
and must also start and finish with angle observations to other known points.
Working in the direction A to B to C etc is the FORWARD DIRECTION
This gives two possible angles at each station.
LEFT HAND ANGLES
RIGHT HAND ANGLES

7. Using a theodolite we can measure all the internal angles.B C D E F
Consider the POLYGON traverse
The L.H.Angles are also the
INTERNAL ANGLES
Using a theodolite we can measure all the internal angles.
Σ (Internal Angles) =
( 2 N ) * 900
The difference between
Σ Measured Angles and Σ Internal Angles
is the Angular Misclosure
(or 3)
Maximum Alloable Angular Misclosure =
Theodolite *  (No. of Angles)
2 * Accuracy of
(Rule of thumb)

8. A F B C Standing at A looking towards F - looking BACK ΘAB ΘAF
Hence ΘAF is known as a Azimuth A B C F
ΘBA
ΘBC
LH angle ABC
Angle FAB
(LH angle)
Standing at A looking towards B - looking FORWARD
Hence ΘAB is known as a FORWARD Azimuth
BACK Azimuth (ΘAF ) L.H.ANGLE (<FAB)
= NEXT FORWARD Azimuth (ΘAB)
Reminder: every line has two Azimuth
ΘBA =
ΘAB
1800
FORWARD Azimuth ( )
BACK Azimuth (

9. Observed Clockwise Horizontal Angle
Traverse Example
12” / 4 = 3”
Observations, using a Zeiss O15B, 6” Theodolite, were taken in the field for an
anti - clockwise polygon traverse, A, B, C, D.
Traverse Station
Observed Clockwise Horizontal Angle
‘ “ C N B A
- 3” A B C D D
Σ (Internal Angles) =
Line
Horizontal
Distance
Σ (Internal Angles) should be
(2N-4)*90 =
Allowable = 3 * 6” * N= 36” AB
638.57
OK - Therefore distribute error BC
The bearing of line AB is to be
assumed to be 00 and the
co-ordinates of station A are
( mE ; mN) CD DA

10. LINE
BACK AZIMUTH
STATION
ADJUSTED LEFT
HAND ANGLE
FORWARD
AZIMUTH
WHOLE
CIRCLE
Azimuth q
HORIZONTAL
DISTANCE D + = + =
Use Distance and Azimuth to go from
POLAR to RECTANGULAR to get
Delta E and Delta N values.
Check 1 AD A
+or-
1800
638.57 AB BA B C D
+or-
1800 BC CB CD DC DA
+or-
1800 AD

11. LATITUDES AND DEPARTURES
FIGURE 6.5:
LOCATION OF A POINT.
A)POLAR TIES
B)RECTANGULAR TIES
LATITUDE = NORTH(+) SOUTH (-)=distance(H) x cos
DEPARTURE = EAST(+) WEST (-)= distance(H) x sin

12. WHOLE
CO-ORDINATE DIFFERENCES
HORIZONTAL
CIRCLE
DISTANCE
CALCULATED
BEARING q D D E D N
638.57
0.000 G
-0.094
-0.654

13. =+931.227m =+638.570m =-3677.764m =+2107.313m D EBC D NCD D NBC D ECD
NAB
= m
= m A D
NDA D
EDA
= m D

14. e is the LINEAR MISCLOSURE
e =  (eE2 + eN2 ) B eE A eN e A’ D

15. Fractional Linear Misclosure (FLM) = 1 in G D / e
WHOLE
CO-ORDINATE DIFFERENCES
HORIZONTAL
CIRCLE
DISTANCE
CALCULATED
BEARING q D D E D N
638.57
0.000 G G
-0.094
-0.654 eE eN
e =  (eE2 + eN2) =  ( ) = 0.661m
Fractional Linear Misclosure (FLM) = 1 in G D / e
= 1 in ( / 0.661) = 1 in 13500
[To the nearest 500 lower value]

16. Acceptable FLM values :-
1 in for most engineering surveys
1 in for control for large projects
1 in for major works and monitoring for
structural deformation etc.

17. Fractional Linear Misclosure (FLM) = 1 in G D / e
WHOLE
CO-ORDINATE DIFFERENCES
HORIZONTAL
CIRCLE
DISTANCE
CALCULATED
BEARING q D D E D N
638.57
0.000 G G
-0.094
-0.654 eE eN
e =  (eE2 + eN2) =  ( ) = 0.661m
Fractional Linear Misclosure (FLM) = 1 in G D / e
= 1 in ( / 0.661) = 1 in 13500
Check 2
If not acceptable Ex. 1 in 3500 then we have an error in fieldwork

18. If the misclosure is acceptable then distribute it by: -
a) Bowditch Method - proportional to line distances
b) Transit Method - proportional to D
N values E
and
c) Numerous other methods including Least Squares
Adjustments

19. a) Bowditch Method - proportional to line distances
The eE and the eN have to be distributed
For any line IJ the adjustments are dE IJ and dN IJ
dE IJ = [ eE / SD ] x D IJ
Applied with the opposite
sign to eE
dN IJ = [ eN / SD ] x D IJ
Applied with the opposite
sign to eN

20. Fractional Linear Misclosure (FLM) = 1 in SD / e
CO-ORDINATE DIFFERENCES
CALCULATED
WHOLE
CIRCLE
BEARING q
HORIZONTAL
DISTANCE D E N
638.57
0.000
-0.094
-0.654 eE eN
e =  (eE2 + eN2) =  ( ) = 0.661m 3
Fractional Linear Misclosure (FLM) = 1 in SD / e
= 1 in / = 1 in 13500
Check 2

21. dE IJ = [ eE / SD ] x D IJ
Applied with the opposite
sign to eE
eE = m SD = m
dE IJ = [ / ] x D IJ
= …... x D IJ
Store this in the memory
For line AB
dE AB = …x D AB
= …x
dE AB = m
For line BC
dE BC = …x D BC
= …x
dE BC = m
For line CD
dE CD = m
For line DA
dE DA = m

22. CO-ORDINATE DIFFERENCES
CALCULATED
ADJUSTMENTS
ADJUSTED
CO-ORDINATES D E N d S T A I O
0.000
-0.094
-0.654 eE eN
+0.007
+0.016
+0.039
+0.032

23. dN IJ = [ eN / SD ] x D IJ
Applied with the opposite
sign to eN
eN = m
dN IJ = [ / ] x D IJ
= …... x D IJ
Store this in the memory
dN AB = … x D AB
= …x
dN AB = m
dN BC = m
dN CD = m
dN DA = m

24. CO-ORDINATE DIFFERENCES
CO-ORDINATES A
CALCULATED
ADJUSTMENTS
ADJUSTED T I O N D E D N d E d N D E D N E N A
0.000
+0.007
+0.046
+0.007 B C D A
+0.016
+0.112
+0.039
+0.273
680.61
Check 3
+0.032
+0.223
-0.094
-0.654
S= 0
S= 0 eE eN

25. 6.13 SUMMARY OF TRAVERSE COMPUTATIONS
Balance the field angles (1st step)
Correct (if necessary) the field distances (2nd step)
Compute the bearings and/or azimuths (3rd step)
Compute the linear error of closure and the precision ratio of the traverse (4th step)
Compute the balances latitudes (y) and balanced departures (x) (5th step)
Compute coordinates (6th step)
Compute the area (7th step)

26. 6.14 AREA OF A CLOSED TRAVERSE BY THE COORDINATE METHOD
The double area of a closed traverse is the algebraic sum of each X coordinate by the difference between the Y values of the adjacent stations.
The final area can result in a positive or negative number, reflecting only the direction of computation (either clockwise or anti clockwise). However there area is POSITIVE…THERE ARE NO NEGATIVE AREAS.

27. 6.14 AREA OF A CLOSED TRAVERSE BY THE COORDINATE METHOD

30. 6.14 AREA OF A CLOSED TRAVERSE BY THE COORDINATE METHOD

31. THANK YOU FOR YOUR ATTENTION
END OF CHAPTER 6
THANK YOU FOR YOUR ATTENTION