Chapter 6 TRAVERSE

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.

Grass N A C D E B

Horizontal Control Co-ordinates of CONTROL or STATIONS.

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

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

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 - 4 ) * 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)

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 (

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 0 ‘ “ C N B A 132 15 30 - 3” A B 126 12 54 C 69 41 18 D D 31 50 30 Σ (Internal Angles) = 360 00 12 Line Horizontal Distance Σ (Internal Angles) should be (2N-4)*90 = 360 00 00 Allowable = 3 * 6” * N= 36” AB 638.57 OK - Therefore distribute error BC 1576.20 The bearing of line AB is to be assumed to be 00 and the co-ordinates of station A are (3000.00 mE ; 4000.00 mN) CD 3824.10 DA 3133.72

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 227 44 33 A 132 15 27 +or- 1800 638.57 1576.20 3824.10 3133.72 AB 00 00 00 00 00 00 BA 180 00 00 B C D +or- 1800 126 12 51 69 41 15 31 50 27 BC CB CD DC DA 306 12 51 306 12 51 195 54 06 47 44 33 126 12 51 +or- 1800 195 54 06 15 54 06 47 44 33 AD 227 44 33

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

WHOLE CO-ORDINATE DIFFERENCES HORIZONTAL CIRCLE DISTANCE CALCULATED BEARING q D D E D N 00 00 00 638.57 0.000 +638.570 306 12 51 1576.10 -1271.701 +931.227 195 54 06 3824.10 -1047.754 -3677.764 47 44 33 3133.72 +2319.361 +2107.313 G -0.094 -0.654

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

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

Fractional Linear Misclosure (FLM) = 1 in G D / e WHOLE CO-ORDINATE DIFFERENCES HORIZONTAL CIRCLE DISTANCE CALCULATED BEARING q D D E D N 00 00 00 638.57 0.000 +638.570 -1271.701 +931.227 306 12 51 1576.10 195 54 06 3824.10 -1047.754 -3677.764 47 44 33 3133.72 +2319.361 +2107.313 9172.59 G G -0.094 -0.654 eE eN e =  (eE2 + eN2) =  (0.0942 + 0.6542) = 0.661m Fractional Linear Misclosure (FLM) = 1 in G D / e = 1 in (9172.59 / 0.661) = 1 in 13500 [To the nearest 500 lower value]

Acceptable FLM values :- 1 in 5000 for most engineering surveys 1 in 10000 for control for large projects 1 in 20000 for major works and monitoring for structural deformation etc.

Fractional Linear Misclosure (FLM) = 1 in G D / e WHOLE CO-ORDINATE DIFFERENCES HORIZONTAL CIRCLE DISTANCE CALCULATED BEARING q D D E D N 00 00 00 638.57 0.000 +638.570 306 12 51 1576.10 -1271.701 +931.227 195 54 06 3824.10 -1047.754 -3677.764 3133.72 +2319.361 +2107.313 47 44 33 9172.59 G G -0.094 -0.654 eE eN e =  (eE2 + eN2) =  (0.0942 + 0.6542) = 0.661m Fractional Linear Misclosure (FLM) = 1 in G D / e = 1 in (9172.59 / 0.661) = 1 in 13500 Check 2 If not acceptable Ex. 1 in 3500 then we have an error in fieldwork

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

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

Fractional Linear Misclosure (FLM) = 1 in SD / e CO-ORDINATE DIFFERENCES CALCULATED WHOLE CIRCLE BEARING q HORIZONTAL DISTANCE D E N 00 00 00 306 12 51 195 54 06 47 44 33 638.57 1576.20 3824.10 3133.72 0.000 +638.570 -1271.701 +931.227 -1047.754 -3677.764 +2319.361 +2107.313 -0.094 -0.654 eE eN e =  (eE2 + eN2) =  (0.0942 + 0.6542) = 0.661m 3 9172.59 Fractional Linear Misclosure (FLM) = 1 in SD / e = 1 in 9172.59 / 0.661 = 1 in 13500 Check 2

dE IJ = [ eE / SD ] x D IJ Applied with the opposite sign to eE eE = -0.094m SD = 9172.59 m dE IJ = [+0.094 / 9172.59 ] x D IJ = +0.0000102479…... x D IJ Store this in the memory For line AB dE AB = +0.0000102479…x D AB = +0.0000102479…x 638.57 dE AB = +0.007m For line BC dE BC = +0.0000102479…x D BC = +0.0000102479…x 1576.20 dE BC = +0.016m For line CD dE CD = +0.039m For line DA dE DA = +0.032m

CO-ORDINATE DIFFERENCES CALCULATED ADJUSTMENTS ADJUSTED CO-ORDINATES D E N d S T A I O 0.000 +638.570 -1271.701 +931.227 -1047.754 -3677.764 +2319.361 +2107.313 -0.094 -0.654 eE eN +0.007 +0.016 +0.039 +0.032

dN IJ = [ eN / SD ] x D IJ Applied with the opposite sign to eN eN = -0.654m dN IJ = [+0.654 / 9172.59 ] x D IJ = +0.000071299…... x D IJ Store this in the memory dN AB = + 0.000071299… x D AB = + 0.000071299…x 638.57 dN AB = +0.046m dN BC = +0.112m dN CD = +0.273m dN DA = +0.223m

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 3000.00 4000.00 A 0.000 +638.570 +0.007 +0.046 +0.007 +638.616 B C D A 3000.01 4638.62 1728.32 5569.96 -1271.701 +931.227 +0.016 +0.112 -1271.685 +931.339 -1047.754 -3677.764 +0.039 +0.273 -1047.715 -3677.491 680.61 1892.46 Check 3 +2319.361 +2107.313 +0.032 +0.223 +2319.393 +2107.536 3000.00 4000.00 -0.094 -0.654 S= 0 S= 0 eE eN

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)

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.

6.14 AREA OF A CLOSED TRAVERSE BY THE COORDINATE METHOD

6.14 AREA OF A CLOSED TRAVERSE BY THE COORDINATE METHOD

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