C hapter 6 TRAVERSE Horizontal Control Horizontal control

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C hapter 6 TRAVERSE

C hapter 6 TRAVERSE

Horizontal Control • Horizontal control is required for initial survey work (detail surveys) and

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.

N A E D B Grass C

N A E D B Grass C

Horizontal Control Co-ordinates of CONTROL or STATIONS.

Horizontal Control Co-ordinates of CONTROL or STATIONS.

Horizontal Control • There are two types : - POLYGON or LOOP TRAVERSE LINK

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

X A B F A B C E D E C D • Both

X A B F A B C E D E C D • Both types are closed. F G Y 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

A F Consider the POLYGON traverse The L. H. Angles are also the INTERNAL

A F Consider the POLYGON traverse The L. H. Angles are also the INTERNAL ANGLES B E C D 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 2 * Accuracy of Misclosure = Theodolite * (No. of Angles) (Rule of thumb)

Standing at A looking towards F - looking BACK A ΘAF Hence ΘAF is

Standing at A looking towards F - looking BACK A ΘAF Hence ΘAF is known as a Azimuth F ΘBA B ΘAB Angle FAB (LH angle) ΘBC C LH angle ABC 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 BACK Azimuth ( ΘBA ) = FORWARD Azimuth ( ΘAB ) 1800

Traverse Example 12” / 4 = 3” Observations, using a Zeiss O 15 B,

Traverse Example 12” / 4 = 3” Observations, using a Zeiss O 15 B, 6” Theodolite, were taken in the field for an anti - clockwise polygon traverse, A, B, C, D. Observed Clockwise Traverse Station C N Horizontal Angle 0 B A D Line Horizontal Distance AB 638. 57 BC 1576. 20 CD 3824. 10 DA 3133. 72 ‘ “ A 132 15 30 - 3” B 126 12 54 - 3” C 69 41 18 - 3” D 31 50 30 - 3” Σ (Internal Angles) = 360 00 12 Σ (Internal Angles) should be (2 N-4)*90 = 360 00 00 Allowable = 3 * 6” * N= 36” OK - Therefore distribute error The bearing of line AB is to be assumed to be 00 and the co-ordinates of station A are (3000. 00 m. E ; 4000. 00 m. N)

LINE BACK AZIMUTH STATION LINE AD A AB BA B BC CB C CD

LINE BACK AZIMUTH STATION LINE AD A AB BA B BC CB C CD DC D DA AD WHOLE + ADJUSTED LEFT + HAND ANGLE FORWARD == AZIMUTH Check 1 227 132 00 0180 126 306 0126 69 195 0 15 31 47 227 +or 180 44 15 00 00 12 12 12 41 54 54 50 44 44 33 27 00 00 51 51 51 15 06 06 27 33 33 HORIZONTAL CIRCLE DISTANCE Azimuth Use Distance and Azimuth to. Dgo from q POLAR to RECTANGULAR to get Delta E and Delta N values. 00 00 00 638. 57 306 12 51 1576. 20 195 54 06 3824. 10 47 44 33 3133. 72

LATITUDES AND DEPARTURES FIGURE 6. 5: LOCATION OF A POINT. A)POLAR TIES B)RECTANGULAR TIES

LATITUDES AND DEPARTURES FIGURE 6. 5: LOCATION OF A POINT. A)POLAR TIES B)RECTANGULAR TIES LATITUDE = NORTH(+) SOUTH (-)=distance(H) DEPARTURE = EAST(+) WEST (-)= x cos distance(H) x sin

CO-ORDINATE DIFFERENCES WHOLE HORIZONTAL CIRCLE DISTANCE CALCULATED BEARING D q E N 00 00

CO-ORDINATE DIFFERENCES WHOLE HORIZONTAL CIRCLE DISTANCE CALCULATED BEARING D q E N 00 00 00 638. 57 306 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 0. 000 G -0. 094 +638. 570 -0. 654

 EBC C NBC =+931. 227 m B NAB =+638. 570 m NCD =-3677.

EBC C NBC =+931. 227 m B NAB =+638. 570 m NCD =-3677. 764 m A NDA =+2107. 313 m D ECD EDA

e is the LINEAR MISCLOSURE C e = (e E 2 + e N

e is the LINEAR MISCLOSURE C e = (e E 2 + e N 2 ) B e. E e. N e A’ D A

CO-ORDINATE DIFFERENCES WHOLE HORIZONTAL CIRCLE DISTANCE CALCULATED BEARING D q E N 00 00

CO-ORDINATE DIFFERENCES WHOLE HORIZONTAL CIRCLE DISTANCE CALCULATED BEARING D q E N 00 00 00 638. 57 306 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 9172. 59 0. 000 G -0. 094 e. E +638. 570 -0. 654 e. N e = (e. E 2 + e. N 2) = (0. 0942 + 0. 6542) = 0. 661 m 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

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.

CO-ORDINATE DIFFERENCES WHOLE HORIZONTAL CIRCLE DISTANCE CALCULATED BEARING D q E N 00 00

CO-ORDINATE DIFFERENCES WHOLE HORIZONTAL CIRCLE DISTANCE CALCULATED BEARING D q E N 00 00 00 638. 57 306 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 9172. 59 0. 000 G -0. 094 e. E +638. 570 -0. 654 e. N e = (e. E 2 + e. N 2) = (0. 0942 + 0. 6542) = 0. 661 m 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

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

a) Bowditch Method - proportional to line distances The e. E and the e.

a) Bowditch Method - proportional to line distances The e. E and the e. N have to be distributed For any line IJ the adjustments are d. E IJ and d. N IJ d. E IJ = [ e. E / SD ] x D IJ d. N IJ = [ e. N / SD ] x D IJ Applied with the opposite sign to e. E Applied with the opposite sign to e. N

CO-ORDINATE DIFFERENCES WHOLE HORIZONTAL CIRCLE DISTANCE CALCULATED BEARING D q E N 00 00

CO-ORDINATE DIFFERENCES WHOLE HORIZONTAL CIRCLE DISTANCE CALCULATED BEARING D q E N 00 00 00 638. 57 306 12 51 1576. 20 -1271. 701 +931. 227 195 54 06 3824. 10 -1047. 754 -3677. 764 47 44 33 3133. 72 +2319. 361 +2107. 313 3 9172. 59 3 0. 000 +638. 570 -0. 094 e. E -0. 654 e. N e = (e. E 2 + e. N 2) = (0. 0942 + 0. 6542) = 0. 661 m Fractional Linear Misclosure (FLM) = 1 in SD / e = 1 in 9172. 59 / 0. 661 = 1 in 13500 Check 2

d. E IJ = [ e. E / SD ] x D IJ e.

d. E IJ = [ e. E / SD ] x D IJ e. E = -0. 094 m Applied with the opposite sign to e. E SD = 9172. 59 m d. E IJ = [+0. 094 / 9172. 59 ] x D IJ = +0. 0000102479…. . . x D IJ For line AB Store this in the memory d. E AB = +0. 0000102479…x D AB = +0. 0000102479…x 638. 57 For line BC d. E AB = +0. 007 m d. E BC = +0. 0000102479…x D BC = +0. 0000102479…x 1576. 20 For line CD d. E CD = +0. 039 m d. E BC = +0. 016 m For line DA d. E DA = +0. 032 m

CO-ORDINATE DIFFERENCES CO-ORDINATES CALCULATED E ADJUSTMENTS N d. E 0. 000 +638. 570 +0.

CO-ORDINATE DIFFERENCES CO-ORDINATES CALCULATED E ADJUSTMENTS N d. E 0. 000 +638. 570 +0. 007 -1271. 701 +931. 227 +0. 016 -1047. 754 -3677. 764 +0. 039 +2319. 361 +2107. 313 +0. 032 -0. 094 -0. 654 e. E e. N d. N ADJUSTED E N E N S T A T I O N

d. N IJ = [ e. N / SD ] x D IJ e.

d. N IJ = [ e. N / SD ] x D IJ e. N = -0. 654 m Applied with the opposite sign to e. N d. N IJ = [+0. 654 / 9172. 59 ] x D IJ = +0. 000071299…. . . x D IJ Store this in the memory d. N AB = + 0. 000071299… x D AB = + 0. 000071299…x 638. 57 d. N AB = +0. 046 m d. N BC = +0. 112 m d. N CD = +0. 273 m d. N DA = +0. 223 m

CO-ORDINATE DIFFERENCES CO-ORDINATES CALCULATED E ADJUSTMENTS N d. E d. N ADJUSTED E 0.

CO-ORDINATE DIFFERENCES CO-ORDINATES CALCULATED E ADJUSTMENTS N d. E d. N ADJUSTED E 0. 000 +638. 570 +0. 007 +0. 046 +0. 007 -1271. 701 +931. 227 +0. 016 +0. 112 -1271. 685 N N E S T A T I O N 3000. 00 4000. 00 A 3000. 01 4638. 62 B +931. 339 1728. 32 5569. 96 C +638. 616 -1047. 754 -3677. 764 +0. 039 +0. 273 -1047. 715 -3677. 491 680. 61 1892. 46 D +2319. 361 +2107. 313 +0. 032 +0. 223 +2319. 393 +2107. 536 3000. 00 4000. 00 A -0. 094 -0. 654 e. E e. N S= 0 Check 3

6. 13 SUMMARY OF TRAVERSE COMPUTATIONS 1. 2. 3. 4. 5. 6. 7. Balance

6. 13 SUMMARY OF TRAVERSE COMPUTATIONS 1. 2. 3. 4. 5. 6. 7. Balance the field angles (1 st step) Correct (if necessary) the field distances (2 nd step) Compute the bearings and/or azimuths (3 rd step) Compute the linear error of closure and the precision ratio of the traverse (4 th step) Compute the balances latitudes ( y) and balanced departures ( x) (5 th step) Compute coordinates (6 th step) Compute the area (7 th step)

6. 14 AREA OF A CLOSED TRAVERSE BY THE COORDINATE METHOD The double area

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

6. 14 AREA OF A CLOSED TRAVERSE BY THE COORDINATE METHOD

6. 14 AREA OF A CLOSED TRAVERSE BY THE COORDINATE METHOD

END OF CHAPTER 6 THANK YOU FOR YOUR ATTENTION

END OF CHAPTER 6 THANK YOU FOR YOUR ATTENTION