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Introduction to Traversing using distances and directions of lines between Traversing is the method of using distances and directions of lines between.

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Presentation on theme: "Introduction to Traversing using distances and directions of lines between Traversing is the method of using distances and directions of lines between."— Presentation transcript:

1 Introduction to Traversing using distances and directions of lines between Traversing is the method of using distances and directions of lines between points to determine positions of the points. CDE F Example of Typical traverse

2 Types of Traverses I. Loop Traverse II. Link Traverse Distance and angles measured

3 A B CDE F G H Types of Link Traverses Special case of the traverse

4 1.Adjustment of Angles 2.Compute Azimuths 3.Adjustment of distances (Latitudes and Departures) 4.Compute Coordinates Typical sequence of computing Procedures

5 A Left angle  B Compute direction 1 2 11 Traverse leg d 1-2 B Left angle  1 1 orientation point 2 orientation point d 1-2 d B-1 Orientation Point with known X,Y Traverse Station with unknown X,Y

6 Adjustment of Angles in a Loop Traverse N51 0 22’00”E 101 0 24’00” 149 0 13’00” 80 0 58’30” 116 0 19’00” 92 0 04’30” 579.03 350.10 401.58 382.20 368.28 A B C D E Check Interior Angle Closure ObservedAdjusted A = 101 0 24’ 00” 101 0 24’ 12” B = 149 0 13’ 00” 149 0 13’ 12” C = 80 0 58’ 30” 80 0 58’ 42” D = 116 0 19’ 00” 116 0 19’ 12” E = 92 0 04’ 30” 92 0 04’ 42” Total = 539 0 59’ 00”= 540 0 00’ 00” Should = 540 0 00’ 00” = (n-2)*180 Misclosure = 01’ 00” = 60” Adjustment = 60/5 = +12” per angle 1.Adjustment of Angles

7 B 180  1 ApAp AnAn ApAp   Compute next azimuths A n A A n = A p +  – 180  A n = A p –  + 180  1.Compute Azimuths

8 Compute Azimuths in Loop traverse  AB = 51 0 22’ 00” (given)  BA = 231 0 22’ 00”  = 149 0 13’ 12”  BC = 82 0 08’ 48”  CB = 262 0 08’ 48” C = 80 0 58’ 42”  CD = 181 0 10’ 06”  DC = 1 0 10’ 06” D = 116 0 19’ 12”  DE = 244 0 50’ 54”  ED = 64 0 50’ 54” E = 92 0 04’ 42”  EA = 332 0 46’ 12”  AE = 152 0 46’ 12” A = 101 0 24’ 12”  AB = 51 0 22’ 00” Check 51 0 22’00” 101 0 24’12” 149 0 13’12” 80 0 58’42” 116 0 19’12” 92 0 04’42” B C D E A 82° 08’ 48” 181° 10’ 06” 244° 50’ 54” 332° 46’ 12” 1.Compute Azimuths

9 d B-1 d 1-2 d2-3d2-3 d 3-C Diredctin of the calculations 1 3 2 A D B C BB 11 22 33 CC Bilateral traverse typical A B-1 = A AB +  B – 180  A 1-2 = A B-1 +  1 – 180  A 2-3 = A 1-2 +  2 – 180  A 3-C = A 2-3 +  3 – 180  A CD = A 3-C +  C – 180  A CD = A AB + [ left ] – n  180  A CD = A AB - [ right ] + n  180  A CD = A AB - [ right ] + n  180  [ left] = A K – A P + n  180  [ right ] = A P – A K + n  180  Adjustment of Angles in a Link Traverse Should = A K – A P + n 180  or (A P – A K + n  180  ) Misclosure = [  left ] – (A K – A P + n 180  ) or [  right ] - (A P – A K + n  180  ) 1.Adjustment of Angles II. Link Traverse

10 Sum angles Should be = A K – A P + n 180  or A P – A K + n  180  Misclosure = [ angles left ] – (A K – A P + n 180  ) or [angles right ] - (A P – A K + n  180  ) Correction to angles = - Misclosure /number of angles Adjustment of Angles in a Link Traverse CDE F 1.Adjustment of Angles

11 Link (open) Traverse 355-40-40 75-20-30 255-20-36 110-36-42 290-36-40 80-22-26 260-22-30 111-25-29 291-25-24 82-25-20 262-25-26 Orientation Point with known X,Y Traverse Terminal with known X,Y Traverse Station with unknown X,Y A B O1 T1 T2 T3 T4 355-40-45 170-26-20 O2 170-26-15 500.364 600.556 625.484 480.253 674.374 170-26-10 O2 170-26-15 1.Compute Azimuths

12 Latitudes (  Y) and Departures (  X) T1 T2 YY XX Latitude = L Cos  Y Departure = L Sin  X L  Latitudes and Departures computed for each leg of a traverse

13 Compute Lats (D Cos  ) and Deps (D Sin  ) Leg Azimuth Distance Lat Dep AB51 0 22’ 00” 401.58 250.720 313.697 BC 82 0 08’ 48”382.20 52.222 378.615 CD 181 0 10’ 06” 368.28 -368.203 -7.509 DE 244 0 50’ 54” 579.03 -246.097-524.130 EA 332 0 46’ 12” 350.10311.301-160.193 Total -0.057 0.480 Total Traverse Distance = 2081.19 Linear Misclosure =  (0.057) 2 + (0.480) 2 = 0.483 Precision = 0.483/2081.19 = 1/4305 …… 1/4300 Adjustment of distances (Lats and Deps)

14 Correction to Lats = -Traverse leg distance * Lat Misclosure Total traverse distance Correction to Deps = -Traverse leg distance * Dep Misclosure Total traverse distance LegLats Deps Corrn LatCorrn Dep Adj Lats Adj Deps AB250.720 313.6970.011 -0.093 250.731 313.604 BC 52.222 378.6150.010 -0.088 52.232 378.527 CD -368.203 -7.5090.010 -0.085 -368.193 -7.594 DE -246.097 -524.1300.016 -0.134 -246.081 -524.264 EA 311.301 -160.1930.010 -0.081 311.311 -160.274 0.000 0.001 Total 0.057 -0.481 0.000 0.001 Adjustment of distances (Lats and Deps)


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