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PPT - 2

PPT - 2. Errors in Chaining. Cumulative Error Compensating Error (Balance) Personal Error. Errors occurs due to :. Bad Ranging ( Cumulative , +) Variation in Temperature ( Cumulative , + or –) Variation in Pull ( Cumulative , + or – or Compensating (Balance) , + )

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PPT - 2

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  1. PPT - 2

  2. Errors in Chaining • Cumulative Error • Compensating Error (Balance) • Personal Error

  3. Errors occurs due to : • Bad Ranging (Cumulative, +) • Variation in Temperature (Cumulative, + or –) • Variation in Pull (Cumulative, + or – orCompensating(Balance), +) • Incorrect Length of Tape or Chain (Cumulative, + or –) • Sag in Chain (Cumulative +) • Errors in Reading the tape and marking (Compensating, +) etc.,

  4. Correction • If Error is + vethen Correction is – ve • If Error is – vethen Correction is + ve

  5. Correction for Absolute Length Measured Length of Line with Designated Tape or Chain Χ True Length of Line = Where, • L’ = The Actual Length of a Chain or Tape • L = Designated Length of aChainor Tape (e.g. 20 m or 30 m) Examples L’ L

  6. 1. The Length of a Chain line measured with a 20 m Chain was found to be 250 m calculate the true length of the line if the chain was 10 cm too Long. Measured Length of Line with Designated Tape or Chain Χ True Length of Line = Where, • L’ = The Actual Length of a Chain or Tape • L = Designated Length of aChainor Tape (e.g. 20 m or 30 m) L’ L True Length of Line = Χ 250 = 251.25 m

  7. 2. The Length of a survey line was measured with a 20 m chain and was found to be equal to 1200 m. As a check, the length was again measured with a 25 m chain and was found to be 1212 m on comparing the 20 m chain with the test gauge, it was found to be 1 decimeter too long. Find the actual length of the 25 m chain used. Where, • L’ = The Actual Length of aChain or Tape • L = Designated Length of aChainor Tape (e.g. 20 m or 30 m) L’ L L’ 25 L’ L 20 + 10-1 20 Measured Length of Line with Designated Tape or Chain Χ True Length of Line = Χ 1200 With 20 m chain True Length of Line = = 1206 m Χ 1212 With 25 m chain True Length of Line = Χ 1212 .˙. 1206 = .˙. L’ = 24.876 m .˙. 25 – 24.876 = 0.124 m = 12.4 cm too short (25 m chain)

  8. 3. A 20 m chain was found to be 8 cm too long at the end of day’s work after measuring 4000 m. If the chain was correct before the commencement of the work, find the correct length of the line. Where, • L’ = The Actual Length of aChain or Tape • L = Designated Length of aChainor Tape (e.g. 20 m or 30m ) L’ L 20.04 20 20 + 20.08 2 .˙. 20.08 m after the work 8 cm long after the work, & 20 m in the starting. .˙. Mean actual length of the chain = = 20.04 m Measured Length of Line with Designated Tape or Chain Χ True Length of Line = Χ 4000 True Length of Line = = 4008 m

  9. Corrections for Temperature CT = α(Tm – Ts) L • α = Coefficient of thermal expansion • Tm = meanTemperature of the field • Ts = Temperature during standardization of the tap • L = Designated length of achain or tape

  10. 4. A line was measured with a steel tape which was exactly 30 m long at 18ºc and found to be 450 m the temperature during measurement was 31ºc. Find the true length of the line. Take the coefficient of expansion of tape per ºc = 0.0000117 Where, • L’ = The Actual Length of aChain or Tape • L = Designated Length of aChainor Tape (e.g. 20 m or 30m ) L’ L 30.004563 30 CT = α(Tm – Ts) L Temperature correction tape length = = 0.004563 m = 0.0000117 x (31-18) x 30 .˙. Actual Length of the Tape at 31ºc = 30 + CT = 30+0.004563 = 30.004563 m Measured Length of Line with Designated Tape or Chain Χ True Length of Line = Χ 450 True Length of Line = = 450.068 m

  11. Corrections for Pull or Tension CP = (Pm – Ps) L AE • Pm = Pull applied during measurement (N, kg, etc.,) • Ps = StandardPull (N, kg, etc.,) • L = Designated length of a chain or tape • A = Cross-sectional Area of the tape or chain (cm2,m2 , etc.,) • E = Young’s Modulus of Elasticity (N/cm2, kg/cm2, N/m2, kg/m2, etc., )

  12. 5. A line was measured with a steel tape which was exactly 30 m at a pull of 5 kg and the measured length was 230 m. The pull applied during measurement was 12 kg and the tape was uniformly supported. Find the true length of the line if cross-sectional area of tape was 0.02 cm2and the Modulus of Elasticity = 2.1 x 106 kg/ cm2. (Pm – Ps) L AE Where, • L’ = The Actual Length of aChain or Tape • L = Designated Length of aChainor Tape (e.g. 20 m or 30m ) L’ L (12 – 5 ) x 30 0.02 x 2.1 x 106 30.005 30 The Correction for Pull per 30 m, CP = = = 0.005 m (+ Ve) .˙. Actual Length of the Tape L’ = L+ CP =30+ 0.005 = 30.005 m Measured Length of Line with Designated Tape or Chain Χ True Length of Line = Χ 230 = 230.03833 m True Length of Line =

  13. Corrections for Slope CSL = – h2 2Lm • h = Difference in elevation between the ends • Lm = Measured Length on Slope

  14. 6. The ends of the tape are held at an elevation difference of 1.5 m in a vertical plane because of some obstruction. What is the horizontal distance of the line if measured length is 30 m. h2 2Lm 1.52 2 (30) – The Corrections for Slope , CSL = – = = – 0.0375 m Or = 0.0375 (–Ve) .˙. Horizontal Length = 30 – 0.0375 = 29.9625 m Ans.

  15. 7. A chain was measuring exactly 20 m at the beginning of survey. At the end, it was found to be 20 cm shorter. The area surveyed and drawn to a scale of 1 cm = 100 m was found to be 100 cm2. Find the true area in hectares. ( ) 2 Where, • L’ = The Actual Length of aChain or Tape • L = Designated Length of aChainor Tape (e.g. 20 m or 30 m) L’ L 19.9 20 20 + 19.8 2 Χ True Area of Field = Measured Area = 19.9 m Actual Length = L’ = & L= 20 m ( ) 2 Actual Area = Χ 100 cm2 = 99.0025 cm2 100 Now Scale, 1 cm = ______ m 2 = 990025 m2 .˙. Actual Area of the Field= 99.0025 x (100 m) 99.0025 hectares .˙. Actual Area of the Field =

  16. Conventional Symbols • North, Bridge, etc.……………..,

  17. Angular Measurement: A Compass Instrument used for Angular Measurement • Surveyor's Compass • Prismatic Compass Instruments used for the direct measurements of directions of survey lines. B. Instruments used for the measurement of horizontal angles between survey lines. 1. Abox-sextant 2. A Theodolite 3. A Total Station, etc.,

  18. Bearing of a Line The horizontal angle made by a survey line with reference to North (or with South) is known bearing of a line. • Types of Bearing System • 1. W.C.B. • 2. R.B. • (Whole Circle Bearing) • (Reduced Bearing Or • Quadrantal Bearing)

  19. (Whole Circle Bearing) W.C.B. When the Bearing of a line is measured with respect to North in Clockwise Direction is known as W.C.B. (Whole Circle Bearing). • R.B. (Reduced Bearing Or • Quadrantal Bearing) TheBearing of a line is measured with respect to North or Southin the direction towards East or West whichever is Nearer.

  20. W.C.B. • (Whole Circle Bearing)

  21. R.B. (Reduced Bearing Or • Quadrantal Bearing)

  22. (Convert WCB to RB) Examples 1. Find the Reduced bearing of a line having W.C.B. 280 ° • Figure. • AnsReduced bearing =N 80° W 2. Find the Reduced bearing of a line having W.C.B. 200 ° • AnsReduced bearing =S 20° W 3. Find the Reduced bearing of a line having W.C.B. 145 ° • AnsReduced bearing =S 35° E 4. Find the Reduced bearing of a line having W.C.B. 30°45’ • AnsReduced bearing =N 30°45’ E 5. Find the Reduced bearing of a line having W.C.B. 100°20’30’’ • AnsReduced bearing =S 79°39’30’’ E 6. Find the Reduced bearing of a line having W.C.B. 160° • AnsReduced bearing =S20° E 7. Find the Reduced bearing of a line having W.C.B. 250° • AnsReduced bearing =S 70° W

  23. (Convert RB to WCB) Examples 1. Find the WCBof a line having R.B. N 40°15’ W • AnsWCB =319°45’ 2. Find the WCBof a line having R.B.S 43°30’ E • AnsWCB =136°30’ 3. Find the WCBof a line having R.B.S36°30’30’’ W • AnsWCB =216°30’30’’ 4. Find the WCBof a line having R.B.N 26°45’ E • AnsWCB =26°45’ 5. Find the WCBof a line having R.B.S 80° W • AnsWCB =260° • Figure.

  24. Equations if Required For W.C.B. to R.B. • .˙.R.B. = NαE • .˙.R.B. = S 180°-αE • .˙.R.B. = Sα-180° W • .˙.R.B. = N 360°-αW • 0°< α < 90°→ W.C.B. = α • 90°< α < 180°→ W.C.B. = α • 180°< α < 270°→ W.C.B. = α • 270°< α < 360°→ W.C.B. = α For R.B. to W.C.B. • R.B.=NβE • R.B.=S βE • R.B.=S βW • R.B.=N βW • .˙.W.C.B=β • .˙.W.C.B= 180°-β • .˙.W.C.B= 180°+β • .˙.W.C.B= 360°-β

  25. (Convert WCB to RB) Examples 1. Find the Reduced bearing of a line having W.C.B. 280 ° • AnsReduced bearing =N 80° W • .˙.R.B. = N 360°- αW 2. Find the Reduced bearing of a line having W.C.B. 200 ° • 270°< α < 360°→ W.C.B. = 280° • 180°< α < 270°→ W.C.B. = 200° • AnsReduced bearing =S 20° W • .˙.R.B. = Sα - 180° W 3. Find the Reduced bearing of a line having W.C.B. 145 ° • 90°< α < 180°→ W.C.B. = 145° • AnsReduced bearing =S 35° E • .˙.R.B. = S 180°- αE 4. Find the Reduced bearing of a line having W.C.B. 30°45’ • 0°< α < 90°→ W.C.B. = 30°45’ • AnsReduced bearing =N 30°45’ E • .˙.R.B. = NαE 5. Find the Reduced bearing of a line having W.C.B. 100°20’30’’ • 90°< α < 180°→ W.C.B. = 100°20’30’’ • AnsReduced bearing =S 79°39’30’’ E • .˙.R.B. = S 180°- αE 6. Find the Reduced bearing of a line having W.C.B. 160° • 90°< α < 180°→ W.C.B. = 160° • AnsReduced bearing =S20° E • .˙.R.B. = S 180°- αE 7. Find the Reduced bearing of a line having W.C.B. 250° • 180°< α < 270°→ W.C.B. = 250° • AnsReduced bearing =S 70° W • .˙.R.B. = Sα - 180° W

  26. (Convert RB to WCB) Examples 1. Find the WCBof a line having R.B. N 40°15’ W • R.B.=N βW • AnsWCB =319°45’ • .˙.W.C.B= 360°-β 2. Find the WCBof a line having R.B.S 43°30’ E • R.B.=S βE • AnsWCB =136°30’ .˙.W.C.B= 180°-β 3. Find the WCBof a line having R.B.S36°30’30’’ W • R.B.=S βW • AnsWCB =216°30’30’’ • .˙.W.C.B= 180°+ β 4. Find the WCBof a line having R.B.N 26°45’ E • R.B.=NβE • AnsWCB =26°45’ • .˙.W.C.B=β 5. Find the WCBof a line having R.B.S 80° W • R.B.=S βW • AnsWCB =260° • .˙.W.C.B= 180°+β

  27. Differentiate between W.C.B. and R.B. • W.C.B.R.B. • The Bearing of a line is measured with respect to Northor Southin the direction towards East or West whichever is Nearer. • The value of R.B. varies from 0° to 90° • 3.With respect to North or South. • 4.It is measuredClockwise and Anticlockwise Direction depends upon the situation. • 5.R.B. = Reduced Bearing Or Quadrantal Bearing • When the Bearing of a line is measured with respect to North in Clockwise Direction is known as W.C.B. (Whole Circle Bearing). • The value of W.C.B. varies from 0° to 360° 3.With respect to Northonly. 4.Measured alwaysClockwise Direction. 5. W.C.B. = Whole Circle Bearing

  28. Types of Compass • Surveyor's Compass • Prismatic Compass Above Instruments are used for Angular Measurement

  29. Differentiate between Prismatic Compass and Surveyor's Compass • Prismatic CompassSurveyor's Compass • It gives R.B. • Prism is notprovided to take reading. • 3.Bearings are measured 0° to 90° • 4.The graduated ring and needle are free to move independently with respect to each other. • 5.The instrument cannot be used without a Tripod. • 6.It gives lessaccuracy than Prismatic compass. • 7.Less Costly. • It gives W.C.B. 2.Prism is provided to take reading. 3.Bearings are measured 0° to 360° 4.The graduated ring is attached with the magnetic needle. It (graduated ring) does not rotate along with the line of site 5.Tripodmay or may not be provided. 6.It gives moreaccuracy than Surveyor’s compass. 7.More Costly.

  30. (Fore Bearing or • Forward Bearing) F.B. If the Bearing of a line AB is measured from AtowardsB, it is known as F.B. (Fore Bearing or Forward Bearing). • B.B. (Back Bearing or Backward Bearing) • If the Bearing of a line AB is measured from BtowardsA, it is known as B.B. (Back Bearing orBackward Bearing). For W.C.B. B.B. = • F.B. • ± 180º B.B. = • F.B. • + 180º • (F.B. < 180º) B.B. = • F.B. • – 180º • (F.B. > 180º) For R.B. F.B. to B.B. OR B.B. to F.B. • N ↔ S& E ↔W

  31. The following are observed fore-bearings of the lines. Find their back bearing. • 1. AB 12°24’ • F.B. < 180º • F.B. • +180º • AnsB.B. =192°24’ • .˙. B.B. = 2. BC 119°48’ • F.B. < 180º • F.B. • +180º • AnsB.B. =299°48’ • .˙. B.B. = • 3. CD 260°30’ • F.B. > 180º • F.B. • – 180º • AnsB.B. =80°30’ • .˙. B.B. = 4.DE354°18’ • F.B. > 180º • F.B. • – 180º • AnsB.B. =174°18’ • .˙. B.B. = • 5. PQ N18°0’E • AnsB.B. =S18°0’W 6. QR S12°24’E • AnsB.B. =N12°24’W 7. RS S59°18’W • AnsB.B. =N59°18’E 8. ST N86°12’W • Figure. • AnsB.B. =S86°12’E

  32. Easy way to convertFB to BB & BB to RB in both the Bearing System (WCB & RB) For W.C.B. B.B. = • F.B. • ± 180º B.B. = • F.B. • + 180º • (F.B. < 180º) B.B. = • F.B. • – 180º • (F.B. > 180º) For R.B. F.B. to B.B. OR B.B. to F.B. • N ↔ S& E ↔W • .˙. B.B. = • S • α • W • F.B. = N αE • .˙. B.B. = • S • α • E • F.B. = N αW • .˙. B.B. = • N • α • W • F.B. = S αE • .˙. B.B. = • N • α • E • F.B. = S αW

  33. 1.The following bearing were observed with a compass. Calculate the interior angles. Solution : - Step : - 1 Calculate Back Bearing (if it is not Given)

  34. 240°30’ • 302° • 226° • 25°30’ • 120° B.B. = • F.B. • ± 180º B.B. = • F.B. • + 180º • (F.B. < 180º) B.B. = • F.B. • – 180º • (F.B. > 180º) Step : - 2 Draw Traverse and FindTraverse is Clockwise or Anticlockwise

  35. Step : - 3 Make Angle calculation Table Traverse is Clockwise. • .˙. F.B.–B.B. • +Ve • -Ve Angle Line FB - BB BB Exterior Interior • 59°30’ • EA • 120° • -59°30’ • AB • 240°30’ • 118°30’ • -118°30’ ` 360° – 274°30’ • BC • 302° • -256° • 256° • CD • 226° • -20°30’ • 20°30’ • DE • 85°30’ • 25°30’ • 274°30’ • 274°30’ • Total • 540° Check, • (2n-4) x 90 = • (2x5-4) x 90 = 540° .˙. O.K. • n = Number of Angles or Number of Sides

  36. 2.Find Included Angles. Solution : - Step : - 1 Calculate Back Bearing (If it is not Given)

  37. 277°15’ • 192° • 91°30’ • 9°15’ • 304°45’ B.B. = • F.B. • ± 180º B.B. = • F.B. • + 180º • (F.B. < 180º) B.B. = • F.B. • – 180º • (F.B. > 180º) Step : - 2 Draw Traverse and FindTraverse is Clockwise or Anticlockwise

  38. Step : - 3 Make Angle calculation Table Traverse is Anticlockwise. • .˙. B.B.–F.B. • +Ve • -Ve Angle Line BB - FB BB Exterior Interior • 207°30’ • 152°30’ • EA • 304°45’ • 207°30’ • AB • 277°15’ • 265°15’ • 94°45’ • 265°15’ • BC • 192° • -79°30’ • 79°30’ ` 360° – 207°30’ • CD • 91°30’ • -97°45’ ` • 97°45’ 360° – 265°15’ • DE • 9°15’ • -115°30’ • 115°30’ • Total • 540° Check, • (2n-4) x 90 = • (2x5-4) x 90 = 540° .˙. O.K. • n = Number of Angles or Number of Sides

  39. 3. Find Angles, A,B,C,D =? Solution : - Step : - 1 • Convert RB in WCB & Calculate Back Bearing (if it is not Given)

  40. FB (in WCB) Back Bearing • 50° • 230° • 120° • 300° • 195° • 15° • 289°30’ • 109°30’ B.B. = • F.B. • ± 180º For R.B. to W.C.B. • R.B.=NβE • R.B.=S βE • R.B.=S βW • R.B.=N βW • .˙.W.C.B=β • .˙.W.C.B= 180°-β • .˙.W.C.B= 180°+β • .˙.W.C.B= 360°-β B.B. = • F.B. • + 180º • (F.B. < 180º) B.B. = • F.B. • – 180º • (F.B. > 180º) Step : - 2 Draw Traverse and FindTraverse is Clockwise or Anticlockwise

  41. Step : - 3 Make Angle calculation Table Traverse is Clockwise. • .˙. F.B.–B.B. • +Ve • -Ve Angle Line FB - BB BB Exterior Interior • 59°30’ • DA • 109°30’ • -59°30’ ` 360° – 274°30’ • AB • 230° • 110° • -110° • BC • 300° • -105° • 105° • CD • 274°30’ • 15° • 274°30’ • 85°30’ • Total • 360° Check, • (2n-4) x 90 = • (2x4-4) x 90 = 360° .˙. O.K. • n = Number of Angles or Number of Sides

  42. 4.The following bearing were observed on a closed traverse. Calculate the Interior angles and Correct them for observational errors. Solution : - Step : - 1 • {Convert RB in WCB & Calculate Back Bearing (If it is not Given)} Draw Traverse and FindTraverse is Clockwise or Anticlockwise

  43. Then Make Angle calculation Table Traverse is Anticlockwise. • .˙. B.B.–F.B. • +Ve • -Ve Angle Line BB - FB BB Exterior Interior • 64°15’ • DA • 44°30’ • -64°15’ • AB • 289°30’ • 288°30’ • 71°30’ • 288°30’ • BC • 181° • -122°15’ • 122°15’ ` 360° – 288°30’ • CD • 124°30’ • -102° • 102° • Total • 360° Check, • (2n-4) x 90 = • (2x4-4) x 90 = 360° .˙. O.K. • n = Number of Angles or Number of Sides

  44. If F.B. & B.B. are given in Question, Find Difference between F.B. & B.B. Step : 4 • Big-Small • 361° • -360° • =1° Corrected Bearing • =71°30’ Difference BB FB • 289°30’+ • B • 180°45’ • 109°30’ • 289°30’ • =122°15’ • 180° • 181° • 181°+ • 1° • C • =102° • 178°45’ • 303°15’ • D • 123°15’ • 123°15’+ • 182° • 225°15’ • 45°15’+ • A • 45°15’ • 64°15’= • FB = 303°15’ • BB = ? • 303°15’ > 180° • .˙. 303°15’-180° • =123°15’ • FB = 225°15’ • BB = ? • 225°15’ > 180° • .˙. 225°15’-180° • =45°15’ • FB = 109°30’ • BB = ? • 109°30’ < 180° • .˙. 109°30’+180° • =289°30’ .˙. O.K.

  45. Corrected Bearing Difference BB FB • 180°45’ • 109°30’ • 289°30’ • 180° • 181° • 1° • 178°45’ • 303°15’ • 123°15’ • 182° • 225°15’ • 45°15’

  46. 5.The following bearing were taken. Compute the Included angles and do correction if required. Also determine the corrected bearing. Solution : - Step : - 1 • {Convert RB in WCB Calculate Back Bearing (If it is not Given)} Draw Traverse and FindTraverse is Clockwise or Anticlockwise

  47. Step : Make Angle calculation Table Traverse is Anticlockwise. • .˙. B.B.–F.B. • +Ve • -Ve Angle Correction Corrected Angle Line BB - FB BB -1° -1° Exterior Interior • 126° • FA • 234° • 235°30’ • 234° • -0°10’ • 125°50’ • AB • 182° • -106°30’ • 106°30’ • -0°10’ • 106°20’ ` 360° – 234° • BC • 111°30’ • -127° • 127° • -0°10’ • 126°50’ ` • CD • 54°15’ • -138°15’ • 138°15’ • -0°10’ • 138°05’ 360° – 189°30’ • DE • 12°45’ • -52°45’ • 52°45’ • -0°10’ • 52°35’ • EF • 245° • 189°30’ • 189°30’ • 170°30’ • -0°10’ • 170°20’ 720° • 721° • n = Number of Angles or Number of Sides • Total Check, • (2n-4) x 90 = • (2x6-4) x 90 = 720° .˙. O.K. .˙. Total Error = 1° .˙. Total Correction = -1° .˙. Correction for Each Angle = = = -0°10’ n 6

  48. If F.B. & B.B. are given in Question, Find Difference between F.B. & B.B. Then • =125°50’ • Big-Small • A 235°30’+ • -360° Corrected Bearing =361°20’ Difference • B 181°20’+ =106°20’ FB BB =126°50’ • 180°30’ • 1°20’ • 181°20’ • C 107°40’+ =138°05’ • 177° • 287°40’ • 107°40’ 54°30’+ • D =52°35’ • 184°15’ • 234°30’ • 54°30’ • E 12°35’+ =170°20’ • 179°45’ • 192°35’ • 12°35’ 245°10’+ • F =415°30’ -360° • 179°30’ • 65°10’ • 245°10’ • =55°30’ • 180° • 55°30’ • 235°30’ • FB = 1°20’ • BB = ? • 1°20’ < 180° • .˙. 1°20’+180° • =181°20’ • FB = 287°40’ • BB = ? • 287°40’ > 180° • .˙. 287°15’-180° • =107°40’ • FB = 234°30’ • BB = ? • 234°30’ > 180° • .˙. 234°30’-180° • =54°30’ • FB = 192°35’ • BB = ? • 192°35’ > 180° • .˙. 192°35’-180° • =12°35’ • FB = 65°10’ • BB = ? • 65°10’ < 180° • .˙. 65°10’+180° • =245°10’ .˙. O.K.

  49. Meridian True Meridian2.Magnetic Meridian3.Arbitraryor assumed Meridian Meridians are reference direction with reference to which the bearings are taken. Types of Meridian Types of Bearing • True Bearing2.Magnetic Bearing3.Arbitraryor assumed Bearing

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