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- 1. Compass Surveying Unit-II
- 2. Compass Surveying • Chain surveying can be used when the area to be surveyed is comparatively small and is fairly flat. • But when the area is large, undulated and crowded with many details, triangulation (which is the principle of chain survey) is not possible. In such an area, the method of traversing is adopted.
- 3. Traversing • In Traversing, the framework consist of a number of connected lines. The length are measured by a chain or a tape and the directions measured by angle measuring instruments. In one of the methods, the angle (direction) measuring instrument is the compass. Hence, in compass surveying directions of survey lines are determined with a compass and the length of the lines are measured with a tape or a chain. This process is known as Compass Traversing.
- 4. Traversing
- 5. Principle of Compass Surveying • The Principle of Compass Survey is Traversing; which involves a series of connected lines the magnetic bearing of the lines are measured by prismatic compass and the distance (lengths) of the are measured by chain. Such survey does not require the formulation of a network of triangle. • Compass surveying is recommended when the area is large, undulating and crowded with many details. • Compass surveying is not recommended for areas where local attraction is suspected due to the presence of magnetic substances like steel structures, iron ore deposits, electric cables conveying currents, and so on.
- 6. Traversing
- 7. Types and Uses of Compass • Compass: A compass is a small instrument essentially Consisting of magnetic needle, a graduated circle, and a line of sight. The compass can not measure angle between two lines directly but can measure angle of a line with reference to magnetic meridian at the instrument station point is called magnetic bearing of a line. The angle between two lines is then calculated by getting bearing of these two lines. • There are two forms of compass available: • The Prismatic Compass • The Surveyor’s Compass
- 8. Compass Surveying The Prismatic Compass The prismatic compass is a magnetic compass which consists of the following parts. Cylindrical Metal Box Cylindrical metal box is having diameter of 8 to 12 cm. It protects the compass and forms entire casing or body of the compass. It protects compass from dust, rain etc.
- 9. The Prismatic Compass Pivot • Pivot is provided at the centre of the compass and supports freely suspended magnetic needle over it. Lifting Pin and Lifting Lever • A lifting pin is provided just below the sight vane. When the sight vane is folded, it presses the lifting pin. The lifting pin with the help of lifting lever then lifts the magnetic needle out of pivot point to prevent damage to pivot head. Spring Brake or Brake Pin • To damp the oscillation of the needle before taking a reading and to bring it to rest quickly, the light spring brake attached to the inside of the box is brought in contact with edge of the ring by gently pressing inward the brake pin.
- 10. The Prismatic Compass • Magnetic Needle: Magnetic needle is the heart of the instrument. This needle measures angles of a line from magnetic meridian a the needle always remains pointed towards north and south pole at the two ends of the needle when freely suspended on any support. • Graduated Circle or Ring: This is an aluminium graduated ring marked with 0 0 to 360 0 to measure all possible bearings of lines, and attached with the magnetic needle. The ring is graduated to half a degree.
- 11. The Prismatic Compass • Prism Prism is used to read graduations on ring and to take exact reading by compass. It is placed exactly opposite to object vane. The prism hole is protected by prism cap to protected by prism cap to protect it from dust and moisture. • Object Vane Object Vane is diametrically opposite to the prism and eye vane. The object vane is carrying a horse hair or black thin wire to sight object in line with eye sight. • Eye Vane Eye Vane is a fine slit provided with eye hole at bottom to bisect the object from the slit and to take reading simultaneously from the eye hole. This eye vane is provided with prism and can be lifted up and down by the stud to bisect the object of higher level.
- 12. The Prismatic Compass • Glass Cover: It covers the instrument box from the top such that needle and graduated ring is seen from the top. • Sun Glasses: These are used when some luminous objects are to be bisected. These are placed in front of the eye slit and in bunch of three or four shades of different colours to give sharp picture of the object only. • Reflecting Mirror: It is used to get image of an object located above or below the instrument level while bisection. It is placed on the object vane.
- 13. Working of the Prismatic Compass • When the needle of the compass is suspended freely. It always points towards the north. Therefore, all the angles measured with prismatic compass are with respect to north (magnetic meridian). • “The horizontal angle made by a survey line with reference to magnetic meridian in clockwise direction is called the bearing of a line.’ • While using the compass, it is usually mounted on a light tripod which is having vertical spindle in the ball and socket arrangement to which the compass is screwed.
- 14. Temporary Adjustment of a Prismatic Compass • The following procedure should be adopted after the prismatic compass on the tripod for measuring the bearing of a line: Centering Centering is the operation in which compass is kept exactly over the station from where the bearing is to be determined. The centering is checked by dropping a small pebble from the underside of the compass. If the pebble falls on the top of the peg then the centering I correct, if not then the centering is corrected by adjustment the legs of the tripod. Levelling Levelling of the compass is done with the aim to freely swing the graduated circular ring of the prismatic compass. The ball and socket arrangement on the tripod will help to achieve a proper lever of the compass. This can be checked by rolling round pencil on glass cover. Focusing The prism is moved up or down in its slide till the graduations on the aluminium ring are seen clear, sharp and perfect focus. The position of the prism will depend upon the vision of the observer.
- 15. The Prismatic Compass
- 16. Observing Bearing of a line • Consider a line AB of which the magnetic bearing is to be observed. • Let the ranging rod be fixed at B in line AB and the compass is centered on A. • Turn the compass in the direction of line AB. • When B is bisected by the vertical hair. i.e. when ranging rod at B comes in line with the slit of eye vane and the vertical hair. i.e. when ranging rod at B comes in line with the slit of eye vane and the vertical hair of the object vane, the reading, under the vertical hair through prism is taken, which gives the bearing of line AB. The enlarged portion gives actual pattern of graduations marked on ring.
- 17. Bearing
- 18. The Surveyor’s Compass • It is similar to a prismatic compass except that it has only plain eye slit instead of eye slit with prism and eye hole. • This compass is having pointed magnetic needle in place of broad form needle as in case of prismatic compass.
- 19. The Surveyor’s Compass Working of Surveyor’s Compass: • Centering • Levelling • Observing the Bearing of a Line • First two operations are similar to that of prismatic compass but the method of taking observation differs from that. • Observing the bearing of a line. In this type of compass, the reading is taken from the top of glass and under the tip of north end of the magnetic needle directly. No prism is provided here.
- 20. Difference between Prismatic Compass and Surveyor’s Compass
- 21. Bearing • The bearing of a line is the horizontal angle which it makes with a reference line (meridian) depending upon the meridian, there are four types of bearings. True Bearing The true bearing of a line is the horizontal angle between the true meridian and the survey line. The true bearing is measured from the true north in the clockwise direction. Magnetic Bearing The magnetic bearing of a line is the horizontal angle which the line makes with the magnetic north.
- 22. Bearing Grid Bearing The grid bearing of a line is the horizontal angle which the line makes with the grid meridian. Arbitrary Bearing The arbitrary bearing of a line is the horizontal angle which the line makes with the arbitrary meridian
- 23. Bearing
- 24. Bearing
- 25. Designation of Bearings • The bearing are designated in the following two systems. • Whole Circle Bearing System (W.C.B) • Quadrantal Bearing System ( Q.B.)
- 26. Whole Circle Bearing System (W.C.B) • The bearing of a line measured with respect to magnetic meridian in clockwise direction is called magnetic bearing and its value varies between 0 0 to 360 0. • The Quadrants start from North and Progress in a clockwise direction as the first quadrant is 0 0 to 90 0 in clockwise direction, 2nd 90 0 to 180 0 , 3 rd 180 0 to 270 0 , and up to 360 0 is 4th one.
- 27. Whole Circle Bearing System (W.C.B)
- 28. Whole Circle Bearing System (W.C.B)
- 29. Quadrant Bearing System (Q.B.) • In this system, the bearing of survey lines are measured with respect to north line or south line which ever is the nearest to the given survey line and either in clockwise direction or in anticlockwise direction. • The bearing of lines which fall in Ist and IV th Quadrant are measured with respect to north line is nearer than south line, and bearing of lines fall in II nd and IIIrd quadrants are measured from south line as south is the nearer line. The surveyor’s compass measures the bearing of lines in the quadrant system.
- 30. Reduced Bearing (RB) • When the whole circle bearing of a line is converted into quadrantal bearing it is termed as ‘Reduced Bearing’. Thus, the reduced bearing is similar to the quadrantal bearing. It’s value lies between 0 0 to 90 0, but the quadrants should be mentioned for proper designation.
- 31. Reduced Bearing (RB)
- 32. Reduced Bearing (RB)
- 33. The Following Table Should be Remembered for Conversion of WCB to RB
- 34. The Following Table Should be Remembered for Conversion of RB to WCB
- 35. Fore Bearing and Back Bearing • The bearing of a line measured in the forward direction of survey line is called the ‘Fore Bearing’ (FB) of that line. • The bearing of the line measured in the direction opposite to the direction of the progress of survey is called the ‘Back Bearing’ (BB) of the line.
- 36. Fore Bearing
- 37. Fore Bearing Fore Bearing • The bearing of a line measured in the forward direction (i.e. along the progress of survey) is known as fore bearing. • ForeBearing = Back Bearing ± 180°
- 38. Back Bearing • The bearing of a line measured in the Backward direction is known as Back Bearing. • BB= FB 180 0 • + sign is applied when FB is < 180 0 • - sign is applied when FB is > 180 0 • In the quadrantal bearing (i.e. reduced bearing) system the FB and BB are numerically equal but the quadrant are just opposite. • For example if the bearing of AB is N 60 0 E, then its BB is S 60 0 W.
- 39. Examples • Convert the following WCB into Reduced Bearing. • 49 0 • 240 0 • 133 0 • 335 0
- 40. Examples 49 0 • Since the line falls in the first quadrant therefore the nearer pole is the north pole and is measured from North towards E as 49 0 • There fore RB = N 49 0 E
- 41. Examples 240 0 • Since the line falls in the third quadrant therefore the nearer pole is the north pole and is measured from North towards S as 0 • RB = WCB- 180 0 • RB = 240 0 - 180 0 = 60 0 • RB= S 60 0 W
- 42. Examples 240 0 • Since the line falls in the third quadrant therefore the nearer pole is the north pole and is measured from South towards W as 0 • RB = WCB- 180 0 • RB = 240 0 - 180 0 = 60 0 • RB= S 60 0 W
- 43. Examples 133 0 • Since the line falls in the second quadrant therefore the nearer pole is the south pole and is measured from South towards E as 0 • RB = 180 0 - Ө • RB = 180 0 – 133 0 = 47 0 • RB= S 47 0 E
- 44. Examples 335 0 • Since the line falls in the third quadrant therefore the nearer pole is the north pole and is measured from North towards W as 0 • RB = 360 0 – WCB • RB = 360 0 – 335 0 • RB= N 25 0 W
- 45. Examples Convert the following WCB into RB • 190 0 • 260 0 • 315 0
- 46. Examples Soln 190 0 • RB= WCB – 180 0 • RB = 190 0- 180 0 • RB = S 10 0 W 260 0 • RB = WCB-180 0 • RB= 260 0 – 180 0 • RB = S 80 0 W
- 47. Examples Soln 315 0 • RB = 360 0 – WCB • RB = 360 0 – 315 0 • RB = N45 0 W
- 48. Examples • Convert the following reduced bearings into whole circle bearings: • N 65° E • S 43° 15′ E • S 52° 30′ W • N 32° 42′ W
- 49. Examples Let ‘θ’ be whole circle bearing. (i) Since it is in NE quadrant, θ = α = 65 Ans. (ii) Since it is in South East quadrant 43 15′ = 180 – θ or θ = 180 – 43 15′ = 136 45′ Ans.
- 50. Examples (iii) Since it is in SW quadrant 52 30′ = θ – 180 or θ = 180 + 52 30′ = 232 30′ (iv) Since it is in NW quadrant, 32 42′ = 360 – θ or θ = 360 – 32 42′ = 327 18′
- 51. Examples • The following fore bearings were observed for lines, AB, BC, CD, DE, EF and FG respectively. Determine their back bearings: • 148 • 65 • 285 • 215 • N 36 W • S 40 E
- 52. Examples Solution: • The difference between fore bearing and the back bearing of a line must be 180 . Noting that in WCB angle is from 0 to 360 , • we find Back Bearing = Fore Bearing 180 • + 180 is used if θ is less than 180 and • – 180 is used when θ is more than 180
- 53. Examples Hence, • BB of AB = 145 + 180 = 325 • BB of BC = 65 + 180 = 245 • BB of CD = 285 – 180 = 105 • BB of DE = 215 – 180 = 35 • In case of RB, back bearing of a line can be obtained by interchanging N and S at the same time E and W. Thus • BB of EF = S 36 E • BB of FG = N 40 W.
- 54. Example The Fore Bearing of the following lines are given Find the Back Bearing. (a) FB of AB= 310 0 30’ (b) FB of BC= 145 0 15’ (c) FB of CD = 210 0 30’ (d) FB of DE = 60 0 45’
- 55. Example Solution (a) BB of AB = 310 0 30’ – 180 0 0’ = 1300 30’ (b) BB of BC = 145 0 15’ + 180 0 0’ = 3250 15’ (c) BB of CD = 210 0 30’ – 180 0 0’ = 30 0 30’ (d) BB of DE = 600 45’ + 180 0 0’ = 240 0 45’
- 56. Example FB of the following lines are given, find the BBs. (a) FB of AB = S 300 30’ E (b) FB of BC = N 400 30’W (c) FB of CD= S 600 15’W (d) FB of DE = N 45030’ E
- 57. Example Solution • (a) BB of AB = N 30 0 30’ W • (b) BB of BC = S 40 0 30’ E • (c) BB of CD = N 60 0 15’ E • (d) BB of DE = S 45 0 30’ W
- 58. Example • The fore bearing of the lines AB, BC, CD and DE are 45 0 30’, 1200 15’, 200 0 30’ and 280 0 45’ respectively, find angles B,C,D
- 59. Example • Interior angle B= BB of AB-FB of BC = (45 0 30’ + 180 0 0’) -120 0 15’ = (225 0 30’ – 120 0 15’) = 105 0 15’ • Interior angle C= BB of BC – FB of CD = (120 0 15’ + 180 0 0’) – 200 0 15’ = 300 0 15’ – 200 0 15’ = 100 0 0’ • Exterior angle D = FB of DE- BB of CD = 280 0 45’ – (200 0 30’ – 180 0 0’) = 280 0 45’ – 20 0 30’ = 260 0 15’ • Interior angle D= 360 0 0’ – 260 0 15’ = 99 0 45’
- 60. Computation Of Angles • Observing the bearing of the line of a closed traverse, it is possible to calculate the included angles, which can be used for plotting the traverse. • At the station where two survey lines meet, two angles are formed, an exterior angle and an interior angle. The interior angle or included angle is generally the smaller angle (< 180 0).
- 61. Computation Of Angles
- 62. Computation Of Angles
- 63. Computation Of Angles • While calculating the interior or included angles, it is strongly recommended that a rough sketch of the traverse must be drawn for the purpose of calculating the interior angles or bearing from included angles. A sketch always gives a better idea for calculations. • At any survey stations generally FB of one line and BB of another line are measured. Then difference of these two bearings will give you either an interior angle or an exterior angle depending upon the station position.
- 64. Computation Of Angles • In a closed traverse the following bearings were observed with a compass. Calculate the interior angles.
- 65. Computation Of Angles We find, Back Bearing = Fore Bearing 180 + 180 is used if θ is less than 180 and – 180 is used when θ is more than 180
- 66. Computation Of Angles
- 67. Computation Of Angles Referring to Figure: ∠A = 150 00′ – 65 00′ = 85 00′ ∠B = 245 00′ – 125 30′ = 119 30′ ∠C = 305 30′ – 200 00′ = 105 30′ ∠D = (360 – 265 15′) + 20 00′ = 114 45′ ∠E = (360 – 330 00′) + 85 15′ = 115 15′
- 68. Computation Of Angles • The angles observed with a surveyor compass in traversing the lines AB, BC, CD, DE and EF are as given below. • Compute the included angles and show them in a neat sketch.
- 69. Computation Of Angles In case of RB, back bearing of a line can be obtained by interchanging N and S at the same time E and W
- 70. Computation Of Angles
- 71. Computation Of Angles • Referring to the figure, we find • ∠B = 55 30′ + 63 30′ = 119 00′. • ∠C = 63 30′ + 70 00′ = 133 30′. • ∠D = 70 00′ + 45 30′ = 115 30′. • ∠E = 45 30′ + 72 15′ = 117 45′.
- 72. Computation Of Angles • The following Bearing were observed for a closed traverse. • Calculate the interior angle of a traverse
- 73. Computation Of Angles
- 74. Computation Of Angles ∠ A= 180 0 – (FB of DA+ BB of DA) = 180 0 – (45 0 30’ +75 0 45’) = 180 0 – 121 0 15’ = 58 0 45’ ∠ B= BB of AB + FB of BC = 45 0 30’ + 60 0 0’ = 105 0 30’
- 75. Computation Of Angles ∠ C = 180 0 – (BB of BC + FB of CD) = 180 0 – ( 60 0 0’ + 10 0 30’) = 180 0 – 70 0 30’ = 109 0 30’ ∠ D= BB of CD + FB of DA = 10 0 30’ + 750 45’ = 86 0 15’ Check Sum of Interior angle (2N-4) 90 0 =360 0 Now, ∠ A + ∠ B + ∠ C + ∠ D = 58 0 45’ + 105 0 30’ + 109 0 30’ + 86 0 15’ = 360 0
- 76. Computation Of Angles • The following bearing were taken in a closed traverse find out the interior angles of a traverse Line FB BB AB 45 0 00’ 225 0 00’ BC 123 0 30’ 303 0 30’ CD 181 0 00’ 1 0 DA 289 0 00’ 109 0 00’
- 77. Computation Of Angles
- 78. Computation Of Angles Calculation of Interior Angle: ∠ A= BB of DA- FB of AB = 109 0 0’ – 45 0 0’ = 64 0 0’ ∠ B= BB of AB – FB of BC = 225 0 0’ – 123 0 30’ = 101 0 30’ ∠ C= B B of BC – FB of CD = 303 0 30’ – 181 0 0’ = 122 0 30’
- 79. Computation Of Angles Exterior angle ∠ D = FB of DA – BB of CD = 289 0 0’ – 1 0 = 288 0 Interior Angle ∠ D = 360 0 – 288 0 = 72 0 Sum of Angles = ∠ A + ∠ B + ∠ C + ∠ D = 64 0 + 101 0 30’ = 122 0 30’ + 72 0 0’ = 360 0 Check = ( 2N -4) x 90 0 = ( 2 N – 4) x 90 0 = 4 x 90 0 = 360 0 O.K.
- 80. Magnetic Declination • The horizontal angle between the magnetic meriadian and true meridian is known as ‘Magnetic declination’ • When the north end of the magnetic needle is pointed towards the west side of the true meridian the position is termed as ‘Declination West (ӨW). • When the north end of the needle is pointed towards east side of the true meridian the position is termed as ‘Declination East (Ө E)
- 81. Magnetic Declination
- 82. Determination of True bearing and and Magnetic Bearing • True Bearing = Magnetic Bearing Declination • Use + sign when declination is towards East • Use – sign when declination is towards West • Magnetic Bearing = True Bearing Declination • Use + sign when declination is towards West • Use – sign when declination is towards East
- 83. Determination of True bearing and and Magnetic Bearing Example • The magnetic bearing of a line AB is 135 0 30’. What will be the true bearing, if the declination is 5 0 15’ W • The true bearing of a line CD is 2100 45’, what will be its be its magnetic bearing of the declination is 8 0 15’ W
- 84. Determination of True bearing and Magnetic Bearing • True Bearing of AB = Magnetic Bearing – Declination = 135 0 30’ – 5 0 15’ = 130 0 15’ • Magnetic Bearing of AB = True bearing – Declination = 210 0 45’ + 80 15’ = 219 0
- 85. Example • True bearing of line AB is 357 and its magnetic bearing is 1 30′. Determine the declination. Also find the true bearing of AC which has magnetic bearing equal to 153 30′.
- 86. Example • Magnetic Declination = 1 30′ + (360 – 357 ) = 4 30′ W. • Magnetic bearing of AC = 153 30′. • ∴ True bearing of AC = 153 30′ – 4 30′ =149
- 87. Local Attraction • North end of a freely suspended magnetic needle will always point towards the magnetic north, if it is not influenced by any other external forces except the earth’s magnetic field. It is common experience that the magnetic needle gets deflected from its normal position, if placed near magnetic rocks, iron ore, cables carrying currents or iron electric poles., therefore , not reliable unless these are checked against the presence of local attraction at each station and their elimination.
- 88. Local Attraction Detection of Local Attraction • The presence of local attraction at any station may be detected by observing the fore and back bearing of the line. If the difference between fore and back bearing is 180 0, both end station are free from local attraction. If not, the discepancy may be due to • An error in observation of either fore and back bearing or both • Presence of Local Attraction at either station • Presence of local Attraction at both the stations
- 89. Local Attraction • It may be noted that local attraction at any station affects all the magnetic bearings by an equal amount and hence, the included angles deduced from the affected bearing are always correct. • In case the fore and back bearing of neither line of traverse differ by the permissible error of reading, the mean value of the bearing of the line least affected may be accepted. The correction to other stations, may be made according to the following methods. • By calculating the Included Angles at the affected stations • By checking the required correction, starting from the unaffected bearing.
- 90. Local Attraction • Methods of elimination of local attraction by included angles. The following steps are followed: • Computing the included angle at each station from the observed bearing, in case of a closed traverse. • Starting from the unaffected line, run down the correct bearing of successive sides.
- 91. Computation Of Angles • The following bearing were taken for a closed traverse compute the interior angle and correct them for observational error assume the bearing of line CD to be correct adjust the bearing of remaining sides.
- 92. Computation Of Angles ∠ A = BB Bearing of AE – FB of AB = 130 0 15’ – 80 0 10’ = 50 0 5’ ∠ B = BB of AB – FB of BC = 259 0 -120 0 20’ = 138 0 40’ ∠ C = BB of BC – FB of CD = 301 0 – 170 0 50’ = 131 0 0’
- 93. Computation Of Angles ∠ D = BB of CD – FB of DE = 350 0 – 230 0 10’ = 120 0 40’ ∠E = BB of DE- FB of EA = 49 0 30’ – 310 0 20’ + 360 0 = 99 0 10’ ∠A + ∠B + ∠ C + ∠D + ∠E = 50 0 5’ + 138 0 40’ + 131 0 0’ + 120 0 40’ + 99 0 10’ = 549 0 35’ Theoritical Sum = (2N-4) 90 0 = 540 0 Therefore Error = - 25’
- 94. Computation Of Angles Error = -25’ Hence a correction of + 5’ is applied to all the angles There fore the corrected angles are ∠ A = 50 0 10’ ∠B = 138 0 45’ ∠C = 131 0 5’ ∠D = 120 0 45’ ∠E = 99 0 15’
- 95. Computation Of Angles Starting with the corrected bearing of CD all other bearings can be calculated as under FB Bearing of DE= Bearing of DC - ∠ D = 350 0 50’ – 120 0 45’ = 230 0 5’ BB of Bearing of DE = 230 0 5’ – 180 0 = 50 0 5’
- 96. Computation Of Angles FB Bearing of EA = BB of Bearing of DE - ∠ E = 50 0 5’ – 99 0 15’ + 360 0 = 310 0 50’ BB Bearing of EA = 310 0 50’ – 180 0 = 130 0 50’ FB Bearing of AB = BB of Bearing of EA- ∠ A = 130 0 50’ – 50 0 10’ = 80 0 40’
- 97. Computation Of Angles BB of Bearing of AB = 80 0 40’ + 180 0 = 260 0 40’ FB Bearing of BC = BB of Bearing of AB- ∠ B = 260 0 40’ – 138 045’ = 1210 55’
- 98. Computation Of Angles BB of Bearing of BC = 1210 55’ + 180 0 = 301 0 55’ FB of Bearing of CD = BB of Bearing of BC - ∠ C = 301 0 55’ – 131 0 5’ = 170 0 50’ BB of Bearing of CD = 170 0 50’ + 180 0 = 350 0 50’ (Check)
- 99. Examples • The following are the observed bearing of the line of a traverse ABCDEA with the compass in a place where local attraction was suspected. • Find the correct bearing of the lines. Line FB BB AB 191 0 45’ 13 0 0’ BC 39 0 30’ 222 0 30’ CD 22 0 15’ 200 0 30’ DE 242 0 45’ 62 0 45’ EA 330 0 15’ 147 0 45’
- 100. Traverse ABCDEA (Anticlockwise) B N N N N A C D N E
- 101. Computation Of Angles Line FB BB Difference AB 191 0 45’ 13 0 0’ 178 0 45’ BC 39 0 30’ 222 0 30’ 183 0 CD 22 0 15’ 200 0 30’ 178 0 15’ DE 242 0 45’ 62 0 45’ 180 0 EA 330 0 15’ 147 0 45’ 182 0 30’
- 102. Computation Of Angles Calculation of Interior Angle Interior ∠ A= FB of AB – BB of EA = 191 0 45’ – 147 0 45’ =44 0 00’ Interior ∠ B= FB of BC – BB of AB = 39 0 30’ – 13 0 00’ = 26 0 30’ Exterior ∠ C= BB of BC – FB of AB = 222 0 30’ – 22 0 15’ = 200 0 15’
- 103. Computation Of Angles Interior ∠ C= 360 0 00’ – 200 0 15’ = 159 0 45’ Interior ∠ D= FB of DE – BB of CD = 242 0 45’ – 200 0 30’ = 42 0 15’ Interior ∠ E= FB of EA – BB of DE = 330 0 15’ – 62 0 45’ = 267 0 30’ Sum of Interior Angle = 44 0 00’ + 26 0 30’ + 159 0 45’ + 42 0 15’ + 267 0 30’ = 540 0 00’ Which is equal to ( 2N-4) x 90 0 = 540 0
- 104. Computation Of Angles Calculation of corrected bearing The Line DE is free from local attraction, So, FB of DE = 242 0 45’ (Correct) FB of EA = 330 0 15’ (Correct) FB of AB = BB of EA + ∠ A = (330 0 15’ + 180 0 0’) + 44 0 00’ = 150 0 15’ + 44 0 00’ = 194 0 15’
- 105. Computation Of Angles FB of BC = BB of EA + ∠ B = (194 0 15’ – 180 0) + 26 0 30’ = 14 0 15’ – 26 0 30’ = 40 0 45’ FB of CD = BB of BC – Exterior ∠ C = (40 0 45’ + 180 0 0’) - 200 0 15’ = 220 0 45’ – 200 0 15’ = 20 0 30’ FB of DE = BB of BC + ∠ D = ( 20 0 30’ + 42 0 15’) = 200 0 30’ + 42 0 15’ = 242 0 45’ (Checked)
- 106. Computation Of Angles Line FB BB Corrected AB 194 0 15’ 14 0 15’ BC 40 0 45’ 220 0 45’ CD 20 0 30’ 200 0 30’ DE 242 0 45’ 62 0 45’ EA 330 0 15’ 147 0 45’
- 107. Computation Of Angles Second method- Directly applying Correction Procedure (a) On verify the observed bearing it is found that the FB and BB of line DE differ by exactly 180 0. So, the station D and E are free from local attraction and the observed FB and BB of DE are correct. (b)The Observed FB of EA is also Correct
- 108. Computation Of Angles (c) The action BB of EA should be 330 0 15’ – 180 0 = 150 0 15’ But the observed bearing is 147 0 45’ So the correction of ( 150 0- 147 0 45’) = + 2 0 30’ at Station A
- 109. Computation Of Angles (d) Correct FB of AB = 191 0 45’ + 2 0 30’ = 194 0 15’ Therefore, the actual correct BB of AB should be 194 0 15’ – 180 0 00’ = 14 0 15’ But observed = 13 0 0’ So a correction of (14 0 15’ – 13 0 0’) = +1 0 15’ At Station B
- 110. Computation Of Angles (e) Correct FB of BC = 39 0 30’ + 1 0 15’ = 40 0 45’ Therefore Correct BB of BC should be = 40 0 45’ + 180 0 = 220 0 45’ So a correction of =( 220 0 45’ – 222 0 30’) = - 1 0 45’At Station C (f) Correct FB of CD = 22 0 15’ – 1 0 45’ = 20 0 30’ which tallies with the observed BB of CD So, D is free from local attraction, which also tallies with remark made at the beginning
- 111. Computation Of Angles Line Observed Correction Corrected Bearing Remark FB BB FB BB AB 191 0 45’ 13 0 00’ + 2 0 30’ at A 194 0 15’ 14 0 15’ BC 30 0 30’ 2220 30’ + 1 0 15’ 40 0 45’ 220 0 45’ CD 22 0 15’ 200 0 30’ -1 0 45’ at C 20 0 30’ 200 0 30’ St D and E are free from local attraction DE 242 0 45’ 62 0 45’ 0 0 at D 242 0 45’ 62 0 45’ EA 330 0 15’ 147 0 45’ 0 0 at E 330 0 15’ 150 0 15’
- 112. Sources of Error in Compass Survey The errors may be classified as (i) Instrumental Error (ii) Error of manipulation and sighting (iii) error due to external influence
- 113. Sources of Error in Compass Survey Instrumental Errors • Needle not being perfectly straight • The pivot being bent, i.e. not being at the centre of the graduated circle. • The needle being sluggish, i.e. the needle having lost its magnetism • The pivot point being dull • The needle neither moving horizontally nor moving freely on the pivot due to the dip of the needle. • The plane of sight not passing through the centre of the graduated ring • The vertical hair being too thick or loose.
- 114. Sources of Error in Compass Survey Error due to Manipulation and Sighting • Inaccurate centring of the compass over the station occupied • Inaccurate leveling of the compass box when the instrument is set up • Imperfect bisection of the ranging rods at station or other objects • Carelessness is reading the needle or in reading the graduate circle through the prism in a wrong direction. • Carelessness in recording the observed reading.
- 115. Sources of Error in Compass Survey Error due to External Influences • Magnetic changes in the atmosphere on a cloudy or stormy day. • Irregular variation due to magnetic storms, earthquakes, sun spots, lunar perturbations etc. • Variation in declination, viz, secular, annual and diurnal. • Local attraction due to proximity of steel structure, electric lines.
- 116. Precaution to be taken in Compass Surveying The following precaution should be taken conducting a compass traverse • The centring should be done perfectly • To stop the rotation of the graduation ring, the break pin should be pressed very gently and not suddenly. • Reading should be taken along the line sight and not from any side. • When the compass has to be shifted from one station to other, the sight vane should be folded over the glass cover. This is done to lift the ring out of the pivot to avoid unnecessary wear of the pivot.
- 117. Precaution to be taken in compass Surveying • The compass box should be tapped gently before taking the reading. This is done to find out whether the needle rotates freely. • The station should not be selected near magnetic substances. • The observer should not carry magnetic substances. • The glass cover should not be dusted with a handkerchief, because the glass may be charged with electricity and the needle may be deflected from its true direction. The glass cover should be cleaned with a moist finger.
- 118. References • “Surveying and Levelling” Vol- I Kanetkar and Kulkarni (2011) • “ Surveying” Vol- I Dr. B.C. Punamia
- 119. Thanks !

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