Compass survey part 2
- 1. COMPASS SURVEYING Arun Ravi.K.R Instructor, Draughtsman (Civil) PART - II
- 3. LEARNING OBJECTIVES Bearing Meridian WCB & QB Fore Bearings and Back Bearings Calculation of angles
- 4. The bearing of a line is the horizontal angle which the line makes with some reference direction or meridian. Bearing of lines Bearing Meridian
- 5. The meridian or reference direction employed in surveying may be. 1. A true meridian 2. A magnetic meridian, or 3. An arbitrary or assumed meridian Meridian
- 6. True Meridians The reference line, which passing through the geographical north pole and the geographical south pole is known as True Meridian. It is known as an azimuth.
- 7. True Bearing The horizontal angle between a line and the true meridian is called true bearings of the line. It is also called as azimuth.
- 8. Magnetic Meridian The direction indicated by a freely suspended magnetic needle, unaffected by local attraction, is called the magnetic meridian or the magnetic north and south line. MAGNATIC MERIDIAN
- 9. Magnetic Bearing The horizontal angle which a line makes with this meridian is called magnetic bearings or bearings of the line.
- 10. Arbitary meridian Arbitary meridian or an assumed meridian or is any convenient direction towards a permanent mark or signal such as a church spire or top of a chimney. Such meridians are used to determine the relative positions of lines in a small area.
- 11. Arbitary bearings of a line is the horizontal angle which it makes with any arbitrary meridian is called arbitrary bearing of the line. Arbitary Bearing
- 13. The meridian or reference direction employed in surveying may be. 1. A true meridian 2. A magnetic meridian, or 3. An arbitrary or assumed meridian The true meridian is usually employed in geodetic surveys, while the magnetic meridian is used in plane surveys. Meridian
- 14. Magnetic declination The horizontal angle between the magnetic meridian and true meridian is known as magnetic declination.
- 15. Designation of a Magnetic bearing The bearings are expressed in the following two ways. Whole Circle Bearing (WCB) Quadrantal Bearing (QB)
- 16. WCB & QB WCB: The magnetic bearing of a line measured clockwise from the North Pole towards the line is known as WCB. Varies 0-360° Quadrantal Bearing: The magnetic bearing of a line measured clockwise or anticlockwise from NP or SP (whichever is nearer to the line) towards the east or west is known as QB. This system consists of 4-quadrants NE, SE, NW, SW. The values lie between 0-90°
- 18. Reduced Bearing When the Whole Circle Bearings exceed 90°, then it is to be converted or reduced to Quadrantal Bearing system which has the same numerical values of the trigonometrical function is known as reduced bearing (R.B)
- 19. Conversion of W.C.B into R.B
- 20. 1° = 60’ 1’ = 60’’ 80°- 60°30’ = ? 79° 60’ – 60° 30’ 19° 30’ 80°+ 60°30’ = ? 140° 30’
- 21. Convert the following W.C.B to Quadrantal Bearings. 1) 12° 30’ 2) 160° 30’ 3) 210° 30’ 4) 285° 30’ 1) W.C.B = 12° 30’ which is less than 90° ∴ R.B = N 12° 30’ E N 12° 30’ E
- 22. Convert the following W.C.B to Quadrantal Bearings. 1) 12° 30’ 2) 160° 30’ 3) 210° 30’ 4) 285° 30’ 2) W.C.B = 160° 30’ The W.C.B is within 90° to 180° ∴ RB = 180° - W.C.B =180° - 160° 30’ = S19° 30’ E S19° 30’ E
- 23. Convert the following W.C.B to Quadrantal Bearings. 1) 12° 30’ 2) 160° 30’ 3) 210° 30’ 4) 285° 30’ 3) W.C.B = 210° 30’ The W.C.B is within 180° to 270° ∴ RB = W.C.B - 180° = 210° 30’ – 180’ = S 30° 30’ W S 30° 30’ W
- 24. Convert the following W.C.B to Quadrantal Bearings. 1) 12° 30’ 2) 160° 30’ 3) 210° 30’ 4) 285° 30’ 4) W.C.B = 285° 30’ The W.C.B is within 270° to 360° ∴ R.B = 360° - W.C.B = N 74° 30’ W N 74° 30’ W
- 25. Convert the following quadrantel bearings to whole circle bearings. 1) N 30° 30’ E 2) S 70° 30’ 3) S 36° 30W 4) N 85° 30W 1) Q.B = N30° 30’E W.C.B = RB = 30° 30’ 30° 30’
- 26. Convert the following quadrantel bearings to whole circle bearings. 1) N 30° 30’ E 2) S 70° 30’ E 3) S 36° 30W 4) N 85° 30W 2) Q.B = S 70° 30’E W.C.B =180° - R.B = 180° - 70° 30’ = 109° 30’ 109° 30’
- 27. Convert the following quadrantel bearings to whole circle bearings. 1) N 30° 30’ E 2) S 70° 30’ 3) S 36° 30W 4) N 85° 30W 3) Q.B = S 36° 30’W W.C.B = 180° + Q.B = 180° + 36° 30’ = 216° 30’ 216° 30’
- 28. Convert the following quadrantel bearings to whole circle bearings. 1) N 30° 30’ E 2) S 70° 30’ 3) S 36° 30W 4) N 85° 30W 4) Q.B = N 85° 30’W W.C.B = 360º - 85º 30’ = 274º 30’ 274° 30’
- 29. Conversion of W.C.B into R.B
- 30. 1° = 60’ 1’ = 60’’ 80°- 60°30’ = ? 79° 60’ – 60° 30’ 19° 30’ 80°+ 60°30’ = ? 140° 30’
- 31. Fore Bearings and Back Bearings The bearing of a line taken in the progress of the survey or in the forward direction is known as Fore Bearing (F.B) of the line. The bearing taken in the reverse or opposite direction is known as reverse or Back Bearing (B.B) Every line has two bearings, observed one at each end of the line. Fore Bearing Back Bearing N N A B
- 32. B.B of a line = F. B ± 180° Use (+) sign if the given F.B is less than 180° and (-) sign if it exceeds 180° Fore Bearing Back Bearing N N Fore Bearings and Back Bearings A B
- 33. Local Attraction The external attractive forces are, magnetic rock, iron ore, steel structures, rails, electric cables, conveying electric current iron pipes etc. While compass surveying, the magnetic needle is sometimes disturbed from its normal position under the influence of external attractive forces. Such a disturbing influence is called as local attraction.
- 34. Magnetic dip Magnetic dip, dip angle, or magnetic inclination is the angle made with the horizontal by the Earth's magnetic field lines. This angle varies at different points on the Earth's surface.
- 35. Magnetic declination & Dip The horizontal angle between the magnetic meridian and true meridian is known as Magnetic Declination. The angle made with the horizontal by the Earth's magnetic field lines is known as Magnetic dip.
- 36. F.B of AB = 63º 30’ ∴ B.B of AB = F.B of AB + 180º = F.B of AB + 180º = 63º 30’ + 180º = 243º 30’ B.B of AB = 243º 30’ F.B of AB = 63º 30’, Find BB B.B of a line = F. B ± 180° 243º 30’
- 37. F.B of BC = 112º 30’ B.B of BC = F. B of BC + 180° = 112º 30’ + 180 = 292º 30’ F.B of BC = 112º 30’ , Find BB B.B of a line = F. B ± 180° 292º 30’
- 38. F.B of CD = 203º 30’ B.B of CD = F.B of CD - 180º = 203º 30’ – 180 = 23º 30’ F.B of CD = 203º 30’ , Find BB B.B of a line = F. B ± 180° 23º 30’
- 39. F.B of DE = 320º 30’ B.B of DE = F.B of DE - 180º = 320º 30’ - 180º B.B of DE = 140º 30’ F.B of DE = 320º 30’ , Find BB B.B of a line = F. B ± 180° 140º 30’
- 40. F.B of DE = N 65º 30’ W ∴ B.B of DE = S 65º 30’E When bearings are expressed on the quadrantal systems, the back bearings of a line is numerically equal to its fore bearings but with opposite letters. F.B OF AB = N 32º 30’ E , ∴ B.B of AB = S 32º 30’ W F.B of BC = S 43º 30’ E ∴ B.B of BC = N 43º 30’ W F.B of CD = S 26º 30’ W ∴ B.B of CD = N 26º 30’ W
- 41. Calculation of angles from Bearing (WCB) Find the angle between the lines OA and OB, if their respective bearings are 32°30’ and 148° 00’ Angle α = 115° 30’ F.B of line OA = 32°30’ F.B of line OB = 148°00’ ∴ angle AOB or α = F.B of line OB – F.B of line OA = 148°00’ - 32°30’ 115° 30’
- 42. Calculation of angles from Bearing (WCB) Find the angle between the lines OA and OB, if their respective bearings are 16°00’ and 332° 30’ Angle α = 43° 30’ F.B of line OA = 16°30’ F.B of line OB = 332°30’ ∴ angle AOB or α = (360° - F.B of line OB) + F.B of line OA = (360° 00’ - 332° 30’) + 16° 00 43° 30’ = 27° 30’ + 16° 00’
- 43. Calculation of angles from Bearing (WCB) Find the angle between the lines OA and OB, if their respective bearings are 126° 30’ and 300° 30’ Angle α = 174° 00’ F.B of line OA = 126°30’ F.B of line OB = 300°30’ ∴ angle AOB or α = F.B of line OB – F.B of line OA = 300° 30’ - 126° 30’ 174° 00’
- 44. The bearings are given in quadrantal systems of lines AB and AC. Calculate the angle BAC Calculation of angles from Bearing (QB) Line AB Line AC N 25° 30 E S 85° 30’ E Angle α1 or BAC = 69° 0’ Angle α1 = 180° - (θ1+ θ2) = 180° - (25° 30’ + 85° 30’) = 180° - 111° 69° 0’
- 45. The bearings are given in quadrantal systems of lines AB and AC. Calculate the angle BAC Calculation of angles from Bearing (QB) Line AB Line AC N 25° 30 E S 85° 30’ E Angle α1 or BAC = 69° 0’ Included Angle α1 = 180° - (θ1+ θ2) = 180° - (25° 30’ + 85° 30’) = 180° - 111° 69° 0’
- 46. The bearings are given in quadrantal systems of lines AB and AC. Calculate the angle BAC Calculation of angles from Bearing (QB) Line AB Line AC N 20° 30 E N 85° 30’ E Angle α2 or BAC = 65° 0’ Included Angle α2 = FB of AC – FB of AB = (θ2 - θ1) = 85° 30’ - 20° 30’ = 65° 0’ 65° 0’
- 47. The bearings are given in quadrantal systems of lines AB and AC. Calculate the angle BAC Calculation of angles from Bearing (QB) Line AB Line AC S 70° 00 E S 10° 00 W Angle α3 or BAC = 80° 0’ Included Angle α3 = FB of AC + FB of AB = (θ1 + θ2) = 70° 00’ + 10° 00’ = 80° 0’ 80° 0’
- 48. The bearings are given in quadrantal systems of lines AB and AC. Calculate the angle BAC Calculation of angles from Bearing (QB) Line AB Line AC N 50 ° 30’ E S 20° 30’ W Angle α4 or BAC = 150° 0’ Included Angle α4 = (180° + θ2)- θ1 = (180° + 20° 30’) - 50° 30’ = 200°30’ - 50° 30’ 150° 0’
- 49. The bearings are given in quadrantal systems of lines AB and AC. Calculate the angle BAC Calculation of angles from Bearing (QB) Line AB Line AC N 40° 30, W N 46° 0’E Angle α5 or BAC = 86° 30’ Included Angle α5 = FB of AC + FB of AB = (θ1 + θ2) = 40° 30’ + 46° 00’ = 86° 30’ 86° 30’
- 50. The bearings are given in quadrantal systems of lines AB and AC. Calculate the angle BAC Calculation of angles from Bearing (QB) Line AB Line AC S 45° 30’ W N 60° 0’ W Angle α1 or BAC = 74° 30’ Angle α6 = 180° - (θ1+ θ2) = 180°- (45° 30’ + 60° 00’) = 180° 00’ - 105° 30’ 69° 0’ 74° 30’
- 51. The following bearings are observed in a triangular plot with a compass, calculate in the interior angles. Line Fore bearing AB 50° 30’ BC 125° 30’ CA 270° 30’
- 52. Included angle at A 40° Angle BAC = θ1 = 40° = (270° 30’ - 180°0’) - 50° 30 = 90° 30’ - 50° 30’ Angle BAC (θ1) = BB of CA – FB of AB
- 53. Included angle at B 105°00’ Angle BAC = θ2 = 105°00’ = (50° 30’ + 180° 00) -125° 30’) = 230° 30’ - 125° 30’ Angle ABC (θ2) = BB of AB –FB of BC
- 54. Included angle at C 35°00’ Angle BAC = θ3 = 35° 00’ = (125° 30 + 180° 00) - 270° 30’ = 305°30’ - 270° 30’ Angle ABC (θ3) = B.B OF BC – FB OF CA
- 55. 35°00’ 105°00’ 40° Check: ⎣A + ⎣B+ ⎣C =180° 40° + 105° +35° = 180°
- 56. ARUN RAVI.K.R Instructor, Draughtsman (Civil)