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- 1. Computation of Area Unit IV
- 2. Syllabus • Area of a irregular figure by Trapezoidal rule, Mid-ordinate rule, average- ordinate rule, Simpson’s rule, various coordinate methods, planimeter. • Computation of volume by trapezoidal and prismoidal formula
- 3. Compution of Area • The term area in the context of surveying refers to the area of a tract of land projected upon the horizontal plane, and not to the actual area of the land surface • Area may be expressed in the following units • Square metre • Hectares • Square feet • Acres
- 4. Various methods of Computation of Area Area Graphical Method From field Notes Instrumental Method From Plotted Plan Entire Area Mid Ordinate Avg Ordinate Boundary Area Trapezoidal Simpson’s Rule
- 5. Computation of Area from Field Notes This is done in two steps Step-I • In cross-Staff survey, the area of field can be directly calculated from the field notes. During survey work the whole area is divided into some geometrical figures, such as triangle, rectangles, square, and trapeziums, and then the area is calculated as follows • Area of Triangle= √s (s-a) (s-b) (s-c)
- 6. Computation of Area from Field Notes Where a, b and c are the sides, s = a+ b+ c 2 Or Area of the triangle = ½ x b x h Where, b= base h= Altitude Area of square= a2 Where a is the side of the square Area of trapezium= ½ (a+b) x d Where a and b are the parallel sides, and d is the perpendicular distance between them.
- 7. Computation of Area from Field Notes • The Area along the boundaries is calculated as follows • O1, O2= Ordinates • X1, X2, = Chainages • Area of Shaded Portion= O1 + O2 x (X1 + X2 ) 2
- 8. Computation of Area from Field Notes
- 9. Computation of Area from Field Notes • Similarly, the areas between all pairs of ordinates are calculated and added to obtain the total boundary area. • Hence, Total area of the field = area of geometrical figure + Boundary areas (Step-1 + Step-2) = Area of ABCD + Area of ABEFA
- 10. Computation of Area from Plotted Plan Case-I Considering the entire area • The entire area is divided into regions of a convenient shape, and calculated as follows: (a) By dividing the area into triangles • The triangle are so drawn as to equilize the irregular boundary line. • Then the bases and altitude of the triangles are determined according to the scale to which the plan was drawn. After this, the area of these triangles are calculated • (area= ½ x base x altitude) • The area are then added to obtain the total area.
- 11. Computation of Area from Plotted Plan
- 12. Computation of Area from Plotted Plan (b) By dividing the area into Squares • In this method, squares of equal sizes are ruled out on a piece of tracing paper. Each square represents a unit area, which could be 1 cm2 or 1 m2. The tracing paper is placed over the plan and the number of full squares are counted. The total area is then calculated by multiplying the number of squares by the unit area of each square.
- 13. Computation of Area from Plotted Plan
- 14. Computation of Area from Plotted Plan (c) By drawing parallel lines and converting them to rectangles In this method, a series of equidistant parallel lines are drawn on a tracing paper. The constant distance represents a metre or centimetre. The tracing paper is placed over the plan in such a way that the area is enclosed between the two parallel lines at the top and bottom. Thus the area is divided into a number of strips. The curved end of the strip are replaced by perpendicular lines and a number of rectangles are formed. The sum of the length of the rectangles is then calculated. Then, Required area= ∑ Length of Rectangle x constant distance
- 15. Computation of Area from Plotted Plan Case II • In this method, a large square or rectangle is formed within the area in the plan. Then ordinates are drawn at regular intervals from the side of the square to the curved boundary. The middle area is calculated according to one of the following rules: • The Mid-Ordinate rule • The Average Ordinate Rule • The Trapezoidal Rule • Simpson’s Rule.
- 16. Computation of Area from Plotted Plan • The Mid-Ordinate Rule
- 17. Computation of Area from Plotted Plan • Let, O1, O2, O3….. On= Ordinates at equal Intervals • l= length of base line, • d= common distance between ordinates, • h1, h2, …hn= mid ordinates • Area of Plot= h1 x d+ h2 x d + .. + hn x d • = d(h1+ h2+ … hn) • i.e. Area= Common Distance x Sum of MidOrdinates.
- 18. The Average-Ordinate Rule
- 19. The Average-Ordinate Rule • Let, O1, O2, O3….. On= Ordinates Regular Intervals • l= length of base line, • n= number of divisions, • n+1= number of ordinates • Area= O1 + O2+ O3….. On x l On+1 • Area= Sum of Ordinates x length of base line No of Ordinates
- 20. The Trapezoidal Rule • While applying the trapezoidal rule, boundaries between the ends of the ordinates are assumed to be straight. Thus the area enclosed between the base line and the irregular boundary lines are considered as trapezoids
- 21. The Trapezoidal Rule
- 22. The Trapezoidal Rule Let, O1, O2, O3….. On = Ordinate at equal intervals d = common distance 1st Area= O1+ O2 x d 2 2nd Area= O2+ O3 x d 2 3 rd Area= O3+ O4 x d 2 Last Area= On-1+ On x d 2 Total Area = d [O1 + 2O1 +2O2….. 2 On-1 + On ] 2 = Common distance (1st ordinate + Last Ordinate+ 2 (sum of other Ordinate) 2
- 23. The Trapezoidal Rule Thus, the trapezoidal rule may be stated as follows: • To the sum of the first and the last ordinate, twice the sum of intermediate ordinates is added. This total sum is multiplied by the common distance, Half of this product is the required area
- 24. Simpson’s Rule • In this rule, the boundaries between the ends of ordinate are assumed to form an arc of a parabola. Hence Simpson’s rule is sometimes called the parabolic rule.
- 25. Simpson’s Rule Let, O1, O2, O3 = Three Consecutive Ordinate D= common distance between the ordinates Area= AFeDC= Area of trapezium AFDC + area of Segment FeDEF Here, Area of Trapezium = O1 + O2 x 2d 2 Area of Segment= 2 x area of parallelogram FfdD 3 = 2 x Ee x 2d = 2 x {O2- O1 + O3 } x 2 d 3 3 2
- 26. Simpson’s Rule So, the area between the first two divisions, ∆ = O1 + O2 x 2d + 2 {O2- O1 + O3 } x 2 d 2 3 3 2 = d ( O1 + 4 O2 + O3) 3 Similarly, the area between next two divisions, ∆2 = d ( O3 + 4 O4 + O5) 3 Total Area= d ( O1 + On + 4 (O2 + O4+ …+ 2 (O3+ O5+…) = Common distance ( 1st ordinate + last Ordinate) 3 + 4 (sum of even ordinates) + 2 (sum of remaining odd Ordinates)
- 27. Simpson’s Rule Thus, the rule may be stated as follow: • To the sum of the first and the last ordinate, for times the sum of even ordinates and twice the sum of the remaining odd ordinates are added. This total sum is multiplied by the common distance. One-third product is the required area.
- 28. Limitation • To apply this rule, the number of ordinates must be odd. That is, the number of divisions must be even.
- 29. The Trapezoidal rule and Simpson rule Trapezoidal Rule Simpson’s Rule The boundary between the ordinates is The Boundary between the ordinates is considered to be straight. considered to be an arc of a parabola. There is no limitation. It can be applied To apply this rule, the number of ordinates for any number of ordinates. must be odd. That is, the number of divisions must be even. It Gives an approximate Result. It gives a more accurate result.
- 30. Examples The following offset were taken from a chain line to an irregular boundary line at an interval of 10 m; • 0, 2.5, 3.5, 5.0, 4.6, 3.3, 0 m • Compute the area between the chain line, the irregular boundary line and the end offset by; • • • • (a) The mid-Ordinate Rule (b) The average- Ordinate Rule (c) The trapezoidal rule (d) Simpson’s Rule
- 31. Examples By mid-ordinate rule: The mid- Ordinate are H1= 0 + 2.5 = 1.25. 2 H2= 2.5 + 3.5 = 3.00 2 H3= 3.5 + 5.00 = 4.25 m 2 H4= 5.00+ 4.6 = 4.8 m 2 H5 = 4.6 + 3.2 = 3.9 m 2 H6= 3.2 + 0 = 1.6 m 2
- 32. Examples Required Area • = 10 (1.25 + 3.00 + 4.25 + 4.8 + 3.9 + 1.6) • = 10 x 18.8 = 188 m2 • By average-ordinate rule: • Here d= 10 m and n= 6 (no of div) • Base length= 10 x 6 = 60 m • Number of ordinates= 7 • Required Area= 60 x (0+ 2.5 +3.5+5+4.6+3.2+0) • 7 • = 60 x 18.80 = 161.14 m 2
- 33. Examples By Trapezoidal Rule: • Here d= 10 • Required Area • = 10/2 ( 0 + 0 + 2 ( 2.5 + 3.5+ 5.0+ 4.6+ 3.2)) • = 5 x 37.60= 188 m2 • By Simpson’s Rule • d= 10 • Required Area • = 10 (0 + 0 + 4 (2.5+ 5.0+ 3.2) + 2 ( 3.5+ 4.6) • 3 • 10 x 59.00 • 3 • 196.66 m2
- 34. Examples • The following offset were taken at 15 m interval from a survey line to an irregular boundary line: • 3.5, 4.3, 6.75, 5.25, 7.5, 8.8, 7.9, 6.4, 4.4, 3.25 m • Calculate the area enclosed between the survey line, the irregular boundary line, and the first and last offset by: • (a) The Trapezoidal Rule • (b) Simpson’s Rule
- 35. Examples By Trapezoidal Rule • Required Area • = 15 (3.5 + 3.25 + 2 (4.3+ 6.75+ 5.25 + 7.5+ 8.8 + 7.9 + 6.4+ 4.4) • 2 • = 15 (6.75 + 102.60) = 820.125 m2 • 2
- 36. Examples Simpson’s Rule • If this rule is to be applied, the number of ordinates must be odd but here the number of ordinate is even (ten) • So, Simpson’s rule is applied from O1 to O 9 and the area between O9 and O 10 is found out by the trapezoidal rule. • A1= 15 (3.5 + 4.4+4(4.3+5.25+8.8+6.4) + 2 (6.75 + 7.5 + 7.9) • 3 • = 15 (7.9 + 99+ 44.3)= 756 m2 • 3 • A2= 15 (4.4 + 3.25) = 57.38 m2 • 2 • Total Area= A1 + A2 = 756 + 57.38 = 813.38 m2
- 37. Examples • The perpendicular offsets taken at 10 m intervals from a survey line to an irregular • boundary are 2.18 m, 3.2 m, 4.26 m, 6.2 m, 4.8 m, 7.20 m, 8.8 m, 8.2 m and 5.2 m. Determine the area • enclosed between the boundary, survey line, the first and the last offsets by • (i) Trapezoidal rule (ii) Simpson’s rule. ANS • Area = 463.5 m2 • = 474.333 m2
- 38. Co-ordinate Method of Finding Area • When Offset are taken at very irregular intervals, then the application of the trapezoidal rule and Simpson’s rule is very difficult. In such a case, the coordinate method is the best.
- 39. Co-ordinate Method of Finding Area
- 40. Co-ordinate Method of Finding Area
- 41. Co-ordinate Method of Finding Area
- 42. Co-ordinate Method of Finding Area Sum of Products along the Solid line, • ∑ P = (y0x1 + y1 x2 + … 0.0) Sum of Products, along the dotted Line • ∑ Q = ( 0.y1 + x1 y2 + …+ 0. y0) • Required Area= ½ (∑ P - ∑ Q)
- 43. Examples • The following perpendicular offset were taken from a chain line to a hedge Chainage (m) 0 – 5.5- 12.7- 25.5- 40.5 Offset (m) 5.25- 6.5- 4.75-5.2-4.2 • Taking g as the origin, the coordinates are arranged as follows:
- 44. Examples
- 45. Examples • Sum of products along the solid line, • ∑ P = (5.25 x 5.5 + 6.5 x 12.7 + 4.75 x 25.5 + 5.2 x 40.5 + 4.2 x 40.5 + 0x 0 + 0 x 0). • = 28.88 + 82.55 + 121.13 +210.6+ 170.1= 613.26 m2 • Sum of Products along dotted line, • ∑ Q= (0 x 6.5 + 5.5 x 4.75 + 12.7 x 5.2 + 25.5 x 4.2 + 40.5 x 0 + 40.5 x 0 + 0 x 5.25). • = 26.13 +66.04 +107.10= 199.27 m2 • Required Area= ½ (∑ P - ∑ Q) • = ½ (613.26-199.27) = 206.995 m2
- 46. Instrument Method • The Instrument used for computation of area from a plotted map is the planimeter. The area obtained by planimeter is more accurate than obtained by the graphical method. There are various types of planimeter in use. But the Amsler Polar Planimeter is the most common used now.
- 47. Amsler Polar Planimeter
- 48. Instrument Method
- 49. Instrument Method • The Constructional Details of the planimeter are Shown; • It consists of two arms. The arm A is known as the tracing arm. Its length can be adjusted and it is graduated. The tracing arm carries a tracing point D which is moved along the boundary line of the area. There is an adjustable Support E which always keep the tracing point just clear of the surface.
- 50. Instrument Method • The other arm F is known as the pole arm or anchor arm, and carries a needle pointed weight or fulcrum K at one end. The weight forms the centre of rotation. The other end of the pole arm can be pivoted at point P by a ball and socket arrangement. • There is a carriage B which can be set at various points of the tracing arm with respect to the vernier of the index mark I.
- 51. Instrument Method
- 52. Instrument Method • The carriage consists of a measuring wheel W and a vernier V. The wheel is divided into 100 divisions and the vernier into 10 divisions. The wheel and the vernier measures reading up to 3 places of decimal (i.e. 0.125, 0.174) • The wheel is geared to a counting disc which is graduated into 10 divisions for ten complete revolutions of the wheel, the disc shows a reading of one divisions
- 53. Instrument Method • Thus the planimeter shows a reading of four digits (i.e. 1.125, 1.174 etc.) • The counting disc shows units • The wheel shows tenth and hundredth and vernier shows thousandths • The planimeter rests on the tracing point, anchor point and the periphery of the wheel.
- 54. Amsler Polar Planimeter
- 55. Procedure of finding area with a Planimeter • Procedure of finding the area with a planimeter • The vernier of the index mark is set to the exact graduation marked on the tracing arm corresponding to the scale as obtained from the table. • The anchor point is fixed firmly in the paper outside or inside the figure. It should be ensured that the tracing point is easily able to reach every point on the boundary line.
- 56. Procedure of finding area with a planimeter • But it is always preferable to set the anchor point outside the figure. If the area is very large. It can be divided into a number of sections. • A good starting point is marked on the boundary line. A good starting point is one at which the measuring wheel is dead. i.e. a point where the wheel does not revolve even for a small movement of the tracing point.
- 57. Procedure of finding area with a Planimeter • By observing the disc, wheel and vernier, the initial reading (IR) is recorded. • The tracing point is moved gently in a clockwise direction along the boundary of the area. • The number of times the zero mark of the dial passes the index mark in a clockwise or anticlockwise direction should be observed. • Finally, by observing the disc, wheel and vernier the final reading (FR) is recorded.
- 58. Instrument Method
- 59. Procedure of finding area with a planimeter • Then, the area of the figure may be obtained from the following expression. • Area A= M (FR- IR 10 N + C) • Where, • M= Multiplier given in the table • N= Number of times the zero mark of the dial passes the index mark, • C= the constant given in the table • FR= Final Reading • IR= Initial Reading.
- 60. Zero Circle in a Planimeter • When the tracing point is moved along a circle without rotation of the wheel (i.e. when the wheel slides without any change in reading), the circle is known as ‘ Zero Circle’ or ‘Circle of Correction’. The zero circle is obtained by moving the tracing arm point in such a way that the tracing point makes an angle of 90 0 with the anchor arm. • The anchor point A is the centre of rotation and AT (R’) is known as the radius of the zero circle. • When the anchor point is inside the figure the area of the zero circle is added to the area computed by planimeter.
- 61. Finding Radius of Zero Circle • • • • • • • • • • • When the wheel is outside of the pivot point, A= Anchor Point. P= Pivot point T= Tracing Point W= Wheel Let TP= L PW= L1 AP= R AT= R’ (Radius of Zero Circle) From right-angled triangle AWT ( Angle AWT= 90 0) AT 2 = AW 2 + TW 2
- 62. Finding Radius of Zero Circle Or • R’ 2= (AP2- PW2) + TW2 • = (R2-L12) + (L + L1)2 • = R2- L12+ L2 + 2 LL1 + L12 • = R2 + L2 + 2LL1 • R’= R2 + 2 LL1 + L2
- 63. Finding Radius of Zero Circle When the wheel is inside the pivot, Let, TP= L PW= L1 AP= R AT= R1 Angle AWT= 90 0 From the right-angled triangle AWT, • AT2= TW2 + AW2 • = (TP-PW) 2 + (AP2- PW2) • R12= (L-L1)2 + ( R2- L12) • = L2- 2 LL1 + L12 + R2 – L1 2 • L2- 2 L L1 + R2 • R’= R2 - 2 LL1 + L2
- 64. Finding Area of Zero Circle By measuring radius of Zero Circle • Area of Zero Circle= ∏ R’2 When the wheel is placed beyond the pivot point • Area of Zero Circle= A= ∏ R’2 = ∏ (R2 + 2LL1 + L2 ) • When the Wheel is placed between the pivot and the tracing point • Area of Zero Circle, • A = ∏ R’ • = ∏ (R2 + 2LL1 + L2 )
- 65. Finding Area of Zero Circle From multiplier and Constant • Area of Zero Circle • A= M x C Where, • M= multiplier value given in table • C= Constant given in table By Planimeter • A geometrical figure is considered whose actual area is known. After this, the area of the figure is computed by the Planimeter • Then, • Area of Zero Circle • = Actual Area- Area Computed by Planimeter.
- 66. Examples • The following reading were recorded by a planimeter with the anchor point inside the figure: • • • • IR= 9.377. FR= 3.336 M= 100 cm2, C= 23.521 • Calculate the area of the figure when it is observed that the zero mark of the dial passed the index mark once in the anticlockwise direction.
- 67. Examples Solution • IR= 9.377 M= 100 cm2 • FR= 3.336 C= 23.521. • N= -1 (for anticlockwise rotation) • From the expression, A= M (FR-IR 10 N + C). • We get, • A= 100 (3.336- 9.377- 10 x 1 + 23.521) = 748 cm2
- 68. Examples • The area of an irregular figure was measured with a planimeter having the anchor point outside the figure. The initial and final reading were 4.855 and 8.754 respectively. The tracing arm was set to the natural scale. The scale of map was 1 cm = 5 m. Find the area of the figure.
- 69. Examples Given • IR= 4.855 • FR= 8.754 • M= 100 cm2 • N= 0 • Scale 1 cm= 5 m • C= 0 • So, • Area, A = M (FR- IR) • = 100 (8.754- 4.855) • = 389.9 cm2 • Required area= 389.9 x 25 = 9747.5 m2
- 70. Examples • To determine the constants of a planimeter a 20 cm 20 cm area was measured with anchor point outside the plan area. The zero mark of disc crossed the index in clockwise direction once. • The observed readings are • Initial reading = 7.422 • Final reading = 1.422 • Determine the multiplying constant M. • Solution: Area = M (F – I + 10 N) • Now Area = 20 20 = 400 cm2 • F = 1.422, I = 7.422 • ∴ 400 = M (1.422 – 7.422 + 10 1) • ∴ M = 100 cm2
- 71. Examples A planimeter was used to measure the area of a map once keeping anchor point outside the figure and second time keeping it inside the figure. The observations are as follows: (i) When the anchor point was outside the figure: • Initial reading = 1.486 • Final reading = 7.058 • The zero of the dial did not pass the index at all. (ii) When the anchor point was inside the map: • Initial reading = 3.486 • Final reading = 8.844
- 72. Examples • Zero mark of the dial passed the fixed index mark twice in anticlockwise direction. • Find the (i) area of the map (ii) area of the zero circle. • Take multiplier constant M = 100 cm2.
- 73. Examples • Solution: (i) When the anchor point was outside the plan • I = 1.486 F = 7.058, N = 0, M = 100 cm2. • ∴ A = M (F – I + 10 N) • = 100 (7.058 – 1.486 + 0) • ∴ A = 557.2 cm2 Ans. (ii) When the anchor point was inside the plan • I = 3.486 F = 8.844 N = – 2, M = 100 cm 2. • A = M (F – I + 10 N + C) • 557.2 = 100 (8.844 – 3.486 – 10 2 + C) • ∴ C = 20.214
- 74. Thanks !

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