1 / 24

Area calculation

Area calculation. Area is divided into triangles, rectangles, squares or trapeziums Area of the one figure (e.g. triangles, rectangles, squares or trapeziums) is calculated and multiplied by total number of figures. Area along the boundaries is calculated as

catherine
Télécharger la présentation

Area calculation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Area calculation

  2. Area is divided into triangles, rectangles, squares or trapeziums • Area of the one figure (e.g. triangles, rectangles, squares or trapeziums) is calculated and multiplied by total number of figures. • Area along the boundaries is calculated as • Total area of the filed=area of geometrical figure + boundary areas

  3. Problem-1

  4. Problem 1-Result

  5. Computation of area from plotted plan • Boundary area can be calculated as one of the following rule: • The mid-ordinate rule • The average ordinate rule • The trapezoidal rule • Simpson’s rule

  6. Mid-ordinate rule l

  7. Average ordinate rule

  8. Trapezoidal rule

  9. Simpson’s rule

  10. Problems • The following perpendicular offsets were taken from chain line to an irregular boundary: • Chainage 0 10 25 42 60 75 m • offset 15.5 26.2 31.8 25.6 29.0 31.5 Calculate the area between the chain line, the boundary and the end offsets. • The following perpendicular offsets were taken from a chain line to a hedge: Calculate the area by mid ordinate and Simpson’s rule.

  11. Area by double meridian distances • Meridian distance of any point in a traverse is the distance of that point to the reference meridian, measured at right angle to the meridian. • The meridian distance of a survey line is defined as the meridian distance of its mid point. • The meridian distance sometimes called as the longitude.

  12. d4/2 d3/2 d4/2 d3/2 D d m4 m3 4 3 Mid points Meridian distance of points (d1, d2, d3, d4) A C c 1 B 2 b d2/2 d2/2 d1/2 d1/2 m1 m2

  13. Meridian distances of survey line: • m1=d1/2 • m2= m1+d1/2+d2/2 • m3=m2+d2/2-d3/2 • m4=m3-d3/2-d4/2 • Area by latitude and meridian distance • Area of ABCD=area of trapezium CcdD + area of trapezium CcbB – area of triangle AbB – area of triangle AdD • = m3*L3+ m2*L2-1/2*2*m4*L4-1/2*2*m1*L1 • =m3*L3+m2*L2-m4*L4-m1*L1

  14. Double meridian distance: • M1= meridian distance of point A + meridian distance of point B • M1=0+d1 • M2=meridian distance of point B + meridian distance of point C • =d1+(d1+d2) • =M1+(d1+d2) • M3=(d1+d2)+(d1+d2-d3)

  15. Area of the traverse ABCD = M3*L3+M2*L2-M1*L1-M4*L4 • Area by Co-Ordinates

  16. The following table gives the corrected latitudes and departures (in m) of the sides of a closed traverse ABCD. Compute the area by (a) M.D. method (b) co-ordinate method

  17. Volume calculation • From cross sections • From spot levels • From contours

  18. Measurement of volume • 3 methods generally adopted for measuring the volume are • (i) from cross sections • (ii) from spot levels • (iii) from contours

  19. Methods of volume calculation • Prismoidal method D2 A2 C2 B2 D2 D1 A2 A1 C2 C1 B1 B2 D1 A1 B1 C1

  20. Also called Simpson’s rule for volume. • Necessary to have odd number of cross sections. • What if there are even number of C/S? • Trapezoidal method: • Assumption mid area is mean of end areas. • Volume =d{(A1+An)/2+A2+A3+…+An-1}

  21. A railway embankment is 10 m wide with side slopes 1.5:1. assuming the ground to be level in a direction transverse to the centre line, calculate the volume contained in a length of 120 m, the centre heights at 20 m intervals being in metres 2.2, 3.7, 3.8, 4.0, 3.8, 2.8, 2.5. A railway embankment 400 m long is 12 m wide at the formation level and has the side slope 2:1. The ground levels across the centre line are as under: The formation level at zero chainage is 207.00 and the embankment has a rising gradient of 1:100. The ground is level across the centre line. Calculate the volume of earthwork.

More Related