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SURVEYING-II

SURVEYING-II. TRIANGULATION. TRIANGULATION. A triangulation system consists of a series of triangles in which one or more sides of each triangle are also sides of adjacent triangles, as illustrated in fig.

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SURVEYING-II

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  1. SURVEYING-II

  2. TRIANGULATION

  3. TRIANGULATION A triangulation system consists of a series of triangles in which one or more sides of each triangle are also sides of adjacent triangles, as illustrated in fig.

  4. The lines of a triangulation system form a network tying together the triangulation stations at the triangles. • If it were possible to measure one side and all angles in a triangulation system with absolute precisions, no further linear measurement would be necessary. • Unavoidable errors in the field measurement, however, make it describe that the length of two or more lines in each system be measured as a means of checking the computed distances. The lines whose lengths are measured are called base lines.

  5. The arrangements of triangles in most system afford many different geometrical figures, each of which theoretical value of included angles is know. Also, the sum of angles about any station should be equal 360o, and in any triangle the length of the sides should be proportional to the sines of the angle opposite. These known conditions serve as a measure of the precision of the angle measurements and as a means of adjusting the errors so as to secure the most probable values of the measured quantities.

  6. If two angles of any triangle are measured, the value of the third can be readily computed. This procedure however does not permit the application of the known conditions as a measure of the precisions of the measurements or as a means of adjusting the errors; therefore, it is customary to measure all angles.

  7. Triangulation may be used for a simple topographic survey covering but a few acres, or it may be used to extend control of the highest order across the continent. The relative merits of the triangulation method are based upon the character of the area to be surveyed and not upon the degree of precision to be attained. If favorable routes are available, the method of traversing is superior to the method of triangulation, but if the terrain offers many obstacles to traverse work (such as hills, vegetation, or marsh) triangulation is superior.

  8. STEPS OF TRIANGULATION SURVEY The work of triangulation consists of the following steps: • Reconnaissance, (To select the location of station). • Erection of signals and, in some cases, tripods or towers for elevating the signals or the instrument. • Measurement of angles between the sides of triangles.

  9. STEPS OF TRIANGULATION SURVEY • In the cases, astronomical observations at one or more triangulation stations, in order to determine the true meridian to which azimuths are referred. 5. Measurement of base line. 6. Computation, including the adjustment of the observations, the computation of the length of each triangle side, and the computation of the coordinates of stations.

  10. ROUTINES OF TRIANGULATION SURVEYING The entire triangulation surveying may b divided into two stages i.e, (i). Field works. (ii). Computation of triangulation data.

  11. Keeping in mind the above stages the triangulation survey can be consisted of the following operations. • Reconnaissance. • Erection of signals and towers. • Measurement of base lines. • Measurement of horizontal angles. • Measurement of vertical angles. • Astronomical observation to find azimuth of the sides of triangles at regular intervals. • Computation of triangulation data.

  12. RECONNAISSANCE Preliminary field inspection of the area to be surveyed is known as “Reconnaissance”. The reconnaissance survey is very important in all types of surveys and required great skill, experience and judgment for the survey party chief. As the economy and accuracy of the whole triangulation system depends upon the efficiency of reconnaissance. The following operations are required in reconnaissance.

  13. Pre examination of country terrain to be surveyed. • Selection of best site for base line. • Selection of suitable positions of triangulation station. • Determination of indivisibility and heights of stations. • Selection of well defined natural points. • Collection of information for communication of water, food, laborers and guides etc.

  14. When the topographical maps of the area are available the reconnaissance can be done effectively in the office. Moss iced vertical aerial photograph can also be used to make the general study of the area. If maps and photographs are not available then a rapid preliminary survey for reconnaissance is carried out to ascertain the general location of schemes of triangulation suitable for that country. By selecting well conditioned triangles, proper identification of points can be judged by the accurate bisection of intersected points. The essential features of the topography should be sketched. The relative strength and cast of various triangulation schemes are studies and a final scheme is selected.

  15. INSTRUMENTS USED As the reconnaissance, is a rapid survey the instruments are used as follows: • Small theodolite and sextant for measurement of angles. • Prismatic compass for measurement of bearing. • A barometer for ascertaining the elevation. • A steel tape. • A powerful filed glass or a good telescope. • Miscellaneous items such as ladders, ropes & creepers etc, for climbing trees.

  16. Examination of country Terrain The country terrain should be properly examined by the chief of the survey party to know the several topography of the country terrain. The places of importance such as undulation, hills, valleys, mineral quarries etc, the nature of people, their traditions, labor available, sources of drinking water, food, etc, the means of transportation should be studied.

  17. ELECTION OF BEST SITE FOR BASE LINE The site, selected for base line, should be best from the all consideration of survey work. The site should be free from undulations, woods, etc. Means of transportation should be available.

  18. SELECTION OF TRIANGULATION STATION The following points should be considered while selecting the stations for triangulation work: • The triangulation stations should be inter-visible. The inter-visibility of stations can be made by placing the stations upon the most elevated ground such as tops of hills etc., so that they should not be disturbed by atmosphere. • The stations should be accessible to instruments. The supplies of food, water etc., should be available. Camping site should be available.

  19. SELECTION OF TRIANGULATION STATION 3. The triangles formed by the stations should be well conditioned as far as possible. No angle should be less than 30o and more than 120o. • The stations selected should provide good intersection for various points and detailed survey of the area. 5. Extensive distance stations should be avoided. The station’s distance should be neither too small nor too lager. The small sight will have errors due to centering & bisection while large line of sight (on distance station) will make the signal too indistinct for accurate bisections.

  20. SELECTION OF TRIANGULATION STATION • The station should be commanding stations so that they can be used for controlling subsidiary stations and for future extension of survey work. The subsidiary triangulation stations should be selected such that they are most useful for detailed survey work. • Grazing lines of sights should be avoided. No line of sight should pass over industrial areas to be Avoid irregular atmosphere refraction. 8. In a wooded country, stations should be so located that the cost of clearing, cutting of wood should be minimum.

  21. MEASUREMENT OF BASE LINE Equipment required: • Theodolite. • Ranging Rod. • Steel tape. • Thermometer. • Spring Balance. • Trusses. • Levels. • Staff Rod.

  22. Correction applied to the measurement of base line with a steel tape: Correction due to absolute length: Ca = Lc / l Where; Ca = correction due to absolute length. L = measured length. C = error in one length of the tape. l = nominal length of the tape.

  23. Correction due to temperature: CT = α (T – To) L Where; CT = correction due to change in temperature. α = coefficient of thermal expansion. T = Temperature during the measurement. To = Temperature at which the tape is standardized. L = measured length of line.

  24. Correction due to steep of slope (negative): C slope = L – l = L – (L2 – h2)1/2 = L – L (1 – h2/L2)1/2 = L – L [1 – (h/L) 2]1/2 = L –L [1 – h2/2L] C slope = h2/2L Therefore; Slope = h2/2L

  25. If θ is given then: Slope = L –l = L -Lcos θ = L (1-cos θ) = L (2sin2 θ/2) = 2L sin2 θ/2

  26. Problem A line is measured with the help of a steel tape which is standardized at 65oF and was found to be 1571ft. Find the length of the line. Even that the tape during measurement was 85oF. α = change in unit length per degree change or rise in temperature.

  27. Solution: Given data: To = 65o. T = 85oF L = 1571ft α = 0.0000065 /o F Calculations: Since; CT = α (T – To) L = 0.0000065 × (85 o – 65 o) × 1571 = 0.204 True length = CT + L = 0.204 + 1571 = 1571.204ft as required.

  28. Problem A line measured with the help of a steel tape which is standardized at 88o F and was found to be 1571ft. Find the true length of line. Given that the temperature during measurement was 75o F and coefficient of thermal expansion is 0.00000625/o F.

  29. Solution Given Data: To = 88o F L = 1571 ft T = 75o F α = 0.00000625/o F Calculations: Since CT = α (T – To) L = 0.00000625 × (75o – 88o) × 1571 = -0.1276 True length = L + CT = 1571 + (-0.1276) = 1570.8724ft as required.

  30. Problem A line is measured with the help of a steel tape which is standardized at 30 lbs and was found to be 280ft. Find true length of the line? Given that the pull applied during measurement was 20 lbs and A =0.004 inch2 and E = 30×106 psi

  31. Solution: Given Data: Po = 30 lbs P =20 lbs L = 1280ft • A = 0.004 inch2 E = 30 ×106 psi Calculations: Cp = [(P – Po) × L] /AE = [(20 – 30) × 1280] / [0.004 × 30×106] = -0.106ft True length = L + Cp = L + Cp = 1280 + (– 0.106) = 1279.893ft

  32. Problem A line was measured with the help of a 100ft steel tape at 65o F and pull of 20 lbs and was found to be 875ft. Find the true length of the line? Given that standard temperature and pull for the tape is 75o F and 10 lbs respectively. α = 0.0000065/oF A = 0.005 inch2 E = 29 ×106 psi Also given that weight of the tape is 1.5 lbs

  33. Solution: Given Data: T = 65o F P = 20 lbs To = 75o F Po = 10 lbs L = 875ft α = 0.0000065/oF A = 0.005 inch2 E = 29 × 106 psi Weight of tape is 1.5 lbs.

  34. Calculations: (a). CT = α (T – To) L = 0.0000065 × (65o – 75o) × 875 = -0.057 (b). Cp = [(P – Po) ×L]/AE = [(20 – 10) × 75] / [0.005 × 29×106] = 0.060 (c). C sag = (nw2l3) / 24×P2 = [8.75 × (0.015)2 × (100)3] / [24 × 202] = 0.205

  35. **NOTE: n = 875 / 100 = 8.75 w = W / l = 1.5 / 100 = 0.015 lbs/ft True length = L + CT + Cp – C sag = 875 + (-0.057) + 0.60 – 0.205 = 874.798ft **Units = (lbs × ft) ÷ (inch2 × lbs / inch2) = ft If area is given in ft2 then it will have to be converted into inch2. In above case it is completely justified without converting the units as shown above.

  36. Problem A steel tape was exactly 100ft long at 64oF when supported through its length under a pull of 100 lbs. A line was measured with this tape under pull of 200 lbs at 84oF and found to be 2406 ft. Assuming the tape to be supported at every 100ft. Compute the true length of the line; given that weight of one cubic inch of steel is 0.0000065/oF. Modulus of elasticity is 30×106 & psi & total weight of the tape is 4.0 lbs.

  37. Solution: Given Data: P = 20 lbs Po = 10 lbs T o = 64oF T = 84oF l = 100ft E = 30×106 psi α = 0.0000065/oF Specific weight of steel = γ = 0.28 lbs/inch3 Total weight of tape = 40 lbs True length =?

  38. Calculations: Now; CT = α (T –To) L = 0.0000065 × (84 – 64) × 2406 = 0.313 Cp = [(P –Po) × L]/AE = 0.0674 C sag = nw2l3 / 24×P2 = 4.01

  39. **Note: n = 2406 / 100 = 24.06 w = 4 lbs / 100 ft = 0.04 lbs/ft **To calculate ‘A’: Suppose X-section area = A inch2 Length of tape = 100 ft = 1200 inch Volume = 1200 × A × 0.28 A = 4 / (0.28 × 1200) inch2 A = 0.0119 inch2 Now; True length = 2406 + 0.313 + 0.0674 – 4.01 = 2402.37 ft

  40. Problem During the measurement of base line of four bays, the following information was obtained.

  41. The tape weigh 0.0088 lbs/ft run and has X-sectioned area 0.0025 inch2. It was standardized 64oF under a pull of 10 lbs. α = 0.00000625 /o F; E = 30×106 psi. Compute the true length of base line.

  42. Solution: Given Data: W = 0.0088 /ft A = 0.0025 in2 To = 64oF Po = 10 lbs α = 0.00000625 /oF E = 30×106 psi.

  43. Also L = ∑l + ∑CT + ∑CP + -∑CS L = True length of base line L = 329.308 ft L = 329.308 ft

  44. Thanks

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