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# EDM and Total Stations

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### EDM and Total Stations

1. 1. Project Surveying 20/01/2013 1
2. 2. Electronic Distance Measurement(EDM) 20/01/2013 2
3. 3. • 1950s.• Large, heavy, complicated and expensive (old EDM).• Lighter, simpler and less expensive (new EDM).• EDM Instruments use infrared light, laser light ormicrowaves.• Receiver/Transmitter at one end and Prism at the other end.• EDM Instruments come in long range (10 – 20) km, mediumrange (3 – 10) km and short range (0.5 – 3) km.• Some laser EDM Instruments measure short distances (100– 350) m without reflecting prism.•Microwave instruments are often used in hydrographicsurveys 50 km. Currently GPS. 20/01/2013 3
4. 4. Electronic Angle Measurement (Total Stations)Combine the function of Theodolite and EDM 20/01/2013 4
5. 5. Principles of EDM y x   = c/f c = 299,792.458 km/s (in vacuum): wavelength (m)c: velocity (km/s)f: frequency (Hz, hertz; one cycle per second) 20/01/2013 5
6. 6. Principles of EDM  1 2 3 4 2L = (n + )  n n-1c, the velocity of light through the atmosphere can be affected by:• Temperature• Atmospheric Pressure• Water Vapour Content 20/01/2013 6
7. 7. EDM Characteristics • The more expensive EDM instruments have longer distance range and higher precision.Distance Range:• 800 m – 1 km (single prism).• Short-range EDM can be extended to 1,300 m using 3 prisms.• Long-range EDM can be extended to 15 km using 11 prisms.Accuracy Range: (5 mm + 5 ppm) for short-range EDM. (2 mm + 1 ppm) for long-range EDM. 20/01/2013 7
8. 8. EDM CharacteristicsMeasuring Time:1.5 seconds for short-range EDM.3.5 seconds for long-range EDM. Battery Capability: 1,400 – 4200 measurements depending on the size and condition of battery and the temperature. Temperature Range: - 20c + 50c Non-prism measurements: 100 – 350 m 20/01/2013 8
9. 9. PrismsPrisms are surveying tools used with EDM and Total Stations toreflect the transmitted signals. • Prisms must be tribrach-mounted if a higher level of accuracy is required (Control Surveys). • Prisms mounted on adjustable-length prism poles are portable and suited particularly for (Stakeout and Topographic Surveys) 20/01/2013 9
10. 10. EDM Instrument AccuraciesAccuracy Range: (2 mm + 1 ppm) to  (10 mm + 10 ppm)Constant Instrumental Error e.g. 5 mm represents an accuracy of1/2,000 at 10 m, 1/20,000 at 100 m, 1/200,000 at 1,000 m.Measuring error is proportional to the distance being measurede.g. 10 ppm. 20/01/2013 10
11. 11. Field Method for Determining the Instrument-Reflector ConstantA B C AC – AB – BC = Instrument/Present Constant 20/01/2013 11
12. 12. EDM Operation• Set Up• Aim• Measure• Record 20/01/2013 12
13. 13. Geometry of EDM• Optical target and reflecting prism are at the same height• EDM (S), theodolite ()• Adjustable-length prism pole (HR = hi)• Elev. (B) = Elev. (A) +hi  h - HR 20/01/2013 13
14. 14. Problem 1Refer to Figure below. A top-mounted EDM instrumentis set up at station A, its elevation, (HA = 650.000 m).Using the following values, compute the horizontaldistance from A to B and the elevation of B, (HB). Theoptical centre of the theodolite is hi = 1.601 m abovestation and a vertical angle of + 4◦ 18 30 is measuredto the target, which is 1.915 m (HR) above station B.The EDM instrument centre is 0.100 m (hi) above thetheodolite and the reflecting prism is 0.150 m (HR)above the target. The slope distance is measured tobe 387.603 m. 20/01/2013 14
15. 15. Prism S HR Target  EDM 2.065 m  HR hi hi B1.701 m X cos  S X A S  k  20/01/2013 15
16. 16. Solution X cos sin   SX  HR  hi 20/01/2013 16
17. 17. X = 0.15 – 0.10 = 0.05 m = 4◦ 18 30 = 4◦.308S = 387.603 m = 26.53k =  +   =4◦ 18 30 + 26.53 = 4◦ 18 56.53 = 4◦.3157 20/01/2013 17
18. 18. Computation of the horizontal distance from A to B (H):H = S cos (k) = 386.504 m 20/01/2013 18
19. 19. Computation of the elevation of B, (HB): H B  H A  hi  h  HR , 20/01/2013 19
20. 20. HA = 650.000 mh = S sin (k) = 29.168 mHB = 650.000 + (1.601 + 0.10) + 29.168 – (1.915 + 0.15) = 678.804 mHB = 678.804 m , 20/01/2013 20
21. 21. Problem 2A line AB is measured at both ends as follows:At A, slope distance = 1458.777 m,zenith angle = 91 26 50.At B, slope distance = 1458.757 m,zenith angle = 88 33 22.The heights of the instrument, reflector and target are equal for each observation.• Compute the horizontal distance AB.• If the elevation at A is 590.825 m, what is the elevation at B?• Calculate the gradient between A and B. 20/01/2013 21
22. 22. H = S cos  -1 26 44 h = S sin  - 2.5% S = 1458.767 mA B 20/01/2013 22
23. 23. SolutionS1 = 1458.777 m, S2 = 1458.757 mS = (S1 + S2)/2 = 1458.767 m1 = 90 00 00 - 91 26 50 = - 1 26 502 = 90 00 00 - 88 33 22 = 1 26 38 = (1 + 2)/2 = 1 26 44 20/01/2013 23
24. 24. SolutionH = S cos  = 1458.303 mHA = 590.825 m,h = S sin  = 36.800 m.HB = HA - h = 590.825 - 36.800 = 554.025 m.Gradient between A and B= (h/H)  100 = - 2.5% 20/01/2013 24
25. 25. Problem 3Refer to Figure below; angles 1, 2,…....8 of a quadrilateral of a triangulation network were adjusted. If, DA = 213.36 m, XD = 171,719.32 m, YD = 114,056.00 m and the bearing of the line DA is equal to N 25 00 00 W.• Calculate the distances AB, BC and CD.• Calculate the bearings of the lines AB, BC and CD.• Calculate the coordinates of points A, B and C. 20/01/2013 25
26. 26. Quadrilateral Adjustment A 1 2 B 3 4 8 7 5D 6 C 20/01/2013 26
27. 27. Quadrilateral Adjustment Angles i Adjusted values 1 38 44 08 2 23 44 33 3 42 19 08 4 44 51 57 5 69 04 22 6 39 37 47 7 26 25 54 8 75 12 11 20/01/2013 27
28. 28. Solution• Distances:Applying the law of sines by solving triangles:ABD for AB, ABC for BC, ACD for DC:AB = AD sin8/sin3 = 306.40 mBC = AB sin2/sin5 = 132.07 mCD = AD sin1/sin6 = 209.31 m 20/01/2013 28
29. 29. SolutionBearings:If you redraw the figure below, you can calculatethe bearings of the lines AB, BC and CD as follows: 20/01/2013 29
30. 30. SolutionBearings: W E A S B D C 20/01/2013 30
31. 31. SolutionBearings: Brg of AB = (1 + 2) – Brg of AD = (38 44 08 + 23 44 33) - 25 00 00 = 37 28 41 The complete bearing of line AB is S37 28 41E Brg of BC = 180 00 00 - (3 + 4 + Brg of BA) = 179 59 60 - (42 19 08 + 44 51 57 +37 28 41) = 55 20 14 The complete bearing of line BC is S55 20 14W Brg of CD = (5 + 6) – Brg of CB = (69 04 22 + 39 37 47) - 55 20 14 = 53 21 55 The complete bearing of line CD is N53 21 55W 20/01/2013 31
32. 32. Coordinates: SolutionXD = 171,719.32 m, YD = 114,056.00 m XA = XD + DA sin (25 00 00) = 171,809.49 m YA = YD + DA cos (25 00 00) = 114,249.37 m XB = XA + AB sin (37 28 41) = 171,995.92 m YB = YA - AB cos (37 28 41) = 114,006.22 m XC = XB - BC sin (55 20 14) = 171,887.29 m YC = YB - BC cos (55 20 14) = 113,931.10 m 20/01/2013 32
33. 33. Total Stations• Background• Main Characteristics• Applications 20/01/2013 33
34. 34. Main Characteristics• Parameter Input1. Angle Units: degrees or gon.2. Distance Units: ft or m.3. Pressure Units: inches HG or mm HG.4. Temperature Units: F or C.5. Prism constant (- 0.03 m). 20/01/2013 34
35. 35. Main Characteristics• Parameter Input6. Offset distance.7. Face 1 or face 2 selection.8. Automatic point no. incrementation.9. Hi10. HR 20/01/2013 35
36. 36. Main Characteristics• Parameter Input11. Point numbers and code numbers for occupied and sighted stations.12. Date and time settings. 20/01/2013 36
37. 37. Main Characteristics• Capabilities1. Monitor: battery status, signal attenuation, horizontal and vertical axes status, collimation factors.2. Compute coordinates (N, E, Z).3. Traverse closure and adjustment, and areas.4. Topography reductions.5. Object heights.6. Distances between remote points. 20/01/2013 37