Assignment 2023 Solution updated

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McGill University *

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210

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Civil Engineering

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Jan 9, 2024

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McGill UNIVERSITY Civil Engineering and Applied Mechanics CIVE 210 Surveying, Summer 2023 Assignment Solutions May 3 rd , 2023 Due: Saturday May 6 th , 2023 8:00 am EST PART A: (25 points) 1. Discuss why the term geomatics is being used to identify the profession of surveying (4 points) Geomatics engineers design, develop, and operate systems for collecting and analyzing spatial information about the land , the oceans, natural resources, and manmade features. /2 Surveying, which has recently also been called geomatics, has traditionally been defined as the science, art, and technology of determining the relative positions of points above, on, or beneath the Earth’s surface, or of establishing such points. /2 2. Explain how aerial photographs and satellite images can be valuable in surveying. (6 points) Photogrammetry has many applications in surveying: It can be used in land surveying to compute coordinates of section corners, boundary corners or points of evidence to locate these corners. /2 Photogrammetry is also used to map shorelines in hydrographic surveying , to determine precise ground coordinates of points in control surveying and to develop maps and cross sections for engineering surveys. /2 Photogrammetry also plays an important role in Land and Geographic Information Systems . /2 3. State the number of significant digits in each of the following values: (2points – 0.5 each) 520 (2 sig figs), 0.0125 (3 sig figs), 0.0009010 (4 sig figs), 2023.429 (7 sig figs) /2 4. Give answers to the following problems in the correct number of significant figures: (3points – 1 each) 12.2021 + 5.19 + 1902.0 = 1919.4 /1 123 – 4.5 = 119 /1 945.60 x 4.15 = 3920 /1 5. Identify the accuracy and precision of Figures a, b and c. Identify which one shows a systematic error. (4points) Figure a: precise, not accurate /1 Figure b: not precise or accurate /1 Figure c: precise and accurate /1 Figure a shows a systematic error. /1

6. Describe how a very precise measurement can be inaccurate. (2 points) Precision is a measurement of repeatability. If the equipment has certain errors (systematic) associated with a particular measurement, that measurement will not be accurate but easily may be repeatable. /2 7. Assume that five observations of a line are recorded as follows: 602.55, 620.59, 602.51, 602.60, and 602.57. What type of error may have happened in the second observation? (2points) A gross/random error occurred. /2 8. Define the term systematic error and give two surveying examples of a systematic error? (2points) A systematic error is a consistent and repeatable error attributed to faulty equipment or experimental procedure. This error can be calibrated. /1 Examples /1 PART B: (77 points) Raw data: /0.5 Correct values in the table (0.5 x 62) Table 1: Level loop field book table BS HI IS FS ELEV NOTES 0.675 100.675 100.000 BM1 0.855 97.795 3.735 96.940 TP1 0.450 94.790 3.455 94.340 TP2 1.380 92.080 4.090 90.700 TP3 0.250 89.110 3.220 88.860 TP4 3.210 88.620 3.700 85.410 TP5 2.555 90.590 0.585 88.035 TP6 1.555 92.060 0.085 90.505 TP7 3.450 94.560 0.950 91.110 TP8 3.335 97.325 0.570 93.990 TP9 3.695 100.385 0.635 96.690 TP10 1.450 101.420 0.415 99.970 TP11 1.250 100.170 BM1 Sample calculations for the first BM1: /2 ELEV + BS = HI 100.000 + 0.675 = 100.675 Sample calculation for TP1: /2 HI − FS = ELEV 100.675 − 3.735 = 96.940 Closure error: Closure error calculation /2 Closure error = |∑ BS − ∑ FS| ??????? ????? = 22.860 − 22.60 = 0.170

Adjusted elevations: /2 Correct number of readings /2 Correct adjustment 𝑁????? ?? ???????? ?? ?? ???????? = 12 𝐴????????? = ??????? ????? / 𝑁????? ?? ???????? ?? ?? ???????? = 0.170 / 12 = 0.014 /2 Explanation for how the error was adjusted Because the final BM1 value was higher than the first, the adjustment will be subracted to each elevation equally. /0.5 Correct values in the table (0.5 x 52) /2 No adjustment on the first benchmark reading /2 The adjustment is cumulative If calculations are done on Excel (with no rounding): Table 2: Adjusted elevations Location Elevation Adjustments Adjusted Elevations BM1 100.000 0.000 100.000 TP1 96.940 0.014 96.926 TP2 94.340 0.028 94.312 TP3 90.700 0.043 90.658 TP4 88.860 0.057 88.803 TP5 85.410 0.071 85.339 TP6 88.035 0.085 87.950 TP7 90.505 0.099 90.406 TP8 91.110 0.113 90.997 TP9 93.990 0.128 93.863 TP10 96.690 0.142 96.548 TP11 99.970 0.156 99.814 BM1 100.170 0.170 100.000 If calculations are done by hand (with rounding): Table 3: Adjusted elevations Location Elevation Adjustments Adjusted Elevations BM1 100.000 0.000 100.000 TP1 96.940 0.014 96.926 TP2 94.340 0.028 94.312 TP3 90.700 0.042 90.658 TP4 88.860 0.056 88.804 TP5 85.410 0.070 85.340 TP6 88.035 0.085 87.950 TP7 90.505 0.099 90.406 TP8 91.110 0.113 90.997 TP9 93.990 0.127 93.863 TP10 96.690 0.141 96.549 TP11 99.970 0.155 99.815 BM1 100.170 0.170 100.000 /2 Ensured that the loop was balanced

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