Surveying 6h5sx

SURVEYING REVIEWER 1. A line is recorded as 472.90m long. It is measured with a 0.65 kg. tape which is 30.005m long at 20:C under a 50N pull ed at both ends. During measurement, the temperature is 5:C and the tape is suspended under a 75N pull. The line is measured on 3% grade. E = 200GPa, cross-sectional area of tape is 3mm2 and the coefficient of linear expansion is 11.6x10-6m/:C. a. Compute the actual length of tape during measurement. – 29.992m b. Compute the total error to be corrected for the inclined distance. – 0.126m c. What is the true horizontal distance? – 472.561m 2. Compute the normal tension which will be applied to a tape ed over two s in order to make the tape equal to its nominal length when ed only at ends points. The steel tape is 30m long and weighs 0.84kg when ed throughout its length under a standard pull of 5.6kg, with the modulus of elasticity is 2x10^6kg/cm2 and area of 0.06 cm2. – 17.33kg 3. The following data observed are the difference in between BM1 and BM2 by running a line of levels over four different routes. Route Diff. In Elevations Probable Error 1 340.22 +-2 2 340.30 +-4 3 340.26 +-6 4 340.32 +-8 Determine the most probable value of difference in elevation. – 340.242m 4. From the measured values of distance AB, the following trials were recorded. Trials Distance 1 120.68 2 120.84 3 120.76 4 120.64 a. Find the probable error. – 0.03 b. Find the standard deviation. – 0.089 c. Find the standard error. – 0.044 5. Compute the adjusted value of B. Angle Observed Value Repetitions A 86:15’20’’ 6 B 134:44’35’’ 2 C 75:48’50’’ 6 D 167:02’05’’ 6 E 76:08’50’’ 4 - 134:44’43’’

6. From the given data of a differential levelling as shown in the tabulation: STA. BS FS ELEV 1 5.87 392.25 2 7.03 6.29 3 3.48 6.25 4 7.25 7.08 5 10.19 5.57 6 9.29 4.45 7 4.94 a. Find the difference in elevation of station 7 and 5. – 10.09m b. Find the difference in elevation of station 7 and 4. – 11.77m c. Find the elevation of station 3. – 392.61m 7. In the plan below shows a differential levelling from bench mark to another bench mark, along each line represents a sight in the actual rod reading. The direction of the field work is indicated by the number of turning points.

a. Compute the elevation of TP2. – 39.644m b. Compute the elevation of BM2. – 40.938m c. Compute the elevation of TP3. – 34.762m 8. The figure shows a schematic arrangement of a profile level route from BM1 and BM2. The values indicated represent backsight, foresight, and intermediate foresight reading taken on stations along the route. Elevation of BM1 = 328.70m. a. Difference in elevation between stations 5 and 9. – 2.24m b. Find the elevation of TP2. – 327.30m c. Find the elevation of BM2. – 327.07m

9. For the given profile levelling notes: STA BS FS IFS ELEV BM1 0.95 225.50 1 3 2 2.3 TP1 3.13 0.64 3 2.7 4 2.8 5 3.1 6 0.5 7 0.8 TP2 2.16 1.28 8 0.9 9 1.2 10 1.7 11 2.8 TP3 0.82 2.37 TP4 1.35 3.50 12 3.0 BM2 1.24 a. Difference in elevation bet. 5 and 2. – 1.69m b. Elevation of TP2. – 227.66m c. Elevation of BM2. – 224.88m

10. In the two peg-test method of a dumpy level the following observations were taken. Instrument set up near A Instrument set up near B Rod reading on A 1.505m 0.938m Rod reading on B 2.054m 1.449m a. What is the difference in elevation between A and B? – 0.53m b. If the line of sight is not in adjustment, determine the correct rod reading on A with the instrument still set up at B. – 0.919m c. Determine the error in the line of sight. – 0.019m 11. In a two peg test using a dumpy level, the following observations were taken: Instrument set up at C Instrument set up at D Rod reading on A 1.103 0.568 Rod reading on B 0.991 0.289 Point C is equidistant from A and B and D is 12m from A and 72m from B. a. What is the true difference in elevation between A and B? – 0.112m b. With the level in the same position at D, to what rod reading on B should the line of sight be adjusted? – 0.4894m c. What is the corresponding rod reading on A for a horizontal line of sight with instrument still at D? – 0.6014m 12. a. The observed com bearing of a line in 1981 was S. 37:30’ E. And the magnetic declination of the place then was known to be 3:10’W. It has also discovered that during the observation local attraction of the place at that moment of 5: E existed. Find the true azimuth of the line. 324:20’ b. The bearing of a line from A to B was measured as S. 16:30’ W. It was found that there was local attraction at both A and B and therefore a forward and a backward bearing were taken between A and a point C at which there was no local attraction. If the bearing of AC was S. 30:10’ E. and that of CA was N. 28:20’ W., what is the corrected bearing of AB? – S 18:20’ W c. In a particular year, the magnetic declination was 1:10’ E and the magnetic bearing of line DE was N. 16:30’ W. If the secular variation per year is 0:03’ E, determine the magnetic bearing of line DE 5 years later? – N 16:45’ W 13. The following bearings taken on a closed com. Assuming the observed bearing of line AB to be correct. Line Forward Bearing Backward Bearing AB S 37:30’ E N 37:30’ W BC S 43:15’ W N 44:15’ E CD N 73:00’ W S 72:15’ E DE N 12:45’ E S 13:15’ W EA N 60:00’ E S 59:00’ W a. Compute the bearing of line BC. – S 43:30 W b. Compute the bearing of line CD. – N 73:30 W c. Compute the bearing of line DE. – N 11: 45’ E

14. Given the following deflection angles of a closed traverse. Assume bearing of line AB is S 40: E. Station Deflection Angles Station Deflection Angles A 85:20’L E 34:18’L B 10:11’R F 72:56’L C 83:32’L G 30:45’L D 63:27’L a. Compute the total error of the deflection angle. - 0:07’ too big b. Compute the bearing of line DE. – N 3:15’ E c. Compute the bearing of line GA. – S 45:19 W 15. From the field notes of a closed traverse shown below, adjust the traverse. LINES BEARING DISTANCES AB Due North 400.00m BC N 45: E 800.00m CD S 60: E 700.00m DE S 20: W 600.00m EA S 86:59’ W 966.34m a. Compute the correction of latitude on line CD using transit rule. – 0.18 b. Compute the linear error of closure. – 1.97740 c. Compute the relative error of precision. – 1/1753 16.