REFRESHER MODULE 1 - SURVEYING

Room 206 JPD Building 1955 CM Recto Avenue, ManilaTelephone Number: (02) 516 7559 E-Mail: megareview_2008 @yahoo.com

PROBLEM 1A 0.65kg, 50m tape was standardized and supported throughout its whole length and found to be 0.00205m longer at an observed temperature of 31.8°C and a pull of 10KN. This tape was used to measure a 4% grade line which was found to be 662.702m. During measurement, the temperature is 15°C and the tape is suspended under a pull of 20KN. E=200GPa, cross-sectional area of tape is 3mm2 and the coefficient of linear expansion is 0.0000116m/°C.1. Compute the standard temperature.a. 28.27°C b. 30.24°C c. 26.27°C d. 32.24°C2. Compute the total correction per tape length.a. -0.2378 b. +0.2378 c. -0.2187 d. +0.2187

3. Compute the true horizontal distance.a. 659.46m b. 659.27m c. 659.20m d. 659.35m

PROBLEM 2A line measures 7800m at elevation 900m. The average radius of the curvature in the area is 6400km. 1. Compute the sea level distance.a. 7768.90m b. 7778.90m c. 7788.90m d. 7798.90m2. Compute the reduction factor.a. 0.99886 b. 0.99986 c. 0.99686 d. 0.99786

PROBLEM 3A line 125 m. long was paced by a surveyor for four times with the following data 161,165, 159 and 158. Then another line was paced for five times with the following results, 520, 525, 524, 522 and 518.1. Determine the pace factor. a. 0.7756 b. 0.7776 c. 0.7796 d. 0.7806 2. Determine the number of paces for the new line. a. 521.8 b. 519.8 c. 523.8 d. 524.8 3. Determine the distance of the new line.a. 405.75m b. 402.75m c. 407.75m d. 408.75m

PROBLEM 4From the measured values of distance AB, the following data were recorded.

DATA DISTANCE No. OF MEASUREMENTS1 120.68 12 120.84 43 120.76 64 120.64 8

1. Find the probable weight of data 2.a. 4 b. 6 c. 8 d. 12. Find the probable error of the mean. a. +0.0106 b. +0.0126 c. +0.0146 d. +0.01663. Find the standard deviation.a. +0.0856 b. +0.0836 c. +0.0816 d. +0.0796 4. Find the standard error. a. +0.0187 b. +0.0167 c. +0.0147 d. +0.01375. Find the probable distance of AB. a. 120.72 b. 120.92 c. 120.52 d. 120.32

PROBLEM 5The following interior angles of a triangle traverse were measured with the same precision.

ANGLE VALUE (DEGREES) No. OF MEASUREMENTSA 39 5B 65 4C 75 3

1. Determine the most probable value of angle A. a. 39° 15.32’ b. 39° 19.15’ c. 39° 25.53’ d. 39° 13.35’2. Determine the most probable value of angle B. a. 65° 13.35’ b. 65° 15.32’ c. 65° 19.15’ d. 65° 25.53’3. Determine the most probable value of angle C. a. 75° 19.15’ b. 75° 13.35’ c. 75° 25.53’ d. 75° 19.15’

PROBLEM 6From the given data of a different leveling as shown in the tabulation:

STA. B.S. F.S. ELEV.1 5.87 392.252 7.03 6.293 3.48 6.254 7.25 7.085 10.19 5.576 9.29 4.457 4.94

1. Find the diff. in elevation of station 7 and station 5. a. 10.09m b. 8.09m c. 12.09m d. 14.09m2. Find the height of instrument at station 4. a. 400.26m b. 396.26m c. 392.26m d. 386.26m

PROBLEM 7The table shows the values of backsight, foresight & intermediate foresight reading taken from BM1 to BM2. Elevation of BM1 = 328.70m.

STA. B.S. F.S. I.F.S. ELEVATIONBM1 2.32 328.70

1 1.72 2.23 1.24 0.9

TP1 2.77 3.435 2.26 3.77 1.6

TP2 2.22 3.068 2.89 3.610 2.011 1.1

BM2 2.45

1. Find the difference in elevation between stations 5 & 9. a. 2.64m b. 2.44m c. 2.24m d. 2.74m2. Find the elevation of TP2. a. 325.30m b. 327.30m c. 329.30m d. 323.30m

PROBLEM 8In a two peg – test of a dumpy level, the following observations were taken.

Instrument at C Instrument at DRod reading on A 0.296 1.563 Rod reading on B 0.910 2.140

Point C is equidistant from A and B. D is 2.5m from A and 72.5m from B.

1. What is the difference in elevation between A and B? a. 0.614 b. 0.624 c. 0.634 d. 0.6442. Determine the error in the rod reading at B with the instrument at D. a. 0.0393 b. 0.0383 c. 0.0373 d. 0.03633. What is the corresponding rod reading on A for a horizontal line of sight with instrument still at D. a. 1.574 b. 1.564 c. 1.554 d. 1.544

PROBLEM 9Considering the effects of curvature and refraction, the difference in elevation of points B & C is found out to be 111.356m. From point A, the angle of elevation of B is 18⁰30’ & that of C is 8⁰15’. A is between B & C. 1. If C is 2000m from A, how far is B from C? a. 1.6km b. 1.4km c. 1.2km d. 1.0km2. Find the elevation of A if elevation of B is 450m. a. 48.79m b. 48.59m c. 48.19m d. 48.39m

PROBLEM 10Two hills A & C have elevations of 600m & 800m respectively. In between A & C, is hill B which has an elevation of 705m. B is located 12km from A & 10km from C. Determine the clearance or obstruction of the line of sight if an observer is @ A so that C will be visible from A. Ans. 3.95ma. 3.95m (obstruction) b. 3.95m (clearance)c. 3.75m (obstruction) d. 3.75m (clearance)

REFRESHER MODULE 1 - SURVEYING

Room 206 JPD Building 1955 CM Recto Avenue, ManilaTelephone Number: (02) 516 7559 E-Mail: megareview_2008 @yahoo.com

PROBLEM 11A field is in the form of a regular pentagon. The directions of the bounding sides were surveyed with an assumed meridian 5⁰ to the right of the true north & south meridian. As surveyed, the bearing of one side AB is N 33⁰20’ W. 1. Compute the true bearing of Line CD.a. S65⁰20’E b. S66⁰20’E c. S64⁰20’E d. S63⁰20’E2. Compute the true azimuth of Line BC.a. 220⁰40’ b. 221⁰40’ c. 222⁰40’ d. 223⁰40’

PROBLEM 12The bearing of a line from A to B was measured as S16°30’W. It was found that there was a local attraction at both A & B and therefore a forward and a backward bearing were taken between A & a point C at which there was no local attraction. If the bearing of AC was S30°10’E & that of CA was N28°20’W, what is the corrected bearing of AB?a. S18°20’W b. S16°20’W c. S17°20’W d. S19°20’W

PROBLEM 13From the given closed traversed shown.

LINES BEARING DISTANCESA - B S. 35 30’ W 44.37 m.B - C N. 57 15’ W 137.84 m.C - D N. 1 45’ E 12.83 m.D - E ? 64.86 m.E - A ? 106.72

1. Find the Bearing of line D-E.a. N 66.17°E b. N 68.17°E c. N 70.17°E d. N 72.17°E2. Find the Bearing of line E-A.a. S54.2°E b. S52.2°E c. S50.2°E d. S48.2°E

PROBLEM 14From the data below:

LINE LATITUDE DEPARTUREAB -36.13 -25.77BC +74.56 -115.93CD +12.82 +0.39DE +19.90 +61.74EA -68.40 +69.57

1. Solve for area by Transit Rule.a. 6753.29m2 b. 6726.62m2 c. 6726.29m2 d. 6753.62m2

2. Solve for area by Compass Rule. a. 6753.29m2 b. 6726.62m2 c. 6726.29m2 d. 6753.62m2