Fundamentals of Surveying

 

  REVIEW EXERCISES: 

1.  Determine the most probable value of the measurements having different values based on the number of measurements as tabulated in the table shown below. Distance No. of Measurement 520.14 1 520.20 3 520.18 6 520.24 8 2.  The following data are the observed elevation of a point by running a line of levels over four different routes. It is required to determine the most  probable value of the elevation. Route

Elevations

Probable Error

5.  Three independent line of level are run from BM 1 to BM2. Route A is 6 km long, route B is 4 km long and route C is 8 km long. By route A, BM 2  is 82.27 m above BM 1, by route B, BM2 is 82.40 m above BM 1 and by route C, BM2 is 82.10 m above BM 1. If the elevation of BM1 is 30.69 m, compute the elevation of BM2 by weighted mean. Route Distance Diff. in Elev. Weight A 6 82.27 4 B 4 82.40 6 C 8 82.10 3 6.  From the measured values shown determine if the measurements are within the required precision. Show all computations. Allowable precision is 1/5000. Trials Length (m) 1 106.87

1 340.22 ±02 2 340.30 ±04 3 340.26 ±06 4 340.32 ±08 3.  A baseline measured with an invar tape, and with a steel tape as follows:

2 106.90 3 106.93 4 106.89 5 106.81 7.  Given the following data in measuring a distance of a certain line.

Invar Tape Steel Tape 571.185 571.193 571.186 571.190 571.179 571.185 571.180 571.189 571.183 571.182 a.  What are the most probable values under each set and their corresponding probable errors?  b.  What is the most probable value of the two sets and the probable error of the general mean? 4.  Two angles AOB and BOC and a single angle AOC are measured at the same point O. Determine the most probable value. Angle Observed Value No. of Measurement AOB 33°46’00”   33°46’00” 1

Distance No. of Measurements 740.53 4 740.59 3 740.57 6 740.53 7 a.  Determine the most probable value of the measurement.  b.  Calculate the standard deviation of any single observation. c.  Calculate the standard error of the mean. d.  Calculate the probable error of any single measurement. e.  Calculate the probable error of the mean. f.  Calculate the relative error or precision of the mean. 8.  The following data are the observed elevation of a point by running a line of levels over four different routes.

BOC AOC

63°14’00”  63°14’00”  97°00’30”   97°00’30”

3 6

Route 1 2

Elevation 521.22 m 520.80 m

Probable Error ±0.01 ±0.02

 

  3 521.36 m ±0.03 4 521.32 m ±0.06 a.  Using the method of least square, determine the weight of elevation taken from route 4.  b.  Determine the corresponding relative weight of the elevation taken from route 2 if the relative weight of route 4 is set as 1.0. c.  Determine the most probable value of the elevation of the observed point. 9.  The two sides of a rectangular lot were measured with certain estimated  probable errors errors as follows: W = 312.755 312.755 ±0.050 ±0.050 m and L = 721.550 721.550 ±0.0 ±0.025 25 m. Calculate the probable error in the area of the rectangle. 10.  The sides of a container in a form of a rectangular prism were measured with certain estimated probable errors as follows: L = 23.575 ±0.030 m, W = 12.455 ±0.015 m and H = 5.025 ±0.010 m. Calculate the probable 11. 

3.  A rectangular lot was being measured using a 30 m metallic tape which was 6 mm too long. The recorded dimensions where 70.50 m long by 37.10 m wide. a.  What is the error i ntroduced due to the erroneous length of tape?  b.  What are the actual dimensions of the lot? 4.  A 50 meter steel tape was standardized and supported throughout its whole length and found to be 0.00205 m longer at an observed temperature of 31.8°C and a pull of 10 kilos. This tape was used to measure a line which was found to be 662.702 m at an average temperature of 24.6°C using the same pull. What is the correct length of the line? Use coefficient of expansion of 0.0000116 m per degree centigrade. 5.  A line is recorded as 472.90 m long. It is measured with a 0.65 kg tape which is 30.005 m long at 20°C under a 50 N pull supported at both ends. During measurement the temperature is 5°C and the tape is suspended

error in the volume of the container. 6. 

REVIEW EXERCISE:  1.  A 50 m tape was standardized and was found to be 0.0042 m, too long then the standard length at an observed temperature of 58°C and a pull of 15 kilos. The same tape was used to measure a certain distance and was recorded to be 673.92 m long at an observed temperature of 68°C and a

 pull of 15 kilos. Determine the standard temperature an and d the true length of the line. Coefficient of li near expansion is 0.0000116 m/°C. 2.  A civil engineer used a 30 m tape in measuring an inclined distance. The measured length on the slope was recorded to be 459.20 m long. The difference in elevation between the initial point and the end point was found to be 1.25 m. The 30 m tape is of standard length at a temperature of 10°C and a pull of 50 N. During measurement the t emperature reading was 15°C and the tape was supported at both ends with an applied pull of 75 N. The cross-sectional area of the tape is 6.50 mm2 and the modulus of elasticity is 200 GPa. The tape has a mass of 0.075 kg/m. Determine the true horizontal distance. K = 0.0000116 m/°C.

7. 

8. 

9. 

under a 75 N pull. The line is measured on 3% grade. What is the true2 horizontal distance? E = 200 GPa, cross-sectional area of tape is 3 mm   and the coefficient of linear expansion is 0.0000116 m/°C. A 30 m steel tape i s 2 mm long at 20°C with a pull of 55 N. A rectangle is measured with this tape. The sides are recorded as 144.95 m and 113.00 m. The average temperature during the measurement is 30°C with a pull of 55 N. a.  If the recorded measurement are used, will the computed area of the field be too small, or too big, and why?  b.  What is the error in area in square meter? Use coefficient of expansion of steel tape a s a 0.0000116 m/°C. A baseline was measured using a 100 m tape which is standardized at 15°C with a standard pull of 10 kg. The recorded distance was found out to be 430.60 meters. At the time of measurement, t he temperature was 20°C and the pull exerted was 16 kg. Determine the true length of the base if the weight of one cubic cm of steel is 7.86 grams weight of tape is 2.67 kg. E = 2 x 10° kg/cm2, K = 7 x 10-7 m/°C. A rectangular field was measured using a 100 m tape which was actually 10 cm too short. The recorded area was 2500 sq.m. What is the true area of the field? A 100 m steel tape standardized at 20°C had a l ength of 100.600 m. It was used to measure a line A to B on slightly sloping ground. The recorded

 

  tape distance was 622.70 m. The temperature at the time of measurement  being 29°C. The stadia distance for the same line was only only 560.00 meters. meters. Upon investigations the discrepancy was found out to be due to the fact that a 10 meter length (between 75 m and 85 m) was cut off. a.  What is the true length of the sloping line?  b.  What is the horizontal distance of the same line if the difference of elevation between A and B is 0.72 meters? (Coefficient of expansion of the tape per °C per meter is 0.0000116). 10.  A civil engineer recorded 51, 52, 53 and 54 paces in walking along a 45m course to determine his pace factor. He then t ook 320, 323, 322 and 319  paces in walking an unknown distance. Compute the the distance based on his  pace factor. 11.  A distance was measured ten times and the average distance was found to  be 554.215 m. If two measurements 559.125 m and 550 550.234 .234 m are deleted from the data as being inconsistent with the other measurements, then the average of the remaining eight measurement is? 12.  A 100 m tape weighing 5.08 kg was used to measure a line. It was supported at end points, midpoints and quarter points and the tension applied is 60 Newtons. If the total measured distance is 2345.76 m, what is the correct distance of the line? 13.  In every measurement, a 100 m tape is suspended at the ends under a pull of 15kg. It is also supported at 30 m and 75 m marks. If the tape is used to measure a 543.25 long line, determine the total correction due to sag? The tape weighs 5 kg. 14.  It is required to lay out a distance of 687.78 m with a 50 m tape that is 0.030 too long. Compute the distance measured with the tape to make the  points the proper distance apart. 15.  A civil engineer used a 30 m tape in measuring an inclined distance. The measured length on the slope was recorded to be 459.20 m long. The difference in elevation between the initial point and the end point was found to be 1.25. The 30 m tape is of standard length at a temperature of 10°C and a pull of 50 N. During measurement the temperature reading was 15°C and the tape was supported at both ends with an applied pull of 75 2

 N. The cross-sectional areatape of the is 6.50 mm  and the modulus elasticity is 200 GPa. The hastape a linear density of 0.075 kg/m, α of = 0.0000116/°C.. Determine the horizontal distance. 0.0000116/°C

REVIEW EXERCISES:

1.  The observed compass 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. Fin the true azimuth of the line. 2.  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? AB?   3.  The interior angles of a five-side five-side traverse are as follows: A = 117°30’, B = 96°32’, C = 142°54’, and D = 132°18’. The angle at E is not measured.  measured.  a.  Compute the deflection angles.  b.  Calculate the bearings of the lines assuming AB due North. 4.  A triangular lot for one of its boundaries a line 1500 m long which runs due East and West. The eastern boundary is 900 m long and the western  boundary is 1200 m long. long. A straight line cutes the wester boundary at the middle point and meets the e asterly boundary 600 m from SE corner. Find the bearings and length of the line of the south portion of this triangular lot and give its technical description starting from the SW corner going counter clockwise.

 

  5.  The side AB of an equilateral field ABC with an area of 692.80 sq.m. has a magnetic bearing of N 48°45’ E in 1930 when the magnetic declination was 0°52’ E. Find the length and true bearing of the side AB. Find also the length and true bearing of line AD connecting corner A and point D on the line BC and making the area of the triangle ABD one third of the whole area. 6.  The following are bearings taken on a closed compass traverse. Compute the interior angles and correct them for observational errors. Assuming the observed bearing of line AB to be correct, adjust the bearings of the remaining sides. Line Forward Bearing Backward Bearing AB S 37°30’ E  E   N 37°30’ W  W  BC S 43°15’ W  W   N 44°15’ E  E  CD  N 73°00’ W  W  S 72°15’ E  E  DE  N 12°45’ E  E  S 13°15’ W  W  EA  N 60°00’ E  E  S 59°00’ W  W  7.  Given the following deflection angles of a closed traverse. Compute the  bearing of all the lines if the bearing of AB is 5.40° E. STATION A B C D

DEFLECTION ANGLES 85°20’ L  L  10°11’ R   83°32’ L  L  63°27’ L  L 

E 34°18’ L  L  F 72°56’ L  L  G 30°45’ L  L  8.  In 1925, the magnetic bearing of a line OA was N 15°45’ W, the magnetic declination at that time is 1°15’ E. The secular variation per year is 03’ E. What will be the declination of the needle and the magnetic bearing of the line in 1938? 9.  A field is in the form of a regular pentagon. The direction of the bounding sides was surveyed with an assumed meridian 5° to the right of the true north and south meridian. As surveyed with an assumed meridian, the  bearing of one side AB is N 33°20’ 33°20’ W. Find the true bearing and azimuth of all sides of the field.

10.  From the given data of a compass survey, compute the c orrected bearings of all the lines. Line AB

Forward Bearing  N 30°30’ W  W 

Backward Bearing S 32°15’ E  E 

BC S 80°45’ W  W   N 82°45’ E  E  CD S 53°00’ W  W   N 50°15’ E  E  DE S 13°00’ W  W   N 11°30’ E  E  EA  N 66°30’ E  E  S 69°30’ W  W  11.  Given the magnetic bearing AB = N 72° E, and the magnetic declination 3° W. Find the t rue bearing of AB. 12.  Given the bearings, OA = N 62°15’ E, and OB = N 81°30’ W. Find the angle AOB. 13.  Given the bearing of OC = S 10°14’ W and the clockwise angle COD = 83°17’.. Find the calculated bearing of OD. 83°17’ 14.  At a given place in 1875 the magnetic bearing of a line was N 89°15’ W, and the declination of the needle 5° W. At the present time the declination is 2°30’ W. What is the present magnetic bearing of the line? What is the t he true bearing of the line? 15.  The magnetic bearings of the sides of a field have been observed as follows: AB = S 25°30’ E; BC = S 12°00’ W; CD = S 68°15’ W; DA = N 18°45’ E. Find the interior angles.  angles.   16.  The interior angles of a field are as follows: A = 73°08’; B = 132°22’; C = 88°47’; and D = 65°43’. The magnetic bearing of AB = N 65°30’ E. If the direction of the taken to field? be clockwise, what are the calculated bearings of courses the otherissides of the 17.  In a survey the following magnetic bearings have been observed. AB = N 62°15’ E; BC = S 81°00’ E; CD = N 75°45’ E; DE = S 13°00’ W; and EF = S 0°30’ E. Find the deflection angles.  angles. 

 

  A –   B B S 35° 30’ W  W  44.37 m B –   C C  N 57° 15’ W  W  137.84 m C –   D D  N 1° 45’ E  E  12.83 m D –  E 64.86 m E –  A 106.72 m 4.  From the data shown below, determine t he value of the unknown bearing and distance.

REVIEW EXERCISES:

1.  From the field notes of a closed traverse shown below, adjust the transverse using. a.  Transit Rule

traverse, the field notes of which are the following:

 b.  Compass Rule c.  Compute the linear error of closure d.  Compute the relative error or precision Sta. Occ. A B C D E

Sta. Obs. B C D E A

Bearings Due North N 45° E S 60° E S 20° W S 86°59’ W  W 

Distances 400.00 m 800.00 m 700.00 m 600.00 m 966.34 m

2.  Find the bearing of line 4-5 and the missing side 5-1 of the closed traversed shown in the field notes s hown. Lines Bearings Distances 1 –   2 2 S 70° 15’ E  E  32.20 m 2 –   3 3 S 36° 30’ W  W  31.20 m 3 –   4 4  N 66° 30’ W  W  17.40 m 4 –  5 36.30 m 5 –   1 1  N 60° 00’ E  E  3.  Determine the bearings of lines 4-5 and 5-1 of the closed traverse shown the technical description of which is as follows: Lines

Bearing

Lines Bearings Distances AB  N 32° 27’ E  E  110.8 m BC 83.6 m CD S 8° 51’ W  W  126.9 m DE S 73° 31’ W  W  EA  N 18° 44’ W  W  90.2 m 5.  Determine the lengths of the two non-adjacent missing sides of a closed

Distances

Line Azimuth Distance 1 –   2 2 250° 55’  55’  437.26 m 2 –   3 3 354° 30’  30’  299.08 m 3 –   4 4 30° 44’  44’  4 –   5 5 86° 40’  40’  185.85 m 5 –   1 1 156° 19’  19’  6.  Find the unknown values of the foll owing notes of a transit survey. Linea Bearings Distances AB  N 48° 20’ E  E  529.60 m BC 592.00 m CD S 7° 59’ E  E  563.60 m DE 753.40 m EA  N 48° 12’ W  W  428.20 m 7.  In the survey of a closed lot with five sides, the following data are given where in all the bearings and distances of all sides except the lengths of lines 4-5 and 5-1 were omitted. Find the lengths of these two missing lines. Lines 1 –   2 2 2 –   3 3

Bearing S 73° 21’ E  E  S 40° 10’ E  E 

Distance 247.20 m 154.30 m

 

  3 –   4 4 4 –  5  5 5 –   1 1

S 26° 42’ W  W  N 14° 20’ W  W   N 12° 20’ E  E 

611.90 m -

REVIEW EXERCISES: 

1.  A differential leveling was run from BM 1 to BM2 which is approximately 4200 m from each other and closing the circuit on the same rout. The elevation of BM1 is 100 m above sea level and that of BM2 was found out to have an elevation of 142.53 m above sea l evel. However, in closing the circuit the computed elevation of BM 1  was 99.96 m only. What is the corrected elevation of BM 2? Supposing the computed elevation of BM 1 is 100.06 m, what is the corrected elevation of BM2? 2.  Compute the differential level notes shown and show the usual arithmetic check. STATION BS FS ELEVATION BM5  1.04 186.38 m 1 3.95 4.93 2 6.56 6.78 3 8.48 1.35 4 9.22 0.91 5 8.37 0.35 6 4.92 2.84 7 7.77 8.75 8 9.25 6.82 BM6  5.50 3.  In the plan below shows a differential leveling from benchmark to another  benchmark, along each l ine represents a sight in the actual rod reading. The direction of the fieldwork is indicated by the numbering of the TPs. Place the data in the form of level notes, compute the elevations, show the arithmetical check and record the error if there’s any.  any. 

4.  Below are the field notes of a profile leveling work where some data are missing accidentally. Go over the arithmetical computations and determine beck the missing data on broken lines. Station

Backsight

HI

BM1  1+000 1+100 1+200 1+300 1+400 PT1  1+500 1+600 1+700

1.56

--------

------

566.00

1+800 1+900 2+000 2+100 2+200 2+300 2+400 ------2+500 2+600

6.10

------

Foresight

Elevation

5.67 2.00 3.40 -----6.70 8.00 3.00 4.10 5.40

569.44 565.33 ------567.60 568.00 564.30 -----563.00 561.90 560.60

-----5.00 3.50 2.80 2.00 0.60 0.80 -----1.72 2.78

559.00 561.00 -----563.20 564.00 565.40 565.20 563.90 568.28 567.22

5.  Arrange the following description in the form of profile level notes complete the elevation and sketch the profile. A level is set up and reading

 

  of 2.995 m is taken on a benchmark the elevation of which is 12.135. At the beginning of the line to be profiled, the rod reading is 2.625 m; 30 m from the beginning, it is 1.617 m; at 60 m, it is 0.702 m; at 66 m and 81 m, the rod readings are 1.281 m and 0.762 m, respectively. On a rock t hat REVIEW EXERCISES: 

is thereading rod reading is 0.555 The level then ahead, setnot up on andline, a rod of 1.952 m is m. observed, theisrod stiremoved ll being held on the rock. The readings along the profile are then resumed: 90 m from the  beginning of the line, the rod reading is 1.159 m; 120 m from the beginning of the line the rod reading is 1.434 m; finally 150 m from the beginning of the line the rod reading is 2.196 m. 6.  A line of levels was run from BLLM No. 1 to BLLM No. 2 covering a route of approximately 5 km. Backsight and foresight distances every set up averages 100 m long. If at every TP the rod settl es about 0.04 m, what would be the corrected elevation of BLLM No. 2. The computed

1.  If the vertical angle from one station to another 100 m apart is 60°, the staff intercept for a tacheometer with K = 100 and C = 0, would be? 2.  With the transit at point B and the line of sight horizontal, the stadia intercept at C is 1.15 m. If the stadia interval factor is 100.32 and the stadia constant is 0.3, find the distance. 3.  The length intercepted on the stadia rod is 2.83 m and the line of sight makes an angle of 4°30’ with the horizontal. Find the vertical distance, from the center of the instrument to the rod, if the stadia constant is 0.3 m

difference in elevation between BLLM No. 1 and BLLM No. 2 is 100.00 m and the elevation of BLLM No. 1 is 60.00 m. 7.  Differential leveling are run from BM1  elevation 103.05 m to BM2, a distance of 30 km. The backsight distances are averages 150 m in length and the foresight distances averages 100 m in length. The elevation of BM 2  as deduced from the level notes was 420.50 m. If the level used is out of adjustment so that when the bubble was centered the line of sight was inclined 0.003 m upward in a dista nce of 100 m. What would be t he total error and the calculated value of the elevation of BM 2?

the stadia intervalthe factor is 100. 4.  and In order to determine stadia interval factor of a transit on the field, the transit was set up at a certain point on the ground. Observations were made on the rod placed at a distance of 240 m from the instrument and the rod readings were 4.505 and 2.105 for the upper and lower stadia hairs respectively. If instrument interval used has an interior focusing telescope, what would be the stadia interval factor of the instrument? 5.  The constant K and C for a certain instrument were 100 and 0 respectively. The ground makes a uniform slope of 12% from point A to point B. With the instrument at A and staff at B, readings were taken but due to obstruction in the line of sight, only t he upper reading was recorded to be 1.915 m. If the vertical angle of the instrument was 6°43’ and height of instrument above A was 1.82 m, determine the horizontal distance  between A and B. 6.  Given the following stadia level notes: K = 100 and C = 0 Station Observation Stadia Intercept Vertical Angle A 1.11 +3° on 1.50 m B 1.36 -5° on 1.70 m The instrument was set up at point C (point along line AB) with elevation 59 m. The instrument m. A Compute horizontal distance  between Aheight and B,ofthe elevation is of1.5 point and the the difference in elevation  between A and B.

 

  7.  The slope distance and vertical angle between points A and B were measured with a total station instrument as 9585.26 ft and 8°17’40”, 8°17’40” , respectively. The height of instrument and rod reading were equal. If the elevation of A is 1238.42 ft above the mean sea level, compute the elevation of B. 8.  The following tachemetric observations were made on two points P and Q from station A. The height of the tacheometer at A a bove the ground was 1.55 m. Elevation of A is 75.5 m. The stadia interval factor is 100 and the stadia constant is 0. Staff Reading Vertical Staff at Angle Upper Middle Lower P -5°12’ 5°12’   1.388 0.978 0.610 Q +27°35’   +27°35’ 1.604 1.286 0.997 Determine the elevation of P and Q.