SOLUTION MANUAL
, CONTENTS
SECTION A: Text Problem Solutions
Chapter 1 ....................................................................................... Page 1
Chapter 2 ....................................................................................... Page 2
Chapter 3 ....................................................................................... Page 8
Chapter 4 ....................................................................................... Page 12
Chapter 6 ....................................................................................... Page 16
Chapter 7 ....................................................................................... Page 24
Chapter 8 ....................................................................................... Page 27
Chapter 9 ....................................................................................... Page 30
Chapter 10 ..................................................................................... Page 34
Chapter 11 ..................................................................................... Page 37
Chapter 12 ..................................................................................... Page 39
Chapter 13 ..................................................................................... Page 42
Chapter 14 ..................................................................................... Page 49
,CHAPTER 01 – Basics of Surveying
1.1 How do plane surveys and geodetic surveys differ?
Plane surveying assumes all horizontal measurements are taken on a single plane and all vertical
measurements are relative to that plane, thereby allowing for all calculations to be done using plane
trigonometry. Geodetic surveys take into account the shape of the earth and generally bases positions on
ellipsoidal models of the earth (Datum).
1.2 Are preliminary (or data-gathering) surveys plane surveys or geodetic surveys? Explain your response.
Generally preliminary surveys are based on plane surveys. However some preliminary surveys are based
on geodetic surveys, because of the size of the area being surveyed and/or the equipment being used.
1.3 What kinds of data are collected during preliminary surveys?
The data being collected is positions of objects or points along objects. The objects may be man-made
features, such as edge of roads, building corners or water values; or they may be natural features such as
creeks, top of banks or grade breaks.
1.4 How is a total station different from an electronic theodolite?
An electronic theodolite only measures horizontal and vertical angles, while a total station also measures
slope distances and integrates that with the vertical angle to compute horizontal and vertical distances
1.5 Describe an integrated survey system(geomatics data model).
See Figure 1.7 on how an integrated survey system works.
1.6 Why is it that surveyors must measure or determine the horizontal distances, rather than just the slope
distances, when showing the relative locations of two points?
Horizontal distances are always determined so that all measurements are on the same plane. This allows
for a frame of reference (coordinate system) that can be used to check and adjust distance
measurements.
1.7 Describe how a very precise measurement can be inaccurate.
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.
1.8 Describe the term error; how does this term differ from mistake?
An error is any difference between the true value and the measured value. A mistake is one type of error
that is caused by human blunders.
1.9 Describe several different ways of locating a physical feature in the field so that it later can be plotted in its
correct position on a scaled plan.
A base line may be established and any feature may be located by measuring along and perpendicular to
the base line (Stationing). Distances and directions may be measured and then plotted from control points
(radial survey). Positions may be located using survey grade GNSS receivers and then those points plotted
based on their positions.
1
,Chapter 02 – Leveling
2.1 Compute the error due to curvature and refraction for the following distances:
(a) 500 ft (c+r) = 0.0206 x (500/1000)2 = 0.005 ft.
(b) 4,000 ft (c+r) = 0.0206 x (4)2 = 0 ft.
(c) 300 m (c+r) = 0.0675 x 0.3002 = 0.006 m.
(d) 2.2 mi (c+r) = 0.574 x 2.22 = 2.78 ft.
(e) 2,800 m (c+r) = 0.0675 x (2.8)2 = 0.529 m
(f) 3 km (c+r) = 0.0675 x (3)2 = 0.608m.
2.2 Determine the rod readings indicated on the foot and metric rod illustrations in Figure 2-32. The foot
readings are to the closest 0.01 ft, and the metric readings are to the closest one-half or one-third cm.
Rod A i 1.90 ii 1.73 iii 1.57 iv 1.21 v 1.03 (1.04)
Rod B i 1.185 ii 1.150 iii 1.040 iv 1.000 v 0.930
Rod C i 3.06 ii 2.85 (2.84) iii 2.57 (2.56) iv 2.21 v 1.92
Rod D i 1.145 ii 1.065 iii 1.000 iv 0.935 v 0.880
2.3 An offshore drilling rig is being towed out to sea. What is the maximum distance away that the navigation
lights can still be seen by an observer standing at the shoreline? The observer’s eye height is 5’ 0” and the
uppermost navigation light is 147 ft. above the water.
2
5.00 = .574 K1 , K1 = √5.00/.574 = 2.95 miles
2
147 = .574 K2 , K2 = √147/.574 = 16.00 miles
Maximum visibility distance = 18.95 miles
2.4 Prepare a set of level notes for the survey in Figure 2-33. Show the arithmetic check.
STATION BS HI IS FS ELEVATION
BM #50 1.27 390.34 389.07
TP #1 2.33 387.76 4.91 385.43
TP #2 6.17 381.59
BS = 3.60 FS = 11.08
389.07 +3.60 = 392.67 - 11.08 = 381.59 check
2.5 Prepare a set of profile leveling notes for the survey in Figure 2-34. In addition to computing all elevations,
show the arithmetic check and the resulting error in closure.
STATION BS HI IS FS ELEVATION
BM #61 4.72 401.46 396.74
0+00 4.42 397.04
0+50 4.30 394.16
TP #1 5.11 404.56 2.01 399.45
1+00 4.66 399.90
1+50 3.98 400.58
1+75 1.20 403.36
TP #2 1.80 402.76
BS = 9.83 FS = 3.81
E = - 0.02m [small error – no need for adjustments]
396.74 + 9.83 = 406.57 - 3.81 = 402.76 check
2
,2.6 Complete the set of differential leveling notes in Table 2-5, and perform the arithmetic check.
STATION BS HI FS ELEVATION
BM 100 2.71 317.59 314.88
TP 1 3.62 316.33 4.88 312.71
TP 2 3.51 315.87 3.97 312.36
TP 3 3.17 316.23 2.81 313.06
TP 4 1.47 316.08 1.62 314.61
BM 100 1.21 314.87
BS = 4.48 FS = 14.49
314.88 + 14.48 - 14.49 = 314.87, check
2.7 If the loop distance in Problem 2.6 is 1,000 ft,at what order of survey do the results qualify? Use Table 2-1 or
Table 2-2.
Error of closure = 0.01 ft.; for 1000 ft., second order (see Table 2.2) permits .035 √1000/5280 = 0.015;
therefore, results qualify for second order accuracy.
2.8 Reduce the set of differential leveling notes in Table 2-6, and perform the arithmetic check
STATION BS HI IS FS ELEVATION
BM 20 8.27 186.04 177.77
TP 1 9.21 192.65 2.60 183.44
0+00 11.3 181.4
0+50 9.6 183.1
0+61.48 8.71 246.65
1+00 6.1 249.3
TP 2 7.33 195.32 4.66 187.99
1+50 5.8 252.2
2+00 4.97 253.06
BM 21 3.88 191.44
BS =24.81 FS = 11.14
177.7 + 24.81 - 11.14 = 191.44 Check!
2.9 If the distance leveled in Problem 2.8 is 1,000 ft, for what order of survey do the results qualify if the
elevation of BM 21 is known to be 191.40? See Tables 2-1 and 2-2.
Error of closure = 0.04 ft.; for 1000 ft., third order (see Table 2.2) permits ± 0.10√1000/5280 = 0.044;
therefore results qualify for third order accuracy.
3
,2.10 Reduce the set of profile notes in Table 2-7, and perform the arithmetic check.
STATION BS HI IS FS ELEVATION
BM 22 1.203 182.425 181.222
0+00
CL 1.211 181.214
10M LT., 1.430 180.995
10M RT., 1.006 181.419
0+20
10M LT., 2.93 179.50
7.3M LT. 2.53 179.90
4M LT. 2.301 180.124
CL 2.381 180.044
4M RT. 2.307 180.118
7.8M RT. 2.41 180.02
10M RT. 2.78 179.65
0+40
10M LT. 3.98 178.45
6.2M LT. 3.50 178.9
4M LT. 3.103 179.322
CL 3.187 179.238
4M RT. 3.100 179.325
6.8M RT. 3.37 179.06
10M RT. 3.87 178.56
TP 1 2.773 179.65
2.11 Reduce the set of municipal cross-section notes in Table 2-8.
STATION BS HI IS FS ELEVATION
BM 41 4.11 307.104 302.994
TP 13 4.10 310.314 0.89 306.214
12+00
50 ft. lt. 3.9 306.4
18.3 ft. lt. 4.6 305.7
CL 6.33 303.98
20.1 ft. rt. 7.9 302.4
50 ft. rt. 8.2 302.1
13+00
50 ft. lt. 5.0 305.3
19.6 ft. lt 5.7 304.6
CL 7.54 302.77
20.7 ft. rt. 7.9 302.4
50 ft. rt. 8.4 301.9
TP 14 7.39 316.584 1.12 309.194
BM S.22 2.41 314.174
BS = 15.60 FS = 4.42
302.994 + 15.60 - 4.42 = 314.174 check!
4
, 2.12 Complete the set of highway cross-section notes in Table 2-9.
STATION BS HI FS ELEV. LEFT CL RIGHT
BM 37 7.20 385.17 377.97
50 26.7 28.4 50
5+50 4.6 3.8 3.7 3.0 2.7
380.6 381.4 381.5 382.2 382.5
50 24.1 25.0 50
6+00 4.0 4.2 3.1 2.7 2.9
381.2 381.0 382.1 382.5 382.3
50 26.4 23.8 50
6+50 3.8 3.7 2.6 1.7 1.1
381.4 381.5 382.6 383.5 384.1
TP 1 6.71 378.46
2.13 Complete the set of highway cross-section notes in Table 2-10.
STATION BS HI FS ELEV. LEFT CL RIGHT
BM 107 7.71 406.87 399.16
60 28 32 60
80+50 9.7 8.0 5.7 4.3 4.0
397.2 398.9 401.2 402.6 402.9
60 25 30 60
81+00 10.1 9.7 6.8 6.0 5.3
396.8 397.2 400.1 400.9 401.6
60 27 33 60
81+50 11.7 11.0 9.2 8.3 8.0
395.2 395.9 397.7 398.6 398.9
TP 1 10.17 396.70
2.14 A level is set up midway between two wood stakes that are about 300 ft apart. The rod reading on stake A is
8.72 ft, and it is 5.61 ft on stake B. The level is then moved to point B and set up about 6 ft or 2 m away. A
reading of 5.42 ft is taken on the rod at B. The level is then sighted on the rod held on stake A, where a
Reading of 8.57 ft is noted.
(a) What is the correct difference in elevation between the tops of stakes A and B?
(b) If the level had been in perfect adjustment, what reading would have been observed at A from the second
setup?
(c) What is the line-of-sight error in 300 ft?
(d) Describe how you would eliminate the line-of-sight error from the telescope.
a) True difference = 8.72 - 5.61 = 3.11 ft.
b) Correct rod reading = 5.42 + 3.11 = 8.53 ft.; on A
c) Error is +0.04 in 300 ft., or .00001 ft/ft
d) Cross hair adjusted downward from 8.57 to read 8.53, on A
5
, 2.15 A pre-engineering baseline was run down a very steep hill (see Figure 2-35). Rather than measure horizontally
downhill with the steel tape, the surveyor measures the vertical angle with a theodolite and the slope
distance with a 200-ft steel tape. The vertical angle is -21° 26’ turned to a point on a plumbed range pole
that is 4.88 ft above the ground. The slope distance from the theodolite to the point on the range pole is
148.61 ft. The theodolite’s optical center is 4.66 ft above the upper baseline station at 110 + 71.25.
(a) If the elevation of the upper station is 318.71, what is the elevation of the lowerstation?
(b) What is the stationing chainage of the lower station?
a) V=148.61 Sin 21O26' = 54.30 ft Elevation of lower station = 318.71 +4.66 -54.30 -4.88 = 264.19 ft.
b) H = 148.61 Cos (21° 26') = 138.33 ft lower station at 110+71.25 +138.33 = 112+09.58
2.16 You must establish the elevation of point B from point A (elevation 216.612 m). A and B are on opposite sides
of a 12-lane highway. Reciprocal leveling is used, with the following results:
Setup at A side of highway:
Rod reading on A = 0.673 m
Rod readings on B = 2.416 and 2.418 m
Setup at B side of highway:
Rod reading on B = 2.992 m
Rod readings on A = 1.254 and 1.250 m
(a) What is the elevation of point B?
(b) What is the leveling error?
a) First elevation difference = 2.417- 0.673 = 1.744
Second elevation difference = 2.992 - 1.252 = 1.740
Average elevation difference = 1.742
Elevation B = 216.612-1.742 = 214.870
b) The leveling error is 0.004m
2.17 Reduce the set of differential leveling notes inTable 2-11, and perform the arithmetic check.
(a) Determine the order of accuracy (seeTable 2-1 or Table 2-2).
(b) Adjust the elevation of BM K110. The length of the level run was 780 m, with setups that are equally spaced.
The elevation of BM 132 is 187.536 m.
STATION BS HI FS ELEVATION
BM 130 0.702 189.269 188.567
TP 1 0.970 189.128 1.111 188.158
TP 2 0.559 189.008 0.679 188.449
TP 3 1.744 187.972 2.780 186.228
BM K110 1.973 188.277 1.668 186.304
TP 4 1.927 188.416 1.788 186.489
BM 132 0.888 187.528
BS = 7.875 FS = 8.914
188.567 + 7.875 - 8.914 = 187.528, check
a) error = 187.536 – 187.528 = - 0.008 m.
Using specifications from Table 2.1, Third order accuracy, allowable error = .012√.780 = 0.011 m. This
error of 0.008 thus qualifies for third order accuracy (in both Tables 2.1 and 2.2)
6
, CONTENTS
SECTION A: Text Problem Solutions
Chapter 1 ....................................................................................... Page 1
Chapter 2 ....................................................................................... Page 2
Chapter 3 ....................................................................................... Page 8
Chapter 4 ....................................................................................... Page 12
Chapter 6 ....................................................................................... Page 16
Chapter 7 ....................................................................................... Page 24
Chapter 8 ....................................................................................... Page 27
Chapter 9 ....................................................................................... Page 30
Chapter 10 ..................................................................................... Page 34
Chapter 11 ..................................................................................... Page 37
Chapter 12 ..................................................................................... Page 39
Chapter 13 ..................................................................................... Page 42
Chapter 14 ..................................................................................... Page 49
,CHAPTER 01 – Basics of Surveying
1.1 How do plane surveys and geodetic surveys differ?
Plane surveying assumes all horizontal measurements are taken on a single plane and all vertical
measurements are relative to that plane, thereby allowing for all calculations to be done using plane
trigonometry. Geodetic surveys take into account the shape of the earth and generally bases positions on
ellipsoidal models of the earth (Datum).
1.2 Are preliminary (or data-gathering) surveys plane surveys or geodetic surveys? Explain your response.
Generally preliminary surveys are based on plane surveys. However some preliminary surveys are based
on geodetic surveys, because of the size of the area being surveyed and/or the equipment being used.
1.3 What kinds of data are collected during preliminary surveys?
The data being collected is positions of objects or points along objects. The objects may be man-made
features, such as edge of roads, building corners or water values; or they may be natural features such as
creeks, top of banks or grade breaks.
1.4 How is a total station different from an electronic theodolite?
An electronic theodolite only measures horizontal and vertical angles, while a total station also measures
slope distances and integrates that with the vertical angle to compute horizontal and vertical distances
1.5 Describe an integrated survey system(geomatics data model).
See Figure 1.7 on how an integrated survey system works.
1.6 Why is it that surveyors must measure or determine the horizontal distances, rather than just the slope
distances, when showing the relative locations of two points?
Horizontal distances are always determined so that all measurements are on the same plane. This allows
for a frame of reference (coordinate system) that can be used to check and adjust distance
measurements.
1.7 Describe how a very precise measurement can be inaccurate.
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.
1.8 Describe the term error; how does this term differ from mistake?
An error is any difference between the true value and the measured value. A mistake is one type of error
that is caused by human blunders.
1.9 Describe several different ways of locating a physical feature in the field so that it later can be plotted in its
correct position on a scaled plan.
A base line may be established and any feature may be located by measuring along and perpendicular to
the base line (Stationing). Distances and directions may be measured and then plotted from control points
(radial survey). Positions may be located using survey grade GNSS receivers and then those points plotted
based on their positions.
1
,Chapter 02 – Leveling
2.1 Compute the error due to curvature and refraction for the following distances:
(a) 500 ft (c+r) = 0.0206 x (500/1000)2 = 0.005 ft.
(b) 4,000 ft (c+r) = 0.0206 x (4)2 = 0 ft.
(c) 300 m (c+r) = 0.0675 x 0.3002 = 0.006 m.
(d) 2.2 mi (c+r) = 0.574 x 2.22 = 2.78 ft.
(e) 2,800 m (c+r) = 0.0675 x (2.8)2 = 0.529 m
(f) 3 km (c+r) = 0.0675 x (3)2 = 0.608m.
2.2 Determine the rod readings indicated on the foot and metric rod illustrations in Figure 2-32. The foot
readings are to the closest 0.01 ft, and the metric readings are to the closest one-half or one-third cm.
Rod A i 1.90 ii 1.73 iii 1.57 iv 1.21 v 1.03 (1.04)
Rod B i 1.185 ii 1.150 iii 1.040 iv 1.000 v 0.930
Rod C i 3.06 ii 2.85 (2.84) iii 2.57 (2.56) iv 2.21 v 1.92
Rod D i 1.145 ii 1.065 iii 1.000 iv 0.935 v 0.880
2.3 An offshore drilling rig is being towed out to sea. What is the maximum distance away that the navigation
lights can still be seen by an observer standing at the shoreline? The observer’s eye height is 5’ 0” and the
uppermost navigation light is 147 ft. above the water.
2
5.00 = .574 K1 , K1 = √5.00/.574 = 2.95 miles
2
147 = .574 K2 , K2 = √147/.574 = 16.00 miles
Maximum visibility distance = 18.95 miles
2.4 Prepare a set of level notes for the survey in Figure 2-33. Show the arithmetic check.
STATION BS HI IS FS ELEVATION
BM #50 1.27 390.34 389.07
TP #1 2.33 387.76 4.91 385.43
TP #2 6.17 381.59
BS = 3.60 FS = 11.08
389.07 +3.60 = 392.67 - 11.08 = 381.59 check
2.5 Prepare a set of profile leveling notes for the survey in Figure 2-34. In addition to computing all elevations,
show the arithmetic check and the resulting error in closure.
STATION BS HI IS FS ELEVATION
BM #61 4.72 401.46 396.74
0+00 4.42 397.04
0+50 4.30 394.16
TP #1 5.11 404.56 2.01 399.45
1+00 4.66 399.90
1+50 3.98 400.58
1+75 1.20 403.36
TP #2 1.80 402.76
BS = 9.83 FS = 3.81
E = - 0.02m [small error – no need for adjustments]
396.74 + 9.83 = 406.57 - 3.81 = 402.76 check
2
,2.6 Complete the set of differential leveling notes in Table 2-5, and perform the arithmetic check.
STATION BS HI FS ELEVATION
BM 100 2.71 317.59 314.88
TP 1 3.62 316.33 4.88 312.71
TP 2 3.51 315.87 3.97 312.36
TP 3 3.17 316.23 2.81 313.06
TP 4 1.47 316.08 1.62 314.61
BM 100 1.21 314.87
BS = 4.48 FS = 14.49
314.88 + 14.48 - 14.49 = 314.87, check
2.7 If the loop distance in Problem 2.6 is 1,000 ft,at what order of survey do the results qualify? Use Table 2-1 or
Table 2-2.
Error of closure = 0.01 ft.; for 1000 ft., second order (see Table 2.2) permits .035 √1000/5280 = 0.015;
therefore, results qualify for second order accuracy.
2.8 Reduce the set of differential leveling notes in Table 2-6, and perform the arithmetic check
STATION BS HI IS FS ELEVATION
BM 20 8.27 186.04 177.77
TP 1 9.21 192.65 2.60 183.44
0+00 11.3 181.4
0+50 9.6 183.1
0+61.48 8.71 246.65
1+00 6.1 249.3
TP 2 7.33 195.32 4.66 187.99
1+50 5.8 252.2
2+00 4.97 253.06
BM 21 3.88 191.44
BS =24.81 FS = 11.14
177.7 + 24.81 - 11.14 = 191.44 Check!
2.9 If the distance leveled in Problem 2.8 is 1,000 ft, for what order of survey do the results qualify if the
elevation of BM 21 is known to be 191.40? See Tables 2-1 and 2-2.
Error of closure = 0.04 ft.; for 1000 ft., third order (see Table 2.2) permits ± 0.10√1000/5280 = 0.044;
therefore results qualify for third order accuracy.
3
,2.10 Reduce the set of profile notes in Table 2-7, and perform the arithmetic check.
STATION BS HI IS FS ELEVATION
BM 22 1.203 182.425 181.222
0+00
CL 1.211 181.214
10M LT., 1.430 180.995
10M RT., 1.006 181.419
0+20
10M LT., 2.93 179.50
7.3M LT. 2.53 179.90
4M LT. 2.301 180.124
CL 2.381 180.044
4M RT. 2.307 180.118
7.8M RT. 2.41 180.02
10M RT. 2.78 179.65
0+40
10M LT. 3.98 178.45
6.2M LT. 3.50 178.9
4M LT. 3.103 179.322
CL 3.187 179.238
4M RT. 3.100 179.325
6.8M RT. 3.37 179.06
10M RT. 3.87 178.56
TP 1 2.773 179.65
2.11 Reduce the set of municipal cross-section notes in Table 2-8.
STATION BS HI IS FS ELEVATION
BM 41 4.11 307.104 302.994
TP 13 4.10 310.314 0.89 306.214
12+00
50 ft. lt. 3.9 306.4
18.3 ft. lt. 4.6 305.7
CL 6.33 303.98
20.1 ft. rt. 7.9 302.4
50 ft. rt. 8.2 302.1
13+00
50 ft. lt. 5.0 305.3
19.6 ft. lt 5.7 304.6
CL 7.54 302.77
20.7 ft. rt. 7.9 302.4
50 ft. rt. 8.4 301.9
TP 14 7.39 316.584 1.12 309.194
BM S.22 2.41 314.174
BS = 15.60 FS = 4.42
302.994 + 15.60 - 4.42 = 314.174 check!
4
, 2.12 Complete the set of highway cross-section notes in Table 2-9.
STATION BS HI FS ELEV. LEFT CL RIGHT
BM 37 7.20 385.17 377.97
50 26.7 28.4 50
5+50 4.6 3.8 3.7 3.0 2.7
380.6 381.4 381.5 382.2 382.5
50 24.1 25.0 50
6+00 4.0 4.2 3.1 2.7 2.9
381.2 381.0 382.1 382.5 382.3
50 26.4 23.8 50
6+50 3.8 3.7 2.6 1.7 1.1
381.4 381.5 382.6 383.5 384.1
TP 1 6.71 378.46
2.13 Complete the set of highway cross-section notes in Table 2-10.
STATION BS HI FS ELEV. LEFT CL RIGHT
BM 107 7.71 406.87 399.16
60 28 32 60
80+50 9.7 8.0 5.7 4.3 4.0
397.2 398.9 401.2 402.6 402.9
60 25 30 60
81+00 10.1 9.7 6.8 6.0 5.3
396.8 397.2 400.1 400.9 401.6
60 27 33 60
81+50 11.7 11.0 9.2 8.3 8.0
395.2 395.9 397.7 398.6 398.9
TP 1 10.17 396.70
2.14 A level is set up midway between two wood stakes that are about 300 ft apart. The rod reading on stake A is
8.72 ft, and it is 5.61 ft on stake B. The level is then moved to point B and set up about 6 ft or 2 m away. A
reading of 5.42 ft is taken on the rod at B. The level is then sighted on the rod held on stake A, where a
Reading of 8.57 ft is noted.
(a) What is the correct difference in elevation between the tops of stakes A and B?
(b) If the level had been in perfect adjustment, what reading would have been observed at A from the second
setup?
(c) What is the line-of-sight error in 300 ft?
(d) Describe how you would eliminate the line-of-sight error from the telescope.
a) True difference = 8.72 - 5.61 = 3.11 ft.
b) Correct rod reading = 5.42 + 3.11 = 8.53 ft.; on A
c) Error is +0.04 in 300 ft., or .00001 ft/ft
d) Cross hair adjusted downward from 8.57 to read 8.53, on A
5
, 2.15 A pre-engineering baseline was run down a very steep hill (see Figure 2-35). Rather than measure horizontally
downhill with the steel tape, the surveyor measures the vertical angle with a theodolite and the slope
distance with a 200-ft steel tape. The vertical angle is -21° 26’ turned to a point on a plumbed range pole
that is 4.88 ft above the ground. The slope distance from the theodolite to the point on the range pole is
148.61 ft. The theodolite’s optical center is 4.66 ft above the upper baseline station at 110 + 71.25.
(a) If the elevation of the upper station is 318.71, what is the elevation of the lowerstation?
(b) What is the stationing chainage of the lower station?
a) V=148.61 Sin 21O26' = 54.30 ft Elevation of lower station = 318.71 +4.66 -54.30 -4.88 = 264.19 ft.
b) H = 148.61 Cos (21° 26') = 138.33 ft lower station at 110+71.25 +138.33 = 112+09.58
2.16 You must establish the elevation of point B from point A (elevation 216.612 m). A and B are on opposite sides
of a 12-lane highway. Reciprocal leveling is used, with the following results:
Setup at A side of highway:
Rod reading on A = 0.673 m
Rod readings on B = 2.416 and 2.418 m
Setup at B side of highway:
Rod reading on B = 2.992 m
Rod readings on A = 1.254 and 1.250 m
(a) What is the elevation of point B?
(b) What is the leveling error?
a) First elevation difference = 2.417- 0.673 = 1.744
Second elevation difference = 2.992 - 1.252 = 1.740
Average elevation difference = 1.742
Elevation B = 216.612-1.742 = 214.870
b) The leveling error is 0.004m
2.17 Reduce the set of differential leveling notes inTable 2-11, and perform the arithmetic check.
(a) Determine the order of accuracy (seeTable 2-1 or Table 2-2).
(b) Adjust the elevation of BM K110. The length of the level run was 780 m, with setups that are equally spaced.
The elevation of BM 132 is 187.536 m.
STATION BS HI FS ELEVATION
BM 130 0.702 189.269 188.567
TP 1 0.970 189.128 1.111 188.158
TP 2 0.559 189.008 0.679 188.449
TP 3 1.744 187.972 2.780 186.228
BM K110 1.973 188.277 1.668 186.304
TP 4 1.927 188.416 1.788 186.489
BM 132 0.888 187.528
BS = 7.875 FS = 8.914
188.567 + 7.875 - 8.914 = 187.528, check
a) error = 187.536 – 187.528 = - 0.008 m.
Using specifications from Table 2.1, Third order accuracy, allowable error = .012√.780 = 0.011 m. This
error of 0.008 thus qualifies for third order accuracy (in both Tables 2.1 and 2.2)
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