Instructor’s Manual 25
Elementary Surveying: An Introduction to Geomatics
5 LEVELING — FIELD PROCEDURES AND
COMPUTATIONS
Asterisks (*) indicate problems that have answers given in Appendix G.
5.1 Explain the left-thumb rule when centering a level.
5.2 What is the difference between a benchmark and a turning point in differential leveling?
5.3 Discuss how stadia can be used to determine the plus and minus sight distances in
differential leveling.
From Section 5.4: “The stadia method determines the horizontal distance to points
5.4 What is the collimation error, and how can it be removed from the differential leveling
process.
5.5 Discuss how errors due to Earth curvature and refraction can be eliminated from the
differential leveling process.
From Section 5.4: “Balancing plus and minus sight distances will eliminate errors due
5.6 When is it appropriate to use the reciprocal leveling procedure?
5.7 List four considerations that govern a rodperson’s selection of TPs and BMs.
*5.8 What error is created by a rod leaning 10 min from plumb at a 12.51-ft reading on the
leaning rod?
5.9 Similar to Problem 5.6, except for a 5-m reading.
5.10 What error results on a 30-m sight with a level if the rod reading is 2.865 m but the top of
the 3 m rod is 0.3 m out of plumb?
32.865 = 2.8506 m
Error = 0.014 m
5.11 What error results on a 200-ft sight with a level if the rod reading is 6.307 ft but the top of
the 7-ft rod is 0.2 ft out of plumb?
76.307 = 6.3044
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or by any means, without permission in writing from the publisher.
5.12 Prepare a set of level notes for the data listed. Perform a check and adjust the misclosure.
Elevation of BM 7 is 2303.45 ft. If the total loop length is 2400 ft, what order of leveling
is represented? (Assume all readings are in feet)
POINT
S (BS)
S (FS)
BM 7
5.68
TP 1
9.42
7.58
TP 2
9.26
5.81
BM 8
6.45
4.59
TP 3
9.59
8.50
BM 7
13.95
Plus
HI
Minus
ELEV
5.68
2303.45
2309.13
(0.006)
(2301.556)
9.42
7.58
2301.55
2310.97
(0.012)
(2305.172)
9.26
5.81
2305.16
2314.42
(0.018)
(2309.848)
6.45
4.590
2309.83
2316.28
(0.024)
(2307.804)
9.59
8.500
2307.78
2317.37
(0.03)
(2303.45)
13.950
2303.42
40.40
40.43
2303.45+40.4-40.43 = 2303.42
Misclosure = 2303.42 – 2303.45 = –0.03
Correction = – –0.03/5 = 0.006
2400 ft ≈ 0.7315 km and 0.03 ft ≈ 9.1 mm; From Equation 5.3: 𝑚 = 12 √0.7315
⁄=
10.7 mm, Third Order
*5.13 Similar to Problem 5.12, except the elevation of BM 7 is 132.05 ft and the loop length 2400
ft. (Assume all readings are in feet)
STA
Plus
HI
Minus
ELEV
BM7
5.68
132.05
137.73
(0.006)
(130.156)
TP1
9.420
7.58
130.15
139.57
(0.012)
(133.772)
BM 8
9.26
5.81
133.76
143.02
(0.018)
(138.448)
TP2
6.45
4.590
138.43
© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all
5.24 Prepare a set of three-wire leveling notes for the data given and make the page check. The
elevation of BM X is 83.097 m. Rod readings (in meters) are (U denotes upper cross-wire
readings, M middle wire, and L lower wire): +S on BM X: U = 1.076, M = 0.829, L = 0.573;
−S on TP 1: U = 1.131, M = 0.858, L = 0.586; +S on TP 1: U = 0.989, M = 0.767, L = 0.555;
−S on BM Y: U = 1.089, M = 0.851, L = 0.611.
Sta
Plus
stadia
HI
Minus
stadia
Elev (m)
BM X
1.076
83.097
0.829
0.247
0.573
0.256
2.478
0.503
0.8260
83.923
TP1
0.989
1.131
83.0647
0.767
0.222
0.858
0.273
0.555
0.212
0.586
0.272
2.311
0.434
2.575
0.545
0.7703
0.8583
83.835
BM Y
1.089
82.9847
0.851
0.238
0.611
0.24
2.551
0.478
0.8503
1.5963
0.937
1.7087
1.023
Page check:
83.097 + 1.5963 – 1.7087 =
82.9847
5.25 Similar to Problem 5.24, except the elevation of BM X is 543.56 ft, and rod readings (in
feet) are: +S on BM X: U = 4.898, M = 4.146, L = 3.396; −S on TP 1: U = 5.875, M = 4.948,
L = 4.023; +S on TP 1: U = 6.504, M = 5.487, L = 4.472; −S on BM Y: U = 5.874, M =
5.026, L = 4.176.
Sta
Plus
stadia
HI
Minus
stadia
Elev (ft)
BM X
4.898
543.56
4.146
0.752
3.396
0.750
12.44
1.502
4.1467
547.7067
TP1
6.504
5.875
542.7580
5.487
1.017
4.948
0.927
4.472
1.015
4.023
0.925
© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all
copyright laws as they currently exist. No portion of this material may be reproduced, in any form
or by any means, without permission in writing from the publisher.
Instructor’s Manual 35
Elementary Surveying: An Introduction to Geomatics
*5.31 What is the percent grade between stations 11+00 and 16 + 00 in Problem 5.28?
1600 −1100 100% = +𝟐.𝟎%
5.32 Differential leveling between BMs A, B, C, D, and A gives elevation differences (in meters)
of −15.632, +32.458, +38.214 and −55.025, and distances in km of 2.0, 3.0, 1.0, and 4.0,
respectively. If the elevation of A is 634.597, compute the adjusted elevations of BMs B,
C, and D, and the order of leveling.
*5.35 A line of levels with 42 setups (84 rod readings) was run from BM Rock to BM Pond with
readings taken to the nearest 3.0 mm; hence any observed value could have an error of
1.5 mm.
For reading errors only, what total error would be expected in the elevation of
BM Pond?
5.36 Same as Problem 5.35, except for 28 setups and readings to the nearest 0.01 ft with possible
error of
0.005 ft
each.
5.37 Compute the permissible misclosure for the following lines of levels: (a) a 10-km loop of
third-order levels (b) a 20-km section of second-order class I levels (c) a 50-km loop of
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