# Elementary Surveying: An Introduction to Geomatics, 15th Edition Solution Manual

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PART I:
SOLUTIONS TO PROBLEMS
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Instructorโs Solution Manual
Elementary Surveying: An Introduction to Geomatics
1
INTRODUCTION
NOTE: Answers for some of these problems, and some in later chapters, can be obtained by
consulting the bibliographies, later chapters, websites, or professional surveyors.
1.1
List 10 uses for surveying in areas other than boundary surveying.
Answers may vary many are included in Section 1.6, which lists control, topographic
hydrographic, alignment, construction, as-built, mine, solar, optical tooling, ground,
aerial, and satellite surveys. This list is not complete and could also include other types of
surveys such as hydrographic surveys, for example.
1.2
Explain the difference between geodetic and plane surveys.
From Section 1.4:
In geodetic surveys the curved surface of the earth is considered by performing the
computations on an ellipsoid (curve surface approximating the size and shape of the
earth). In plane surveys, except for leveling, the reference base for fieldwork and
computations is assumed to be a flat horizontal surface. The direction of a plumb line
(and thus gravity) is considered parallel throughout the survey region, and all measured
angles are presumed to be plane angles.
1.3
Describe some surveying applications in:
(a) Archeology
There are many different uses of surveying in archeology. Some include using sonar
to identify possible underground or underwater archeology sites, LiDAR to help
identify possible ancient human settlements in unexplored forest and jungles, and
traditional surveying and laser scanning to help locate artifacts in site excavations.
(b) Gas exploration
There are several stages of surveying in gas exploration, which include but are not
limited to determining anomalies in the gravity field, which identify possible gas
deposits, boundary surveys identifying properties that have mineral rights to the gas
deposits, alignment surveys for placement of pipelines to transport extracted gas.
(c) Agriculture
In agriculture, surveying is used to determine the acreage of fields, to locate lines of
constant elevation for strip farming, to track harvesting machinery to enable the size
of the harvest, and to track the position of the planting equipment to allow for precise
applications of seeds and fertilizers. The field is known as high-precision agriculture.
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1.4
List some application of surveying in geology, forestry, and archeology.
Applications in each are multiple. For some in geology and archeology see the answer to
Problem 1.3 (a) and (b). Some uses of surveying in forestry identifying forest boundaries,
locating spread of diseases and insects through remote sensing, using GIS to help
inventory and keep records on resources in forested regions.
1.5
Why is it important to make accurate surveys of underground utilities?
To provide an accurate record of the locations of these utilities so they can be found if
repairs or servicing is needed, and to prevent their accidental destruction during
excavation for other projects.
1.6
Discuss the uses for topographic surveys.
Topographic surveys are used whenever elevation data is required in the end product.
Some examples include (1) creating maps for highway design; (2) creating maps for
construction surveys; (3) creating maps for flood plain delineation; (4) creating maps for
site location of buildings; and so on.
1.7
What are hydrographic surveys, and why are they important?
From Section 1.6, hydrographic surveys define shorelines and depths of lakes, streams,
oceans, reservoirs, and other bodies of water. Sea surveying is associated with port and
offshore industries and the marine environment, including measurements and marine
investigations made by ship borne personnel.
1.8
Print a view of your location using Google Earth.ยฎ
Answers will vary but should be an image in your region.
1.9
Briefly explain the procedure used by Eratosthenes in determining the Earthโs
circumference.
From Section 1.3, paragraph 8 of text: His procedure, which occurred about 200 B.C., is
illustrated in Figure 1.3. Eratosthenes had concluded that the Egyptian cities of
Alexandria and Syene were located approximately on the same meridian, and he had also
observed that at noon on the summer solstice, the sun was directly overhead at Syene.
(This was apparent because at that time of that day, the image of the sun could be seen
reflecting from the bottom of a deep vertical well there.) He reasoned that at that moment,
the sun, Syene, and Alexandria were in a common meridian plane, and if he could
measure the arc length between the two cities, and the angle it subtended at the earth’s
center, he could compute the earth’s circumference. He determined the angle by
measuring the length of the shadow cast at Alexandria from a tall vertical staff of known
length. The arc length was found from multiplying the number of caravan days between
Syene and Alexandria by the average daily distance traveled. From these measurements
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Elementary Surveying: An Introduction to Geomatics
Eratosthenes calculated the earth’s circumference to be about 25,000 mi. Subsequent
precise geodetic measurements using better instruments, but techniques similar
geometrically to Eratosthenes’, have shown his value, though slightly too large, to be
amazingly close to the currently accepted one.
1.10 Describe the steps a land surveyor would need to do when performing a boundary survey.
Briefly, the steps should include (1) preliminary walking of property with owner; (2)
courthouse research to locate deed of property and adjoiners to determine ownership,
possible easements, right-of-ways, conflicts of interest, and so on; (3) location survey of
property noting any encroachments; conflicting elements; and so on; (4) resolution of
conflicting elements between deed and survey; (5) delivery of surveying report to owner.
1.11 What is the name of the state-level professional surveying organization in your state or
region?
Answer will vary by location.
1.12 What organizations in your state furnish maps and reference data to surveyors and
engineers?
Responses will vary but some common organizations are the (1) county surveyor, (2)
register of deeds, (3) county engineer or county highway department (4) Department of
Transportation, (5) Department of Natural Resources of its equivalent, and so on.
1.13 List the legal requirements for registration as a land surveyor in your state.
Responses will vary. Contact with your licensing board can be found on the NCEES
website at http://www.ncees.org/licensure/licensing_boards/.
1.14 Briefly describe an Earth-Centered, Earth-Fixed coordinate system.
From Section 1.4 and 13.4.3, a ECEF coordinate system is an Earth-based threedimensional coordinate system with its origin a the mass-center of the Earth, it Z axis
aligned with the semi-minor (spin) axis of the Earth defined at some epoch, it X axis in
the plane of the equator passing through mean Greenwich meridian, and it Y axis in the
plane of the equator and creating a right-handed coordinate system. At this stage of their
introduction to surveying it should be sufficient for students to simply know that it is an
Earth-based three-dimensional coordinate system.
1.15 List the professional societies representing the geospatial industry in the
(a) United States.
There are several including AAGS, ASCE, ASPRS, GLIS, NSPS, and SaGES.
(b) Canada.
Canadian Institute of Geomatics (CIG)
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(c) International.
International Federation of Surveyors (FIG)
1.15 Explain how aerial photographs and satellite images can be valuable in surveying.
Photogrammetry presently has many applications in surveying. It is used, for example, in
land surveying to compute coordinates of section corners, boundary corners, or point of
evidence that help locate these corners. Largeโscale maps are made by photogrammetric
procedures for many uses, one being subdivision design. Photogrammetry is used to map
shorelines, in hydrographic surveying, to determine precise ground coordinates of points
in control surveying, and to develop maps and cross sections for route and engineering
surveys. Photogrammetry is playing an important role in developing the necessary data
for modern Land and Geographic Information Systems.
1.16 Search the Internet and define a Very Long Baseline Interferometry (VLBI) station.
Discuss why these stations are important to the geospatial industry.
VLBI stands for Very Long Baseline Interferometry. Responses will vary. These stations
provide extremely accurate locations on the surface of the Earth. The stations are used to
develop world-wide reference frameworks such as ITRF08 and thus provide a worldwide
coordinate system that links continents. They also may provide tracking information for
satellites.
1.17 Describe how a GIS can be used in flood emergency planning.
Responses will vary but may mention the capabilities of a GIS to overlay soil type and
their permeability with slopes, soil saturation, and watershed regions. A GIS can also be
used to provide a list of business and residences that will be affected by possible flooding
for evacuation purposes. It can provide โbestโ routes out of a flooded region.
1.19 Visit one of the surveying websites listed in Table 1.1, and write a brief summary of its
contents. Briefly explain the value of the available information to surveyors.
Responses will vary with time, but below are brief responses to the question
ยท NGS โ control data sheets, CORS data, surveying software
ยท USGS โ maps, software
ยท BLM โ cadastral maps, software, ephemerides
ยท U.S. Coast Guard Navigation Center – GPS information
ยท U.S. Naval Observatory โNotice Advisory for NAVSTAR Users (NANU) and other
GPS related links
ยท National Society of Professional Surveyors โ professional organization for boundary
and construction
ยท American Association for Geodetic Surveying โ professional organization for control
surveying
ยท Geographic and Land Information Society โ professional organization for developers
and users of geographic and land information systems
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Instructorโs Solution Manual
Elementary Surveying: An Introduction to Geomatics
ยท American Society for Photogrammetry and Remote Sensing โ professional
organization for photogrammetry and remote sensing
ยท The Pearson Prentice Hall publishers access to software and support materials that
accompany this book.
ยท SaGES โ An organization to advance surveying/geomatics education
1.20 Read one of the articles cited in the bibliography for this chapter, or another of your
choosing, that describes an application where satellite surveying methods were used.
Write a brief summary of the article.
Answer will vary.
1.21 Same as Problem 1.20, except the article should be on safety as related to surveying.
Answers will vary but should be related to safety issues in surveying.
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1.20 Read one of the articles cited in the bibliography for this chapter, or another of your
choosing, that describes an application where satellite surveying methods were used.
Write a brief summary of the article.
Answers will vary. Students should be told to look in trade journals for articles.
1.21 Same as Problem 1.20, except the article should be on safety as related to surveying.
Answers will vary. Students should be told to look in trade journals for articles.
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2 UNITS, SIGNIFICANT FIGURES, AND FIELD NOTES
2.1
List the five types of measurements that form the basis of traditional plane surveying.
From Section 2.1, they are (1) horizontal angles, (2) horizontal distances, (3) vertical
(altitude or zenith) angles, (4) vertical distances, and (5) slope (or slant) distances.
2.2
Give the basic units that are used in surveying for length, area, volume, and angles in
(a) The English system of units.
From Section 2.2:
length (U.S. survey ft or in some states international foot), area (sq. ft. or acres),
volume (cu. ft. or cu. yd.), angle (sexagesimal)
(b) The SI system of units.
From Section 2.3:
length (m), area (sq. m. or hectare), volume (cu. m.), angle (sexagesimal, grad, or
radian)
2.3
The easting coordinate for a point is 725,316.911 m. What is the coordinate using the
(a) Survey foot definition?
(b) International foot definition?
(c) Why was the survey foot definition maintained in the United States?
39.37
(a) 2,379,643.90 sft; 725,316.911 ( 12 ) = 2,379,643.899 sft
(b) 2,379,648,66 ft; 725,316.911/0.3048 = 2,379,648.658 ft
(c) From Section 2.2: โBecause of the vast number of surveys performed prior to 1959,
it would have been extremely difficult and confusing to change all related
documents and maps that already existed. Thus the old standard, now called the U.S.
survey foot, is still used.โ
2.4
Convert the following distances given in meters to U.S. survey feet:
*(a) 4129.574 m 13,548.44 sft
(b) 588.234 m 1929.90 sft
(c) 102,302.103 m 335,636.15 sft
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2.5
2.6
2.7
2.8
2.9
Convert the following distances given in survey feet to meters:
*(a) 537.52 sft
163.836 m
(b)
2,405,687.82 sft
733,255.114 m
(c)
5783.12 sft
1762.699 m or 1762.70 m
Compute the lengths in survey feet corresponding to the following distances measured
with a Gunterโs chain:
*(a) 10 ch 13 lk
668.6 sft
(b) 56 ch 83 lk
3750.8 sft
(c) 124 ch 35 lk
8207.1 sft
Express 5,377,700 sft2 in:
*(a) acres
123.46 ac
(b)
hectares
49.961 ha
(c)
square Gunterโs chains
1234.6 sq. ch.
Convert 23.4587 ha to:
(a) square survey feet
2,525,070 sft2
(b)
acres
57.9676 ac
(b)
square Gunterโs chains
579.676 sq. ch
What are the lengths in feet and decimals for the following distances shown on a
building blueprint:
(a) 12 ft 6-1/4 in. 12.5 ft
601/4/12
(b) 10 ft 6-1/2 in.
2.10
10.5 ft
253/2/12
What is the area in acres of a rectangular parcel of land measured with a Gunterโs chain
if the recorded sides are as follows:
*(a) 9.17 ch and 10.64 ch
9.76 ac
(b) 16 ch 78 lk and 52 ch 49 lk
2.11
Compute the area in acres of triangular lots shown on a plat having the following
recorded right-angle sides:
(a) 335.36 ft and 804.02 ft
3.0945 ac
(b)
2.12
93.064 m and 30.346 m
0.69785 ac
A distance is expressed as 9756.12 sft. What is the length in
*(a) international feet?
9756.14 ft
(b)
2.13
88.08 ac
meters?
2973.67 m
What are the radian and degree-minute-second equivalents for the following angles
given in grads:
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2.14
2.15
2.16
(a)*
136.000 grads
???ยฐ??โฒ??”; 2.13628 rad
(b)
115.089 grads ???ยฐ??โฒ??โณ; 1.80781 rad
(c)
363.809 grads ???ยฐ??โฒ??โณ; 5.71469 rad
Give answers to the following problems in the correct number of significant figures:
*(a) sum of 23.15, 0.984, 124, and 12.5
161
(b)
sum of 2.115, 23.04, 13.8, and 199.66
238.6
(c)
product of 127.08 and 13.1
1660
(d)
quotient of 4466.83 divided by 35.61
125.4
Express the value or answer in powers of 10 to the correct number of significant figures:
(a) 4586.49
4.58679 ร 103
(b)
2450
2.45 ร 103
(c)
square of 199.99
3.9996 ร 104
(d)
sum of (32.087 + 1.56 + 206.44) divided by 2.3
1.95 ร 101
Convert the angles of a triangle to radians and show a computational check:
*(a) 39ยฐ41โฒ 54, 91ยฐ30’16”, 48ยฐ47โฒ50″ 0.692867, 1.59705, and 0.851672
0.6928666 + 1.597054 + 0.8516721 = 3.14159 check
(b) 96ยฐ23’18, 44ยฐ56โฒ53″, 38ยฐ39โฒ 49″ 1.68229, 0.784492, and 0.674807
1.682294 + 0.784492 + 0.674807 = 3.14159 check
2.17
Why should a ball point pen not be used in field notekeeping?
From Section 2.7: “Books so prepared (with 3h or higher pencil) will withstand damp
weather in the field (or even a soaking) and still be legible, whereas graphite from a
soft pencil, or ink from a pen or ballpoint, leaves an undecipherable smudge under
such circumstances.”
2.18
Explain why one number should not be superimposed over another or the lines of
sketches.
From Section 2.7: This can be explained with the need for integrity since it would
raise the issue of what are you hiding, legibility since the numbers are often hard to
interpret when so written, or by clarity since the notes are being crowded.
2.19
Explain why data should always be entered directly into the field book at the time
measurements are made, rather than on scrap paper for neat transfer to the field book
later.
From Section 2.7: Data should always be entered into the field book directly at the
time of the measurements to avoid loss of data.
2.20
Why should the field notes show the precision of the measurements?
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Field notes should show the precision of the measurements made to indicate the
accuracy of the measurements.
2.21
Explain the reason for item 7 in Section 2.11 when recording field notes.
In general a sketch will show more than a table of numbers. As the saying goes, โA
picture is worth a thousand words.โ
2.22
Explain the reason for item 13 in Section 2.11 when recording field notes.
A standard set of symbols and signs improve the clarity of drawings.
2.23
Explain the reason for item 18 in Section 2.11 when recording field notes.
A zero should be placed before a decimal point for the sake of clarity.
2.24
When should sketches be made instead of just recording data?
Sketches should be made instead of recording data anytime observations need to be
clarified so that the personnel interpreting the notes can have a clear understanding of
the field conditions. This also serves as a reminder of the work performed and any
unusual conditions in later references to the project.
2.25
Justify the requirement to list in a field book the makes and serial numbers of all
instruments used on a survey.
Listing the makes and serial numbers of the instruments used in the survey may help
isolate instrumental errors later when reviewing the project.
2.26
Discuss the advantages of survey controllers that can communicate with several
different types of instruments.
The ability of survey controllers to communicate with several different types of
instruments allows the surveyor to match the specific conditions of the project with the
instrument that this is ideally suited for the job. Thus total station, digital levels, and
GNSS receivers can all be used in a single project.
2.27
Discuss why data should always be backed up at regular intervals.
From Section 2.13, paragraph 1: โAt regular intervals, usually at lunchtime and at the
end of a dayโs work, or when a survey has been completed, the information stored in
files within a data collector is transferred to another device. This is a safety precaution
to avoid accidentally losing substantial amounts of data.โ
2.28
Search the Internet and find at least two sites related to
(a) Manufacturers of survey controllers.
(b)
Manufacturers of total stations.
(c)
Manufacturers of global navigation satellite system (GNSS) receivers.
Answers should vary with students.
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2.29
Why do many survey controllers contain digital cameras?
From Section 2.15: โMany modern survey controllers also contain digital cameras that
allow field personnel to capture a digital image of the survey.โ
2.30
What are the dangers involved in using a survey controller?
From Section 2.15: โAlthough survey controllers have many advantages, they also
present some dangers and problems. There is the slight chance, for example, the files
could be accidentally erased through carelessness or lost because of malfunction or
damage to the unit.โ
2.31
Describe what is meant by the phrase โfield-to-finish.โ
From Section 2.15, “The field codes can instruct the drafting software to draw a map
of the data complete with lines, curves and mapping symbols. The process of
collecting field data with field codes that can be interpreted later by software is known
as a field-to-finish survey. This greatly reduces the time needed to complete a project.”
2.32
Why are sketches in field books not usually drawn to scale?
This is true since this would require an overwhelming amount of time. The sketches
are simply to provide readers of the notes an approximate visual reference to the
measurements.
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3
THEORY OF ERRORS IN OBSERVATIONS
3.1
Discuss the difference between an error and a residual.
From Section 3.3, an error is the difference between the observation and its true value,
or ? = ? โ ?ฬ
whereas a residual, which is defined in Section 3.11 is the difference
ฬ
โ?
between the mean of a set of observations and the observation or ? = ?
3.2
Give two examples of (a) direct and (b) indirect measurements.
From Section 3.2: A direct observation is made by applying a measurement instrument
directly to a quantity to be measured and an indirect observation is made by computing
a quantity from direct observations.
Examples should vary by student response.
3.3
Define the term systematic error, and give two surveying examples of a systematic error.
See Section 3.6
3.4
Define the term random error, and give two surveying examples of a random error.
See Section 3.6
3.5
Discuss the difference between accuracy and precision.
From Section 3.7, accuracy is the nearness of the observed quantities to the true value,
which is never known. Precision is the degree of refinement or consistency of a group
of observations and is evaluated on the basis of discrepancy size.
3.6
The observations of 124.53, 124.55, 142.51, and 124.52 are obtained when taping the
length of a line. What should the observer consider doing before a mean length is
determined from the set of observations?
It appears that the observation 142.51 is an outlier and a possible mistake in the data
set. The observer should collect another tape observation of the line and discard the
offending observation(s).
A distance AB is observed repeatedly using the same equipment and procedures, and the results,
in meters, are listed in Problems 3.7 through 3.10. Calculate (a) the lineโs most probable length,
(b) the standard deviation and (c) the standard deviation of the mean for each set of results.
*3.7
65.401, 65.400, 65.402, 65.396, 65.406, 65.401, 65.396, 65.401, 65.405, and 65.404
(a)
65.401
โ654.012
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3.8
3.9
3.10
(b)
ยฑ0.003
(c)
ยฑ0.001
โฮฝ2 = 0.000091
Same as Problem 3.7 but discard only one 65.396 observation.
(a)
65.402
โ588.616
(b)
ยฑ0.003
โฮฝ2 = 0.000064
(c)
ยฑ0.0009
Same as Problem 3.7, but discard both 65.396 observations.
(a)
65.402
โ523.220
(b)
ยฑ0.002
โฮฝ2 = 0.000030
(c)
ยฑ0.0007
Same as Problem 3.7, but include two additional observations, 65.402 and 65.405.
(a)
65.402
โ784.819
(b)
ยฑ0.003
โฮฝ2 = 0.000115
(c)
ยฑ0.0009
In Problems 3.11 through 3.14, determine the range within which observations should
fall (a) 90% of the time and (b) 95% of the time. List the percentage of values that actually fall
within these ranges.
3.11
3.12
3.13
3.14
For the data of Problem 3.7.
*(a)
65.4012ยฑ0.0052 (65.3960, 65.4064), 100%
(b)
65.4012ยฑ0.0062 (65.3950, 65.4074), 100%
For the data of Problem 3.8.
(a)
65.4018ยฑ0.0046 (65.3971, 65.4064), 90%, 65.396 outside of range
(b)
65.4018ยฑ0.0055 (65.3963, 65.4073), 90%, 65.396 outside of range
For the data of Problem 3.9.
(a)
65.4025ยฑ0.0034 (65.3991, 65.4059), 90%, 65.406 outside of range
(b)
65.4025ยฑ0.0040 (65.3985, 65.4065), 100%
For the data of Problem 3.10.
(a)
65.4016ยฑ0.0053 (65.3963, 65.4069), 83.3%, both 65.396 outside of range
(b)
65.4016ยฑ0.0063 (65.3952, 65.4079), 100%
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In Problems 3.15 through 3.17, an angle is observed repeatedly using the same
equipment and procedures. Calculate (a) the angleโs most probable value, (b) the standard
deviation, and (c) the standard deviation of the mean.
*3.15
23ยฐ30โฒ 00โณ , 23ยฐ30โฒ 10โณ , 23ยฐ30โฒ 10โณ , and 23ยฐ29โฒ 55โณ .
(a) 23ยฐ30โฒ04โณ
(b) ยฑ7.5โณ
(c) ยฑ3.8โณ
3.16
Same as Problem 3.15, but with three additional observations, 23ยฐ29โฒ 55โณ , 23ยฐ29โฒ 50โณ
and 23ยฐ30โฒ 05โณ .
(a) 23ยฐ30โฒ01โณ
(b) ยฑ7.9โณ
(c) ยฑ3.0โณ
3.17
Same as Problem 3.15, but with two additional observations, 23ยฐ30โฒ05โณ and
23ยฐ29โฒ 55โณ .
(a) 23ยฐ30โฒ02โณ
(b) ยฑ6.9โณ
(c) ยฑ2.8โณ
3.18*
A field party is capable of making taping observations with a standard deviation of ยฑ0.02
ft per 100 ft tape length. What standard deviation would be expected in a distance of
400 ft taped by this party?
By Equation (3.12): ยฑ0.04 ft ๏ฝ 0.020 400 /100
3.19
Repeat Problem 3.18, except that the standard deviation per 30-m tape length is ยฑ3mm
and a distance of 60 m is taped. What is the expected 95% error in 60 m?
S = by Equation (3.12): ยฑ0.004 m = 0.003โ60/30
S95 = by Equation (3.8): ยฑ0.008 m = 0.0042(1.9599)
3.20
A distance of 200 ft must be taped in a manner to ensure a standard deviation smaller
than ยฑ0.04 ft. What must be the standard deviation per 100 ft tape length to achieve the
desired precision?
ยฑ0.028 ft = ๏ฑ0.04
3.21
200 /100 by Equation (3.12) rearranged.
Lines of levels were run requiring n instrument setups. If the rod reading for each
backsight and foresight has a standard deviation ฯ, what is the standard deviation in each
of the following level lines?
(a) n = 12, ฯ = ยฑ0.005 ft; By Equation (3.12): ยฑ0.017 ft = 0.005โ12.
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(b) n = 32, ฯ = ยฑ3 mm; By Equation (3.12): ยฑ17.0 mm = 3โ32.
3.22
A line AC was observed in 2 sections AB and BC, with lengths and standard deviations
listed below. What is the total length AC, and its standard deviation?
*(a) ?? = 60.00 ยฑ 0.015 ft; ?? = 86.13 ยฑ 0.018 ft; 146.13ยฑ0.023 ft by Equation
(3.11)
(b) ?? = 30.000 ยฑ 0.004 m; ?? = 23.150 ยฑ 0.003 m; 53.150ยฑ0.005 m by Equation
(3.11)
3.23
Line AD is observed in three sections, AB, BC, and CD, with lengths and standard
deviations as listed below. What is the total length AD and its standard deviation?
(a) ?? = 456.78 ยฑ 0.03 ft; ?? = 524.56 ยฑ 0.04 ft; ?? = 692.35 ยฑ 0.05 ft
1673.69 ยฑ 0.071 ft by Equation (3.11)
(b) ?? = 229.090 ยฑ 0.005 ?; ?? = 336.447 ยฑ 0.006 m; ?? = 465.837 ยฑ 0.008 m
1031.374 ยฑ 0.011 m by Equation (3.11)
3.24
The difference in elevation between A and B was observed four times as 32.05, 32.03,
32.08, and 32.01 ft. The observations were given weights of 2, 1, 3 and 2, respectively,
by the observer. *(a) Calculate the weighted mean for distance AB. (b) What difference
results if later judgment revises the weights to 2, 3, 1, and 1, respectively?
By Equation (3.17):
3.25
*(a)
32.036 ft; mw =
32.05(2)+32.03(1)+32.08(3)+32.01(2)
(b)
32.040 ft; mw =
32.05(2)+32.03(3)+32.08(1)+32.01(1)
2+1+3+2
2+3+1+1
Determine the weighted mean for the following angles:
By Equation (3.17):
(a) 222ยฐ12โฒ 36โณ , wt 2; 222ยฐ12โฒ 42โณ , wt 1; 222ยฐ12โฒ 34โณ , wt 3; 222ยฐ12โฒ40.2โณ; mw =
36(2)+42(1)+34(3)
2+1+3
(b) 96ยฐ14โฒ 20โณ ยฑ 3โณ ; 96ยฐ14โฒ 24โณ ยฑ 2โณ ; 96ยฐ14โฒ 18โณ ยฑ 1โณ ; 96ยฐ14โฒ19.3โณ; mw =
1 2
3
1 2
2
1
1
1
+
+
32 22 1 2
1
1
20( ) +42( ) +34( )^2
3.26
Specifications for observing angles of an n-sided polygon limit the total angular
misclosure to E. How accurately must each angle be observed for the following values
of n and E?
By rearranged Equation (3.12):
(a) n = 6, E = ยฑ10โณ; ยฑ4.1โณ; 10/โ6
(b) n = 10, E = ยฑ10โณ; ยฑ3.2โณ; 10/โ10
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3.27
What is the area of a rectangular field and its estimated error for the following recorded
values:
By Equation (3.13):
*(a) 243.89 ยฑ 0.05 ft by 208.65 ยฑ 0.04 ft; 50,888 ยฑ 14 ft2 or 1.1682 ยฑ 0.0003 ac;
โ[243.89(0.04)]2 + [208.65(0.05)]2
(b) 1203.45 ยฑ 0.08 ft by 906.78 ยฑ 0.06 ft; 1,091,300 ยฑ 100 ft2 or 25.052 ยฑ 0.002 ac;
โ[1203.45(0.06)]2 + [906.78(0.08)]2
(c) 344.092 ยฑ 0.006 m by 180.403 ยฑ 0.005 m; 62,075.0 ยฑ 2.0 m2 or 6.2075 ยฑ 0.00020
ha; โ[344.092(0.005)]2 + [180.403(0.006)]2
3.28
Adjust the angles of triangle ABC for the following angular values and weights:
By Equation (3.17):
*(a)
? = 49ยฐ24โฒ 22โณ , wt 2; ? = 39ยฐ02โฒ 16โณ , wt 1; ? = 91ยฐ33โฒ 00โณ , wt 3
Misclosure = โ22โณ
A
B
C
Obs. Ang.
49ยฐ24โฒ22โณ
39ยฐ02โฒ16โณ
91ยฐ33โฒ00โณ
179ยฐ59โฒ38โณ
Wt
2
1
3
6
Corr.
3x
6x
2x
11x
11x = 22โณ
Num. Cor.
6โณ
12โณ
4โณ
Rnd. Cor.
6โณ
12โณ
4โณ
Adj. Ang.
49ยฐ24โฒ28โณ
39ยฐ02โฒ28โณ
91ยฐ33โฒ04โณ
x = 2โณ
(b) ? = 81ยฐ06โฒ 44โณ , wt 2; ? = 53ยฐ33โฒ 56โณ , wt 2; ? = 45ยฐ19โฒ 20โณ , wt 3
Misclosure = โ10โณ
A
B
C
3.29
Obs. Ang.
81ยฐ06โฒ44โณ
53ยฐ33โฒ56โณ
45ยฐ19โฒ10โณ
179ยฐ59โฒ53โณ
Wt
2
2
3
7
Corr.
21x
21x
14x
56x
56x = 10โณ
Num. Cor.
3.8โณ
3.8โณ
2.5โณ
x = 0.178โณ
Rnd Cor.
4โณ
4โณ
2โณ
10โณ
Adj. Ang.
81ยฐ06โฒ48โณ
53ยฐ34โฒ00โณ
45ยฐ19โฒ12โณ
Determine relative weights and perform a weighted adjustment (to the nearest second)
for angles A, B, and C of a plane triangle, given the following four observations for each
angle:
Angle A
Angle B
Angle C
44ยฐ28โฒ16โณ 65ยฐ56โฒ13โณ 69ยฐ35โฒ20โณ
44ยฐ28โฒ12โณ 65ยฐ56โฒ10โณ 69ยฐ35โฒ24โณ
44ยฐ28โฒ17โณ 65ยฐ56โฒ06โณ 69ยฐ35โฒ18โณ
44ยฐ28โฒ11โณ 65ยฐ56โฒ08โณ 69ยฐ35โฒ24โณ
A = 44ยฐ28โฒ14.0โณ ยฑ 2.9โณ; B = 65ยฐ56โฒ09.3โณ ยฑ 3.0โณ; C = 69ยฐ35โฒ21.5โณ ยฑ 3โณ
Misclosure = โ15.3โณ
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A
B
C
3.30
Obs. Ang.
Wt
Corr. Multiplier
44ยฐ28โฒ14.0โณ
65ยฐ56โฒ09.3โณ
69ยฐ35โฒ21.5โณ
179ยฐ59โฒ44.7โณ
0.115385
0.11215
0.111111
0.338645
0.338645/wt = 2.93
0.338645/wt = 3.02
0.338645/wt= 3.05
9.00x
9.0 = 15.3โณ
Num.
Cor.
4.97โณ
5.12โณ
5.16โณ
Rnd.
Cor.
5.0โณ
5.1โณ
5.2โณ
Adj. Ang.
44ยฐ28โฒ19โณ
65ยฐ56โฒ14โณ
69ยฐ35โฒ27โณ
x = 1.69โณ
A line of levels was run from benchmarks A to B, B to C, and C to D. The elevation
differences obtained between benchmarks, with their standard deviations, are listed
below. What is the difference in elevation from benchmark A to D and the standard
deviation of that elevation difference?
(a) BM A to BM B = +37.78 ยฑ 0.12 ft; BM B to BM C = โ73.50 ยฑ 0.16 ft; and BM C to
BM D = โ84.09 ยฑ 0.08 ft
By Equation (3.11): โ119.81 ยฑ 0.22 ft
(b) BM A to BM B = โ60.821 ยฑ 0.015 m; BM B to BM C = +94.378 241 ยฑ 0.020 m; and
BM C to BM D +56.805 ยฑ 0.015 m
By Equation (3.11): 90.362 ยฑ 0.029 m
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4
LEVELING THEORY, METHODS, AND EQUIPMENT
4.1
Define the following leveling terms: (a) vertical line, (b) level surface, and (c)
benchmark.
From Section 4.2:
(a) Vertical line: โA line that follows the local direction of gravity as indicated by a
plumb lineโ
(b) Level surface: โ. A curved surface that at every point is perpendicular to the local
plumb line (the direction in which gravity acts).โ
(c) Benchmark: โA relatively permanent object, natural or artificial, having a marked
point whose elevation above or below a reference datum is known or assumed.โ
*4.2
How far will a horizontal line depart from the Earthโs surface in 1 km? 5 km? 10 km?
(Apply both curvature and refraction)
1 km? Cm = 0.0675(1)2 = 0.068 m
5 km? Cm = 0.0675(5)2 = 1.688 m
10 km? Cm = 0.0675(10)2 = 6.750 m
4.3
Visit the website of the National Geodetic Survey, and obtain a data sheet description of
a benchmark in your local area.
Solutions should vary.
Create plot of the curvature and
refraction correction for sight lines
going from 0 ft to 10,000 ft in 500 ft
increments.
Dist. (ft)
1000
1500
2000
2500
3000
3500
4000
4500
5000
5500
CR (ft)
0.02
0.05
0.08
0.13
0.19
0.25
0.33
0.42
0.52
0.62
tionacfreRduCv
tionecrC
1.50
1.0
0.5
0.
tionCrec)(f
4.4
tanDisce)(f
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6000
6500
7000
7500
8000
Create a plot of curvature and refraction corrections for sight lines going from 0 m to
10,000 m in 500 m increments.
m)Dist.(
0
05
01
0 15
02
0 25
03
0 35
04
4.6
tionacfreRduCv
m)CR(
0.
70.1
80.6
20.15
0.27
20.4
80.6
70.82
01.8
tionecrC
1.20
1.0
0.8
0.6
0.4
0.2
tionCrec)(m
4.5
0.74
0.87
1.01
1.16
1.32
0.
tanDisce)(m
Why are elevations today not referred to as mean sea-level heights?
From Section 4.3: โThis adjustment (NAVD88) shifted the position of the reference
surface from the mean of the 26 tidal gage stations to a single tidal gage benchmark
called Father Point, which is in Rimouski, Quebec, Canada, near the mouth of the St.
Lawrence Seaway. Thus, elevations in NAVD88 are no longer referenced to mean sea
level. Benchmark elevations that were defined by the NGVD29 datum have changed by
relatively small, but nevertheless significant amounts in the eastern half of the
continental United States (see Figure 19.7). However, the changes are much greater in
the western part of the country and reach 1.5 m in the Rocky Mountain region. It is
therefore imperative that surveyors positively identify the datum to which their
elevations are referred. After the adjustment, adjusted elevations are properly known as
orthometric heights.โ
*4.7
On a large lake without waves, how far from shore is a sailboat when the top of its 30-ft
mast disappears from the view of a person lying at the waterโs edge?
30
38,160 ft or 7.3 mi; ? = 1000โ0.0206 = 38,161 ?? = ?. ??? ??
4.8
Similar to Problem 4.7, except for a 5-m mast and a person whose eye height is 1.5 m
above the waterโs edge.
13.3 km; ? = โ1.5โ0.0675 + โ5โ0.0675 = ??. ?? ??
4.9
Readings on a line of differential levels are taken to the nearest 0.2 mm. For what
maximum distance can the Earthโs curvature and refraction be neglected?
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? = 1000โ0.0002/0.0675 = ??. ? ?
4.10
Similar to Problem 4.9 except readings are to the 0.001 ft.
? = 1000โ0.001โ0.0206 = ???. ? ??
4.11
Describe how readings are determined in a digital level when using a bar coded rod.
From Section 4.11: “At the press of a button, the image of bar codes in the telescopeโs
field of view is captured and processed. This processing consists of an on-board
computer comparing the captured image to the rodโs entire pattern, which is stored in
memory. When a match is found, which takes about 4 sec, the rod reading is displayed
digitally.”
Successive plus and minus sights taken on a downhill line of levels are listed in Problems 4.12
and 4.13. The values represent the horizontal distances between the instrument and either the
plus or minus sights. What error results from curvature and refraction?
*4.12 20, 225; 50, 195; 40, 135; 30, 250 ft.
4.13
Plus
20
50
40
30
Sum
CR (ft)
0.00000824
0.0000515
0.00003296
0.00001854
0.00011124
Combined
โ0.003 ft
CR (ft)
0.001043
0.000783
0.000375
0.001288
0.003489
20, 70; 25, 60; 20, 55; 15, 60 m.
Plus
20
25
20
15
CR (mm)
0.027
0.042188
0.027
0.015188
0.111375
Minus
70
60
55
60
Combined: โ0.91 mm
4.14
Minus
225
195
135
250
CR (mm)
0.33075
0.243
0.204188
0.243
1.020938
What error results if the curvature and refraction correction is neglected in trigonometric
leveling for sights: (a) 3000 ft long (b) 1200 m long (c) 4500 ft long?
3000 2
(a) โ? = 0.0206 (1000) = ?. ?? ??
1200 2
(b) โ? = 0.0675 (1000) = ?. ??? ?
4500 2
(c) โ? = 0.0206 (1000) = ?. ?? ??
*4.15 The slope distance and zenith angle observed from point P to point Q were 2406.787 m
and 84ยฐ13โฒ07โณ respectively. The instrument and rod target heights were equal. If the
elevation of point P is 30.245 m, above datum, what is the elevation of point Q?
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????? = 30.245 + 2406.787 cos(84ยฐ13โฒ 07โณ ) + 0.0675 (
2406.787 sin(84ยฐ13โฒ 07โณ )
)
1000
2
????? = 30.245 + 242.443 + 0.387 = ???. ??? ?
4.16
The slope distance and zenith angle observed from point X to point Y were 2907.45 ft
and 97ยฐ25’36โณ. The instrument and rod target heights were equal. If the elevation of
point X is 6547.89 ft above datum, what is the elevation of point Y?
2907.45 sin(97ยฐ25โฒ 36โณ )
????? = 6547.89 + 2907.45 cos(97ยฐ25โฒ 36โณ ) + 0.0206 (
1000
2
)
= 6547.89 โ 375.809 + 0.171 = ????. ?? ??
4.17
Similar to Problem 4.15, except the slope distance was 1543.853 m, the zenith angle
was 83ยฐ44’08โณ and the elevation of point P was 1850.567 m above datum.
2
โฒ
????? = 1850.567 + 1543.853 cos(83ยฐ44 08
โณ)
1543.853 sin(83ยฐ44โฒ 08โณ )
+ 0.0675 (
)
1000
= 1850.567 + 168.462 + 0.159 = ????. ??? ?
4.18
In trigonometric leveling from point A to point B, the slope distance and zenith angle
measured at A were 5462.46 ft and 94ยฐ08’36โณ. At B these measurements were 5462.58
ft and 85ยฐ51 ’47 โณ, respectively. If the instrument and rod target heights were equal,
calculate the difference in elevation from A to B.
???? =
???? =
94ยฐ08โฒ 36โณ +180ยฐโ85ยฐ51โฒ 47โณ
2
5462.58+5462.46
2
= 94ยฐ08โฒ 24โณ
= 5462.52
โ???? = 5462.52 cos(94ยฐ08โฒ 24โณ ) = โ???. ?? ??
4.19
Describe how parallax in the viewing system of a level can be detected and removed.
From Section 4.7:
“After focusing, if the cross hairs appear to travel over the object sighted when the eye
is shifted slightly in any direction, parallax exists. The objective lens, the eyepiece, or
both must be refocused to eliminate this effect if accurate work is to be done.”
4.20
What is the sensitivity of a level vial with 2-mm divisions for: (a) a radius of 13.75 m
(b) a radius of 10.31 m?
2
(a)
? = [13.75(1000)] 206264.8 = ??โณ
(b)
? = [10.31(1000)] 206264.8 = ??โณ
2
*4.21 An observer fails to check the bubble, and it is off two divisions on a 500-ft sight. What
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error in elevation difference results with a 10-sec bubble?
angular error = 2(10) = 20 sec
Error = 250 tan(20) = 0.048 ft
4.22
An observer fails to check the bubble, and it is off two divisions on a 300-m sight. What
error results for a 20-sec bubble?
angular error = 2(20) = 40 sec
Error = 200 tan(40) = 0.058 m
4.23
Similar to Problem 4.22, except a 30-sec bubble is off three divisions on a 300-ft sight.
angular error = 3(30) = 90 sec
Error = 300 tan(90) = 0.13 ft
4.24
With the bubble centered, a 100-m sight gives a reading of 1.352 m. After moving the
bubble three divisions off center, the reading is 1.396 m. For 2-mm vial divisions, what
is: (a) the vial radius of curvature in meters (b) the angle in seconds subtended by one
division?
โrdg = 1.410 โ 1.352 = 0.058 m
0.058
4? = atan (
) = 120″
100
(a) ? = 0.002/tan(120″) = ?. ??? ?
(b)
4.25
120โณ/4 = 30โณ
Similar to Problem 4.24, except the sight length was 300 ft, the initial reading was 5.132
ft, and the final reading was 5.176 ft.
โrdg = 5.176 โ 5.132 = 0.0.044 ft
0.044
3? = atan (
) = 3?”
300
(a) ? = 0.002/tan(30″) = ??. ?? ??
(b)
4.26
30โณ/3 = 10โณ
Sunshine on the forward end of a 20โณ /2 mm level vial bubble draws it off 2 divisions,
giving a plus sight reading of 4.63 ft on a 250-ft sight. Compute the correct reading.
Correction = 200 tan(2โ20โณ) = 0.048 ft
Correct reading = 4.63 โ 0.048 = 4.58 ft
Note: the correction is subtracted since the bubble was drawn off on the forward end
of the level, thus raising the line of sight.
4.27
List in tabular form, for comparison, the advantages and disadvantages of an automatic
level versus a digital level.
See Section 4.10 and 4.11.
*4.28 If a plus sight of 3.54 ft is taken on BM A, elevation 850.48 ft, and a minus sight of 7.84
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ft is read on point X, calculate the HI and the elevation of point X.
HI = 850.48 + 3.54 = 854.02 ft
Elev = 854.02 โ 7.84 = 846.18 ft
4.29
If a plus sight of 0687 m is taken on BM A, elevation 85.476 m, and a minus sight of
1.564 m is read on point X, calculate the HI and the elevation of point X.
HI = 85.476 + 0.687 = 86.163 m
Elev = 86.163 โ 1.564 = 84.599 m
4.30
Similar to Problem 4.28, except a plus sight of 8.98 ft is taken on BM A, elevation
606.33 ft, and a minus sight of 4.32 ft read on point X.
HI = 606.33 + 8.98 = 615.31 ft
Elev = 615.31 โ 4.32 = 610.99 ft
4.31
Describe the procedure used to test if the level vial is perpendicular to the vertical axis
of the instrument.
See Section 4.15.5
4.32
A horizontal collimation test is performed on an automatic level following the
procedures described in Section 4.15.5. With the instrument setup at point 1, the rod
reading at A was 5.548 ft, and to B it was 5.126 ft. After moving and leveling the
instrument at point 2, the rod reading to A was 5.540 ft and to B was 5.126 ft. What is
the collimation error of the instrument and the corrected reading to A from point 2?
5.126 โ 5.548 โ 5.126 + 5.540
= โ0.004
2
Correct reading at A = 5.540 โ 2(โ0.004) = 5.548 ft
?=
4.33
The instrument tested in Problem 4.32 was used in a survey immediately before the test
where the observed elevation difference between two benchmarks was +44.65 ft. The
sum of the plus sight distances between the benchmarks was 250 ft and the sum of the
minus sight distances was 490 ft. What is the corrected elevation difference between the
two benchmarks?
+44.64 ft; = 44.65โ 0.004/100(250 โ 490) = 44.64 ft
4.34
Similar to Problem 4.32 except that the rod readings are 1.894 m and 1.923 m to A and
B, respectively, from point 1, and 1.083 m and 1.100 m to A and B, respectively, from
point 2. The distance between the points in the test was 100 m.
๏ฅ๏ฝ
1.923 ๏ญ 1.894 ๏ญ 1.100 ๏ซ 1.083
๏ฝ 0.006 m
2
Correct reading at A = 1.083 โ 2(0.006) = 1.071 m
4.35 The instrument tested in Problem 4.34 was used in a survey immediately before the test
where the observed elevation difference between two benchmarks was โ13.068 m. The
sum of the plus sight distances between the benchmarks was 1540 m and the sum of the
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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.
24
Instructorโs Solution Manual
csmeoatiGnrudI:AgyvSEl
minus sight distances was 545 m. What is the corrected elevation difference between the
two benchmarks?
โ13.128 m; = โ13.068 โ 0.006/100(1540 โ 545)
ยฉ 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.
Instructorโs Manual
Elementary Surveying: An Introduction to Geomatics
25
5
LEVELING โ 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.
From Section 5.2, paragraph 3: A simple but useful rule in centering a bubble,
illustrated in Figure 5.1, is: A bubble follows the left thumb when turning the screws.
5.2
What is the difference between a benchmark and a turning point in differential leveling?
A benchmark is a relatively permanent object, natural or artificial, having a marked
point whose elevation above or below a reference datum is known, assumed, or will be
established during the leveling process whereas a turning point is an intermediate,
temporary point between benchmarks which are created to perform the differential
leveling process. They are usually temporary points whose elevations are lost after the
differential leveling process is complete.
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
through the use of readings on the upper and lower (stadia) wires on the reticle. The
method is based on the principle that in similar triangles, corresponding sides are
proportional. โฆ Thus the equation for a distance on a horizontal stadia sight reduces to
D = KI (5.2) โฆ It should be realized by the reader that in differential leveling, the
actual sight distances to the rod are not important. All one needs to balance is the rod
intervals on the plus and minus sights between benchmarks to ensure that the sight
distances are balanced.”
5.4
What is the collimation error, and how can it be removed from the differential leveling
process.
From Section 5.12.1: It is caused by the line of sight not being parallel with the axis of
the level vial. When this condition exists, the line of sight will not be horizontal and
thus result in incorrect readings. This is a systematic error and can be removed by
balancing the backsight and foresight distances between benchmarks.
5.5
Discuss how errors due to Earth curvature and refraction can be eliminated from the
differential leveling process.
From Section 5.4: โBalancing plus and minus sight distances will eliminate errors due
to instrument maladjustment (most important) and the combined effects of the Earthโs
curvature and refraction, as shown in Figure 5.6. Here e1 and e2 are the combined
26
Instructorโs Solution Manual
Elementary Surveying: An Introduction to Geomatics
curvature and refraction errors for the plus and minus sights, respectively. If D1 and D2
are made equal, e1 and e2 are also equal. In calculations, e1 is added and e2 subtracted;
thus they cancel each other.โ
5.6
When is it appropriate to use the reciprocal leveling procedure?
From Section 5.7: โSometimes in leveling across topographic features such as rivers,
lakes, and canyons, it is difficult or impossible to keep plus and minus sights short and
equal. Reciprocal leveling may be utilized at such locations.โ
5.7
List four considerations that govern a rodpersonโs selection of TPs and BMs.
1. From Chapter 4: BMs must be permanent.
2. From Section 5.4: “Turning points should be solid objects with a definite high point.”
3. From Section 5.6: “…it is recommended that some turning points or benchmarks used
in the first part of the circuit be included again on the return run. This creates a multiloop circuit, and if a blunder or large error exists, its location can be isolated to one of
the smaller loops.”
4. From Section 5.12.2: “It (settlement) can be avoided by selecting firm, solid turning
points or, if none are available, using a steel turning pin set firmly in the ground.”
5. Find turning points that aid in the balancing of plus and minus sight distances.
*5.8 What error is created by a rod leaning 10 min from plumb at a 12.51-ft reading on the
leaning rod?
Error = 0.000 ft
Correct reading = 12.51 cos(10′) = 12.50995; So error is 0.00005 ft, or 0.000 ft
Problem is designed to show that even for a high reading and a mislevelment outside
of a typical circular bubble, the resulting error is negligble.
5.9
Similar to Problem 5.6, except for a 5-m reading.
Error is 0.000 m
Correct reading = 5 cos(10′) = 4.9999785, so error is 0.000021. The error is negligible.
Problem is designed to show that even for a high reading and a mislevelment outside
of a typical circular bubble, the resulting error is negligble.
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?
0.3
Correct reading = 3 2.865 = 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?
0.2
Correct reading = 7 6.307 = 6.3044
Error = 0.0026 ft
Instructorโs Manual
Elementary Surveying: An Introduction to Geomatics
27
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
๏ซS (BS)
BM 7
TP 1
TP 2
BM 8
TP 3
BM 7
5.68
9.42
9.26
6.45
9.59
STA
BM 7
Plus
5.68
TP 1
9.42
TP 2
9.26
BM 8
6.45
TP 3
9.59
BM 7
40.40
๏ญS (FS)
7.58
5.81
4.59
8.50
13.95
HI
Minus
2309.13 (0.006)
7.58
2310.97 (0.012)
5.81
2314.42 (0.018)
4.590
2316.28 (0.024)
8.500
2317.37 (0.03)
13.950
40.43
ELEV
2303.45
(2301.556)
2301.55
(2305.172)
2305.16
(2309.848)
2309.83
(2307.804)
2307.78
(2303.45)
2303.42
Page check
2303.45+40.4-40.43 = 2303.42
Misclosure = 2303.42 โ 2303.45 = โ0.03
Correction = โ โ0.03/5 = 0.006
2400 ft โ 0.7315 km and 0.03 ft โ 9.1 mm; From Equation 5.3: ? = 12โโ0.7315 =
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
BM7
Plus
5.68
TP1
9.420
BM 8
9.26
TP2
6.45
HI
Minus
137.73
(0.006)
7.58
(0.012)
5.81
(0.018)
4.590
139.57
143.02
ELEV
132.05
(130.156)
130.15
(133.772)
133.76
(138.448)
138.43

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