Chapter 03 – Distance Measurement
3.1 A Gunter’s chain was used to take the following measurements. Convert these distances to feet and meters.
a. 65 chains, 15 links;
65.15 x 66 = 4,299.9ft.
x 0.3048 = 1,310.610m
b. 105 chains, 11 links;
105.11 x 66 = 6,937.26 ft. x 0.3048 = 2,114.48 m
c. 129.33 chains;
129.33 x 66 = 8,535.78ft x 0.3048 = 2,601.71 m
d. 7 chains, 41 links.
7.41 x 66 = 489.06 ft.
x .3048 = 149.07 m
3.2
A 100-ft “cut” steel tape was used to measure between two property markers. The rear surveyor held 52 ft,
while the head surveyor cut 0.27 ft. What is the distance between the markers?
Distance = 52.00 - 0.27 = 51.73 ft
3.3
The slope measurement between two points is 36.255 m and the slope angle is 1°50′. Compute the
horizontal distance.
H = 36.255 Cos (1° 50') = 9.409 m
3.4
A distance of 210.23 ft was measured along a 3 percent slope. Compute the horizontal distance.
Tan slope angle = .03; slope angle = 1.°
H = 210.23 Cos(1.718358°) = 210.14 ft.
3.5
The slope distance between two points is 21.835 m and the difference in elevation between the points is
3.658 m. Compute the horizontal distance.
H = √ 21.8352 – 3.6582 = 21.522 m
3.6
The ground clearance of an overhead electrical cable must be determined. Surveyor B is positioned directly
under the cable (surveyor B can make a position check by sighting past the string of a plumb bob, held in his
or her outstretched hand, to the cable); surveyor A sets the clinometer to 45° and then backs away from
surveyor B until the overhead electrical cable is on the crosshair of the leveled clinometer. At this point,
surveyors A and B determine the distance between them to be 17.4 ft (see figure). Surveyor A then sets the
clinometer to 0° and sights surveyor B; this horizontal line of sight cuts surveyor B at the knees, a distance of
2.0 ft above the ground. Determine the ground clearance of the electrical cable
Clearance = 17.4 + 2.0 = 19.4 ft. (19.6 looking at the image)
3.7
A 100-ft steel tape known to be only 99.98 ft long (under standard conditions) was used to record a
measurement of 276.22 ft. What is the distance corrected for the erroneous tape?
Error per tape length = -0.02 ft.; tape used 276.22/100 times
error = -.02 x 2.7622 = - 0.06 ft.
corrected distance = 276.22 - 0.06 = 276.16 ft. [or, 276.22 x .9998 = 276.16 ft]
3.8
A 30-m steel tape, known to be 30.004 m (under standard conditions), was used to record a measurement of
129.085 m. What is the distance corrected for the erroneous tape length?
Error per tape length =+0.004 m; tape used 129.085/30 times,
Error = +0.004 x 129.085/30 = +0.017 m
Corrected distance = 129.085 + 0.017 = 129.102 m
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3.9
It is required to lay out a rectangular commercial building 200.00 ft wide and 350.00 ft long. If the steel tape
being is 100.02 ft long (under standard conditions), what distances should be laid out?
Error per tape length = 0.02 ft.
Tape used a) 2.0 times; error = 2.0 x .02 = + 0.04 ft.
b) 3.50 times; error = 3.50 x .02 = + 0.07 ft.
Corrected layout distance
a) 200.00 - 0.04 = 199.96 ft.
b) 350.00 - 0.07 = 344.93 ft.
(ie., these are the values, when pre-corrected for errors, result in the required layout distances)
3.10 A survey distance of 287.13 ft was recorded when the field temperature was 96°F. What is the distance,
corrected for temperature?
CT = .00000645(96 - 68)287.13 = + 0.05 ft.
Corrected distance = 287.13 + 0.05 = 287.18 ft.
3.11 Station 2 + 33.33 must be marked in the field. If the steel tape to be used is only 99.98 ft (under standard
conditions) and if the temperature is 87°F at the time of the measurement, how far from the existing station
mark at 0 + 79.23 will the surveyor have to measure to locate the new station?
[2+33.33] – [0+79.23] = 154.10 ft.
CT = 0.00000645(87 - 68)154.10 = +0.019 ft.
CL = 0.02 x 1.5410 = - 0.031 ft.
C = 0.019 – 0.031 = - 0.012 ft.
Layout 154.10 + 0.01 = 154.11 ft.
3.12 The point of intersection of the center line of Elm Rd. with the center line of First St. was originally recorded
(using a 30-m steel tape) as being at 6 + 11.233. How far from existing station mark 5 + 00 on First St. would
a surveyor have to measure along the center line to reestablish the intersection point under the following
conditions?
Temperature to be −6°C, with a tape that is 29.995 m under standard conditions.
[6+11.233] – [5+00] = 111.233 m
CT = .0000116(- 6 - 20)111.233 = - 0.034m
CL = .005 x 111.233/30 = - 0.019m
C = - 0.053 m
Layout 111.233 + 0.053 = 111.286 m
3.13
Temperature
Tape Length
Slope Data
Slope Measurement
−14°F
99.98 ft
Difference in elevation = 6.35 ft
198.61 ft
CT= .00000645(- 14 -68)198.61 = - 0.105 ft.
CL= - 0.02 x 1.9861 = - 0.040 ft.
C = - 0.105 - 0.040 = - 0.15 ft.
Corrected slope distance = 198.61 - 0.15 = 198.46 ft.
H = √198.462- 6.352 = 198.36 ft.
3.14 46°F
100.00 ft
Slope angle 3°18′
219.51 ft
CT = .00000645(46 - 68)219.51 = - 0.031 ft
Corrected slope distance = 219.51 - 0.02 = 219.49 ft
H = 219.49 Cos (3° 18') = 205.54 ft
3.15 26°C
29.993 m
Slope angle −3°42′
177.032 m
CT = .0000116(26 - 20)177.032 = + 0.012 m
CL = - 0.007 x 177.032/30 = - 0.041 m
C = +0.012 - 0.041 = - 0.029 m
Corrected slope distance = 177.032 - 0.029 = 177.003
H = 177.003 Cos (3° 42') = 176.634 m
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3.16 0°C
30.002 m
Slope at 1.50%
225.000 m
CT = .0000116(0 - 20) 225.000 = - 0.052
CL = .002 x 225/30 = + 0.015
C = - 0.058 + 0.016 = - 0.037
Corrected slope distance = 225.000 - 0.037 = 224.963 m
Tan slope angle = .015; slope angle = 0.8593722°
H = 224.963 Cos (0.8593722°0 = 224.938 m.
3.17 100°F
100.03 ft
Slope at −0.80%
349.65 ft
CT = .00000645(100-68)349.65 = + 0.072 ft CL = + 0.03 x 3.4965 = + 0.104 ft
C = +0.072 +0.105= + 0.177 ft
Corrected slope distance = 349.65 + 0.18 = 349.83 ft
Tan slope angle = .008: slope angle = 0.458356°
H = 349.83 Cos(0.458356°) = 329.82 ft
3.18
Temperature
Tape Length
Design Distance Required Layout Distance
29°F
99.98 ft
366.45 ft
?
CT = .00000645(29 - 68)366.45 = - 0.092 ft
CL = - 0.02 x 3.6645 = - 0.073 ft
C = - 0.092 - 0.073 = - 0.165 ft
Layout 366.45 + 0.17 = 366.62 ft
3.19
26°C
30.012 m
132.203 m
CT = .0000116(26 - 20)132.203 = + 0.009 m.
C = 0.009 + 0.053 = + 0.062
Layout 132.203 - 0.062 = 132.141 m
?
CL = .012 x 156.209/30 = + 0.053
3.20
13°C
29.990 m
400.000 m
CT = .0000116(13 - 20)400.000 = - 0.032 m
C = - 0.032 - 0.133 = - 0.165
Layout 400.000 + 0.165 = 400.165 m
?
CL = 0.010 x 400/30 = - 0.133 m
3.21
30°F
100.02 ft
500.00 ft
CT = .00000645(30 - 68)500.00 = - 0.123 ft.
C = - 0.123 + 0.10 = - 0.02 ft.
Layout 500.00 + 0 .02 = 500.02 ft.
?
CL = +0.02 x 5.00 = +0.10 ft.
3.22
100°F
100.04 ft
88.92 ft
CT = .00000645(100 - 68)88.92 = + 0.018 ft.
C = +0.018 + 0.036 = + 0.05 ft.
Layout 88.92 - 0.05 = 88.87 ft. .
?
CL = +.04 x 0.8892 = + 0.036 ft.
3.23
A 50-m tape is used to measure between two points. The average weight of the tape per meter is 0.320 N. If
the measured distance is 48.888 m, with the tape supported at the ends only and with a tension of 100 N,
find the corrected distance.
2 3
2
2
3
2
CS = -w L / 24p = -0.32 x 48.888 / (24 x 100 ) = -.050 m
Corrected distance = 48.888 -.050 = 48.838 m
3.24
A 30-m tape has a mass of 544 g and is supported only at the ends with a force of 80 N. What is the sag
correction?
30 m tape weighs 0.544 x 9.807 = 5.34 Newtons
2
2
2
2
CS=-W L / 24p = -5.34 x 30 / (24 x 80 ) = -0.006 m
3.25 A 100-ft steel tape weighing 1.8 lb and supported only at the ends with a tension of 24 lb is used to measure
a distance of 471.16 ft. What is the distance corrected for sag?
2
2
2
2
CS = -W L / 24p = -1.8 x 471.16 / (24 x 24 ) = -0.11 ft.
Corrected distance = 471.16 -.11 = 471.05 ft.
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3.26
A distance of 72.55 ft is recorded using a steel tape supported only at the ends with a tension of 15 lb and
weighing 0.016 lb per foot. Find the distance corrected for sag.
2 3
2
2
3
2
CS = w L / 24p = -0.016 x 72.55 /(24 x 15 ) = -0.018 ft.
Corrected distance = 72.55 - 0.02 = 72.53 ft.
3.27 In order to verify the constant of a particular prism, a straight line EFG is laid out. The EDM instrument is first
Set up at E, with the following measurements recorded: EG = 426.224 m, EF = 277.301 m. The EDM
instrument is then set up at F, where distance FG is recorded as FG = 148.953 m. Determine the prism
constant.
prism constant
= EG- EF –FG = 426.224 – 277.301 – 148.953 = - 0.030m
3.28 The EDM slope distance between two points is 2556.28 ft, and the vertical angle is +2°′45′30″ (the vertical
angles were read at both ends of the line and then averaged using a coaxial theodolite/EDM combination). If
the elevation of the instrument station is 322.87 ft and the heights of the theodolite/EDM and the
target/reflector are all equal to 5.17 ft, compute the elevation of the target station and the horizontal
distance to that station.
H = 2556.28 Cos(2°45'30") = 2553.32 ft.
V = 2556.28 Sin(2°45'30") = 123.02 ft.
Elevation of target station = 322.87+123.02 = 445.89 ft.
3.29 A line AB is measured at both ends as follows: Instrument at A, slope distance = 879.209 m; vertical angle =
+1°26′50″ Instrument at B, slope distance = 879.230 m; vertical angle = −1°26′38″ The heights of the
instrument, reflector, and target are equal for each observation.
a. Compute the horizontal distance AB.
b. If the elevation at A is 163.772 m, what is the elevation at B?
Instrument at A
H = 879.209 Cos(1°26'50") = 878.929m
Elevation difference = 879.209 Sin(1°26'50") = 22.205m
Instrument at B
H = 879.230 Cos(-1°26'38") = 878.951m
Elevation difference = 879.230 Sin(-1°26'38") = -22.155m
a) Horizontal distance = (878.929+878.951)/2 = 878.940m
b) Elev. B = 163.772 + (22.205+22.155)/2 = 185.952m
3.30 A coaxial EDM instrument at station K (elevation = 198.92 ft) is used to sight stations L, M, and N, with the
heights of the instrument, target, and reflector being equal for each sighting. The results are as follows:
Instrument at station L, vertical angle = +3°30’, EDM distance = 2200.00 ft
Instrument at station M, vertical angle = -1°30’, EDM distance = 2200.00 ft
Instrument at station N, vertical angle = 0°00’, EDM distance = 2800.00 ft
Compute the elevations of L, M, and N (correct for curvature and refraction).
KL = 2200.00 Sin(3°30') =+134.31 ft.
c+r = .0206 x 2.22 = +0.10 ft.
Elevation difference, K to L, = +134.41 ft
Elevation of L = 198.92+134.41 = 333.33 ft.
KM = 2200.00 Sin(-1°30') = - 57.59 ft.
c+r = .0206 x 2.22 = +0.10 ft.
Elevation difference, K to M, = - 57.49 ft.
Elevation of M = 198.92- 52.49 = 146.43 ft.
KN c+r = .0206 x 2.82 = + 0.16 ft.
Elevation difference, K to N, = +0.16 ft.
Elevation of N = 198.92 + 0.16 = 199.08 ft.
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