Surveying: Principles and Applications Ninth Edition
Barry Kavanagh Tom Mastin
Upper Saddle River, New Jersey Columbus, Ohio
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 a ll calculations to be done using plane trigonometry. Geodetic surveys take into account the shap e 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 must measure or determine the horizontal horizontal distances, distances, rather than just 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 plotted in its correct position on a scaled plan. A base line may be established a nd any feature may be loca ted 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 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 away that the the navigation lights can still be seen by an observer standing at the shoreline? The observer’s eye height is 5’ 0” and 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 arithmetic check. 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
2.5
FS = 11.08 389.07 +3.60 = 392.67 - 11.08 = 381.59 check
Prepare a set of profile leveling notes for the survey in Figure 2-34. In addition to computing computing all elevations, show the arithmetic check and the resulting error in closure. STATION BM #61 0+00 0+50 TP #1 1+00 1+50 1+75 TP #2
BS 4.72
HI 401.46
IS
FS
4.42 4.30 5.11
404.56
2.01 4.66 3.98 1.20 1.80
BS = 9.83 FS = 3.81 E = - 0.02m [small error error – – no no need for adjustments] 396.74 + 9.83 = 406.57 - 3.81 = 402.76 check
2
ELEVATION 396.74 397.04 394.16 399.45 399.90 400.58 403.36 402.76
2.6
Complete the set of differential leveling notes in Table 2-5, and perform the arithmetic check. STATION BM 100 TP 1 TP 2 TP 3 TP 4 BM 100
2.7
BS 2.71 3.62 3.51 3.17 1.47
HI 317.59 316.33 315.87 316.23 316.08
FS
ELEVATION 314.88 312.71 312.36 313.06 314.61 314.87
BS = 4.48
FS = 14.49 314.88 + 14.48 - 14.49 = 314.87, check
4.88 3.97 2.81 1.62 1.21
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 BM 20 TP 1 0+00 0+50 0+61.48 1+00 TP 2 1+50 2+00 BM 21
BS 8.27 9.21
IS
FS 2.60
11.3 9.6 8.71 6.1 7.33
195.32
4.66 5.8 4.97 3.88
BS =24.81
2.9
HI 186.04 192.65
ELEVATION 177.77 183.44 181.4 183.1 246.65 249.3 187.99 252.2 253.06 191.44
FS = 11.14 177.7 + 24.81 - 11.14 = 191.44 Check!
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√1000/5 280 = 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 BM 22 0+00
BS 1.203
HI 182.425
C L
10M LT., 10M RT., 0+20 10M LT., 7.3M LT. 4M LT. C L
4M RT. 7.8M RT. 10M RT. 0+40 10M LT. 6.2M LT. 4M LT. C L
4M RT. 6.8M RT. 10M RT. TP 1 2.11
IS
FS
ELEVATION 181.222
1.211
181.214
1.430 1.006
180.995 181.419
2.93 2.53 2.301 2.381
179.50 179.90 180.124 180.044
2.307 2.41 2.78
180.118 180.02 179.65
3.98 3.50 3.103 3.187
178.45 178.9 179.322 179.238
3.100 3.37 3.87
179.325 179.06 178.56 179.65
2.773
Reduce the set of municipal cross-section notes in Table 2-8. STATION BM 41 TP 13 12+00 50 ft. lt. 18.3 ft. lt.
BS 4.11 4.10
HI 307.104 310.314
C L
20.1 ft. rt. 50 ft. rt. 13+00 50 ft. lt. 19.6 ft. lt C L
20.7 ft. rt. 50 ft. rt. TP 14 BM S.22
7.39
IS
FS 0.89
ELEVATION 302.994 306.214
3.9 4.6 6.33
306.4 305.7 303.98
7.9 8.2
302.4 302.1
5.0 5.7 7.54
305.3 304.6 302.77
7.9 8.4
302.4 301.9 309.194 314.174
316.584
1.12 2.41
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
BM 37
7.20
385.17
FS
ELEV.
6+00
6+50
2.13
6.71
STATION
BS
HI
BM 107
7.71
406.87
FS
26.7 3.8
3.7
28.4 3.0
50 2.7
380.6
381.4
381.5
382.2
382.5
50 4.0
24.1 4.2
3.1
25.0 2.7
50 2.9
381.2
381.0
382.1
382.5
382.3
50 3.8
26.4 3.7
2.6
23.8 1.7
50 1.1
381.4
381.5
382.6
383.5
384.1
ELEV.
TP 1
LEFT
C L
RIGHT
399.16
81+50
(c) (d)
50 4.6
Complete the set of highway cross-section notes in Table 2-10.
81+00
(a) (b)
RIGHT
378.46
80+50
2.14
C L
377.97
5+50
TP 1
LEFT
10.17
60 9.7
28 8.0
5.7
32 4.3
60 4.0
397.2
398.9
401.2
402.6
402.9
60 10.1
25 9.7
6.8
30 6.0
60 5.3
396.8
397.2
400.1
400.9
401.6
60 11.7
27 11.0
9.2
33 8.3
60 8.0
395.2
395.9
397.7
398.6
398.9
396.70
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. What is the correct difference in elevation between the tops of stakes A and B? If the level had been in perfect adjustment, what reading would have been observed at A from the second setup? What is the line-of-sight error in 300 ft? 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)
b)
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 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.
a)
STATION
BS
HI
FS
ELEVATION
BM 130 TP 1 TP 2 TP 3 BM K110 TP 4 BM 132
0.702 0.970 0.559 1.744 1.973 1.927
189.269 189.128 189.008 187.972 188.277 188.416
1.111 0.679 2.780 1.668 1.788 0.888
188.567 188.158 188.449 186.228 186.304 186.489 187.528
BS = 7.875
FS = 8.914 188.567 + 7.875 - 8.914 = 187.528, check
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
b) STATION
CUMULATIVE DISTANCE
ELEVATION
CORRECTION
BM 130 188.567 TP 1 130 188.158 130/780 x.008 = +.001 TP 2 260 188.449 260/780 x.008 = +.003 TP 3 390 186.228 390/780 x.008 = +.004 BM K110 520 186.304 520/780 x.008 = +.005 TP 4 650 186.489 650/780 x.008 = +.007 BM 132 780 187.528 780/780 x.008 = +.008 C = 187.536 – 187.528 = - 0.008 The adjusted elevation of BM K110 is 186.309m
7
ADJUSTED ELEVATION 188.567 188.159 188.452 186.232 186.309 186.496 186.536
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 knee s, a distance of 2.0 ft above the ground. Determine the ground clearance of the electrical cable Clearance = 17.4 + 2.0 = 19.4 f t. (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
8
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 99.98 ft Difference in elevation = 6.35 ft 198.61 ft −14°F 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 219.51 ft Slope angle 3°18′ 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
Slope angle −3°42′ 26°C 29.993 m 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 9
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 349.65 ft Slope at −0.80% 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. 10
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 f t.
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). KL = 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. KM = 2200.00 Sin(-1°30') = - 57.59 ft. c+r = .0206 x 2.22 = +0.10 ft. Elevation difference, K to M, = - 57.49 f t. Elevation of M = 198.92- 52.49 = 146.43 ft. KN 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. 11
Chapter 04 – Angles and Directions 4.1
A closed five-sided field traverse has the following interior angles: A = 121°13′00″; B = 136°44′30″; C = 77 °05′30″; D = 94 °20′ 30″. Find the angle at E. (n-2)180 = 3 x 180 = 540°00' A B C D
= 121° 13’ 00" = 136° 44' 30" = 77° 05' 30" = 94° 20' 30" = 427° 23’ 30
539° 59' 60” - 427° 23' 30” E = 112° 36' 30”
4.2
Convert the following azimuths to bearings. (a) 210° 30′ S 30° 30' W (b) 128° 22′ S 51° 38' E (c) 157 ° 55′ S 22° 05’ E (d) 300° 30′ N 59° 30’ W (e) 0° 08′ N 0° 08’ E (f) 355° 10′ N 4° 50' W (g) 182° 00′ S 2°00’ W
4.3
Convert the following bearings to azimuths. (a) N 20° 20′ E 20° 20' (b) N 1° 33′ W 358° 27' (c) S 89° 28′ E 90°32' (d) S 82° 36′ W 262°36' (e) N 89° 29′ E 89° 29' (f) S 11° 38′ W 191° 38' (g) S 41° 14′ E 138° 46′
4.4
Convert the azimuths given in Problem 4.2 to reverse (back) azimuths. (a) 210° 30′ 30° 30' (b) 128° 22′ 308° 22' (c) 157 ° 55′ 337 55’ (d) 300° 30′ 120° 30’ (e) 0° 08′ 180° 08’ (f) 355° 10′ 175° 50' (g) 182° 00′ 2°00’
4.5
Convert the bearings given in Problem 4.3 to reverse (back) bearings. S 20° 20' W (a) N 20°20′ E S 1° 33' E (b) N 1°33′ W (c) S 89°28′ E N 89° 28' W N 82°36' E (d) S 82°36′ W S 89° 29' W (e) N 89°29′ E N 11° 38' E (f) S 11°38′ W N 41° 14′ W (g) S 41°14′ E
12
4.6
An open traverse that runs from A through H has the following deflection angles: B = 6°25′ R; C = 3°54′ R; D = 11°47′ R; E = 20°02′ L; F = 7°18′ L; G = 1°56′ R. If the bearing of AB is N 19°09′ E compute the bearings of the remaining sides. AB N 19° 09' E +B 6° 25' BC N25° 34'E +C 3° 54 ' CD N29° 28'E +D 11° 47' DE N41° 15'E
4.7
DE N 41° 15'E -E 20° 02' EF N 21° 13'E -F 7° 18' FG N13° 55'E +G 1° 56' GH N15° 51'E
Closed traverse ABCD has the following bearings: AB = N 45° 20′ E; BC = S 47 ° 32′ E; CD = S 14 ° 55′W; DA = N68°00′W Compute the interior angles and provide a geometric check for your work. Angle at A Angle at B Angle at C Angle at D GEOMETRIC CHECK 45° 20′ 45° 20’ 47° 32’ 14° 55’ A= 66° 40' + 68° 00' + 47° 32' + 14° 55’ +68° 00’ B= 92° 52’ 113° 20 ' B = 92° 52’ 62° 27’ D=82° 55’ C= 117° 33’ D= +82° 55’ 179°60' 179°60' Total = 360° 00’ Check - 113°20′ - 62° 27’ A = 66° 40' C=117° 33’
13
4.8
If the bearing of AB is N 42° 11′10″ E, compute the bearings of the remaining sides. Provide two solutions: one solution proceeding clockwise and the other solution proceeding counterclockwise. Clockwise Solution for Bearings AB N42° 11′ 10" E [Given] BC N 81° 42′ 20" E CD S 19° 48′ 00" E DE S 87° 23′ 50" W DA N 74° 01′ 50" W
4.9
If the azimuth of AB is 42°11′10″, compute the azimuths of the remaining sides. Pro -vide two solutions: one solution proceeding clockwise and the other proceeding counterclockwise. Clockwise Solution for Azimuths Azimuth AB 42° 11′ 10" [Given] BC 81° 42′ 20" CD 160° 12′00" DE 267° 23′ 50" EA 285° 58′ 10"
4.10
If the bearing of AB is N 45° 48′56″ E, compute the bearings of the remaining sides proceeding in a clockwis e direction. See Answer to Question 4.11
4.11
If the azimuth of AB is 45° 48′56″, compute the azimuths of the remaining sides pro -ceeding in a counterclockwise direction. Azimuth AB = 45°48'56" [Given] N 45° 48’ 56” E +A 63°47'00" Azimuth AE = 109°35'56" S 70° 24’ 04” E + 180°00’00” Azimuth EA = 289°35'56" N 70° 24’ 04” W +E 161°25'40" Azimuth ED = 451° 01'36" 360°00’00” Azimuth ED = 91° 01'36" S 88° 58’ 24” E + 180°00’00” Azimuth DE = 271° 01'36" N 88° 58’ 24” W +D 72° 48'10" Azimuth DC = 343°49'46" N 16° 10’ 14” W 180°00’00” Azimuth CD = 163°49'46" S 16° 10’ 14” E +C 101°30'20" Azimuth CB = 265° 20'06" S 85° 20’ 06” W 180°00’00” Azimuth BC = 85° 20'06" N 85° 20’ 06” E +B 140° 28'50" Azimuth BA = 225°48'56" S 45° 48’ 56” W 180°00’00” Azimuth AB = 48°48'56" CHECK
14
4.12
On January 2, 2000, the compass reading on a survey line in the Seattle, Washington, area was N 39° 30 ’ E.. Use Figure 4-16(a) to determine the following: (a) The compass reading for the same survey line on July 2, 2004. (b) The geodetic bearing of the survey line. a) The annual change in declination is 8' west [scaled from Figure 4.16 (a)] and on July 2, 2004, the change in declination would be 8 x 4.5(yrs.) = 36'. The compass reading on July 2, 2004 would be 39° 30' + 36' = N 40° 06' E. b) On Jan. 2, 2000, at Seattle, Wash., the declination (scaled from Figure 4.16a) is 18°50'. Therefore, the astronomic bearing = N39°30'E + 18°50' = N58° 20'E.
15
Chapter 06 – Traverse Surveys 6.1
A five-sided closed field traverse has the following angles: A = 106° 28′30″, B = 114°20′00″; C = 90°32′30″; D = 109° 33′30″; E = 119 ° 07′00″. Determine the angular error of closure and balance the angles by applying equal corrections to each angle. STATION FIELD ANGLES CORECTIONS ADJUSTED ANGLES A B C D E
106°28'30" 114°20'00" 90°32'30" 109°33'30" 119°07'00"
-18" -18" -18" -18" -18"
106°28'12" 114°19'42" 90°32′12" 109°33'12" 119°06'42"
Sum
540° 01' 30" + 90” -90" Error = 1'30", = 90"; Correction/angle = 18" 6.2
540°00'00 Balanced
A four-sided closed field traverse has the following angles: A = 81° 53′30″; B = 70° 28′30″; C = 86 ° 09′30″; D =121° 30′30″. The lengths of the sides are as follows: AB = 63 6.45 ft; BC = 654.45 ft; CD = 382.65 ft; DA = 469.38 ft. The bearing of AB is S 17° 17′ W. BC is in the N.E. quadrant. (a) Balance the field angles. (b) Compute the bearings or the azimuths. (c) Compute the latitudes and departures. (d) Determine the linear error of closure and the accuracy ratio. STATION A B C D
FIELD ANGLES
CORRECTIONS
ADJUSTED ANGLES
81° 53'30" 70° 28'30" 86° 09'30" 121° 30'30"
- 30" - 30" - 30" - 30"
81° 53' 70° 28' 86° 09' 121° 30'
360° 02’ 00” 120" Error = 02',or 120"; Correction/angle = -30"
360° 00’ ‘’”
Course
Azimuth
Bearing
Distance
Latitude
Departure
AB BC CD DA
197° 17' 87° 45' 353° 54' 295° 24'
S17° 17'W N87° 45'E N 6° 06'W N64° 36'W
636.45 654.45 382.65 469.38
-607.71 25.69 380.48 201.33
-189.09 653.95 - 40.66 - 424.01
2142.93
- 0.21
+ 0.19
E = √.212 + 0.192 = 0.283 f t. Precision ratio = E/P = 0.283/2142.93 = 1/7572 = 1/7600
16
6.3
Using the data from Problem 6.2, (a) Balance the latitudes and departures by use of the compass rule. (b) Compute the coordinates of A, C, and D if the coordinates of station B are 1,000.00 N, 1,000.00 E.
Course AB BC CD DA
CDep
Adjusted Lat.
-0.06 -0.06 -0.03 -0.04
- 607.64 + 25.76 +380.51 + 201.37
- 189.15 +653.89 - 40.69 - 424.05
+0.21
-0.19
0.00
0.00
NORTH
EAST
B
1000.00 + 25.76 1025.76 +380.51 1406.27 +201.37 1607.64 - 607.64 1000.00
1000.00 +653.89 1653.89 - 40.69 1613.20 - 424.05 1189.15 - 189.15 1000.00
D A B
Adjusted Dep.
+0.07 +0.07 +0.03 +0.04
STATION
C
6.4
CLat
Check
Using the data from Problems 6.2 and 6.3, compute the area method. Area by coordinates XB(YC-YA) = 1000.00(1025.76 –1607.64) = XC(YD-YB) = 1653.89(1406.27 -1000.00) = XD(YA-YC) = 1613.20(1607.64 – 1025.76) = XA(YB-YD) = 1189.15(1000.00 –1406.27) = Twice the area = Area = (1 Acre = 43,560 ft2) or, A =
17
enclosed by the traverse using the coordinate
- 581,880 + 671,926 + 938,689 - 483,116 545,619 2 272,810 ft 6.26 Acres
6.5
A five-sided closed field traverse has the following distances in meters: AB = 51.766; BC = 26.947; CD = 37.070; DE = 35.292; EA = 19.192. The adjusted angles are as follows: A = 101° 03′19″; B = 101 ° 41′49″; C = 102° 22′03″; D = 115 ° 57′20″; E = 118 ° 55′29″. The bearing of AB is N 73° 00′00″ E. BC is in the S.E. quadrant. (a) Compute the bearings or the azimuths. (b) Compute the latitudes and departures. (c) Determine the linear error of closure and the accuracy ratio. ANGLES A B C D E
101° 03' 19" 101° 41' 49" 102° 22' 03" 115° 57' 20" 118° 55' 29" 540°00'00" CHECK
Course
AZIMUTH
BEARING
DISTANCE
LATITUDE
DEPARTURE
AE ED DC CB BA
176° 08' 49" 115° 04' 18" 51° 01' 38" 333° 23' 41" 255° 05' 30"
S 3 ° 51' 11"E S64° 55' 42"E N51° 01' 38"E N26° 36' 19"W S75° 05' 30"W
19.192 35.292 37.070 26.947 51.766
- 19.149 - 14.955 +23.315 +24.094 - 13.318
+ 1.290 +31.967 +28.820 - 12.068 - 50.023
Sums
170.267
- 0.013
- 0.014
E = √0.0132 + 0.0142 = 0.019m Precision ratio = E/P = 0.019 / 170.267 = 1/8961 = 1/9000 6.6
Using the data from Problem 6.5: (a) Balance the latitudes and departures by use of the compass rule. (b) Compute the coordinates of the stations using coordinates of station A as 1,000.000N, 1,000.000E. Balance using Compass Rule COURSE CLAT CDEP Corrected Corrected LAT DEP AE ED DC CB BA
+.001 +.003 +.003 +.002 +.004
+.002 +.003 +.003 +.002 +.004
-19.148 -14.952 +23.318 +24.096 - 13.314
+ 1.292 +31.970 +28.823 - 12.066 - 50.019
Check
+.013
+.014
0.000
0.000
COORDINATES STATION A E D C B A
NORTH
EAST
1000.000 - 19.148 980.852 - 14.952 965.900 + 23.318 989.218 + 24.096 1013.314 - 13.314 1000.000
1000.000 + 1.292 1001.292 + 31.970 1033.262 + 28.823 1062.085 - 12.066 1050.019 - 50.019 1000.000 18
Check
6.7
6.8
6.9
Using the data from Problems 6.5 and 6.6, compute the area method. Area by the Coordinate Method XA(YB-YE) = 1000.000(10 13.314 - 980.852) XE(YA-YD) = 1001.292(1000.000 – 965.900) XD(YE-YC) = 1033.262(980.852 – 989.218) XC(YD-YB) = 1062.085(965.900 – 1013.314) XB(YC-YA) = 1050.019(989.218 - 1000.000) Twice the Area Area
enclosed by the traverse using the coordinate
= = = = = = =
+ 32,462 + 34,144 - 8,540 - 50,358 - 11,321 2 3,713 m 2 1,857 m
The two frontage corners of a large tract of land were joined by the following open traverse: Course Distance (ft) Bearing AB 100.00 N 70° 40′05″ E BC 953.83 N 74° 29′00″ E CD 818.49 N 70° 22′45″ Compute the distance and bearing of the property frontage AD. COURSE
BEARING
AB BC CD
N70°40'05"E N74°29'00"E N70°22'45"E
DISTANCE 100.00 953.83 81 8.49
LATITUDE
DEPARTURE
+ 33.10 +255.17 +274.84
+94.36 +919.07 +770.96
AD +563.11 Distance AD = √563.112 + 1784.392 = 1,871.13 ft. Tan Bearing AD = 1784.39/563.11 = 3.168812 Bearing AD = N72°29'08"E
+1784.39
Given the following data for a closed property traverse: Course Bearing Distance (m) AB N 37° 10′49″ E 537.144 BC N 79° 29′49″ E 1109.301 CD S 18° 56′31″ W ? DE ? 953.829 EA N 21° 10′08″ W 477.705 Compute the missing data (i.e., distance CD and bearing DE), and verify the results by summing the latitudes and departures of the property traverse. Solve for Distance and Direction from E to C COURSE BEARING DISTANCE LATITUDE EA AB BC
N21°10'08"W N37°10'49"E N79°29'49"E
477.705 537.144 1109.301
DEPARTURE
+445.469 +427.963 +202.212
-172.508 +324.609 +1090.715
EC +1075.644 Distance EC = √1075.6442 + 1242.8162 = 1,643.655m Tan Bearing EC = 1242.816/1075.644; Bearing EC = N49°07'27"E
+1242.816
Using Triangle ECD Find angle at C, = 49° 07'27" - 18°56'31" Angle @ C = 30° 10'56"
19
Solve for angle at D Sin D = Sin 30°10'56"/953.829 x 1643.655 D = 60° 02'14" or 119°57’46" From the given data, D = 119°57'46" Therefore E = 180° - (119°57'46"+30°10'56") = 29°51'18" CD = Sin 29°51'18"(1643.655/Sin 119°57'46") = 944.448 m Bearing DE = S78°58'45"W Verification: Course Bearing Distance Latitude Departure EA AB BC CD DE
N21°10'08"W N37°10'49"E N79°29'49"E S18°56’31”W S78°58’45”W
477.705 537.144 1109.301 944.448 953.829
+445.469 +427.963 +202.212 - 893.304 - 182.340
-172.508 +324.609 +1090.715 -306.577 - 936.238
0.000
+ 0.001≈ 0.000 (rounding errors)
6.10
A six-sided traverse has the following station coordinates: A (559.319 N, 207.453 E); B (738.562 N, 666.737 E); C (541.742 N, 688.350 E); D (379.861 N, 839.008 E); E (296.099 N, 604.048 E); F (218.330 N, 323.936 E). Compute the distance and bearing of each side. 2 2 AB = √ (738.562-559.319) +(666.737-207.453) = 493.021m Tan Brg AB = 459.284/179.243; Brg. AB = N68°40'52"E 2 2 BC = √ (541.742-738.562) +(688.350-666.737) = 198.003m Tan Brg BC = 21.613/-196.820; Brg. BC = S 6°16'00"E 2 2 CD = √ (379.861-541.742) +(839.008-688.350) = 221.141m Tan Brg CD = 150.658/-161.881; Brg. CD = S42°56'36"E 2 2 DE = √ (296.099-379.861) +(604.048-839.008) = 249.444m Tan Brg DE = - 234.960/-83.762; Brg. DE = S70°22'45"W 2 2 EF = √ (218.330-296.099) +(323.936-604.048) = 290.707m Tan Brg EF = -280.112/-77.769; Brg. EF = S74°29'00"W 2 2 FA = √ (559.319-218.330) +(207.453-323.936) = 360.336m Tan Brg FA = -116.483/340 .989; Brg. FA = N18°51'37"W
6.11
Using the data from Problem 6.10, compute the area enclosed by the traverse. XA(YB-YF) = 207.453(738.562-218.330) = +107,924 XB(YC-YA) = 666.737(541.742-559.319) = - 11,719 XC(YD-YB) = 688.350(379.861-738.562) = -246,912 XD(YE-YC) = 839.008(296.099-541.742) = -206,096 XE(YF-YD) = 604.048(218.330-379.861) = - 97,572 XF(YA-YE) = 323.936(559.319-296.099) = + 85,266 2 Twice the Area = 369,109m Area = 184,555m2 Area = 18.455 hectares
20
6.12
Use the data from Problem 6.10 to solve the following. If the intersection point of lines AD and BF is K, and if the intersection point of lines AC and BE is L, compute the distance and bearing of line KL. LINE AD Y-559.319 = (379.861-559.319)/(839.008-207.453)x (X-207.453) (1) 0.28415X + Y = 618.2673 LINE FB Y-218.330 = (738.562-218.330)/(666.737-323.936)x (X-323.936) (2) 1.51759X - Y = 273.2726 1.80174X = 891.5399 (1) + (2) X = 494.8204 Y = 477.6628 Coordinates of K = 477.633N, 494.820E LINE EB Y-296.099 = (738.562-296.099)/(666.737-604.048)x (X-604.048) (3) 7.05806X - Y = 3967.3106 LINE AC Y-559.319 = (541.742-559.319)/(688.350-207.453)x (X-207.453) (4) 0.03655X + Y = 566.9015 7.0946X = 4534.212 (3) + (4) X = 639.106 Y = 543.542 Coordinates of L = 543.542N, 639.106E Tan Brg KL = (639.106-494.820)/(543.542-477.663); Brg KL = N65°27'33"E KL = √ 144.2862 + 65.8792 = 158.614m
6.13
A five-sided field traverse has the following balanced angles: A = 101° 28′26″; B = 102 ° 10′42″; C = 104 ° 42′06″; D = 113° 04′42″; E = 118° 34′04″. The lengths of the sides are as follows: AB = 50 .276 m; BC = 26.947 m; CD = 37.090 m; DE = 35.292 m; EA = 20.845 m. The bearing of EA is N 20° 20′20″ W, and AB is oriented northeasterly. (a) Compute the bearings. (b) Compute the latitudes and departures. (c) Compute the linear error of closure and the accuracy ratio. COURSE
AZIMUTH
BEARING
DISTANCE
LATITUDE
DEPARTURE
AE ED DC CB BA
159°39'40" 98°13'44" 31°18'26" 316°00'32" 238°11'14"
S20°20'20"E S81°46'16"E N31°18'26"E N43°59'28"W S58°11'14"W
20.845 35.292 37.090 26.947 50.276
-19.545 - 5.051 +31.689 +19.387 -26.503
+ 7.245 +34.929 +19.273 -18.716 -42.723
170.450 E = √ 0.0232 + 0.0082 = 0.0244m Precision ratio = E/P = 0.0244/170.450 = 1/7000
- 0.023
+ 0.008
21
6.14
From the data in Problem 6.13, balance the latitudes and departures employing the compass rule. COURSE CLAT CDEP Corrected Corrected LAT. DEP. AE ED DC CB BA CHECK
6.15
+0.003 +0.005 +0.005 +0.004 +0.006
-0.001 -0.002 -0.002 -0.001 -0.002
-19.542 - 5.046 +31.694 +19.391 -26.497
+ 7.244 +34.927 +19.271 - 18.717 - 42.725
+0.023
-0.008
0.000
0.000
From the data in Problems 6.13 and 6.14, if the coordinates of station A are (1,000.000 N, 1,000.000 E), compute the coordinates of the other stations. Calculate COORDINATES STATION NORTH A E D C B A
6.16
EAST
1000.000 - 19.542 980.458 - 5.046 975.412 + 31.694 1007.106 + 19.391 1026.497 - 26.497
1000.000 + 7.244 1007.244 + 34.927 1042.171 + 19.271 1061.442 - 18.717 1042.725 - 42.725
1000.000
1000.000
CHECK
From the data in Problems 6.14 and 6.15, use the coordinate method to compute the area of the enclosed figure. XA(YB-YE) = 1000.000(1026.497-980 .458) = +46,039 XE(YA-YD) = 1007.244(1000.000-975 .412) = +24,766 XD(YE-YC) = 1042.171(980.458-1007 .106) = - 27,772 XC(YD-YB) = 1061.442(975.412-1026 .497) = - 54,224 XB(YC-YA) = 1042.725(1007.106-100 0.000) = + 7,410 2 Twice the area = 3,781 m 2 AREA = 1,890 m
22
6.17
See Figure 6.25. A total station was set up at control station K that is within the limits of a five-sided property. Coordinates of control station K are 1,990.000 N, 2,033.000 E. Azimuth angles and distances to the five property corners were determined as follows: Direction Azimuth Horizontal distance (m) KA 286° 51′00″ 34.482 KB 37° 35′28” 31.892 KC 90° 27′56″ 38.286 KD 166° 26′49″ 30.916 KE 247° 28′43″ 32.585 Compute the coordinates of the property corners A, B, C, D, and E. STATION
NORTH
EAST
K
1990.000 + 10.000
2033.000 - 33.000
A K
2000.000 1990.000 + 25.271
2000.000 2033.000 + 19.455
B K
2015.271 1990.000 - 0.311
2052.455 2033.000 + 38.285
C K
1989.689 1990.000 - 30.055
2071.285 2033.000 + 7.245
D K
1959.945 1990.000 - 12.481
2040.245 2033.000 - 30.100
E
1977.519
2002.900
6.18
Using the data from Problem 6.17, compute the area of the property. XA(YE-YB) = 2000.000(1977.519-201 5.271) = - 75,504 XB(YA-YC) = 2052.455(2000.000-198 9.689) = + 21,163 XC(YB-YD) = 2071.285(2015.271-195 8.945) = +114,596 XD(YC-YE) = 2040.245(1989.689-197 7.519) = + 24,830 XE(YD-YA) = 2002.900(1959.945-200 0.000) = - 80,226 2 Twice the Area = 4,859 m 2 AREA = 2,430 m
6.19
Using the data from Problem 6.17, compute the bearings and distances of the five sides of the property. COURSE
BEARING
DISTANCE (m)
AB BC CD DE EA
N73°46'06"E S36°21'20"E S46°13'17"W N64°47'56"W N 7°20'52"W
54.633 31.765 42.991 41.273 22.675
23
Chapter 07 – Satellite Positioning Systems 7.1
Why is it necessary to observe a minimum of four GPS satellites to solve for position? GPS pseudoranging requires the receiver clock to be synchronized with satellite clock. As the receiver clock chip is not synchronized to GPS time, there are basically four unknowns that must be solved for, the three positions (X,Y,Z) and time.
7.2
How does the United Sta tes’ GPS constellation compare with the GLONASS constellation and with the proposed Galileo constellation? At the time of this writing (June 2011), the GPS system, is the only system that has a full constellation of 24 satellites operational. (A full constellation provides full coverage all the time). The GPS system currently has 31 operational satellites. The GPS system primary design is for 60 orbit planes inclined 55° from the equator, with 4 satellites in each plane. The Russian GLONASS system currently has 22 operational satellites, but should soon have the full constellation of 24 (It had a full constellation when it was operated by USSR, but went into d isrepair during the breakup of the USSR). The system has 3 orbital planes with 8 satellites each and has a 64.8° incline. This system gives better coverage in high northern latitudes than the GPS system. The European Galileo system currently has only two test satellites launched. It anticipates having its first 4 operational satellites up by the end of 2011. It has a planned constellation of 30 satellites in three orbital planes with a 56° inclination.
7.3
How does differential positioning work? Differential GPS (DGPS) works by using two GPS receivers. One is set as a base station at a known locat ion, which may be a precise location or just location determined by autonomous GPS positioning. The second receiver, the rover, is the position that is trying to be determined. The position of the rover is determined relative to the “known” position of th e base station. At any given epoch the difference between the known position at the base station and the GPS computed position can be used to determine the error correction to be applied to the rover for that epoch. This can give relative horizontal position accuracy of the rover of 1 meter. The further the rover is from the base station the larger the error due to differences in atmosphere. However in GPS positioning a distance 100 kilometers is a short distance.
7.4
What is the difference between range and pseudorange? Range in GPS is a term for determining the distance from the satellite to the receiver. G enerally GPS determines the distance from the satellite to the receiver by ca lculating the time shift in the code. This is actually measuring time and then computing distance using the speed of light, so it is known as a “pseudorange”. In carrier phase differential solutions, time is not used, but the count of wavelengths and the partial wavelength, so it is considered a direct di stance measurement or “range”.
7.5
What are the chief sources of error in GPS measurements? How can you minimize or eliminate each of these errors? Time Difference: This can be minimized by doing differential GPS or eliminated by doing Ca rrier Phase Differential GPS. Atmosphere: Again this can be minimized by differential GPS over relatively short distances (100km) or the error can be minimized by an error model when using dual frequency GPS systems over large distances. Ephemeris: The Ephemeris is the information on the satellite orbit and is part of the signal the satellite sends, known as the broadcast ephemeris. This is the predicted orbit and is not necessarily its true orbit. This broadcast ephemeris gets worse over time. Orbits based on the monitoring stations tracking of the satellites can be downloaded after the observations. There is the rapid orbit available about 24 hours after the observation and the precise orbit available about 12 days after the observation. Receiver noise: The power used to broadcast the GPS sig nal is minimal and by the time it gets to the receiver it is even lower. Natural and human sources generate energy in and around the frequency used by the GPS signals. The receiver must isolate out the GPS signal from all that noise. Static surveys collect the signal information over a long period of time, which minimizes the error caused by the receiver noise. 24
Multipath: Multipath is the error caused by the receivers antenna not only picking up the signal directly from the satellite, but also picking up the signal as it bounces off surfaces near the antenna. This can be minimalized by not setting points near smooth surfaces, such as roads and buildings. Luckily there are other ways to minimize it, such a s specially designed antennas and more sophisticated s oftware design. 7.6
What are the factors that must be analyzed in GPS planning? GPS Planning has two purposes, minimizing potential errors and ensuring successful occ upation. Planning for a static or rapid static GPS survey is usually more rigorous than planning for a RTK survey. For an RTK Survey factors that must be considered are; What to use for a base station if you are not using your own What coordinate system is the survey going to be based on Are there areas that may have significant obstructions Are there features such as buildings that must be located What to do if communications fail For a static or rapid static survey the process is; Determine the location of control to be set Determine existing control to be located If elevations are a factor determine what benchmarks to use Determine and map all obstructions Determine the best time for occupations on each site Plan order of observations
7.7
Describe RTK techniques used for a layout survey. For RTK layout or construction surveys, a two person crew is more efficient than a one person crew. Once the rover has resolved its position relative to the base and “fixed”, the rover must get on the project coordinate system by doing a transformation or localization onto existing control. Verify the solution by staking out to control not used for the transformation. When setting stakes, have one person using the RTK system and software to walk to the desired point and determine the location, close enough for setting the wood. The second person should drive in the wood (or whatever is used as the mark er) and then the final measurement is taken with the RTK. The RTK person calls out the information to be marked on the lath, and the second person mark s the lath and then repeats the markings to be verified by the RTK person. Depending on the type of the layout, the RTK person may be able to estimate the location of the next stakeout by pacing. During the survey if there is a ny loss of power, or a software issue arises, as soon as the problem is resolved, a check on a previously set point should be made to verify the integrity of the solution.
7.8
Which positioning method is best for precise control surveys and why. For large distances a Static or Rapid Static GPS survey process is best. The solution errors involved in G PS positioning when doing these types of surveys do not change much whether the points a re 100 feet or 100 miles apart. However there are a number of potential systematic errors and blunders that can occur. These types of surveys lend themselves to redundant measurements which first allows for blunder detection and second provides for a rigorous least squares solution for the random errors.
7.9
Explain why station visibility diagrams are used in survey planning. A visibility diagram is a diagra m that shows what obstructions there are at each monument that is going to be occupied. An obstruction is considered anything that blocks open sky above the elevation mask being used (usually around 10°). This would include buildings, trees and mountains. With this information, GPS planning software can determine which satellites are avai lable for observation over the complete time span of the project. This allows for the planning of the best times to occupy any given station.
25
7.10
Explain the difference between orthometric heights and ellipsoid heights. Orthometric heights are what people think of as elevations and are heights above a vertical datum that is based on the geoid. These heights are relative to gravity. Ellipsoid heights are heights above the ellipsoid used for a horizontal datum. This height is measured perpendicular to the ellipsoid surface. GPS measures ellipsoid height above the WGS84 datum. Often your GPS receiver has the ability to provide you with the Orthometric height based on a Geoid model. A Geoid model is a file that contains positions and the geoid separation at each position. The geoid separation is the vertical difference between the geoid and ellipsoid measured perpendicular to the ellipsoid. The software can then interpolate to estimate what the separation is at the position of the G PS. This is just estimation, although the most current models are generally providing high quality estimations.
7.11
Explain the CORS system The CORS (Continuously Operation Reference Station) system is a series of GPS stations managed by NGS that continually operate. These stations are what are currently being used a s the backbone of the horizontal control system for the United States. NGS can determine the relative rate of positional change between these stations and the NAD83 datum, so that a velocity and direction are provided for each station. The GPS data from these stations are freely available from the NGS website. Besides the CORS system there are thousands more of these types of stations commonly called CGPS ( Continuous GPS) that are managed by other groups, primarily for monitoring earth movement. However this da ta is also freely available.
7.12
For RTK to work, what do we need besides 2 or more receivers collecting data from a sufficient number of satellites simultaneously? RTK (Real Time Kinematic) surveys require communication between the base station and the rover. This communication can be via radio or cell phone, however it must be uninterrupted. This is one more component that makes RTK surveys susceptible to malfunctions.
7.13
Assume a horizontal positional accuracy for the determination of a point using GPS to be 2 cm, and the vertical positional accuracy to be 3 cm. How many meters would two points have to be apart before I could be assured that the error in slope is less than 1.0%? Assume the slope is flat, with a worst case scenario for the vertical positional a ccuracy. This would mean that one point would be 0.03 m to low and the other would be 0.03 meter too high. This gives a worst case vertical error of 0.06 meters. So any horizontal distance that is 6.00 meters or greater would give a 1% or less error. However the horizontal positional accuracy has not been taken into account. Worst case for this example, would be the each point is 0.02 m closer to each other than indicated. This means at 6.00 meters they would only be 5.96 meters apart. Therefore the minimum distance to have a 1% error or less would require the GPS positions to be 6.04 meters a part.
7.14
What type of GPS process gets positions within 1 to 2 meters? This would be Differential GPS where two receivers a re used, with one receiver at a known position and the other determining relative position by noting the error at the base station (see a nswer to Question 7.3). This still uses pseudoranging to determine position.
7.15
What is the geometric indicator of precision used by the GPS system? The geometric indicator is called Dilution of Precision (DOP ).It is a measure of the positional solution based solely on the geometry between the receiver and the satellites. The lower the DOP the better the geometry and therefore the solution. There are a number of types of DOP measurements, Horizontal (HDOP), Vertical (VDOP), Time (TDOP), relative (RDOP) and then the two most commonly used in surveying Geometric (GDOP) and Position (PDOP).
26
Chapter 08 – Topographic and Hydrographic Surveying and Mapping 8.1
Establish the grid, plot the elevations (the decimal point is the plot point from Table 8-5), and interpolate the data to establish contours at 1-m (1-ft) intervals. Scale at 1:500 for metric units, or 1 in. = 10 ft or 15 ft for foot units. Use pencil. TRAVERSE COMPUTATIONS COURSE BEARING AB BC CD DE EF FG GA
8.2
DISTANCE
N. 3°30'E. N. 0°30'W. N.88°40'E. S.23°30'E. S.28°53'W. SOUTH N.83°37'W.
56.05 61.92 100.02 31.78 69.11 39.73 82.67
LATITUDE
DEPARTURE
+ 55.95 + 61.92 + 2.33 - 29.14 - 60.51 - 39.73 + 9.18
+ 3.42 - 0.54 + 99.99 + 12.67 + 33.38 0 - 82.16
Compute the interior angles of the traverse and check for geometric closure, that is (n − 2)180° (see Table 8 5). INTERIOR ANGLES A 92°53' B 184°00' C 90°50' D 112°10' E 127°37' F 208°53' G 83°37' Sum 900°00' CHECK; (n-2)x180 = 5 x 180 = 900°
8.3
Plot the traverse, using the interior angles and the given distances. Scale as in Problem 8.1 (see Table 8-7).
8.4
Plot the detail using the plotted traverse as control. Scale as in Problem 8.1. See Table 8-7.
27
8.5
Determine the area enclosed by the traverse in m2 (ft2) by using one or more of the following methods. a. Use grid paper as an overlay or underlay. Count the squares and partial squares enclosed by the traverse, and determine the area represented by one square at the chosen scale. From that relationship, determine the area enclosed by the traverse. b. Use a planimeter to determine the area. c. Divide the traverse into regular-shaped figures (squares, rectangles, trapezoids, triangles) and use a scale to determine the figure dimensions. Calculate the areas of the individual figures and add them to produce the overall traverse area. d. Use the balanced traverse data in Table 8-4 and the technique of coordinates to compute the area enclosed by the traverse. e. Use the software program in the total station (or its controller) to compute the area.
AREA COMPUTATIONS First, compute coordinates - assume coordinates of 1,000.00 N and 1,000.00 E for station A Station
Northing
Easting
A
1,000.00 + 55.95 1,055.95 61.92 1,117.87 + 2.33 1,120.20 - 29.14 1,091.06 - 60.51 1,030.55 - 39.73 990.82 + 9.18
1,000.00 + 3.42 1,003.42 0.54 1,002.88 + 99.99 1,102.87 + 12.67 1,115.54 - 33.38 1,082.16 0.00 1,082.16 - 82.16
1,000.00
1,000.00
B C D E F G A
Check
Second, compute the area: XA(YB - YG) = 1,000.00(1,055.95 - 990.82) = XB(YC - YA) = 1,003.42(1,117.87 - 1,000.00) = XC(YD - YB) = 1,002.88(1,120.20 - 1,055.95) = XD(YE - YC) = 1,102.87(1,091.06 - 1,117.87) = XE(YF - YD) = 1,115.54(1,030.55 - 1,120.20) = XF(YG - YE) = 1,082.16(990.82 - 1,091.06) = XG(YA - YF) = 1,082.16(1000.00 - 1,030.55) = Double area = Area =
28
65,130 118,273 64,435 - 29,568 -100,008 - 108,476 - 33,060 - 23,274 2 2 11,637 ft or m
8.6
Draw profile A to E showing both the existing ground and the proposed CL of highway. Use the following scales: metric units —horizontal is 1:500, vertical is 1:100; foot units —horizontal is 1 in. = 10 ft or 15 ft, vertical is 1 in. = 2 ft or 3 ft.
8.7
A highway is to be constructed to pass through points A and E of the traverse. The proposed highway CL grade is +2.30 percent rising from A to E (CL elevation at A = 68.95). The proposed cut and fill sections are shown in Figure 8-36(a) and (b). Plot the highway CL and 16 m (ft) width on the plan (see Problems 8.1, 8.3, and 8.4). Show the limits of cut and fill on the plan.
29
Chapter 09 – Geographic Information Systems 9.1
How do scale and resolution affect the creation and operation of a GIS? Scale pertains to GIS design at two levels; first deals with at what display scale should certain information be displayed. This affects the detail of the information shown and the look of the data. The second issue with scale, is what resolution can values be stored at. If the scale is going to be such that the whole country or continent needs to be shown, there is a limit as to the precision that i nformation can be shown at. There is a limit as to how many significant figures can be stored. It is not practical to show the whole continent but then to have resolutions to the nearest millimeter. So, when designing a G IS some thought as to the coverage and quality of information must be taken into consideration.
9.2
What are the various ways that scale can be shown on a map? Directly scale can be shown by a scale bar, which shows ground distances along a bar at specific intervals. Scale can be shown by text, such as a representative fraction 1:24,000 which is unit less. The value on the left “1” indicates the measure on the map, and the value on the right “24, 000” indicates the measure on the ground. 1 cm on the map equals 24,000 cm on the ground. Text scales can also have units, such as 1inch=2000feet. In this case 1inch on the map equals 2000 feet on the ground. Note this is the sa me as 1:24,000 as there are 24,000 inches in 2000 feet. Additionally scale of a map ca n be determined by having a k nown distance on the ground and measuring that distance on the map. Say the distance between two fence corners was ½ mile ( 2,640 feet). The distance on a map measures 33.53 mm. The scale can be calculated as shown below; Scale = (33.53MM/2440ft) x (1ft/304.8mm) Scale = 33.53/804492 ≈ 1/24000 Therefore the scale = 1:24,000
9.3
When is it advantageous to use cylindrical projections? To use conical projections? (See also Chapter 10) Projections, which are two dimensional models of a three dimensional surface, always h ave some distortion. A cylindrical projection tends to minimize that distortion in a north south direction while a conical projection minimizes the distortion in an east west direction. Therefore if the area being covered is long in a north south direction a cylindrical projection would work best while a conical projection works best in long east west direction.
9.4
Which features are best modeled using vector techniques? Using raster techniques? Vectors work best for features that can be defined as a point, line or polygon, while raster works best for features with vast amount of information such as an image or a surface. Vector data has specific geometry associated with it, so at any scale vector data will look the same (except for change in scale), while raster data has a specific cell size and each cell has a single value associated with it. So as the scale gets larger (more detail) at some point the edges of the cells will become apparent creating what is called a “pixelated” image, havin g jagged edges along cell boundary. Generally attributes are associated with vector data, a s a single feature has specific inf ormation tied to it. However there are thematic raster images where there are a finite number of values associated with the cells. These are considered attributes.
30
9.5
List and describe as many GIS applications as you can. This list is endless these days, but some common ones are: • Car Direction systems • Google Earth • Weather predictions • Survey Cadastre • Census • Fire management • Urban and Regional Planning • Facilities management • Military operations • Precision Agriculture • Forestry management • Logistics • Utility Mapping • Environmental Management • Disaster Management
9.6
Accepting that GIS is difficult to define, in your own words describe GIS, as you see it.
ESRI Website defines GIS as: A geographic information system (GIS) integrates hardware, software, and data for capturing, managing, analyzing, and displaying all forms of geographically referenced information. GIS allows us to view, understand, question, interpret, and visualize data in many ways that reveal relationships, patterns, and trends in the form of maps, globes, reports, and charts. A GIS helps you answer questions and solve problems by looking at your data in a way that is quickly understood and easily shared. GIS technology can be integrated into any enterprise information system framework. 9.7
What are the similarities and differences between mapping and GIS? The similarities are that both are presenting information in a graphical representation. Both require a knowledge of cartography in order to clearly present the information. Both have the ability to graphically display analysis based on the data collected and mapped. GIS is different as there is non-graphical information that is linked to the graphical information that provides for sophisticated analysis and gives the map a n intelligence. GIS also has the a bility to present the data at different scales on the fly and automatically update changes as they are mapped.
9.8
What are the similarities and differences between GIS and CAD? As mentioned in the question above, the similarities are that both are presenting information in a graphical representation. Both require a knowledge of cartography in order to clearly present the information. Both have the ability to graphically display analysis based on the data collected and mapped. Within the CAD environment the ability to edit graphical features is very sophisticated as well as the ability to extract geometric information about the features. GIS has more sophisticated capabilities of working with non-graphical date and in analysis of data both graphical and non-graphical. These days the line between CAD and GI S is getting fuzzier, however most surveying and engineering design is still done with CAD while project planning and management is done with GIS 31
9.9
What modern developments enabled the creation of GIS? The two primary developments were the computer and database software. Modern GIS a lso relies on graphical monitors, servers and high speed communication.
9.10
Why is it important to tie spatial data to a recognized reference system? In surveying it is common to create local or assumed spatial reference systems. This just means that the coordinate system is random, such as giving point #1 a value of North 10,000.00 and East 10,000.00. This works fine for the project that the reference system is being used for, and often for other nearby projects where some overlapping control is surveyed in. With GIS one of the major benefits is the ability to use many different data sources. However, if there is no spatial relationship between the sources, that benefit is lost. Therefore the benefit of surveying in to control based on a known spatial reference system outweighs the cost of the additional survey, especially these days with the tools that are available.
9.11
Why is it that the conversion of raster data to vector data has the potential for locational errors? There are two primary sources of locational error when converting raster data to vector data. The first is because Raster data is cell or pixel based which means that any location is only as precise as that location is to the center of the nearest pixel. If pixel size is 1meter x 1 meter than a ll conversions are going to be to the nearest 0.5 meters. The second source is that ability for the software or digitizer to correctly interpret the position of the feature being converted. Although at a smaller scale, when looking at a raster image the lines of say the corner of a building may look sharp, when zoomed down to the pixel, it is not always clear which pixel represents that corner.
9.12
How do you measure polygon areas in a vector model? In a raster model? Computing the area of a polygon based on a vector model is just the process of using coordinate geometry to compute area. See chapter 6 for methods for area calculations using coordinates. Computing the area of a Polygon based on a raster model requires counting the pixels that a re within the polygon, and then multiplying by the area of each pixel.
9.13
What would the appropriate positional precision be for a GIS showing the following information When determining positional precision, it should always be based on the specifications associated with a project, but the answers below represent general guidelines Watershed limits It is common to use USGS topographic quadrangle maps for determining watershed limits. On those maps the positional accuracy is roughly 100 feet, which should work for large watersheds. Smaller watersheds might require a precision of 20-30 feet. Assessing the change of Redwood Growths in western U.S. This would require very low precision; again the precision of USGS topographic quadrangle map would be sufficient. Mapping Row crops in the Midwest Most row crops in the Midwest are on large plots such as a ¼ square mile. To map these over the entire Midwest, the precision of the Landsat satellite system would be appropriate (30m x 30m) 32
Locating Water and Sewer features in a mid-size city. If the location was just for managing maintenance of the system, the precision would probably by 1 meter, which can be achieved using DGPS units (See Chapter 7). This would allow city crews to quickly find surface features of the systems. It would not be precise enough for digging up a pipeline, but with the surface features such as valves located the underground feature locations could be determined. Determining Legal Street Right of Ways in a mid-size city This requires the highest precision possible, beyond what could be achieved even with a survey grade RTK GPS system. More often RTK GPS is used to determine very good approximations of where the rights of way are and then if there appears to be an issue with the right of way, a survey crew will be sent out to determine the legal boundary.
9.14
For the list in the above question what might be the appropriate tool for collecting the positional information for each item listed. Watershed limits USGS Topographic Quadrangle Maps Assessing the change of Redwood Growths in western U.S. NAIP Aerial Photos (See Chapter 12) Mapping Row crops in the Midwest LANDSAT Images (See Chapter 11) Locating Water and Sewer features in a mid-size city. GIS Grade GPS Determining Legal Street Right of Ways in a mid-size city Survey Crew experienced with Boundaries
9.15
If a digital image covered an area of 2 km. by 2 km. and the pixel size is one meter, how many pixels are there in the total image? If each side is 2 km and the pixel size is one meter, there are 2000 pixels along each side, so the count would be 2000 x 2000 equaling 4,000,000 pixels. This would be for each layer. A color image would have three layers (Red, Blue, Green), so there would be 12,000,000 pixels.
9.16
In the question above, if the pixel size is 0.5 meter, how many pixels are there? If the pixel is half the size it will take 4 pixels to cover 1 of the larger pixels, so the number of pixels per layer would be 4 times larger or 16,000,000 pixels. This shows that there is a large storag e and processing cost with high resolution images
. 9.17
What coordinate system would you expect a Digital Raster Graphic (DRG) from USGS of a 7 ½ minute Quad sheet to be on? The USGS Hardcopy maps are based on NAD27 datum and have both State Plane and UTM coordinate systems on the map. The DRG’s uses the UTM coordinate for their projection.
33
Chapter 10 – Control Surveys 10.1
Draw a representative sketch of the four control points and then determine the grid distances and grid bearings of sides AB, BC, CD, and DA. Monument
Distance
Azimuth
Bearing
A 290.722
254°29'00"
S 74.° 29' 00" W
225.931
260°50'10"
S 80.° 50' 10" W
206.038
89°15'59"
N 89.° 15' 59" E
317.248
69.29'50"
N 69.° 29' 50" E
B C D A
10.2
From the grid bearings computed in Problem 10.1, compute the interior angles (and their sum) of the Traverse A, B, C, D, A, thus verifying the correctness of the grid bearing computations. Angle at A 254°29'00" A->B A 4°59'10" 249°29'50" A->D B 186°21'10" 4°59'10" C 8°25'49" D 160°13'51" Angle at B 260°50'10" B->C Sum = 360°00'00" 74°29'00" B->A 186°21'10" Angle at C -
89°15'59" 80°50'10" 8°25'49"
C->D C->B
-
429°29'50" 269°15'59" 160°13'51"
D->A D->C
Angle at D
34
10.3
Determine the ground distances for the four traverse sides by applying the scale and elevation factors (i.e., grid factors). Use average latitude and longitude. Scale factor at Mo CM East Radius 20906000ft = St a A
Elev
0.9999 304,800 m 6372161.544m
Grid Dist
Ave Elev
Elev. SF
Average East
Dist From CM
Scale Factor
Combined
Ground Dist
290.722
180.594
0.999972
316,963.999
12163.999
0.999902
0.999873
290.759
225.931
185.166
0.999971
316,712.413
11912.413
0.999902
0.999873
225.959
188.214
0.999970
316,703.900
11903.900
0.999902
0.999872
179.832
B
181.356
C
188.976 206.038
D
187.452
A
179.832
317.248
10.4
206.064 183.642
0.999971
316,955.486
12155.486
0.999902
0.999873
17.288
Determine the convergence correction for each traverse side and determine the geodetic bearings for each traverse side. Use average latitude and longitude. Central Meridian Average Latitude Average Longitude Delta Longitude Delta Longitude Sec Station A
79°.30'00" 43°.47'30" 79°.21'02" 0°.08'58" 538 Grid Azimuth
Geodetic Az.
Geodetic Brg
254°.29'00"
254°.22'48"
S 74.° 22' 48" W
260°.50'10"
260°.43'58"
S 80.° 43' 58" W
89°.15'59"
89°.09'47"
N 89.° 09' 47" E
69°.29'50"
69°.23'38"
N 69.° 23' 38" E
B C D
35
10.5
From the geodetic bearings computed in Problem 10.4, compute the interior angles (and their sum) of the traverse A, B, C, D, A, thus verifying the correctness of the geodetic bearing computations.
Angle at A -
254°22'48" 249°23'38" 4°59'10"
A->B A->D
-
260°43'58" 74°22'48" 186°21'10"
B->C B->A
-
89°09'47" 80°43'58" 8°25'49"
C->D C->B
-
429°23'38" 269°09'47" 160°13'51"
D->A D->C
Angle at B
Angle at C
Angle at D
36
A B C D Sum =
4°59'10" 186°21'10" 8°25'49" 160°13'51" 360°00'00"
Chapter 11 – Satellite Imagery 11.1
What are the chief differences between maps based on satellite imagery and maps based on airborne imagery? Maps based on satellite imagery generally have a lower resolution (larger pixel size) than airborne imagery. Satellite imagery, although usually georeferenced, does not have high precision positions, and generally not used for topographic mapping. Airborne imagery often are flown for precise photogrammetric mapping, requiring ground control and overlapping images.
11.2
What are the advantages and disadvantages inherent in the use of hyperspectral sensors over multispectral sensors? The advantage with hyperspectral is the ability to look at a very narrow range on the electromagnetic spectrum. This allows for very detailed spectral signatures and classifications. The disadvantage is the amount of information that needs to be processed. Hyperspectral imagery can contain as much a s 60 times more data than multi-spectral imagery for the same area, requiring much more processing overhead. Determining which set of layers to use to perform classification is more difficult whith Hyperspectral data, although the results can be far more detailed.
11.3
Describe all the types of data collection that can be used to plan the location of a highway/utility corridor spanning several counties? The classical method for initial route planning consisted of using USGS topographic maps a nd field visits. More recently, Aerial photos and orthophotos from USGS f lights were used. These days, it is common to use data such as NAIP images, GIS layers of roads and ownership and even Google Earth.
11.4
Why is radar imaging preferred over passive sensors for Arctic and Antarctic data collection? Polar regions have long periods of darkness which are problematic for most passive sensors. In addition radar has the capability of penetrating cloud cover which can be common in the polar regions. Finally the high reflectance of ice and snow can limit the ability to analyze using passive systems.
11.5
Compare and contrast the two techniques of remote sensing, aerial and satellite imagery, by listing the possible uses for which each technique would be appropriate. Use the comparative examples shown in Figures 11.7 and 11.8 as a basis for your response. Satellite imagery is useful for: • Large area mapping • Planning growth on a state level • Temporal Crop Monitoring • Monitoring changes in ocean temperatures • Monitoring environmental changes across the world • Remote observation Aerial imagery is useful for: • Precision mapping • Disaster recovery • Community Planning • Vegetation classification • Route planning
37
11.6
What is the difference between classification and feature extraction? Classification is the processes of comparing pixels to determine which pixels have similar values within certain bands. Classification creates boundaries just by looking to see if the neighboring pixel is within the same class. Feature extraction is the process of determining edges of features by looking at the neighboring pixels and see if they match. Feature extraction looks a t a single class (buildings, forest) a t a time.
11.7
What type of remote sensing projects would Landsat 7 be appropriate for, and which ones would it not be appropriate for. The projects what benefit from Landsat can be determined by looking at the resolutions. Landsat has a low spatial resolution (30 m), but a high temporal resolution (16 days) and spectral resolution (8 sensors). This means that worldwide crop monitoring would be ideal with Landsat. Mapping of large features such as forest or desert boundaries would be an a ppropriate Landsat Project. Also mapping of thermal change over large areas would be an a ppropriate use. Using Landsat to do urban planning would not be appropriate, nor would the use of Landsat f or topographic mapping or route design.
38
Chapter 12 – Airborne Imagery 12.1
Calculate the flying heights and altitudes, given the following information. a Photographic scale = 1:20,000, lens focal length = 153 mm, and elevation of mean datum = 180 m. b. Photographic scale is 1 in. = 20,000 ft, lens focal length = 6.022 in., and elevation of mean datum = 520 ft. a) b)
12.2
Calculate the photo scales, given the following data. a. Distance between points A and B on a topographic map (scale 1:50,000) = 4.75 cm; distance between the same points on an air photo = 23.07 cm. b. Distance between points C and D on a topographic map (scale 1:100,000) = 1.85 in.; distance between the same points on an air photo = 6.20 in. a) b)
12.3
Photo scale = 23.07 x 1 :50,000/4.75 = 1: 10,295 Photo scale = 6.20 x 1:100,000/1.85 = 1 :29,839
Calculate the approximate numbers of photographs required for stereoscopic coverage (60 percent forward overlap and 25 percent sidelap) for each of the following conditions. a. Photographic scale = 1:30,000; ground area to be covered is 30 km by 45 km. b. Photographic scale = 1:15,000; ground area to be covered is 15 miles by 33 miles. c. Photographic scale is 1 in. = 500 ft; ground area to be covered is 10 miles by 47 miles. a) b) c)
12.4
H = SR.f= 20,000 x .153 3060m Altitude = 3060 + 180 = 3240m SR = 20,000 x 12 = 240,000 H = SR.f= 240,000 x 6.022/12 = 120,440 ft. Altitude = 120,400 + 520 = 120,920 ft.
2
No. of photos required = 30 x 45/1 (10,000) = 150 2 (30,000) 2 No. of photos required = 15x33/0.4 (10,000) = 550 2 (15,000) SR is 1:500 x 12 or 1:6000 2 No. of photos required = 10x47/0.4 (10,000) = 3265 2 (6,000)
Calculate the dimensions of the area covered on the ground in a single stereo model having a 60 percent forward overlap if the scale of the photograph (9 by 9 in. format) is: a. 1:10,000. b. 1 in. = 400 ft.
a.
9 in. ≈ (228mm) Dimension across flight line
Dimension along flight line b.
SR is 1 :400 x 12 or 1 :4,800 Dimension across flight line
Dimension along flight line
= =
10,000 x .228 = 10,000 x 0.75 =
2280m 7,500 ft.
= =
2280 x 0.60 7,500 x 0.60
= =
1368m 4500 ft.
= =
4,800 x .228 4,800 x 0.75
= =
1094m 3,600 ft.
= =
1094 x 0.60 3,600 x 0.60
= =
656 m 2160 ft.
39
12.5
a. Calculate how far the camera would move during the exposure time for each of the following conditions: 1. Ground speed of aircraft = 350 km/h; exposure time = 1/100 s. 2. Ground speed of aircraft = 350 km/h; exposure time = 1/1,000 s. 3. Ground speed of aircraft = 200 miles/h; exposure time = 1/500 s. b. Compare the effects on the photograph of all three situations in part (a).
b)
12.6
1
Ground speed of aircraft `
= =
350 km/hr 350 x 1000 = 97.2 m/s 60 x 60 :. camera would move 97.2 x 1/100 = .972 m during exposure.
2
Ground speed of aircraft = 350 km/hr :. camera would move 97.2 x 1/1000 = .097 m
3
Ground speed of aircraft
=
97.2 m/s
= =
200 miles per hour 200 x 5280 = 293 ft./sec. 60 x 60 :. camera would move 293 x 11500 = 0.59 ft. (018m)
The "image motion" during the exposure is least in situation 2. Therefore, the resolution of an aerial photo taken under these conditions would be higher, assuming that all other conditions (weather, film types) are equal
Does a longer focal-length camera increase or decrease the relief displacement? Briefly explain the reason for your answer. A longer focal length decreases the relief displacement. The explanation for this is evident from examining Figure 12 .3 b). A longer focal length increases the distance from C to the focal plane. Therefore, the relief displacement, such as bb' and cc' is reduced.
12.7
Would a wide-angle lens increase or decrease relief displacement? Briefly explain the reason for your answer. A wide-angle lens increases relief displacement. As illustrated in Figure 12 .3 b),
12.8
Choose the best alternative from the following list for the assemblage of an uncontrolled mosaic, using alternative photographs, over rolling terrain having vertical relief differences of up to 50 ft (15 m). Give the main reasons for your choice. a. Focal length = 150 mm; scale = 1:20,000. b. Focal length = 150 mm; scale = 1:5,000. c. Focal length = 12 in. (300 mm); scale 1 in. = 1,000 ft. d. Focal length = 12 in. (300 mm); scale 1 in. = 2,000 ft. The best alternative is d), for t he following reasons. The major consideration is relief displacement. (i) The focal length is longer than a) and b), thus reducing the radial distortion. (ii) The scale of d) is 1 in. = 2000 ft., a scale ratio (SR) of 1 :24,000. This is the highest scale ratio of all the alternatives. Reference to Figure 12 .13 b) will illustrate that the higher the scale ratio, the greater the distance from C to the ground, a nd the less the radial distortion 40
12.9
Calculate the time interval between air photo exposures if the ground speed of the aircraft is 100 miles/h (160 km/h), the focal length is 50 mm (2 in.), the format is 70 mm (2.8 in.), the photographic scale is 1:500 (approximately 1 in = 40 ft), and the forward overlap = 60 percent. On the ground, the dimensions of the area covered by one photograph are 70mm x 500 = 35,000mm = 350m (1150 ft.). For a forward overlap of 60%, the distance along the direction of the flight line = 350 x 0.60 = 210m. Therefore, the "new" area is to be photographed between successive exposures is 350 - 210 = 140m. Hence, an exposure must be taken every 140m along the flight line. The ground speed of the aircraft = 160 km/hr 160 x 1000/(60 x 60) 44.4 m/s. Time between exposures thus equals 140/44.4 = 3.15, say 3.2 seconds.
41
Chapter 13 – Engineering Surveys 13.1
Given PI at 9 + 27.26, Δ = 29°42′, and R = 700 ft, compute tangent (T) and length of arc (L).
T L
13.2
= =
R tan Δ/2 2πR Δ 360
= =
700 Tan(4°51') = 2π700 x 29.7 = 360
185.60 ft. 362.85 ft.
Given PI at 15 + 88.10, Δ = 7°10′, and D = 8°, compute tangent (T) and length of arc (L).
R
=
T L
= =
5729.58 D R tan Δ/2 100 Δ D
= = =
5729.58 = 716.20 ft. 8 716.20Tan(3°35')= 44.85 ft. 100 x 7.166" = 89.58 ft. 8
13.3
From the data in Problem 13.1, compute the stationing of the BC and EC. PI@ 9 + 27.26 -T 1 85.60 BC = 7 + 41.66 +L 3 62.85 EC= 11 + 04.51
13.4
From the data in Problem 13.2, compute the stationing of the BC and EC. PI@ 15 + 88.10 -T 44.85 BC = 15 + 43.25 +L 89.58 EC = 16 + 32.83
13.5
A straight-line route survey, which had PIs at 3 + 81.27 (Δ = 12°30′) and 5 + 42.30 (Δ = 10°56′), later had 600 ft-radius circular curves inserted at each PI. Compute the BC and EC stationing (chainage) for each curve. Curve #1 Δ = 12°30' R = 600 ft. T1 = 65.71 L1 = 130.90
Curve #2 Δ = 10°56' R = 600 ft. T2 = 57.42 L2 = 114.49
PI1 @ 3 + 81.27 - T1 65.71 BC1 = 3 + 15.56 +L1 1 30.90 EC1 4 + 46.46
Pi2 @ 5 +42.30 PI1 @ 3 + 81.27 Diff. = 161.03 T1 + T2 = -123.13 EC1 to BC 2 = 37.90
EC1 to BC 2 BC2 + L2 EC2
37.90 4+ 84.36 1 14.49 5 +98.85
=
42
13.6
Given PI at 5 + 862.789, Δ = 12°47′, and R = 300 m, compute the deflections for even 20-m stations.
T L
= =
300 Tan(6°23'30") = 33.606 m 2π X 300 x 12.783" = 66.933 m 360 PI @ 5 + 862.789 Deflection angle for 1m = 6.3916° -T 33.606 66.933 BC = 5 + 829.183 = 0.0954935° +L 66.933 EC 5 + 896.116 Deflections:
BC
5 + 829.183 5+840 5+860 5+880 5 + 896.116
EC 13.7
(1) 10.817 x .0959935 (2) 20 x .0954935 (3) 16.116 x .0954935 0°00'00" 1°01'59" 2°56'35" 4°51'11" 6° 3'31" ≈
1°01'59" 1 °54'36" 1°32'20"
Δ/2
Given PI at 8 + 272.311, Δ = 24°24′20″, and R = 500 m, compute E (external), M (mid ordinate), and the stations of the BC and EC.
E
500[ 1 ] -1 [Cos (12°12'10")] M = 500 (1 – Cos(12°12'10")) T = 500 Tan (12°12'10") L = 500 π x 24.40556 180 PI @ 8 + 272.311 -T 108.129 BC = 8 + 164.182 +L 212.979 EC = 8 + 377.161
13.8
= = =
=
=
11.558m
= = =
11.297 m 108.129 m 212.979 m
Given PI at 10 + 71.78, Δ = 36°10′30″ RT, and R = 1, 150 ft, compute the deflections for even 1 00-ft stations.
T L
= =
1150Tan(18°05'15") = 375.60 ft. 2 x π x 1150 x 36.175 = 726.08 ft. 360 PI @ 10 + 71.78 -T 3 75.60 BC = 6 + 96.18 Deflection/ft. = +L 7 +26.08 EC = 14 +22.26
43
Δ/2 L
=
.0249111°
BC
EC 13.9
6 + 96.18 7 + 00 8 + 00 9 + 00 10 + 00 11 + 00 12 + 00 13 + 00 14 + 00 14 + 22.26
0°00'00" 0°05'43" 2°35'11" 5°04'39" 7°34'07" 10°03'35” 12°33'03" 15°02'31" 17°31'59" 18°05'15" = Δ/2
Deflections: (1) 3.82 x .0249111 = 0°05'43" (2) 100 x .0249111 = 2°29'28" (3) 22.26 x.0249111 = 0°33'16"
From the distances and deflections computed in Problem 13.6, compute the three key CL chord layout lengths, (1) BC to first 20-m station, (2) Chord distance for 20-m (arc) stations (3) From the last even 20-m station to the EC. For 10.817m arc, deflection For 20m arc, deflection For 16.116m arc, deflection
= = =
1°01'58.6" 1°54'36" 1°32'20"
(1) C = 2RSin(1°01'58.6") = 10.816 m (2) C = 2RSin(1°54'36") = 19.996 m (3) C = 2RSin(1°32'20") = 16.114 m
13.10 From the distances and deflections computed in Problem 13.8, compute the three key CL chord layouts as in the previous problem. For 3.82 ft. arc, deflection For 100 ft. arc, deflection For 22.26 ft. arc, deflection
= = =
0°05'43" 2°29'28" 0°33'16"
(1) C = 2RSin(0°05'43")= 3.82 ft. (2) C = 2RSin(2°29'28") = 99.97 ft. (3) C = 2RSin(0°33'16") = 22.26 ft.
13.11 From the distances and deflections computed in Problem 13.8, compute the chords (6) required for layout directly on offsets 50-ft right and 50-ft left of CL. Δ Curve offset curve
= = =
36°10'30" RT, therefore radius of right offset 1150 - 50 = 1100 ft. and radius of left 1150 + 50 = 1200 ft.
Interval BC to 7+00
LS, R=1200 ft CL R=1150 RS, R=1100 ft. c=2400Sin(0°05'43") c=2300Sin(0°05'43”) c=2200Sin(0°05'43") = 3.99 =3.82 =3.65 100 ft. c=2400Sin(2°29'28") c=2300Sin(2°29'28") c=2200Sin(2°29'28") = 104.31 = 99.97 = 95.62 14+00 to EC c=2400Sin(0°33'16") c=2300Sin(0°033'16") c=2200Sin(0°33'16") =23.22 =22.26 =21.29 Computation check: At each interval, differences between LS & 4L, and 4L and RS are equal.
44
13.12 Two highway CL tangents must be joined with a circular curve of radius 1,000 ft. See Figure 13-50. The PI is inaccessible, because its location falls in a river. Point A is established near the river on the back tangent and point B is established near the river on the forward tangent. Distance AB is measured to be 615.27 ft. Angle a = 51°31′20″, and angle b = 32°02′45″. Perform the calculations required to locate the BC and the EC in the field. Measured values: AB = 615.27 ft. Δ T
= =
α + β = R TanΔ/2=
α
=
51°31'20"
83°34'05" 1000 Tan 41°47'02"
β = 32°02'45"
=
893.60 ft.
Side A,PI
=
Sin 32°02'45"
x
615.27 Sin 96°25'55"
= 328.53 ft.
Side PI,B
=
Sin 51°31'20"
x
615.27 Sin 96°25'55"
= 484.71 ft.
To locate BC, Layout (893.60 - 328.53); To locate EC, Layout (893.60 - 484.71);
i.e., 565.07 back from point A. i.e., 408.89 forward from point B.
13.13 Two street curb lines intersect with Δ = 71°36′. See Figure 13 -51. A curb radius must be selected so that an existing catch basin (CB) abuts the future curb. The curb-side of the CB CL is located from point V: V to CB = 8.713 m and angle E@V@CB = 21°41′. Compute the radius that will permit the curb to abut the existing CB. Sin (0, CB, V) = OV x Sin (0, V, CB) R
(1)
[triangle 0, CB, V]
R/OV = Cos Δ/2,
(2)
[triangle 0, V, E]
Sin (O,CB,V) = = =
OV = R/Cos Δ/2 R/Cos Δ/2 x Sin (0, V,CB)/R Sin (O,V,CB) Cos Δ/2 Sin 32°31' Cos 35°48'
Angle (O,CB,V)= 138°29'20" Angle (CB,O,V) = 180 - (138°29'20"+32°31') R Sin 32°31’
=
8.713 Sin 8°59'40"
(substitute (2) into (1)
= 0.662765
= 8°59'40" R = 29.958 m
13.14 Given the following vertical curve data: PVI at 7 + 25.712; L = 100 m; g1 = -3.2 percent; g2 = +1.8 percent elevation of PVI = 210.440, compute the elevations of the curve low point and even 20-m stations. 1. 2. 3. 4.
A = 1.8 - (-3.2) BVC @ 7 + 25.712 - 50 EVC @ 7 + 25.712 + 50 Elevation of BVC = 210.440 + (50 x .032) Elevation of EVC = 210.440 + (50 x .018) Distance from BVC to low point: X = -g1L = 3.2 x 100 A 5 Low pt @ (6 + 75.712) + 64.000 45
= = = = =
5.0 6 + 75.712 7 +75.712 212.040 211.340
=
64.000m
=
7 + 39.712
Station BVC
PVI low point EC
6+75.712 6+80 7+00 7+20 7+25.712 7+39.712 7+60 7+75.712
Tangent Elev
2
Tangent Offset [x/CL/2)] d
Curve Elev.
2
212.040 211.902 211.262 210.622 210.440 210.692 211.057 211.340
(0/50) x .625 = 0.000 2 (4.288/50) x .625 =.005 2 (24.288/5) x .625 =.147 2 44.288/50) x .625 =.490 2 ((50/50) x .625 0.62 2 (36/50) x .625 = .324 )2 (15.712/50 x .625 =.062 2 (0/50) x .625 0.000
212.040 211.907 211.409 211.112 211.065 211.016 211.119 211.340
13.15 Given the following vertical curve data: PVI at 19 + 00; L = 500 ft; g1 = +2.5 percent; g2 = -1 percent; elevation at PVI = 723.86 ft, compute the elevations of the curve summit and even full stations (i.e., 100-ft even stations). 1. 2. 3. 4.
A = -1-2.5 BVC @ 19 + 00 - 250 EVC@ 19 +00 + 250 Elevation of BVC = 723.86 - (250 x .025) Elevation of EVC = 723.86 - (250 x .01) Distance from BVC to summit (high point) X = -g1L = -2.5 x 500 A -3.5 Summit @ (16+ 50) + 357.14
BC
PVI Summit EC
= = = = =
- 3.5 16 + 50 21 +50 717.61 ft. 721.36 ft.
=
357.14 ft.
=
20 + 07.14
Station
Tangent Elev.
Tangent offset [X/(L/2)]2d
Curve Elev .
16+50 17+00 18+00 19+00 20+00 20+07.14 21+00 21+50
717.61 718.86 721.36 723.86 722.86 722.78 721.86 721.36
(0/250)2 x 2.19 = 0 (50/250)2 x 2.19 = .09 (150/250)2 x 2.19 = .79 (2501250)2x 2.19 2.19 (150/250)2 x 2.19= .79 (142.86/250)2x 2.19 = .72 (50/250)2x 2.19 = .09 (0/250)2 x 2.19 = 0
717.61 718.77 720.57 721.67 722.07 722.07 721.77 721.36
13.16 A new road is to be constructed beginning at an existing road (CL elevation = 472.70 ft) for a distance of 600 ft. The CL gradient is to rise at 1.32 percent. The elevations of the offset grade stakes are as follows: 0 + 00 = 472.60; 1 + 00 = 472.36 ; 2 + 00 = 473.92; 3 + 00 = 475.58; 4 + 00 = 478.33; 5 +00 = 479 .77; 6 + 00 = 480.82. Prepare a grade sheet (similar to Table 13-9) showing the cuts and fills in feet and inches. Station
CL Grade
Stake elev.
0+00 1+00 2+00 3+00 4+00 5+00 6+00
472.70 474.02 475.34 476.66 477.98 479.30 480.62
472.60 472.36 473.92 475.58 478.33 479.77 480.82
46
Cut
Fill
Cut
0.10 1.66 1.42 1.08 0.35 0.47 0.20
Fill 0' 1 ¼” 1' 8" 1' 5" 1'1"
0' 4 ¼” 0' 5 ½” 0'2 3/8"
13.17 A new road is to be constructed to connect two existing roads. The CL intersection with the east road (0 + 00) is at an elevation of 210.500 m, a nd the CL intersection at the west road (1 + 32.562) is a t an elevation of 209.603 m. The elevations of the off set grade stakes are as follows: 0 + 00 = 2 10.831; 0 + 20 = 210.600 ; 0 + 40 = 211.307; 0 + 60 = 210.114; 0 + 80 = 209.772; 1 + 00 = 209.621; 1 + 20 = 209.308; and 1 + 32.562 = 209.400. Prepare a grade sheet (similar to Table 13-7) showing cuts and fills in meters. Station
CLGrade
Stake elev.
Cut
Fill
0+00 0+20 0+40 0+60 0+80 1+00 1+20 1+32.562
210.500 210.365 210.230 210.094 209.959 209.823 209.688 209.603
210.831 210.600 211.307 21O.114 209.772 209.621 209.308 209.400
0.331 0.235 1.077 0.020 0.187 0.202 0.380 0.203
13.18 In Figure 13-31(a), if the ditch invert is to be 3 ft 6 in. deep (below subgrade), how far from the highway CL would the CL of the ditch be? 24.5 + (3.5 x 3) = 35 ft. from CL 13.19 In Figure 13-31(b), a. If the ditch invert is to be 2.0 m deep (below subgrade), how far from the highway CL would the CL of the ditch be? b. If the original ground is level with the CL of the highway right across the width of the highway, how far from the CL would be the slope stake marking the far side of the ditch at original ground? a. b
7.45 + (2 x 3) = 13.45m from CL Ditch Invert is 2.67 below original ground; slope stake is 13.45 + (3 x 2.67) = 21.46m from CL
13.20 A storm sewer is to be constructed from existing MH 8 (invert elevation = 360.44) at +1.20 percent for a distance of 240 ft to proposed MH 9. The elevations of the offset grade stakes are as follows: 0 + 00 = 368.75; 0 + 50 = 368.81 ; 1 + 00 = 369.00; 1 + 50 = 369.77; 2 + 00 = 370.22; and 2 + 4 0 = 371.91. Prepare a grade sheet (similar to Table 13-11) showing stake-to batter- board distance in feet and inches. Use a 14-ft grade rod. Station MH
MH
80+00 0+50 1+00 1+50 2+00 92+40
Invert Elev.
Stake Elev
360.44 368.75 361.04 368.81 361.64 369.00 362.24 369.77 362.84 370.22 363.32 371.91 Grade rod = 14'
47
Cut 8.31 7.77 7.36 7.53 7.38 8.59
Stake to batterboard Feet Ft./n 5.69 6.23 6.64 6.47 6.62 5.41
5' 8 1/4" 6' 2 3/4" 6' 7 3/4" 6' 5 1/2" 6' 7 1/2" 5' 4 7/8"
13.21 A sanitary sewer is to be constructed from existing MH 4 (invert elevation = 150.666) at +0.68 percent for a distance of 115 m to proposed MH 5. The elevations of the offset grade stakes are as follows: 0 + 00 = 152.933; 0 + 20 = 152.991; 0 + 40 = 153.626; 0 + 60 = 153.725; 0 + 80 = 153.888; 1 + 00 = 153.710; and 1 + 15 = 153.600. Prepare a grade sheet (similar to Table 13-11) showing stake-to batter- board distances in meters. Use a 4-m grade rod. Station MH
MH
40+00 0+20 0+40 0+60 0+80 1+00 51+15
Invert Elev.
Stake Elev.
150.666 152.933 150.802 152.991 150.938 153.626 151.074 153.725 151.210 153.888 151.346 153.710 151.448 153.600 Grade rod = 4m
48
Cut
Stake to Batterboard
2.267 2.189 2.688 2.651 2.678 2.364 2.152
1.733 1.811 1.312 1.349 1.322 1.636 1.848
Chapter 14 – Land Surveys 14.1
Calculate the angular convergency for two meridians 6 miles (9.66 km) apart at latitude 36°30' and 46°30'. The Convergence can be computed using the formula: (seconds)
52.09" /mi d(miles)ta n
Therefore at 36° 30’ (seconds) 52.09" /mi 6tan(36.5) " 52.09" 6 0.739961 231" 03'51"
At 46° 30’ (seconds) 52.09" /mi 6tan(46.5) " 52.09" .05378 329" 05'29" 6 1
14.2
Find the convergency in feet, chains, and meters of two meridians 24 miles apart if the latitude for the southern limit is 40°20' and the latitude for the northern limit is 40°41'. First the approximate distance from 40° 20’ and 40°41’ must be determined. Assume Earth’s radius as 20,906,000 ft, then the approximate distance would be calculated by :
,000 ft Tan(4041 '4020' ) Distance 20,906
Distance 20,906,000 ft xtan(21' ) 127,709 ft 24.19mi Determine the convergency p:
( feet ) 1 .33 ft mi 2 24mi 24.19mi tan(4030'30" ) 660 ft (chains ) 0.0202 ch mi 2 24mi 24.19mi tan(4030'30" ) 10ch (meters ) 0.4064 m mi 2 24mi 24.19mi tan(4030'30" ) 202m
14.3
If the tangent offset is 3.28 ft (1.00 m) at 3 miles (4.83 km) from the meridian, calculate the offsets for 5 miles (8.05 km), 7 miles (11.27 km), and 10 miles (16.10 km) from the same meridian. The tangent offset is proportional to the square of the distance from the starting meridian. With the given information an equation can be set up:
TanOff
3.28 ft
Dist (mi )2
TanOff
3mi 2
1 m
Dist TanOff 3.28 ft 3
2
Dist (km)2
4.83km 2
Dist TanOff 1 m 4.83
2
To solve for 5, 7 and 10 miles:
8 .05 2.78 m 4.83
5 9.11 ft 3
TanOff 1 m
TanOff 3.28 ft
2
7 TanOff 3.28 ft 17 ft .86 3
2
11 .27 TanOff 1 m 5 .44 m 4.83 2
2
16 .10 10 TanOff 3.28 ft .44 ft TanOff 1 m 11 .11 m 36 4.83 3
49
14.4
For a 6-mile-square sectional township, calculate the area of section 6, if the mean latitude through the middle of the township is 46°30'. Take only the effects of convergence into account in this calculation. p(feet) = 1.33 x 6 x 6 Tan 46o3 0' = 50.4 ft. = 0.0096 mi. North boundary of Section 6 = 1.0000 - 0.0096 = 0.9904 mi. Area of Section 6 = 1 mi. x (1.000 + 0.9904)/2 = 0.9952 sq.mi. = 636.9 ac.
14.5
In Figure 14.10 show how the iron monument would be marked for (a) The northwest corner of township T.20N., R.7E. (b) The southeast corner of section 32 of township T.20N., R.6E. Corner (a) is a closing township corner, so it would look something like:
T21N
R6E
S36 S1
S6
R6E
R7E
T 20 N CC 2011
Corner (b) is corner common to 4 sections and two townships, so it would look something like:
T20N
R7E
S32
S33
S5
S4
T 19 N 2011
50
