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

1

CHAPTER 01 Basics of Surveying 1.1 How do plane surveys and geodetic surveys differ?

Plane surveying assumes all horizontal measurements are taken on a single plane and all vertical measurements are relative to that plane, thereby allowing for all calculations to be done using plane trigonometry. Geodetic surveys take into account the shape of the earth and generally bases positions on ellipsoidal models of the earth (Datum).

1.2 Are preliminary (or data-gathering) surveys plane surveys or geodetic surveys? Explain your response. Generally preliminary surveys are based on plane surveys. However some preliminary surveys are based on geodetic surveys, because of the size of the area being surveyed and/or the equipment being used.

1.3 What kinds of data are collected during preliminary surveys? The data being collected is positions of objects or points along objects. The objects may be man-made features, such as edge of roads, building corners or water values; or they may be natural features such as creeks, top of banks or grade breaks.

1.4 How is a total station different from an electronic theodolite?

An electronic theodolite only measures horizontal and vertical angles, while a total station also measures slope distances and integrates that with the vertical angle to compute horizontal and vertical distances

1.5 Describe an integrated survey system(geomatics data model).

See Figure 1.7 on how an integrated survey system works. 1.6 Why is it that surveyors must measure or determine the horizontal distances, rather than just the slope

distances, when showing the relative locations of two points? Horizontal distances are always determined so that all measurements are on the same plane. This allows for a frame of reference (coordinate system) that can be used to check and adjust distance measurements.

1.7 Describe how a very precise measurement can be inaccurate.

Precision is a measurement of repeatability. If the equipment has certain errors (systematic) associated with a particular measurement, that measurement will not be accurate but easily may be repeatable.

1.8 Describe the term error; how does this term differ from mistake?

An error is any difference between the true value and the measured value. A mistake is one type of error that is caused by human blunders.

1.9 Describe several different ways of locating a physical feature in the field so that it later can be plotted in its

correct position on a scaled plan. A base line may be established and any feature may be located by measuring along and perpendicular to the base line (Stationing). Distances and directions may be measured and then plotted from control points (radial survey). Positions may be located using survey grade GNSS receivers and then those points plotted based on their positions.

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Chapter 02 Leveling 2.1 Compute the error due to curvature and refraction for the following distances:

(a) 500 ft (c+r) = 0.0206 x (500/1000)2 = 0.005 ft. (b) 4,000 ft (c+r) = 0.0206 x (4)2 = 0 ft. (c) 300 m (c+r) = 0.0675 x 0.3002 = 0.006 m. (d) 2.2 mi (c+r) = 0.574 x 2.22 = 2.78 ft. (e) 2,800 m (c+r) = 0.0675 x (2.8)2 = 0.529 m (f) 3 km (c+r) = 0.0675 x (3)2 = 0.608m.

2.2 Determine the rod readings indicated on the foot and metric rod illustrations in Figure 2-32. The foot readings are to the closest 0.01 ft, and the metric readings are to the closest one-half or one-third cm.

Rod A i 1.90 ii 1.73 iii 1.57 iv 1.21 v 1.03 (1.04) Rod B i 1.185 ii 1.150 iii 1.040 iv 1.000 v 0.930 Rod C i 3.06 ii 2.85 (2.84) iii 2.57 (2.56) iv 2.21 v 1.92 Rod D i 1.145 ii 1.065 iii 1.000 iv 0.935 v 0.880

2.3 An offshore drilling rig is being towed out to sea. What is the maximum distance away that the navigation

lights can still be seen by an observer standing at the shoreline? The observers eye height is 5 0 and the uppermost navigation light is 147 ft. above the water.

5.00 = .574 K12, K1 = 5.00/.574 = 2.95 miles

147 = .574 K22, K2 = 147/.574 = 16.00 miles

Maximum visibility distance = 18.95 miles 2.4 Prepare a set of level notes for the survey in Figure 2-33. Show the arithmetic check.

STATION BS HI IS FS ELEVATION

BM #50 1.27 390.34 389.07

TP #1 2.33 387.76 4.91 385.43

TP #2 6.17 381.59

BS = 3.60 FS = 11.08 389.07 +3.60 = 392.67 - 11.08 = 381.59 check

2.5 Prepare a set of profile leveling notes for the survey in Figure 2-34. In addition to computing all elevations,

show the arithmetic check and the resulting error in closure.

STATION BS HI IS FS ELEVATION

BM #61 4.72 401.46 396.74

0+00 4.42 397.04

0+50 4.30 394.16

TP #1 5.11 404.56 2.01 399.45

1+00 4.66 399.90

1+50 3.98 400.58

1+75 1.20 403.36

TP #2 1.80 402.76

BS = 9.83 FS = 3.81 E = - 0.02m [small error no need for adjustments] 396.74 + 9.83 = 406.57 - 3.81 = 402.76 check

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2.6 Complete the set of differential leveling notes in Table 2-5, and perform the arithmetic check.

STATION BS HI FS ELEVATION

BM 100 2.71 317.59 314.88

TP 1 3.62 316.33 4.88 312.71

TP 2 3.51 315.87 3.97 312.36

TP 3 3.17 316.23 2.81 313.06

TP 4 1.47 316.08 1.62 314.61

BM 100 1.21 314.87

BS = 4.48 FS = 14.49 314.88 + 14.48 - 14.49 = 314.87, check

2.7 If the loop distance in Problem 2.6 is 1,000 ft,at what order of survey do the results qualify? Use Table 2-1 or Table 2-2.

Error of closure = 0.01 ft.; for 1000 ft., second order (see Table 2.2) permits .035 1000/5280 = 0.015; therefore, results qualify for second order accuracy.

2.8 Reduce the set of differential leveling notes in Table 2-6, and perform the arithmetic check

STATION BS HI IS FS ELEVATION

BM 20 8.27 186.04 177.77

TP 1 9.21 192.65 2.60 183.44

0+00 11.3 181.4

0+50 9.6 183.1

0+61.48 8.71 246.65

1+00 6.1 249.3

TP 2 7.33 195.32 4.66 187.99

1+50 5.8 252.2

2+00 4.97 253.06

BM 21 3.88 191.44

BS =24.81 FS = 11.14 177.7 + 24.81 - 11.14 = 191.44 Check! 2.9 If the distance leveled in Problem 2.8 is 1,000 ft, for what order of survey do the results qualify if the

elevation of BM 21 is known to be 191.40? See Tables 2-1 and 2-2.

Error of closure = 0.04 ft.; for 1000 ft., third order (see Table 2.2) permits 0.101000/5280 = 0.044; therefore results qualify for third order accuracy.

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2.10 Reduce the set of profile notes in Table 2-7, and perform the arithmetic check.

2.11 Reduce the set of municipal cross-section notes in Table 2-8.

STATION BS HI IS FS ELEVATION

BM 41 4.11 307.104 302.994

TP 13 4.10 310.314 0.89 306.214

12+00

50 ft. lt. 3.9 306.4

18.3 ft. lt. 4.6 305.7

CL 6.33 303.98 20.1 ft. rt. 7.9 302.4

50 ft. rt. 8.2 302.1

13+00

50 ft. lt. 5.0 305.3

19.6 ft. lt 5.7 304.6

CL 7.54 302.77 20.7 ft. rt. 7.9 302.4

50 ft. rt. 8.4 301.9

TP 14 7.39 316.584 1.12 309.194

BM S.22 2.41 314.174

BS = 15.60 FS = 4.42 302.994 + 15.60 - 4.42 = 314.174 check!

STATION BS HI IS FS ELEVATION

BM 22 1.203 182.425 181.222

0+00

CL 1.211 181.214 10M LT., 1.430 180.995

10M RT., 1.006 181.419

0+20

10M LT., 2.93 179.50

7.3M LT. 2.53 179.90

4M LT. 2.301 180.124

CL 2.381 180.044 4M RT. 2.307 180.118

7.8M RT. 2.41 180.02

10M RT. 2.78 179.65

0+40

10M LT. 3.98 178.45

6.2M LT. 3.50 178.9

4M LT. 3.103 179.322

CL 3.187 179.238 4M RT. 3.100 179.325

6.8M RT. 3.37 179.06

10M RT. 3.87 178.56

TP 1 2.773 179.65

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2.12 Complete the set of highway cross-section notes in Table 2-9. STATION BS HI FS ELEV. LEFT CL RIGHT

BM 37 7.20 385.17 377.97

50 26.7 28.4 50

5+50 4.6 3.8 3.7 3.0 2.7

380.6 381.4 381.5 382.2 382.5

50 24.1 25.0 50

6+00 4.0 4.2 3.1 2.7 2.9

381.2 381.0 382.1 382.5 382.3

50 26.4 23.8 50

6+50 3.8 3.7 2.6 1.7 1.1

381.4 381.5 382.6 383.5 384.1

TP 1 6.71 378.46 2.13 Complete the set of highway cross-section notes in Table 2-10.

STATION BS HI FS ELEV. LEFT CL RIGHT BM 107 7.71 406.87 399.16

60 28 32 60

80+50 9.7 8.0 5.7 4.3 4.0

397.2 398.9 401.2 402.6 402.9

60 25 30 60

81+00 10.1 9.7 6.8 6.0 5.3

396.8 397.2 400.1 400.9 401.6

60 27 33 60

81+50 11.7 11.0 9.2 8.3 8.0

395.2 395.9 397.7 398.6 398.9

TP 1 10.17 396.70 2.14 A level is set up midway between two wood stakes that are about 300 ft apart. The rod reading on stake A is

8.72 ft, and it is 5.61 ft on stake B. The level is then moved to point B and set up about 6 ft or 2 m away. A reading of 5.42 ft is taken on the rod at B. The level is then sighted on the rod held on stake A, where a Reading of 8.57 ft is noted.

(a) What is the correct difference in elevation between the tops of stakes A and B? (b) If the level had been in perfect adjustment, what reading would have been observed at A from the second

setup? (c) What is the line-of-sight error in 300 ft? (d) Describe how you would eliminate the line-of-sight error from the telescope.

a) True difference = 8.72 - 5.61 = 3.11 ft. b) Correct rod reading = 5.42 + 3.11 = 8.53 ft.; on A c) Error is +0.04 in 300 ft., or .00001 ft/ft d) Cross hair adjusted downward from 8.57 to read 8.53, on A

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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 theodolites optical center is 4.66 ft above the upper baseline station at 110 + 71.25.

(a) If the elevation of the upper station is 318.71, what is the elevation of the lowerstation? (b) What is the stationing chainage of the lower station?

a) V=148.61 Sin 21O26' = 54.30 ft Elevation of lower station = 318.71 +4.66 -54.30 -4.88 = 264.19 ft. b) H = 148.61 Cos (21 26') = 138.33 ft lower station at 110+71.25 +138.33 = 112+09.58

2.16 You must establish the elevation of point B from point A (elevation 216.612 m). A and B are on opposite sides

of a 12-lane highway. Reciprocal leveling is used, with the following results: Setup at A side of highway: Rod reading on A = 0.673 m Rod readings on B = 2.416 and 2.418 m Setup at B side of highway: Rod reading on B = 2.992 m Rod readings on A = 1.254 and 1.250 m

(a) What is the elevation of point B? (b) What is the leveling error?

a) First elevation difference = 2.417- 0.673 = 1.744

Second elevation difference = 2.992 - 1.252 = 1.740 Average elevation difference = 1.742

Elevation B = 216.612-1.742 = 214.870 b) The leveling error is 0.004m

2.17 Reduce the set of differential leveling notes inTable 2-11, and perform the arithmetic check.

(a) Determine the order of accuracy (seeTable 2-1 or Table 2-2). (b) Adjust the elevation of BM K110. The length of the level run was 780 m, with setups that are equally spaced.

The elevation of BM 132 is 187.536 m.

STATION BS HI FS ELEVATION

BM 130 0.702 189.269 188.567

TP 1 0.970 189.128 1.111 188.158

TP 2 0.559 189.008 0.679 188.449

TP 3 1.744 187.972 2.780 186.228

BM K110 1.973 188.277 1.668 186.304

TP 4 1.927 188.416 1.788 186.489

BM 132 0.888 187.528

BS = 7.875 FS = 8.914 188.567 + 7.875 - 8.914 = 187.528, check a) error = 187.536 187.528 = - 0.008 m.

Using specifications from Table 2.1, Third order accuracy, allowable error = .012.780 = 0.011 m. This error of 0.008 thus qualifies for third order accuracy (in both Tables 2.1 and 2.2)

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b) STATION CUMULATIVE

DISTANCE ELEVATION CORRECTION ADJUSTED

ELEVATION

BM 130 188.567 188.567

TP 1 130 188.158 130/780 x.008 = +.001 188.159

TP 2 260 188.449 260/780 x.008 = +.003 188.452

TP 3 390 186.228 390/780 x.008 = +.004 186.232

BM K110 520 186.304 520/780 x.008 = +.005 186.309

TP 4 650 186.489 650/780 x.008 = +.007 186.496

BM 132 780 187.528 780/780 x.008 = +.008 186.536 C = 187.536 187.528 = - 0.008

The adjusted elevation of BM K110 is 186.309m

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Chapter 03 Distance Measurement 3.1 A Gunters 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 150. Compute the

horizontal distance.

H = 36.255 Cos (1 50') = 9.409 m 3.4 A distance of 210.23 ft was measured along a 3 percent slope. Compute the horizontal distance.

Tan slope angle = .03; slope angle = 1. H = 210.23 Cos(1.718358) = 210.14 ft.

3.5 The slope distance between two points is 21.835 m and the difference in elevation between the points is

3.658 m. Compute the horizontal distance.

H = 21.8352 3.6582 = 21.522 m 3.6 The ground clearance of an overhead electrical cable must be determined. Surveyor B is positioned directly

under the cable (surveyor B can make a position check by sighting past the string of a plumb bob, held in his or her outstretched hand, to the cable); surveyor A sets the clinometer to 45 and then backs away from surveyor B until the overhead electrical cable is on the crosshair of the leveled clinometer. At this point, surveyors A and B determine the distance between them to be 17.4 ft (see figure). Surveyor A then sets the clinometer to 0 and sights surveyor B; this horizontal line of sight cuts surveyor B at the knees, a distance of 2.0 ft above the ground. Determine the ground clearance of the electrical cable

Clearance = 17.4 + 2.0 = 19.4 ft. (19.6 looking at the image)

3.7 A 100-ft steel tape known to be only 99.98 ft long (under standard conditions) was used to record a measurement of 276.22 ft. What is the distance corrected for the erroneous tape?

Error per tape length = -0.02 ft.; tape used 276.22/100 times error = -.02 x 2.7622 = - 0.06 ft. corrected distance = 276.22 - 0.06 = 276.16 ft. [or, 276.22 x .9998 = 276.16 ft]

3.8 A 30-m steel tape, known to be 30.004 m (under standard conditions), was used to record a measurement of 129.085 m. What is the distance corrected for the erroneous tape length?

Error per tape length =+0.004 m; tape used 129.085/30 times, Error = +0.004 x 129.085/30 = +0.017 m Corrected distance = 129.085 + 0.017 = 129.102 m

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3.9 It is required to lay out a rectangular commercial building 200.00 ft wide and 350.00 ft long. If the steel tape being is 100.02 ft long (under standard conditions), what distances should be laid out?

Error per tape length = 0.02 ft. Tape used a) 2.0 times; error = 2.0 x .02 = + 0.04 ft. b) 3.50 times; error = 3.50 x .02 = + 0.07 ft. Corrected layout distance a) 200.00 - 0.04 = 199.96 ft. b) 350.00 - 0.07 = 344.93 ft.

(ie., these are the values, when pre-corrected for errors, result in the required layout distances)

3.10 A survey distance of 287.13 ft was recorded when the field temperature was 96F. 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 87F 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 6C, 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

Temperature Tape Length Slope Data Slope Measurement 3.13 14F 99.98 ft Difference in elevation = 6.35 ft 198.61 ft

CT= .00000645(- 14 -68)198.61 = - 0.105 ft. CL= - 0.02 x 1.9861 = - 0.040 ft. C = - 0.105 - 0.040 = - 0.15 ft. Corrected slope distance = 198.61 - 0.15 = 198.46 ft. H = 198.462- 6.352 = 198.36 ft.

3.14 46F 100.00 ft Slope angle 318 219.51 ft

CT = .00000645(46 - 68)219.51 = - 0.031 ft Corrected slope distance = 219.51 - 0.02 = 219.49 ft H = 219.49 Cos (3 18') = 205.54 ft

3.15 26C 29.993 m Slope angle 342 177.032 m

CT = .0000116(26 - 20)177.032 = + 0.012 m CL = - 0.007 x 177.032/30 = - 0.041 m C = +0.012 - 0.041 = - 0.029 m Corrected slope distance = 177.032 - 0.029 = 177.003 H = 177.003 Cos (3 42') = 176.634 m

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3.16 0C 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 (0.85937220 = 224.938 m.

3.17 100F 100.03 ft Slope at 0.80% 349.65 ft

CT = .00000645(100-68)349.65 = + 0.072 ft CL = + 0.03 x 3.4965 = + 0.104 ft C = +0.072 +0.105= + 0.177 ft Corrected slope distance = 349.65 + 0.18 = 349.83 ft Tan slope angle = .008: slope angle = 0.458356 H = 349.83 Cos(0.458356) = 329.82 ft

Temperature Tape Length Design Distance Required Layout Distance 3.18 29F 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 26C 30.012 m 132.203 m ?

CT = .0000116(26 - 20)132.203 = + 0.009 m. CL = .012 x 156.209/30 = + 0.053 C = 0.009 + 0.053 = + 0.062 Layout 132.203 - 0.062 = 132.141 m

3.20 13C 29.990 m 400.000 m ?

CT = .0000116(13 - 20)400.000 = - 0.032 m CL = 0.010 x 400/30 = - 0.133 m C = - 0.032 - 0.133 = - 0.165 Layout 400.000 + 0.165 = 400.165 m

3.21 30F 100.02 ft 500.00 ft ?

CT = .00000645(30 - 68)500.00 = - 0.123 ft. CL = +0.02 x 5.00 = +0.10 ft. C = - 0.123 + 0.10 = - 0.02 ft. Layout 500.00 + 0 .02 = 500.02 ft.

3.22 100F 100.04 ft 88.92 ft ?

CT = .00000645(100 - 68)88.92 = + 0.018 ft. CL = +.04 x 0.8892 = + 0.036 ft. C = +0.018 + 0.036 = + 0.05 ft. Layout 88.92 - 0.05 = 88.87 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.

CS = -w2L

3 / 24p

2 = -0.32

2 x 48.888

3 / (24 x 100

2) = -.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 CS=-W

2L / 24p

2 = -5.34

2 x 30 / (24 x 80

2) = -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? CS = -W

2L / 24p

2 = -1.8

2 x 471.16 / (24 x 24

2) = -0.11 ft.

Corrected distance = 471.16 -.11 = 471.05 ft.

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3.26 A distance of 72.55 ft is recorded using a steel tape supported only at the ends with a tension of 15 lb and weighing 0.016 lb per foot. Find the distance corrected for sag.

CS = w2L

3 / 24p

2 = -0.016

2 x 72.55

3/(24 x 15

2) = -0.018 ft.

Corrected distance = 72.55 - 0.02 = 72.53 ft. 3.27 In order to verify the constant of a particular prism, a straight line EFG is laid out. The EDM instrument is first

Set up at E, with the following measurements recorded: EG = 426.224 m, EF = 277.301 m. The EDM instrument is then set up at F, where distance FG is recorded as FG = 148.953 m. Determine the prism constant.

prism constant = EG- EF FG = 426.224 277.301 148.953 = - 0.030m

3.28 The EDM slope distance between two points is 2556.28 ft, and the vertical angle is +24530 (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(245'30") = 2553.32 ft. V = 2556.28 Sin(245'30") = 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 = +12650 Instrument at B, slope distance = 879.230 m; vertical angle = 12638 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(126'50") = 878.929m Elevation difference = 879.209 Sin(126'50") = 22.205m Instrument at B H = 879.230 Cos(-126'38") = 878.951m Elevation difference = 879.230 Sin(-126'38") = -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 = +330, EDM distance = 2200.00 ft Instrument at station M, vertical angle = -130, EDM distance = 2200.00 ft Instrument at station N, vertical angle = 000, EDM distance = 2800.00 ft Compute the elevations of L, M, and N (correct for curvature and refraction).

KL = 2200.00 Sin(330') =+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(-130') = - 57.59 ft. c+r = .0206 x 2.22 = +0.10 ft. Elevation difference, K to M, = - 57.49 ft. Elevation of M = 198.92- 52.49 = 146.43 ft. 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.

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Chapter 04 Angles and Directions 4.1 A closed five-sided field traverse has the following interior angles: A = 1211300; B = 1364430; C = 770530; D = 942030. Find the angle at E.

(n-2)180 = 3 x 180 = 54000' A = 121 13 00" B = 136 44' 30" C = 77 05' 30" 539 59' 60 D = 94 20' 30" - 427 23' 30 = 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 200 W

4.3 Convert the following bearings to azimuths.

(a) N 2020 E 20 20' (b) N 133 W 358 27' (c) S 8928 E 9032' (d) S 8236 W 26236' (e) N 8929 E 89 29' (f) S 1138 W 191 38' (g) S 4114 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 00 200

4.5 Convert the bearings given in Problem 4.3 to reverse (back) bearings.

(a) N 2020 E S 20 20' W (b) N 133 W S 1 33' E (c) S 8928 E N 89 28' W (d) S 8236 W N 8236' E (e) N 8929 E S 89 29' W (f) S 1138 W N 11 38' E (g) S 4114 E N 41 14 W

13

4.6 An open traverse that runs from A through H has the following deflection angles: B = 625 R; C = 354 R; D = 1147 R; E = 2002 L; F = 718 L; G = 156 R. If the bearing of AB is N 1909 E compute the bearings of the remaining sides.

AB N 19 09' E DE N 41 15'E +B 6 25' -E 20 02' BC N25 34'E EF N 21 13'E +C 3 54 ' -F 7 18' CD N29 28'E FG N13 55'E +D 11 47' +G 1 56' DE N41 15'E GH N15 51'E

4.7 Closed traverse ABCD has the following bearings: AB = N 4520 E; BC = S 4732 E; CD = S 1455W; DA =

N6800W 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 17960' 17960' Total = 360 00 Check - 11320 - 62 27 A = 66 40' C=117 33

14

4.8 If the bearing of AB is N 421110 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 421110, 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 1200" DE 267 23 50" EA 285 58 10"

4.10 If the bearing of AB is N 454856 E, compute the bearings of the remaining sides proceeding in a clockwise

direction. See Answer to Question 4.11

4.11 If the azimuth of AB is 454856, compute the azimuths of the remaining sides pro-ceeding in a

counterclockwise direction. Azimuth AB = 4548'56" [Given] N 45 48 56 E +A 6347'00" Azimuth AE = 10935'56" S 70 24 04 E + 1800000 Azimuth EA = 28935'56" N 70 24 04 W +E 16125'40" Azimuth ED = 451 01'36" - 3600000 Azimuth ED = 91 01'36" S 88 58 24 E + 1800000 Azimuth DE = 271 01'36" N 88 58 24 W +D 72 48'10" Azimuth DC = 34349'46" N 16 10 14 W - 1800000 Azimuth CD = 16349'46" S 16 10 14 E +C 10130'20" Azimuth CB = 265 20'06" S 85 20 06 W - 1800000 Azimuth BC = 85 20'06" N 85 20 06 E +B 140 28'50" Azimuth BA = 22548'56" S 45 48 56 W - 1800000 Azimuth AB = 4848'56" CHECK

15

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 1850'. Therefore, the astronomic bearing = N3930'E + 1850' = N58 20'E.

16

Chapter 06 Traverse Surveys 6.1 A five-sided closed field traverse has the following angles: A = 1062830, B = 1142000; C = 903230; D =

1093330; E = 1190700. Determine the angular error of closure and balance the angles by applying equal corrections to each angle.

STATION FIELD ANGLES CORECTIONS ADJUSTED ANGLES

A 10628'30" -18" 10628'12"

B 11420'00" -18" 11419'42"

C 9032'30" -18" 903212"

D 10933'30" -18" 10933'12"

E 11907'00" -18" 11906'42"

Sum 540 01' 30" 54000'00 + 90 -90" Balanced

Error = 1'30", = 90"; Correction/angle = 18" 6.2 A four-sided closed field traverse has the following angles: A = 815330; B = 702830; C = 860930; D

=1213030. The lengths of the sides are as follows: AB = 636.45 ft; BC = 654.45 ft; CD = 382.65 ft; DA = 469.38 ft. The bearing of AB is S 1717 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 FIELD ANGLES CORRECTIONS ADJUSTED ANGLES

A 81 53'30" - 30" 81 53'

B 70 28'30" - 30" 70 28'

C 86 09'30" - 30" 86 09'

D 121 30'30" - 30" 121 30'

360 02 00 120" 360 00

Error = 02',or 120"; Correction/angle = -30"

Course Azimuth Bearing Distance Latitude Departure

AB 197 17' S17 17'W 636.45 -607.71 -189.09

BC 87 45' N87 45'E 654.45 25.69 653.95

CD 353 54' N 6 06'W 382.65 380.48 - 40.66

DA 295 24' N64 36'W 469.38 201.33 - 424.01

2142.93 - 0.21 + 0.19

E = .212 + 0.192 = 0.283 ft. Precision ratio = E/P = 0.283/2142.93 = 1/7572 = 1/7600

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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 CLat CDep Adjusted Lat. Adjusted Dep.

AB +0.07 -0.06 - 607.64 - 189.15

BC +0.07 -0.06 + 25.76 +653.89

CD +0.03 -0.03 +380.51 - 40.69

DA +0.04 -0.04 + 201.37 - 424.05

+0.21 -0.19 0.00 0.00

STATION NORTH EAST

B 1000.00 1000.00 + 25.76 +653.89

C 1025.76 1653.89 +380.51 - 40.69

D 1406.27 1613.20 +201.37 - 424.05

A 1607.64 1189.15 - 607.64 - 189.15

B 1000.00 1000.00 Check

6.4 Using the data from Problems 6.2 and 6.3, compute the area enclosed by the traverse using the coordinate

method. Area by coordinates XB(YC-YA) = 1000.00(1025.76 1607.64) = - 581,880 XC(YD-YB) = 1653.89(1406.27 -1000.00) = + 671,926 XD(YA-YC) = 1613.20(1607.64 1025.76) = + 938,689 XA(YB-YD) = 1189.15(1000.00 1406.27) = - 483,116 Twice the area = 545,619 Area = 272,810 ft

2

(1 Acre = 43,560 ft2) or, A = 6.26 Acres

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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 = 1010319; B = 1014149; C = 1022203; D = 1155720; E = 1185529. The bearing of AB is N 730000 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 101 03' 19" B 101 41' 49" C 102 22' 03" D 115 57' 20" E 118 55' 29" 54000'00" CHECK

Course AZIMUTH BEARING DISTANCE LATITUDE DEPARTURE

AE 176 08' 49" S 3 51' 11"E 19.192 - 19.149 + 1.290

ED 115 04' 18" S64 55' 42"E 35.292 - 14.955 +31.967

DC 51 01' 38" N51 01' 38"E 37.070 +23.315 +28.820

CB 333 23' 41" N26 36' 19"W 26.947 +24.094 - 12.068

BA 255 05' 30" S75 05' 30"W 51.766 - 13.318 - 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 LAT

Corrected DEP

AE +.001 +.002 -19.148 + 1.292

ED +.003 +.003 -14.952 +31.970

DC +.003 +.003 +23.318 +28.823

CB +.002 +.002 +24.096 - 12.066

BA +.004 +.004 - 13.314 - 50.019

Check +.013 +.014 0.000 0.000

COORDINATES

STATION NORTH EAST

A 1000.000 1000.000 - 19.148 + 1.292

E 980.852 1001.292 - 14.952 + 31.970

D 965.900 1033.262 + 23.318 + 28.823

C 989.218 1062.085 + 24.096 - 12.066

B 1013.314 1050.019 - 13.314 - 50.019

A 1000.000 1000.000 Check

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6.7 Using the data from Problems 6.5 and 6.6, compute the area enclosed by the traverse using the coordinate method.

Area by the Coordinate Method XA(YB-YE) = 1000.000(1013.314 - 980.852) = + 32,462 XE(YA-YD) = 1001.292(1000.000 965.900) = + 34,144 XD(YE-YC) = 1033.262(980.852 989.218) = - 8,540 XC(YD-YB) = 1062.085(965.900 1013.314) = - 50,358 XB(YC-YA) = 1050.019(989.218 - 1000.000) = - 11,321 Twice the Area = 3,713 m

2

Area = 1,857 m2

6.8 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 704005 E

BC 953.83 N 742900 E CD 818.49 N 702245

Compute the distance and bearing of the property frontage AD.

COURSE BEARING DISTANCE LATITUDE DEPARTURE

AB N7040'05"E 100.00 + 33.10 +94.36

BC N7429'00"E 953.83 +255.17 +919.07

CD N7022'45"E 818.49 +274.84 +770.96

AD +563.11 +1784.39

Distance AD = 563.112 + 1784.392 = 1,871.13 ft. Tan Bearing AD = 1784.39/563.11 = 3.168812 Bearing AD = N7229'08"E

6.9 Given the following data for a closed property traverse: Course Bearing Distance (m) AB N 371049 E 537.144 BC N 792949 E 1109.301 CD S 185631 W ? DE ? 953.829 EA N 211008 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 DEPARTURE

EA N2110'08"W 477.705 +445.469 -172.508

AB N3710'49"E 537.144 +427.963 +324.609

BC N7929'49"E 1109.301 +202.212 +1090.715

EC +1075.644 +1242.816

Distance EC = 1075.6442 + 1242.8162 = 1,643.655m Tan Bearing EC = 1242.816/1075.644; Bearing EC = N4907'27"E Using Triangle ECD Find angle at C, = 49 07'27" - 1856'31" Angle @ C = 30 10'56"

20

Solve for angle at D Sin D = Sin 3010'56"/953.829 x 1643.655 D = 60 02'14" or 1195746" From the given data, D = 11957'46" Therefore

E = 180 - (11957'46"+3010'56") = 2951'18" CD = Sin 2951'18"(1643.655/Sin 11957'46") = 944.448 m Bearing DE = S7858'45"W

Verification:

Course Bearing Distance Latitude Departure

EA N2110'08"W 477.705 +445.469 -172.508

AB N3710'49"E 537.144 +427.963 +324.609

BC N7929'49"E 1109.301 +202.212 +1090.715

CD S185631W 944.448 - 893.304 -306.577

DE S785845W 953.829 - 182.340 - 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.

AB = (738.562-559.319)2+(666.737-207.453)

2 = 493.021m

Tan Brg AB = 459.284/179.243; Brg. AB = N6840'52"E BC = (541.742-738.562)

2+(688.350-666.737)

2 = 198.003m

Tan Brg BC = 21.613/-196.820; Brg. BC = S 616'00"E CD = (379.861-541.742)

2+(839.008-688.350)

2 = 221.141m

Tan Brg CD = 150.658/-161.881; Brg. CD = S4256'36"E DE = (296.099-379.861)

2+(604.048-839.008)

2 = 249.444m

Tan Brg DE = - 234.960/-83.762; Brg. DE = S7022'45"W EF = (218.330-296.099)

2+(323.936-604.048)

2 = 290.707m

Tan Brg EF = -280.112/-77.769; Brg. EF = S7429'00"W FA = (559.319-218.330)

2+(207.453-323.936)

2 = 360.336m

Tan Brg FA = -116.483/340.989; Brg. FA = N1851'37"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 Twice the Area = 369,109m

2

Area = 184,555m2 Area = 18.455 hectares

21

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) 0.28415X + Y = 618.2673

(1)

LINE FB Y-218.330 = (738.562-218.330)/(666.737-323.936)x (X-323.936) 1.51759X - Y = 273.2726

(2)

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) 7.05806X - Y = 3967.3106

(3)

LINE AC Y-559.319 = (541.742-559.319)/(688.350-207.453)x (X-207.453) 0.03655X + Y = 566.9015

(4)

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 = N6527'33"E KL = 144.2862 + 65.8792 = 158.614m

6.13 A five-sided field traverse has the following balanced angles: A = 1012826; B = 1021042; C = 1044206;

D = 1130442; E = 1183404. 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 202020 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 15939'40" S2020'20"E 20.845 -19.545 + 7.245

ED 9813'44" S8146'16"E 35.292 - 5.051 +34.929

DC 3118'26" N3118'26"E 37.090 +31.689 +19.273

CB 31600'32" N4359'28"W 26.947 +19.387 -18.716

BA 23811'14" S5811'14"W 50.276 -26.503 -42.723

170.450 - 0.023 + 0.008

E = 0.0232 + 0.0082 = 0.0244m Precision ratio = E/P = 0.0244/170.450 = 1/7000

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6.14 From the data in Problem 6.13, balance the latitudes and departures employing the compass rule.

COURSE CLAT CDEP Corrected LAT.

Corrected DEP.

AE +0.003 -0.001 -19.542 + 7.244

ED +0.005 -0.002 - 5.046 +34.927

DC +0.005 -0.002 +31.694 +19.271

CB +0.004 -0.001 +19.391 - 18.717

BA +0.006 -0.002 -26.497 - 42.725

CHECK +0.023 -0.008 0.000 0.000

6.15 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 EAST

A 1000.000 1000.000 - 19.542 + 7.244

E 980.458 1007.244 - 5.046 + 34.927

D 975.412 1042.171 + 31.694 + 19.271

C 1007.106 1061.442 + 19.391 - 18.717

B 1026.497 1042.725 - 26.497 - 42.725

A 1000.000 1000.000 CHECK

6.16 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(1007.106-1000.000) = + 7,410 Twice the area = 3,781 m

2

AREA = 1,890 m2

23

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 2865100 34.482 KB 373528 31.892 KC 902756 38.286 KD 1662649 30.916 KE 2472843 32.585 Compute the coordinates of the property corners A, B, C, D, and E.

STATION NORTH EAST

K 1990.000 2033.000 + 10.000 - 33.000

A 2000.000 2000.000

K 1990.000 2033.000 + 25.271 + 19.455

B 2015.271 2052.455

K 1990.000 2033.000 - 0.311 + 38.285

C 1989.689 2071.285

K 1990.000 2033.000 - 30.055 + 7.245

D 1959.945 2040.245

K 1990.000 2033.000 - 12.481 - 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(1977.519-2015.271) = - 75,504 XB(YA-YC) = 2052.455(2000.000-1989.689) = + 21,163 XC(YB-YD) = 2071.285(2015.271-1958.945) = +114,596 XD(YC-YE) = 2040.245(1989.689-1977.519) = + 24,830 XE(YD-YA) = 2002.900(1959.945-2000.000) = - 80,226 Twice the Area = 4,859 m

2

AREA = 2,430 m2

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 N7346'06"E 54.633

BC S3621'20"E 31.765

CD S4613'17"W 42.991

DE N6447'56"W 41.273

EA N 720'52"W 22.675

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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 States 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 disrepair 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 location, 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 the 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. Generally GPS determines the distance from the satellite to the receiver by calculating 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 distance 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 Carrier 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 signal 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.

25

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 as specially designed antennas and more sophisticated software 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 occupation. 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 marker) 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 marks 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 any 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 GPS positioning when doing these types of surveys do not change much whether the points are 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 diagram 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 available for observation over the complete time span of the project. This allows for the planning of the best times to occupy any given station.

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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 GPS. 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 as 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 data 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 accuracy. 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 apart.

7.14 What type of GPS process gets positions within 1 to 2 meters? This would be Differential GPS where two receivers are used, with one receiver at a known position and the other determining relative position by noting the error at the base station (see answer 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).

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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 DISTANCE LATITUDE DEPARTURE

AB N. 330'E. 56.05 + 55.95 + 3.42

BC N. 030'W. 61.92 + 61.92 - 0.54

CD N.8840'E. 100.02 + 2.33 + 99.99

DE S.2330'E. 31.78 - 29.14 + 12.67

EF S.2853'W. 69.11 - 60.51 + 33.38

FG SOUTH 39.73 - 39.73 0

GA N.8337'W. 82.67 + 9.18 - 82.16

8.2 Compute the interior angles of the traverse and check for geometric closure, that is (n 2)180 (see Table 8-

5). INTERIOR ANGLES

A - 9253'

B - 18400'

C - 9050'

D - 11210'

E - 12737'

F - 20853'

G - 8337'

Sum 90000'

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.

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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 1,000.00 + 55.95 + 3.42

B 1,055.95 1,003.42 61.92 - 0.54

C 1,117.87 1,002.88 + 2.33 + 99.99

D 1,120.20 1,102.87 - 29.14 + 12.67

E 1,091.06 1,115.54 - 60.51 - 33.38

F 1,030.55 1,082.16 - 39.73 0.00

G 990.82 1,082.16 + 9.18 - 82.16

A 1,000.00 1,000.00 Check

Second, compute the area:

XA(YB - YG) = 1,000.00(1,055.95 - 990.82) = 65,130 XB(YC - YA) = 1,003.42(1,117.87 - 1,000.00) = 118,273 XC(YD - YB) = 1,002.88(1,120.20 - 1,055.95) = 64,435 XD(YE - YC) = 1,102.87(1,091.06 - 1,117.87) = - 29,568 XE(YF - YD) = 1,115.54(1,030.55 - 1,120.20) = -100,008 XF(YG - YE) = 1,082.16(990.82 - 1,091.06) = - 108,476 XG(YA - YF) = 1,082.16(1000.00 - 1,030.55) = - 33,060 Double area = - 23,274 Area = 11,637 ft

2 or m

2

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8.6 Draw profile A to E showing both the existing ground and the proposed CL of highway. Use the following scales: metric unitshorizontal is 1:500, vertical is 1:100; foot unitshorizontal 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.

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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 information 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 GIS 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 same as 1:24,000 as there are 24,000 inches in 2000 feet. Additionally scale of a map can be determined by having a known distance on the ground and measuring that distance on the map. Say the distance between two fence corners was 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 have 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, having jagged edges along cell boundary. Generally attributes are associated with vector data, as a single feature has specific information 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.

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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 an intelligence. GIS also has the ability 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 GIS is getting fuzzier, however most surveying and engineering design is still done with CAD while project planning and management is done with GIS

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9.9 What modern developments enabled the creation of GIS?

The two primary developments were the computer and database software. Modern GIS also 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 all 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 are 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 square mile. To map these over the entire Midwest, the precision of the Landsat satellite system would be appropriate (30m x 30m)

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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 storage 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 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 DRGs uses the UTM coordinate for their projection.

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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 25429'00" S 74. 29' 00" W

B

225.931 26050'10" S 80. 50' 10" W

C

206.038 8915'59" N 89. 15' 59" E

D

317.248 69.29'50" N 69. 29' 50" E

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 25429'00" A->B A 459'10" - 24929'50" A->D B 18621'10" 459'10" C 825'49" D 16013'51" Angle at B 26050'10" B->C Sum = 36000'00" - 7429'00" B->A 18621'10" Angle at C 8915'59" C->D - 8050'10" C->B 825'49" Angle at D 42929'50" D->A - 26915'59" D->C 16013'51"

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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 0.9999 CM East 304,800 m Radius 20906000ft = 6372161.544m Sta

Elev Grid Dist

Ave Elev

Elev. SF Average East Dist From CM

Scale Factor

Combined Ground Dist

A 179.832

290.722 180.594 0.999972 316,963.999 12163.999 0.999902 0.999873 290.759

B 181.356

225.931 185.166 0.999971 316,712.413 11912.413 0.999902 0.999873 225.959

C 188.976

206.038

188.214 0.999970 316,703.900 11903.900 0.999902 0.999872 206.064

D 187.452

317.248 183.642 0.999971 316,955.486 12155.486 0.999902 0.999873 17.288

A 179.832

10.4 Determine the convergence correction for each traverse side and determine the geodetic bearings for each

traverse side. Use average latitude and longitude.

Central Meridian 79.30'00" Average Latitude 43.47'30" Average Longitude 79.21'02" Delta Longitude 0.08'58" Delta Longitude Sec 538

Station Grid Azimuth Geodetic Az. Geodetic Brg

A

254.29'00" 254.22'48" S 74. 22' 48" W

B

260.50'10" 260.43'58" S 80. 43' 58" W

C

89.15'59" 89.09'47" N 89. 09' 47" E

D

69.29'50" 69.23'38" N 69. 23' 38" E

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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 25422'48" A->B A 459'10" - 24923'38" A->D B 18621'10" 459'10" C 825'49" D 16013'51" Angle at B 26043'58" B->C Sum = 36000'00" - 7422'48" B->A 18621'10" Angle at C 8909'47" C->D - 8043'58" C->B 825'49" Angle at D 42923'38" D->A - 26909'47" D->C 16013'51"

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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 multi-

spectral 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 as 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 and field visits. More recently, Aerial photos and orthophotos from USGS flights 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

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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 at a single class (buildings, forest) at 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 appropriate Landsat Project. Also mapping of thermal change over large areas would be an appropriate use. Using Landsat to do urban planning would not be appropriate, nor would the use of Landsat for topographic mapping or route design.

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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) H = SR.f= 20,000 x .153 3060m

Altitude = 3060 + 180 = 3240m b) 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.

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) Photo scale = 23.07 x 1 :50,000/4.75 = 1: 10,295 b) Photo scale = 6.20 x 1:100,000/1.85 = 1 :29,839

12.3 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) No. of photos required = 30 x 45/1 (10,000)2 = 150

(30,000)2

b) No. of photos required = 15x33/0.4 (10,000)2 = 550

(15,000)2

c) SR is 1:500 x 12 or 1:6000 No. of photos required = 10x47/0.4 (10,000)

2 = 3265

(6,000)2

12.4 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.

9 in. (228mm)

a. Dimension across flight line = 10,000 x .228 = 2280m = 10,000 x 0.75 = 7,500 ft. Dimension along flight line = 2280 x 0.60 = 1368m = 7,500 x 0.60 = 4500 ft.

b. SR is 1 :400 x 12 or 1 :4,800 Dimension across flight line = 4,800 x .228 = 1094m = 4,800 x 0.75 = 3,600 ft. Dimension along flight line = 1094 x 0.60 = 656 m = 3,600 x 0.60 = 2160 ft.

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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).

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 = .972m during exposure. 2 Ground speed of aircraft = 350 km/hr = 97.2 m/s :. camera would move 97.2 x 1/1000 = .097m 3 Ground speed of aircraft = 200 miles per hour = 200 x 5280 = 293 ft./sec. 60 x 60 :. camera would move 293 x 11500 = 0.59 ft. (018m)

b) 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

12.6 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),

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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.

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Chapter 13 Engineering Surveys 13.1 Given PI at 9 + 27.26, = 2942, and R = 700 ft, compute tangent (T) and length of arc (L).

T = R tan /2 = 700 Tan(451') = 185.60 ft. L = 2R = 2700 x 29.7 = 362.85 ft. 360 360

13.2 Given PI at 15 + 88.10, = 710, and D = 8, compute tangent (T) and length of arc (L).

R = 5729.58 = 5729.58 = 716.20 ft. D 8 T = R tan /2 = 716.20Tan(335')= 44.85 ft. L = 100 = 100 x 7.166" = 89.58 ft. D 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 ( = 1230) and 5 + 42.30 ( = 1056), later had 600-

ft-radius circular curves inserted at each PI. Compute the BC and EC stationing (chainage) for each curve.

Curve #1 Curve #2 = 1230' = 1056' R = 600 ft. R = 600 ft. T1 = 65.71 T2 = 57.42 L1 = 130.90 L2 = 114.49 PI1 @ 3 + 81.27 Pi2 @ 5 +42.30 - T1 65.71 PI1 @ 3 + 81.27 BC1 = 3 + 15.56 Diff. = 161.03 +L1 1 30.90 T1 + T2 = -123.13 EC1 4 + 46.46 EC1 to BC2 = 37.90 EC1 to BC2 = 37.90 BC2 4+ 84.36 + L2 1 14.49 EC2 5 +98.85

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13.6 Given PI at 5 + 862.789, = 1247, and R = 300 m, compute the deflections for even 20-m stations.

T = 300 Tan(623'30") = 33.606 m L = 2 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: (1) 10.817 x .0959935 = 101'59" (2) 20 x .0954935 = 1 54'36" (3) 16.116 x .0954935 = 132'20"

BC 5 + 829.183 000'00" 5+840 101'59" 5+860 256'35" 5+880 451'11" EC 5 + 896.116 6 3'31" /2

13.7 Given PI at 8 + 272.311, = 242420, and R = 500 m, compute E (external), M (mid ordinate), and the

stations of the BC and EC.

E = 500[ 1 ] -1 = 11.558m [Cos (1212'10")] M = 500 (1 Cos(1212'10")) = 11.297 m T = 500 Tan (1212'10") = 108.129 m L = 500 x 24.40556 = 212.979 m 180 PI @ 8 + 272.311 - T 108.129 BC = 8 + 164.182 + L 212.979 EC = 8 + 377.161

13.8 Given PI at 10 + 71.78, = 361030 RT, and R = 1, 150 ft, compute the deflections for even 100-ft stations.

T = 1150Tan(1805'15") = 375.60 ft. L = 2 x x 1150 x 36.175 = 726.08 ft. 360 PI @ 10 + 71.78 -T 3 75.60 BC = 6 + 96.18 Deflection/ft. = /2 = .0249111 +L 7 +26.08 L EC = 14 +22.26

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BC 6 + 96.18 000'00" Deflections: 7 + 00 005'43" (1) 3.82 x .0249111 = 005'43" 8 + 00 235'11" (2) 100 x .0249111 = 229'28" 9 + 00 504'39" (3) 22.26 x.0249111 = 033'16" 10 + 00 734'07" 11 + 00 1003'35 12 + 00 1233'03" 13 + 00 1502'31" 14 + 00 1731'59" EC 14 + 22.26 1805'15" = /2

13.9 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) stati