FE REVIEW COURSE SPRING 2017 Surveying 4/19/2017 Surveying Knowledge 4 6 problems Angles, distances, & trigonometry Area computations Earthwork & volume computations Closure Coordinate systems State plane, latitude/longitude Leveling Differential, elevations, percent grades Sections in FE Reference Handbook Civil Engineering 2 ANGLES 3 1
Bearings and Azimuths Bearing - The direction of a line as given by the acute angle between the line and a reference meridian. Azimuth The direction of a line as given by an angle measurement clockwise (usually) from the north end of a reference meridian. (NAD 83 is from the North) (NAD 27 is from the South) 4 Bearings and Azimuths 5 Bearings and Azimuths 6 2
Bearings and Azimuths Reverse Directions Back azimuth (reverse direction) = azimuth + / 180 Back bearing (reverse direction) = same numeric value with opposite directions Convert these Bearings to Back Bearings N 5 51 45 W S 72 12 30 W SOLUTIONS: S 5 51 45 E N 72 12 30 E Convert these North Azimuths to Back Azimuths: 191 58 15 146 11 45 (191 58 15 ) + 180 = 371 58 15 = (371 58 15 ) 360 = 11 58 15 (146 11 45 ) + 180 = 326 11 45 8 Bearings and Azimuths- Problems At point D of a closed traverse, the bearing of side DA is N 33 deg. 15 min. W and the bearing of side DB is S 12 deg. 20 min. E. The clockwise angle BDA is most nearly: A) 20 deg. 55 min. B) 45 deg. 35 min. C) 158 deg. 05 min. D) 159 deg. 05 min. 9 Bearings and Azimuths- Problems Line GH has an azimuth of 310 deg. 16 min. 45 sec. The bearing of this line is most nearly: A) N 31 deg. 16 min. 45 sec. W B) N 49 deg. 43 min. 150 sec. E C) N 49 deg. 43 min. 150 sec. W D) S 49 deg. 43 min. 150 sec. E 10 3
Interior Angles 11 Interior & Exterior Angles - Problems The interior angle at a corner of a closed traverse is 66.292778 degrees. The exterior angle at this corner is most nearly: A) 293 deg. 42 min. 26 sec. B) 293 deg. 35 min. 25 sec. C) 113 deg. 42 min. 26 sec. D) 66 deg. 17 min. 34 sec. 13 DISTANCES 15 4
Interior & Exterior Angles - Problems Distances along a baseline are referred to as stations. English Stations 1000 feet = 10+00.00 (1 sta. = 100 feet) Metric Stations 1000 meters = 1+000.000 Ties to a baseline are made at 90 degrees to the baseline and are called offsets 16 Distances - Problems What is the station at Point B? 2112.22 A B STA 15+27.32 STA?? 17 Distances - Problems The distance (feet) between station 102+64.93 and station 118+43.72 is most nearly: a) 157.88 b) 578.79 c) 1578.79 d) 1678.79 18 5
Distances - Problems A survey had identified the beginning and ending points along an urban collector highway that requires new sidewalks. The first station is located at 8+50.00 and the second station is located at 16+75.00. What is the difference in length between stations in both feet and in stations? a) 82.50 and 82.5 stations b) 182.50 and 18.25 stations c) 925.50 and 9.25 stations d) 825.0 and 8.25 stations 19 TRIGONOMETRY 20 Trigonometry Right Triangle Trig Law of Sines sin cos H O tan A Law of Cosines 21 6
Trigonometry Compute the length of side BA and side AC, if = 68 45 30 A) BA = 247.27, AC = 96.12 B) BA = 96.12, AC = 247.27 C) BA = 247.25, AC = 96.15 D) BA = 96.15, AC = 247.25 22 Trigonometry A reading of 14.1 is taken on a 15 ft. rod that is 3.00 ft. out of plumb at the top. What is the correct reading? A) 13.81 B) 12.81 C) 11.28 D) 14.41 3.00 ft. 23 AREAS 24 7
Area by Coordinates Area by Coordinates: the double area of a closed traverse is the algebraic sum of each X coordinate multiplied by the difference between the Y values of the adjacent station. A=1/2[X 1 (Y 2 -Y 4 )+X 2 (Y 3 -Y 1 )+X 3 (Y 4 -Y 2 )+X 4 (Y 1 -Y 3 )] EQN 7.16 See Examples 7.8, 7.9 on Pg. 173 25 These coordinates can be viewed as elevation and offset to figure end areas 2A= 700(700-300)+650(500-400)+300(300-700)+200(400-500) A= 102500 ft 2 /43560 = 2.35 Acres 26 Area by Coordinates - Problem Given these coordinates, determine the area (in Acres): P (6253, 8273) Q (6684, 8471) R (6598, 8784) S (6186, 8851) T (6381, 8598) a) 289819 acres b) 144910 acres c) 3.3267 acres d) 6.6534 acres 27 8
Areas Bounded by Irregular Boundaries If the irregular side can be divided into a series of cells of width, w, either the trapezoidal rule or Simpson s rule can be used. Trapezoidal = irregular side of each cell is fairly straight Simpson s = irregular side of each cell is curved or parabolic P. 176 28 Trapezoidal Rule Example 29 Simpson s Rule Example 31 9
Earthwork 33 Excavation Volumes Methods for determining the volume of cut or fill along a baseline: Average End Area: V=(A 1 +A 2 )/2 units are ft 3 or m 3 Prismoidal Formula: V=L(A 1 +4A m +A 2 )/6 The Pyramid: V=h(Area of Base)/3 or LA/3 Divide ft 3 by 27 to get yardage 34 End Areas 35 10
Volumes From End Areas 36 Earthwork Volume Calculation Example 37 Volumes of Piles Volume of piles of soil, recycled pavement, and paving materials can be calculated if the pile shapes are assumed Angle of repose 30 for smooth gravel 40 for sharp gravel 25-35 for dry sand 30-45 for moist sand 20-40 for wet sand 37 for cement P. 176 38 11
Volumes of Piles Example 39 Earthwork Volumes Average End Area Method Volume is calculated by averaging the two end areas & multiplying by the prism length P. 176 41 Average End Area Method Example 42 12
Earthwork Volumes Prismoidal Formula Method Preferred when the two end areas differ greatly or when the ground surface irregular Uses the mean area, A m, midway between the two end sections P. 176 44 Prismoidal Formula Method Example 45 Latitude & Departures, Traverses, Closure 47 13
Open Traverses and Closed Traverses: Example of an Open Traverse -If points 1 and 15 are known it would make this figure closed even though it does not form a closed loop. Traverses should always start at a known coordinate and end at a known coordinate if it does not loop into itself so all distances and angles can be geometrically checked! 48 Example of a Closed Traverse with an internal backsight. The figure below is what is known as a Traverse Loop. See Figure 7.4 in book. First half of process report generated by Carlson Survey; Closure Results (Before Angle Balance) Starting Point 1: N 5000.0000 E 5000.0000 Z 100.0000 Closing Reference Point 1: N 5000.0000 E 5000.0000 Z 100.0000 Ending Point 8: N 5000.0906 E 4999.9699 Z 100.0595 Azimuth Of Error: 341 38'22" North Error : 0.09061 East Error : 0.03007 Vertical Error : 0.05953 Hz Dist Error : 0.09547 Sl Dist Error : 0.11251 Traverse Lines : 7 SideShots : 2 Store Points : 1 Horiz Dist Traversed: 2712.2905 Slope Dist Traversed: 2712.6200 Closure Precision: 1 in 28409 Balancing Angles Find the angular misclosure and apply it evenly to all angles to achieve closure 49 Latitude and Departure Latitude = N or Y Departure = E or X 50 14
This is why you should always use azimuths for Latitude and Departure Calculations!! **If the rectangular coordinates for D are N: 5000, E: 5000 the coordinates for A would be N: 5055.180, E: 5031.709. 51 This table is intuitive, but it is also a good reference when finding the direction of a bearing, given two points and their respected coordinates (discussed later in this lecture). 52 Clockwise Latitude and Departure Example along with Coordinates computed from Latitudes and Departures (section 7.12 in text) taken from R.B. Buckner 53 15
X is the Easting and Y is the Northing 54 Computation of Latitudes and Departures The total error is the hypotenuse of the triangle made by the lat and dep The precision of the survey is the Total Error/Perimeter of the traverse The total error is also called the misclosure, error of closure, absolute error 55 Compass Rule Adjustment (Bowditch) Distributes the errors of the latitudes and departures throughout the traverse The sign of the correction will be opposite from that of the error. 56 16
Computing the Adjusted Values from a Balanced Traverse 57 Assigning Coordinates to a Balanced Traverse 58 Latitude & Departures - Problems 59 17
Latitude & Departures - Problems 60 Coordinate Systems 61 Coordinate Systems WGS84 NAD27 NAD83 Approximately 2 meters Earth Mass Center Approximately 236 meters GEOID WGS84 and NAD83 share the GRS80 ellipsoid but the origin differs by about 2m NAD27 uses the Clark spheroid of 1866, the origin is 236 m from WGS84 62 18
State Plane Coordinate Systems (SPCS) Two types: Lambert Conformal Conic Projection Transverse Mercator Projection Projecting a round surface onto a flat surface. 63 Equatorial Mercator Projection (not used for SPCS) 64 Transverse Mercator Projection 65 19
Lambert Conformal Conic Projection 66 Lambert Conformal Conic Projection Intersection of cylinder cone and and ellipsoid SCALE EXACT Cone Ellipsoid 67 Tennessee State Plane Coordinate System 36-25 N 34-20 N False Easting Lambert Conformal Conic Projection 68 20
Kentucky State Plane Coordinate System 2 Zones to Minimize Distortion Lambert Conformal Conic Projection 69 Mississippi State Plane Coordinate System Transverse Mercator Projection 2 Zones to Minimize Distortion 70 Grid to Ground Coordinates 71 21
Scale Factor Example 1 E1 = 10,000.00 ft on ellipsoid Scale = 0.99998237 (average of scale at line endpoints) 10,000.00 ft X 0.99998237 = 9999.82 ft Length of line G1 on grid 72 Scale Factor Example 2 20,000.00 ft X 1.00004234 = 20,000.85 ft Length of line G2 on grid E2 = 20,000.00 ft on ellipsoid Scale = 1.00004234 (average of scale at line endpoints) 73 Coordinate Systems Problems 74 22
Coordinate Systems Problems 75 Coordinate Systems Problems 76 Coordinate Systems Problems 77 23
Leveling 78 Leveling **You can think of a level line as a body of water. 79 Leveling 80 24
Differential Leveling Loop Closure = 2053.21 2053.18 = 0.03 too high Adjustment = 0.03/7 = 0.004 per setup or HI ALWAYS PERFORM ARITHMETIC CHECK!! 81 Leveling Circuit Field Notes Example Known Point New Point Known Point Check 144.30 82 Trigonometric Leveling 83 25
Level Notes - Problem 84 Level Notes - Problem 85 Level Notes - Problem 86 26
Level Notes - Problem 87 Trig Leveling - Problem A transit is set up directly over a benchmark elevation 300.00 m. A reading of 4.5 m is taken on a rod at B. If the instrument height (HI) is 1.5 m, the zenith angle is 75, and the slope distance is 420.0 m, what is the elevation of B? 88 88 Trig Leveling - Problem A transit is set up directly over a benchmark elevation 300.00 m. A reading of 4.5 m is taken on a rod at B. If the instrument height (HI) is 1.5 m, the zenith angle is 75, and the slope distance is 420.0 m, what is the elevation of B? 89 89 27
Horizontal Curves 90 Horizontal Curves Radius for Degree of Curvature1º to 10º Radius 7000 6000 5000 4000 3000 2000 1000 0 1 2 3 4 5 6 7 8 9 10 Degree of Curvature Radius for Degree of Curvature10º to 58º Radius 700 600 500 400 300 200 100 0 10 14 18 22 26 30 34 38 42 46 50 54 58 Degree of Curvature 91 Horizontal Curves 92 28
Horizontal Curves - Problems 93 Horizontal Curves - Problems 94 Vertical Curves 95 29
Vertical Curves 96 Vertical Curves - Problems 97 Vertical Curves - Problems 98 30
THANK YOU 31