Q1. A differential leveling circuit began BM Hydrant (elevation 4823.65) and closed on BM Rock (elevation 4834.47). The plus sight and minus sight distances were kept approximately equal. Readings (in feet) given in the order taken are 2.65 (+S) and BM Hydrant, 3.51 (-S) and 7.23 (+S) and TP1, 5.04 (-S) and 11.41 (+S) on BM1, 8.58 (-S) and 7.65(+S) on BM2, 4.23(-S) and 7.53 (+S) on TP2 and 4.34 (-S) on BM Rock. Prepare, check and adjust the notes. Does it meet third order tolerance if the total distance for the loop is 6000 ft. (Show your calculations)
Q2. A level is set up half way between point x and point y. A three wire reading on point x reads 7.14, 6.39, and 5.64. A three wire reading on point y reads 5.85, 5.10 and 4.35. The instrument is then moved to a point 20 feet from point x and a reading of 5.92 is taken. What does the reading need to be on point y for the instrument to be properly adjusted. What is the distance from point x to point y.
Q3. Course AB of a five-sided traverse runs due North. (Azimuth AB=000-00-00) From the given balanced interior angles to the right, draw a diagram of the traverse then compute and tabulate the bearing and azimuths from north for each side of the traverse. A=82°13´15", B=106°35'18", C=28°45'06", D=205°14'56", E=117°11'25".
Q4. Draw a diagram and then compute and tabulate for the following closed-polygon traverse (a) preliminary azimuth, (b) unadjusted departures and latitudes, (c) linear misclosure, (d) relative precision. (note line BC bears in an easterly in direction), (e) adjust the traverse using the compass rule. (f) if the coordinates of point A are 6521.951 E and 7037.072 N, determine the coordinate of all other points. (g) find the length and bearing of line AC.
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Course
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Azimuth
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Length (m)
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Interior angle (right)
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AB
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179°50'39"
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2862.392
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A = 120°05'50"
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BC
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4189.033
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B = 91°57'50"
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CD
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3815.353
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C = 121°44'06"
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DE
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3645.450
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D = 82°02'08"
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EA
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3490.014
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E = 124°10'11"
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|
|
|
|
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Q5. The (X, Y) coordinates (ft) for a closed-polygon traverse ABCDEFA are as follows. (A) 1000, 1000), (B) 1645.49, 1114.85, (C) 1675.95, 1696.06, (D) 1178.99, 1664.04, (E) 1162.62, 1337.78. Calculate the area by the coordinate method, show all work and express the answer to the nearest sqft.
Q6. Given the following curve data find: 1. The radius, 2. The delta angle, 3. The length of curve 4. The degree of curvature 5. The length of the tangent 6. The long chord length 7. The coordinate of a point on the curve at station 365+00.
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Station
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North
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East
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PC 360+47.96
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608442.42
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2702644.29
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PT 384+02.62
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610508.38
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2701723.21
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Radius Pt
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614927.39
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2712330.71
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Q7. Given the following vertical curve data find 1.) the BVC station, 2.) the EVC station 3.) the station and elevation at the low point or high point of the curve.
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VPI station
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VPI elevation
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G1
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G2
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Length
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247+00
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1324.55
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+3.21
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-1.19
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1600 ft
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Q8. Given the same data as problem 7 calculate the elevation at the profile grade line at 242+00 and then project a grade to 29 ft right using a -2% cross slope.
Q9. You are trying to fit a slope stake on a 4:1 slope. The catch point is at 60 right and the elevation of the catch point is 1150.00. You are at 92 right and measure an elevation of 1142.20. What offset distance out will put you where you need to be to catch the slope.