Graphical Adjustment OF A Closed Traverse PDF

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BATANGAS STATE UNIVERSITY Pablo Borbon Main II College of Engineering, Architecture, and Fine Arts Civil and Sanitary Engineering DepartmentCE 404FUNDAMENTALS OF SURVEYINGLAB EXERCISE NO. 13:GRAPHICAL ADJUSTMENT OF A CLOSED TRAVERSEby:Andal, Van Rouel R. Baldrias, Zyra S. Berania, Mary Kate T. Cante...

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Republic of the Philippines BATANGAS STATE UNIVERSITY Pablo Borbon Main II College of Engineering, Architecture, and Fine Arts Civil and Sanitary Engineering Department

CE 404 FUNDAMENTALS OF SURVEYING

LAB EXERCISE NO. 13: GRAPHICAL ADJUSTMENT OF A CLOSED TRAVERSE

by: Andal, Van Rouel R. Baldrias, Zyra S. Berania, Mary Kate T. Cantero, Louie Roel L. Claveria, Raven Tristan D. Francisco, Kyell Adam G. Mancilla, Eleina Jean Pagcaliwangan, Joshua A. San Andres, Kathleen Lourdes P. Untalan, Joshua M. Zalameda, Rizza Brenda D.

GROUP 3 CE-2201

Submitted to: Engr. Marcelo Teñoso Jr.

Republic of the Philippines BATANGAS STATE UNIVERSITY Pablo Borbon Main II College of Engineering, Architecture, and Fine Arts Civil and Sanitary Engineering Department

LABORATORY EXERCISE NO. 13 GRAPHICAL ADJUSTMENT OF A CLOSED TRAVERSE I.

OBJECTIVE To learn how to plot a closed traverse and adjust the error of closure by graphical application of the compass rule. II.

PARTY MEMBERS ANDAL, Van Rouel R. BALDRIAS, Zyra S. BERANIA, Mary Kate T. CANTERO, Louie Roel L. CLAVERIA, Raven Tristan D. FRANCISCO, Kyell Adam G. MANCILLA, Eleina Jean PAGCALIWANGAN, Joshua A. SAN ANDRES, Kathleen Lourdes P. UNTALAN, Joshua M. ZALAMEDA, Rizza Brenda D.

18-02896 18-55892 18-05591 18-08098 18-03503 18-01488 15-59115 18-54976 18-50486 18-50099 18-59439

III. EQUIPMENTS / INSTRUMENTS Engineer’s Transit or Theodolite, Range Poles, Chaining Pins, Hubs or Pegs, Steel tape, and Plumb Bobs. IV. PROCEDURE: 1. This problem is an indoor lab exercise and it will be assumed that the given data were taken form an actual field observation. 2. The lab instructor has the option to use the prepared data, revise it, or give a new set if data. 3. In order that the student will better understand the procedure involved in working out this lab exercises, the following illustrative problem is given. 4. Illustrative Problem: a.) Constructing the Correction Triangle 1.) Using the same scale used in plotting the given traverse, construct a straight line AA’ representing the total length of the given traverse and mark off distances AB, BC, CD, and EA’ equal to the respective lengths of the different lines on the traverse. (Refer to accompanying sketches) 2.) Measure Bb on the correction triangle and lay out, starting at station B of the plotted traverse, the same length along the line just drawn through the station. Label this line also Bb. 3.) Repeat the foregoing procedure at traverse stations C, D and E to similarly lay out the lengths of Cc, Dd and Ee accordingly 4.) The adjusted is determined by successively connecting the following points now plotted on the traverse: A, b, c, d, e, and back to A. This is shown by dashed lines.

Republic of the Philippines BATANGAS STATE UNIVERSITY Pablo Borbon Main II College of Engineering, Architecture, and Fine Arts Civil and Sanitary Engineering Department

Figure 4-17. Graphical Adjustment of a closed traverse

Figure 4-18. The correction triangle 5. After understanding the above illustrated process, consider the following listed two sets of data for closed traverse. It will be expected that an error of closure exists in both data sets. You are now required to: a.)Plot both traverses separately on 8 ½ “x11” paper using a scale 1:100. b.)Construct the corresponding correction triangle for each traverse. c.) Adjust graphically the error of each traverse. NOTE: Label plotted traverse and their corresponding correction triangles accordingly. All construction lines should be shown. 6. Data for the 1st traverse: The length and bearing of each line are: AB, 69.50 m, S73°30’E; BC, 123.00 m, N79°00’E; CD, 80.00 m, N40°00’W; DE, 104.00 m, N86°40’W; and EA, 90.50 m, S33°00’W. 7. Data for the second traverse: the length of azimuth from south of each line are: AB, 83.10 m, 162°30’; BC, 191.75 m, 265°30’; CD, 116.50 m, 1°15’; DE, 95.00 m, 121°00’; and EA, 99.00 m, 76°30’. V.

DATA Table 1. The mean length of each line of 1st traverse and its observed bearings LINE

LENGTH

OBSERVED BEARINGS

AB

69.50 m

S 73°30’ E

BC

123.00 m

CD

80.00 m

DE

104.00 m

EA

90.50 m

N 79°00’ E

N 40°00’ W N 86°40’ W S 33°00’ W

Republic of the Philippines BATANGAS STATE UNIVERSITY Pablo Borbon Main II College of Engineering, Architecture, and Fine Arts Civil and Sanitary Engineering Department

Table 2. The mean length of each line of 2nd traverse and its observed bearings

VI.

LINE

LENGTH

OBSERVED BEARINGS

AB

83.10 m

162°30’

BC

191.75 m

265°30’

CD

116.50 m,

1°15’

DE

95.00 m,

121°00’

EA

99.00 m,

76°30’

COMPUTATIONS: Since the solution is graphical, there are no mathematical computations to be made. However, the group wished to calculate for the Linear Error of closure following a mathematical computation in order for the group to compare the result on the graphical sketch made in AutoCAD. a. Mathematical Computation FIRST TRAVERSE

Determining the course latitude ( Lat = d cos α) 𝐿𝑎𝑡𝐴𝐵 = 69.50 cos 73°30’ = - 19.739 𝐿𝑎𝑡𝐵𝐶 = 123.00 cos 79°00’ = + 23.469 𝐿𝑎𝑡𝐶𝐷= 80.00 cos 40°00’ = + 61.284 𝐿𝑎𝑡𝐷𝐸= 104.00 cos 86°40’ = + 6.047 𝐿𝑎𝑡𝐸𝐴= 90.50 cos 33°00’ = - 75.899

Determining the precision of measurement ∑NL = 𝐿𝑎𝑡𝐵𝐶 + 𝐿𝑎𝑡𝐶𝐷 + 𝐿𝑎𝑡𝐷𝐸 = 23 .469 + 61.284 +6.047 = 90.8 m ∑SL = 𝐿𝑎𝑡𝐴𝐵 + 𝐿𝑎𝑡𝐸𝐴 = (-19.739) + (-75.899) = - 95.638 m

∑ED = 𝐷𝑒𝑝𝐴𝐵 + 𝐷𝑒𝑝𝐵𝐶 = 66.638 + 120.740 = 187.378 m

∑WD = 𝐷𝑒𝑝𝐶𝐷 + 𝐷𝑒𝑝𝐷𝐸 + 𝐷𝑒𝑝𝐸𝐴 = (-51.423) + (-103.824) + (-49.289) = -204.536 m

Determining the course departure ( Dep = d sin α) 𝐷𝑒𝑝𝐴𝐵 = 69.50 sin 73°30’ = + 66.638 𝐷𝑒𝑝𝐵𝐶 = 123.00 sin 79°00’ = + 120.740 𝐷𝑒𝑝𝐶𝐷 = 80.00 sin 40°00’ = - 51.423 𝐷𝑒𝑝𝐷𝐸 = 104.00 sin 86°40’ = - 103.824 𝐷𝑒𝑝𝐸𝐴 = 90.50 sin 33°00’ = - 49.289

the C𝐿 = ∑NL + ∑SL

= 90.8 + (-95.638) = - 4. 838 m

C𝐷 = ∑ED + ∑WD = 187.378 +(-204.536) = - 17.158 m Linear Error of Closure LEC = √𝐂𝑳 𝟐 + 𝐂𝑫 𝟐

= √(−4.838)𝟐 + (−17.158)𝟐 LEC = 17. 827 m...