S LINEAR ERROR OF CLOSURE Problem 2: For the given data of a transit-tape survey. R 1. Determine the latitudes and departures of the lines. 2. Determine the linear error of closure. 3. Determine the bearing of the line of the side error. V 4. Determine the relative precision. E LINE LENGTH (m) BEARING LATITUDE DEPARTURE +N +E -W Y АВ 495.85 N 05°30'E BC 850.62 N 46°02'E CD 855.45 S 67°38'E DE 1020.87 S 12°25'E EF 1117.26 S 83°44'W FA 660.08 N 55°09'W Total ΣΝL ΣSL ΣED EWD

S LINEAR ERROR OF CLOSURE Problem 2: For the given data of a transit-tape survey. R 1. Determine the latitudes and departures of the lines. 2. Determine the linear error of closure. 3. Determine the bearing of the line of the side error. V 4. Determine the relative precision. E LINE LENGTH (m) BEARING LATITUDE DEPARTURE +N +E -W Y АВ 495.85 N 05°30'E BC 850.62 N 46°02'E CD 855.45 S 67°38'E DE 1020.87 S 12°25'E EF 1117.26 S 83°44'W FA 660.08 N 55°09'W Total ΣΝL ΣSL ΣED EWD

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Q3: A: Draw the diagram of traverse classification. B: In closed traverse ABCD. The partial coordinates were known as shown below. Determine its area by DMD method. 04 N Dofland OWODY dienstl it with (nm) dinostim Mem. AB BC CD DA ΔΧ 40 -80 -50 AY 60 100 -70
30 Y 5OKN 3. 4 40° 7OKN 450 l00 KN z Fy=? ī Fx = ? %3D RF = ? %3D 20
with the given table below, find the following:- Linear Error of Closure- Relative Error of Closure- BSE Bearing Distance Lat Dep cLat cDep AdjLat AdjDep AB S 68 5 35 W 472.68 -176.36 -438.55 45.67633474 45.67926201 -130.68 -392.87 BC N 19 46 0 W 216.3 73.09 -203.40 20.88522093 20.8865594 93.98 -182.51 CD N 45 55 20 E 276.52 192.35 198.65 26.72086842 26.72258088 219.08 225.37 DA S 54 59 15 E 382.24 -219.31 313.06 36.93687525 36.93924243 -182.38 350.00 1347.57 ∈latitude=el=-130.22 ∈departure=ed=-130.23 130.2192993 130.2276447 0.00 0.00
Using the data below, compute for the Linear Error of Closure Also, Determine the adjusted latitude and departure using compass rule: Line Distance Latitude Departure Adjustment (Lat) Adjustment (Dep) Adjusted Lat Adjusted Dep 1-2 388.08 -234.363 -309.322 2-3 732.72 732.665 8.952 3-4 324.78 -22.938 323.969 4-1 477.24 -476.557 -25.531 Express all answer unitless and with 3 decimals for A to J Next
A traverse has the following data: Upload LINE Choose a File AB BC CD DE EA LENGTH 33.48 Use compass rule in adjusting the closed traverse and compute for the Linear Error of Closure and Relative Error of Closure. 8.11 39.42 7.08 22.02 BEARING N 73°23' E $ 39°30' E S43°40' W N 39°50' W N 17°33' W
1 7:46 0.60 4G 49 KB/s 2 A surveying transcript of a closed traverse is defined by the vertices ABCDEFGHIJKL. Engr. Pot is performing Double Meridian Distance Method in order to compute the area of the traverse. Midway in his work, his nephew Ashton accidentally spilled Engr. Pots' coffee onto the notes. Engr. Pot decided to rewrite the data, only to find out that the DMD of the lines CD and DE are unreadable. He figures that he can just solve for the missing data knowing that the DMD of line DE is twice the DMD of the previous line CD. If the departure of line CD is -124.25 and that of line DE is 301.965, determine the DMD of line DE. Your answer A lot has a frontage along a road. The 2 II
The observed lengths of the sides (in feet) of a five-sided closed polygon traverse are as follows: AB= 202.74, BC = 285.87, CD= 499.37, DE = 321.33, and EA = 382.78. (Note: Assume units of feet for all distances.) The preliminary azimuths obtained after balancing the angles are as follows: AB= 138° 16'27", BC = 67°02′46", CD= 354°22'24", DE = 256° 13'01", EA = 185°29/20". Compute the departures of the sides. Express your answers in feet to five significant figures and separated by commas. AB, BC, CD, DE, EA = Part B ||| ΑΣΦ ΑΣΦ AB, BC, CD, DE, EA = vec Compute the latitudes of the sides. Express your answers in feet to five significant figures and separated by commas. [ΠΙ —| ΑΣΦ | w vec ? ? ft, ft, ft, ft, ft ft, ft, ft, ft, ft
I. Determine the interior angles of each stations and azimuths for each surveying lines. AZAB = INTZA = 40 40P 37 AZBC = INTZB = C 30 22 AZCD = INTZC = 66° 40' AZDE = INTZD = 15° 12 1s° 58 AZEA = INTZE =
PROBLEM 1. Given in the accompanying tabulation are the adjusted latitude and adjusted departure of a closed traverse. Calculate the area by the DMD method LINE ADJ ADJ LAT DEP AB - 125.055 + 196.387 BC +20.241 + 241.003 CD +180.107 + 65.326 DE +210.169 - 104.519 EF -74.003 - 260.516 FA - 211.459 - 137.681
Below is a transit and tape survey notes of a lot. a. Compute the angular error of deflection angle. b. Compute the corrected bearing of the lines. Assume that line 1-2 has a corrected bearing of N 13° E. Station Deflection Line Bearing Distance Angle 55°30'R 1 1-2 650 99°30'R 2-3 895 44°00'R 3-4 315 4 92°00'R 4-5 875 68°55'R 5-1 410 23
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Which of the following indicates the correct value of precise closing error if e = 0.54 and lengths of sides are 92.69 m, 119.23 m, 92.64 m, 42.96 m, and 60.96 m? O a. 1/766.445 O b. 1/746.445 O c. 1/756.445 O d. 1/765.445