fundamentals of surveying. transit rule. 10. TRANSIT RULE. Given in the accompanying tabulations areobserved and calculated data for two different traverses.AZIMUTHLATITUDEDEPARTURELINEDISTANCEFROM SOUTH+N+EAB357.20 m99°45'154 00242 30353 15'60.49164.64352.04BC183. 1880.30CD430.05198.57381.46DA425.18422.2349.97Sums1395.6Im+ 423.70-422.23+431.43-432.34(D)(ENL)(ESL)(SED)( WD) LATITUDEDEPARTUREAZIMUTHFROM NORTHCOURSELENGTH+N+E-W5.530°50'296°06'205°54'135 20'33 27'95 39'CD380.38 m380. 34DEEF51 9.30228.46466.35303.08693. 87624.18FG417.98297.27293.83GH403.93337.03222.65HC249.0024.51247.79Sums2664. 46 m+945.83-945.96+769.80 -769.43(D)CENL)(ESL)EEDI(ZWD)balance the latitudes andAlso determine the 1ineartheobservedcalculated values into a convenient format. and prepatea sketch of the adjusted traverse. Use a convenient scale.For each set of traverse data,departures by the transit rule.of closure,bearing of the side of error,'anderrorprecision of the measurements made. Tabulate alland

∑L=∑NL+∑SL=+423.70-422.23=+0.47∑D=∑ED+∑WD=+431.43-422.34=-0.91 ∑l=Sum of all lengths/traverse…

-10. TRANSIT RULE. Given in the accompanying tabulations are observed and calculated data for two diffèrent traverses. AZIMUTH LATITUDE DEPARTURE LINE DISTANCE FROM SOUTH +N -S +E AB вс 357.20 m 99*45' 154 00 242 30 353 15' 60.49 352.04 183. 18 164.64 80.30 CD 430.05 198.57 381,46 DA 425.18 422.23 49,97 Sums 1395.61m + 423. 70 -422.23 +431.43 -432.34 (D) (ENL) (ESL) (SED) (WD)

-10. TRANSIT RULE. Given in the accompanying tabulations are observed and calculated data for two diffèrent traverses. AZIMUTH LATITUDE DEPARTURE LINE DISTANCE FROM SOUTH +N -S +E AB вс 357.20 m 99*45' 154 00 242 30 353 15' 60.49 352.04 183. 18 164.64 80.30 CD 430.05 198.57 381,46 DA 425.18 422.23 49,97 Sums 1395.61m + 423. 70 -422.23 +431.43 -432.34 (D) (ENL) (ESL) (SED) (WD)

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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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
3. The following data if a closed traverse is shown in the technical description. The area of the closed traverse is 63147.05 sq. m. LINES AB BC CD LAT. a -240.20 -160.00 DA 179.70 Solve for the unknown variables. DEP. 120 с e -220 DMD b d f 8 2A 26460.0 X -86480.0 39534.0
8. Compute for the following and show the tabulated solutions: Azimuth LINES Length (S) АВ 1020.87 168d35m ВС 1117.26 263d30m CD 660.08 304d25m DE 495.85 5d30m EF 850.62 46d15m FA 855.45 112d30m (lengths in 3 dec. places; bearings in deg-mm-ss) a. Using Transit Rule, solve for adjusted latitudes, departures, distances, and bearings for all the traverse lines Lines Adj. Latitude Adj. Departure Adj. Length Adj. Bearing AB BC CD DE EF FA
Q3): The following data were collected while running a closed traverse ABCDA. Calculate: 1- Coordinates of B (Easting and Northing). N Line AB BC CD DA 4- 5- Distance BC and CD. 6- Coordinates of C (Easting and Northing). Interior angles A and B Length(m) 330 827 Sum= 2435.15 Azimuth 181°25' 89°50' 355°00' 264°18'20" Point A B C D A (calculated) = Easting 500 B 1323.167 Northing 500 582.059 A= B= Angles
ANSWER KEY=D 108.1502m∆h 22.4596m
Glven the following notes for a closed traverse. Determine the following a) Coordinates of station E (xy). (Blank 1.Blank 2) b) Area of the traversed using coordinate method in m^2. Blank 3 ADJUSTED ADJUSTED LINE DEPARTURE LATITUDE АВ 395.60 352.00 ВС 591.60 23.40 CD 78.30 -558.20 DE -746.30 -102.50 EA -319.20 285.30 Blank 1 Add your answer Blank 2 Add your answer Blank 3 Add your answer
Following are the data gathered from the traverse: Lat Dep DPD 2A 399.790 0 565.388565.277 -350.184605.780 -564.112-205.360 -50.882-965.697 Express all answer unitless and with 3 decimals for A to J No new data to sav
Situation 4] From the given data ofa dosed traverse, The adjusted bearing of BC using compassrule. The adjusted bearing of DA using compass rule. The total area using Coordinates method. The total area using Double Parallel Distance method. i. ii. i. iv. Distance, m 368.76 Bearing N15°18'E S85046'E S18°30W N77035W Line АВ ВС 645.38 CD 467.86 DA 593.00
Q3): The following data were collected while running a closed traverse ABCDA. Calculate: 1- Coordinates of B (Easting and Northing). N Line AB BC CD DA 4- Interior angles A and B 5- Distance BC and CD. Length (m) 330 827 Sum= 2435.15 Azimuth 181°25' 89°50' 355°00' 264°18'20" Point A B C D A (calculated) = Easting 500 B 1323.167 Northing 500 582.059 A= B= Angles
Bearing Angle Traverse Given the tabulated measurement of the closed traverse as shown. Determine the Area of the Traverse. Line Length Bearing AB 589.060 m N 87°30’ E BC 391.730 m N 07°02’ E CD 291.860 m N 30°30’ E DE 656.980 m S 42°45’ E EF 654.060 m Unknown FA 686.660 m Unknown Note: EXPRESS YOUR ANSWER UPTO 3 DECIMAL PLACES.
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
Given the following data for a closed traverse in which the lengths of sides CD and DE are unknown. LINES BEARING LENGTH LAT. DEP. AB S 85° 58' E 541.55 m 38.09 +540.21 BC S 03°05 W 795.62 m -794.47 4280 CD S 520 50' W DE N 12°42 W EA N 40° 15 E 650 25 +496 29 +420.14