3. Compute the missing data in the table shown if the Pace factor of a surveyor is 0.796. Determinealso the relative precision if the Taped distance of CD is 434mLINENo. PacesТарedLINENo. PacesPacedDistancedistance.AB128CD543I.ВА125DC544АВ100mCD543DC546ВАAB124125CD542

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Answered: Compute the missing data in the table…

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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PAGE:. DATE :.. A rectangular lot steei tape which was later found to 5 mm too short. The recorded dimensions was mearured uring a be 121.5m long by 92:5m wide. What ir the Were error in the area
ANSWER KEY:Elev Sta. B= 700.3994mElev Sta. C=243.2945m∆hAB = 458.8394m∆hBC = 457.1049m∆hCA = 1.7345mNOTE: these are the exact answers.
From the field notes of a closed traverse shown below, Lines Bearing AB BC CD DA N 53'27' 0"E S 66 54' 0"E S 29'8' 0"W N 52'0' 0"W Distances 59.82 70.38 76.62 95.75 ma) Determine the Error of closure b) Relative Precision (in fraction form; format: 1/wholenumber)
Find the missing data and complete the table below. DIST (m) 23.324 27.459 ? 69.771 22.361 20.591 20.616 15.811 LINE AB BC CD CD DE EF FG GH HA Missing data: LINE Bearing N 30° 58' E N 56° 53' E ? S 27° 18' W N 26° 34' W S 60° 57' W N 14° 2¹ E N 71° 34' W DISTANCE (M) DIRECTION BEARING ANGLE (°)
Line MN, 250 m long, was paced five Himer by the surveyor and We recorded as follows." TRIAL NO. DISTANCE 121 2 122 3 120 4 123 121 line OP was paced four Hones dgain with the following resuilts'. Ano ther DISTANCE TRIAL NO. 1 321 S 13 3 310 4 312
Three triangulation stations A, C and E are points on the shore whose position were observedfrom X. The angles were measured using a sextant from the boat at X to the three triangulationstations. The following data were recorded during the survey procedure:Distance AC = 700mDistance CE = 680mAngle ACE = 148oAngle AXC = 45oAngle CXE = 35oAngle CAX = 70oDraw the diagrama.) Compute the angle CEX.b.) Compute the distance AX.c.) Compute the distance XE.
Geometric space, i need solutions in 15 minutes please. MCQ/ If Ø = 0.987643 rad. Find Ø in degrees. A-56º 35' 16'' B-45º 42' 16'' C-52º 04' 26'' D-54º 30' 16'' E-35º 25' 25'' F-65º 25' 13''
C. Table: Determination of Stadia Interval Factor (K) D. TRIAL 1 2 3 4 5 6 7 8 DISTANCE STADIA HAIR READINGS UPPER LOWER 15.00 20.00 25.00 30.00 35.00 40.00 45.00 50.00 Where: SKETCH: 1.319 m 1.335 1.358 1.367 1.415 1.435 1.460 1.445 1.169 m 1.135 1.108 1.065 1.035 1.035 1.010 0.935 S = Stadia Intercept K = Stadia Interval Factor STADIA STADIA INTERCEPT INTERVAL FACTOR (K) COMPUTATIONS: 1. Computing Stadia Intercept (S) and Stadia Interval Factor (K). a.) Stadia Intercept = Upper Rod Reading - Lower Rod Reading b.) Stadia Interval Factor (K) = Distance / Stadia Intercept (S) AVERAGE K = (Show computations on the computed stadia intercept and stadia interval factor.) (Draw the resulting figure based on the data given data. Refer to the illustration above.)
From the field notes of a closed traverse shown below, Lines AB BC Bearing Distances N 53° 27' 0"E S 66°54' 0"E CD S 29° 8' 0"W DA N 52'0' 0"W| 59.82 70.38 76.62 95.75 a) Determine the Error of closure b) Relative Precision (in fraction form; format: 1/wholenumber)
4. Find coordinate of H and I in Figure 4.1. NB BEB ΝΑ EA A ANS (Area ABDE)/3, H Station N,m. A B 2(Area ABDE)/3 E NE EE 195.000 155.700 100.250 68.500 ND D ED NOT TRUE SCALE Figure 4.1 Working area E,m. 61.750 100.500 148.000 100.000
ANSWER KEY:Elev Sta. B= 700.3994mElev Sta. C=243.2945m∆hAB = 458.8394m∆hBC = 457.1049m∆hCA = 1.7345m
Line MN, 25Em long, was paced five times by the surveyor and recorded as follows: Trial No. Distance 121 2 122 3 120 4 124 122 Another line OP was paced four times again with the following results: Trial No. Distance 1 122 2 314 311 4 322 Determine the actual length of line OP. 1,

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Compute the missing data in the table shown if the Pace factor of a surveyor is 0.796. Determine also the relative precision if the Taped distance of CD is 434m