A surveyor uses an auto-level instrument to calculate thelength of the sides of a traverse (horizontal distances) basedon stadia method. Table below presents the upper, middle andlower crosshair readings. Determine the horizontal distancesbetween the corners of this traverse.Point Upper stadia Middle stadia Lower stadia Distance (ft.)A1.981.581.18A2.211.410.61C2.191.290.39D2.451.951.45E1.391.190.99

The formula to calculate the horizontal distance is given by, D=KS+C Here, K is the multiplying…

Answered: A surveyor uses an auto-level…

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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to find out the included angles in a closed survey PQRSTP, the following observations were made with the compass. calculate the included angles after correcting the local attractions line fore bearing(FB) back bearing(BB) PQ N 620 45/E S 62015/W QR N 210 00/W S 200 45/E RS N 710 30/W S 710 30/E ST S 390 00/W N 380 00/E TP S 540 30/E N 530 15/E
Elementary Surveying: Measurement of Horizontal Distances, show proper and right solution. 1. A line AC cannot be measured directly because of an obstruction on theline. Accordingly, the two lines AB and BC were measured as 2,357.90mand 1,846.00m, respectively. Point B was set at a perpendicular distanceof 225m from point D on line AC. Using the approximate slope correctionformula for steep slopes, determine the length of line AC. Illustrate.
A 5-sided closed traverse that needs correction has the following distances and azimuths:AB = 285.10 m <26°10'BC = 610.45 m <104°35'CD = 720.48 m <195°30'DE = 203.00 m <358°18'EA = 647.02 m <306°54'
Course AB of a five-sided traverse runs due north. From the given balanced interior angles to the right, compute and tabulate the bearings and azimuths from the north for each side of the traverses. a) A = 77˚23’26’ b) B = 125 ˚58’59” c) C = 105 ˚28’32” d) D = 116 ˚27’02” e) E = 114 ˚42’01”
Problem #1 For the given data of a close traverse, determine the missing bearings of courses DE and EF then solve the AREA using DPD. Sketch the traverse. LINES DISTANCE (m) BEARING AB 60 S 49⁰16’ E BC 59 S 67⁰48’ E CD 95 N 32⁰28’ W DE 110 ? EF 128 ? FA 71 S 18⁰56’ E
A five-sided closed field traverse has the following distances in meters: AB = 51.766; BC = 26.947; CD =37.070; DE = 35.292; EA = 19.192. The adjusted angles are as follows: A = 101°03′19″; B = 101°41′49″; C =102°22′03″; D = 115°57′20″; E = 118°55′29″. The bearing of AB is N 75°05′30″ E. BC is in the S.E. quadrant OR BC runs in the direction of South East. What do the interior angles add up to be and do they meet the interior angle check (n-2)*180? What is the Azimuth of BC? What is the Azimuth of CD? What is the Azimuth of DE? What is the Azimuth of EA? What is the the Latitude of BC? What is the the Latitude of DE What is the the Departure of AB? What is the the Departure of CD? What is the the Departure of EA? What is the the total perimeter distance of the traverse? What is the the total error in the Latitudes? (OR what is the total when you add up all of the latitudes?) What is the the total error in the Departures? (OR what is the total when you add up all of the Departures?)…
Given the closed traverse data. Lines Bearing Distance AB N20o E 20.35 BC N68oE 18.75 CD S22oE 21.30 DE S40oW 13.35 EF S62oW 9.23 FA Compute the following. Compute the bearing and distance of line FA Compute the area of the given traverse
The data in the table below is for a closed-polygon traverse. Balance the traverses by coordinates using the compass rule. Course AB BC CD DA Bearing N8∘17′02"E N87∘02′05"E S14∘47′06"W N68∘43′20"W Length (ft ) 403.73 622.63 653.16 550.84
PROBLEM 1 Given the following interior angles (angles-to-the-right) of a five-sided closedpolygon traverse and if the azimuth of side AB is fixed at 74°31'17", determine the adjusted azimuth of line EA.A =105°13'14"; B = 92°36'06"; C = 67°15'22"; D = 217°24'30"; E = 57°30"38". (Note: line BC bears NW.) NEEDED ANS: Compute the relative precision for the traverse of Problem 1 if the lengths of the sides (in feet) are as follows: AB = 2157.34; BC =1722.58;CD =1318.15; DE =1536.06; and EA =1785.58. (Note: Assume units of feet for all distances.) Round off the denominator to two decimal places.
A closed five-sided field traverse has the following interior angles: A = 121°13′00″; B = 136°44′30″; C = 77°05′30″; D = 94°20′30″. Find the angle at E.
elementary surveying A line AC cannot be measured directly because of an obstruction on line. Accordingly, the two lines AB and BC were measured as 2,357.90m and 1,846.00m, respectively. Point B was set at a perpendicular distance of 225m from point D on line AC. 1.Using the approximate slope correction formula for steep slopes, determine slope correction for AB. 2.Using the approximate slope correction formula for steep slopes, determine slope correction for BC. 3. Determine the length of line AC.
SURVEYING From the following traverse: (Use 3 decimal point) LINE DISTANCE BEARING AB 370m N 15° E BC 648 S 86° E CD 470 S 19° W DA 585 N 78° W Calculate the area of the traverse by DPD method.

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