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CivilAdvanced Survey
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What is said about that traverse when total latitude and departure are zero?

A

Balancing of the traverse

B

Closed traverse

C

Closing error

D

Both A and B

Correct Answer

тЪЩя╕П TE тАв Technical Concept & PrincipleCivilAdvanced Survey
Option D

Both A and B

Quick Summary:

When the sum of all latitudes (тИСL\sum LтИСL) and the sum of all departures (тИСD\sum DтИСD) are both equal to zero, it signifies that the survey starts and ends at the exact same point, defining a closed traverse. Furthermore, this condition means there is no closing error, and the traverse is perfectly balanced mathematically.

тЪЩя╕ПTETechnical SolutionConcept & Principle
ЁЯТб Explanation

When the sum of all latitudes (тИСL\sum LтИСL) and the sum of all departures (тИСD\sum DтИСD) are both equal to zero, it signifies that the survey starts and ends at the exact same point, defining a closed traverse. Furthermore, this condition means there is no closing error, and the traverse is perfectly balanced mathematically.

ЁЯФв Key Formulas

тИСL=0\sum L = 0тИСL=0 тАФ Sum of latitudes for a balanced closed traverse

тИСD=0\sum D = 0тИСD=0 тАФ Sum of departures for a balanced closed traverse

e=(тИСL)2+(тИСD)2e = \sqrt{(\sum L)^2 + (\sum D)^2}e=(тИСL)2+(тИСD)2тАЛ тАФ Closing error magnitude

тЪЩя╕П Working Principle

In surveying, latitude is the north-south component (L=LcosтБб╬╕L = L \cos \thetaL=Lcos╬╕) and departure is the east-west component (D=LsinтБб╬╕D = L \sin \thetaD=Lsin╬╕) of a line. For a closed loop traverse, the algebraic sum of latitudes must equal zero (total northing equals total southing) and the algebraic sum of departures must equal zero (total easting equals total westing). When these conditions are satisfied, the total displacement is zero, confirming both a closed traverse and complete balancing.

ЁЯУМ Key Points
  • тЦ╕

    Latitude (LLL) is the projection of a traverse line on the north-south meridian (L=lcosтБб╬╕L = l \cos \thetaL=lcos╬╕).

  • тЦ╕

    Departure (DDD) is the projection of a traverse line on the east-west reference line (D=lsinтБб╬╕D = l \sin \thetaD=lsin╬╕).

  • тЦ╕

    When тИСL=0\sum L = 0тИСL=0 and тИСD=0\sum D = 0тИСD=0, the closing error e=0e = 0e=0, representing a perfectly balanced closed traverse.

тЬЕ Advantages
  • тЦ╕

    Ensures geometric consistency and closure in field surveying control networks.

  • тЦ╕

    Eliminates linear misclosure, allowing precise boundary mapping and area computation.

тЭМ Disadvantages / Limitations
  • тЦ╕

    Does not guarantee the total absence of systematic or compensating operational errors that cancel each other out.

  • тЦ╕

    Requires high precision in both angular and linear measurements to achieve zero closing error naturally.

ЁЯЫая╕П Applications / Uses
  • тЦ╕

    Establishment of horizontal control networks for major civil engineering projects.

  • тЦ╕

    Cadastral survey, land boundary determination, and topographic mapping.

ЁЯУД Additional Information
  • тЦ╕

    Option A is correct because when total latitude and departure are zero, the traverse requires no further corrections and is considered balanced.

  • тЦ╕

    Option B is correct because a closed traverse forms a closed loop returning to the origin where total latitude and departure algebraically sum to zero.

  • тЦ╕

    Option C is incorrect because closing error exists when тИСLтЙа0\sum L \neq 0тИСLюАа=0 or тИСDтЙа0\sum D \neq 0тИСDюАа=0.

ЁЯУК Diagram / Illustration
Closed Traverse EquilibriumSum of Latitudes (╬гL) = 0Sum of Departures (╬гD) = 0Closing Error (e) = тИЪ(╬гL┬▓ + ╬гD┬▓) = 0тЗТ Traverse is Closed & Fully Balanced
тЬЕ

D is correct тАФ Total latitude and departure being zero defined a perfectly closed traverse with zero closing error, representing a fully balanced condition.

Core Concepts Used
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Latitude and Departure Closing Error Traverse Balancing Methods (Bowditch / Transit Rule)
ЁЯТб EXAM TIP

In exam problems, if тИСLтЙа0\sum L \neq 0тИСLюАа=0 or тИСDтЙа0\sum D \neq 0тИСDюАа=0, the direction (bearing) of the closing error is given by tanтБб╬╕=тИСDтИСL\tan \theta = \frac{\sum D}{\sum L}tan╬╕=тИСLтИСDтАЛ.

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