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Question

In a closed traverse, the sum of the latitudes is 1.39 m and the sum of the departures is 2.17 m. The Length and Whole Circle Bearing of the closing error are

The correct answer is
Length = 2.57 m, Whole Circle Bearing = $57^\circ$

Solving Traverse Closing Error: Length and Bearing

A closed traverse should theoretically end at its starting point. When it doesn't, a closing error occurs, represented by the resultant of the sum of latitudes and departures.

Error Magnitude Calculation

The length of the closing error ($E$) is calculated using the Pythagorean theorem based on the total error in latitude ($\sum \text{Lat}$) and the total error in departure ($\sum \text{Dep}$).

Given:

  • Sum of Latitudes ($\sum \text{Lat}$) = 1.39 m
  • Sum of Departures ($\sum \text{Dep}$) = 2.17 m

The formula for the length of the closing error is:

$ E = \sqrt{(\sum \text{Lat})^2 + (\sum \text{Dep})^2} $

Substituting the given values:

$ E = \sqrt{(1.39 \text{ m})^2 + (2.17 \text{ m})^2} $

$ E = \sqrt{1.9321 \text{ m}^2 + 4.7089 \text{ m}^2} $

$ E = \sqrt{6.641 \text{ m}^2} $

$ E \approx 2.577 \text{ m} $

Rounding to two decimal places, the length is approximately 2.57 m.

Error Bearing Calculation

The Whole Circle Bearing (WCB) of the closing error is determined using the arctangent function. Since the sum of latitudes is positive (error towards North) and the sum of departures is positive (error towards East), the bearing is in the first quadrant.

The formula is:

$ \text{WCB} = \arctan\left(\frac{\sum \text{Dep}}{\sum \text{Lat}}\right) $

Substituting the values:

$ \text{WCB} = \arctan\left(\frac{2.17 \text{ m}}{1.39 \text{ m}}\right) $

$ \text{WCB} = \arctan(1.56115...) $

$ \text{WCB} \approx 57.37^\circ $

Rounding to the nearest degree, the WCB is approximately $57^\circ$.

Conclusion

The calculated length of the closing error is 2.57 m and its Whole Circle Bearing is approximately $57^\circ$.

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Important Questions from Traversing

  1. The forebearing of four sides of a closed quadrilateral ABCDA are
    AB = 60°, BC = 149.87°, CD = 269.5°, DA = 20.17°.

    The calculated value of interior angle D is _____° (Answer in decimal degrees and rounded off to two decimal values).
  2. The angular misclosure of the closed-loop traverse shown in the figure is _____________ ° (Answer in decimal degrees and rounded off to three decimal places)

  3. It is needed to establish the coordinates of a station A. Two stations B and C are available whose coordinates are given in the table.

    Length of line CA = 1212.38 m; ∠BCA = 64°.

    The Easting of station A is _______ m (Rounded off to two decimal places).

    Assume station A lies to the west of the traverse line BC.

    StationEasting (m)Northing (m)
    B11054.099484.37
    C10827.6210112.15
  4. If the departure and latitude of a line are 70 m and -130 m, respectively, then the whole circle bearing of the line in degrees is
  5. While carrying out a traverse survey ABCDA' using a theodolite with the originating station A, the departures and latitudes of the lines, as obtained, are shown in the following figure (not drawn to scale). It is seen that, due to the observational errors, the originating station A and its computed station A' are not the same. For this survey, the 'closing error' in m is 

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