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Chain Surveying (Part 2)

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    Chain surveying:  A type of surveying in which only linear measurements are made in the field. Suitability:    Suitable for surveys of small extent    Secure data for exact description of the boundaries    To take simple details Principle: To provide a skeleton or frame work consisting of a number of connected triangles; as triangle is the only simple figure that can be plotted from the lengths of its sides measured in the field. To get good results in plotting, framework should consist of triangles which are as nearly equilateral as possible. Chain survey is sometimes called “chain triangulation”.   Survey station:      A prominent point on the chain line.    Can be either at the beginning or ending at the chain line.    May be marked by driving pegs (if the ground is soft). Tie station:    A point may be selected anywhere on the chain line.    Tie line may be run through them. Survey lines:      The lines joining the main survey stations.    The biggest of the main survey lines known as base line.    Various survey stations are plotted with reference to base line. Check lines:      Used to check the accuracy of the work.    Length of the check line measured in the field must agree with its length on the plan. Tie lines:      A line joining tie stations on the main line.       Used to take the details of the nearby object.    Also serves the purpose of the check line. Conditions to be fulfilled by the survey lines or survey stations:      Stations must be visible.    Survey lines must be as few as possible.    The framework must have at least one base line. If one base line is used, it must run along the length and through the middle of the area. If 2 base lines are used, the must intersect in the form of „X‟.      The lines must run through level ground.    The lines should form well-conditioned triangles.    Each triangle should have sufficient check line.    The main survey lines should not pass through obstacles.    The lines should fall within the boundaries of the property to be surveyed. Basic problems in Chaining:   1)   To erect a perpendicular to a chain line from a point on it:    Let XY be the chain line. Let it be required to erect a perpendicular to the chain line at a point A on it.    Assume AC, AB and BC being 4, 3, and 5 units respectively.    Then angle   BAC will be 90 0 .    And AB will be perpendicular to the chain line XY.    Because AC 2  + AB 2  = BC 2 .   Y X A C B 4 3 5   90 o      2)   To drop a perpendicular to a chain line from an external point:    Let XY be the chain line and P is the external point.    Keep the zero end of the tape at P and swing the tape along the chain line.    The point of minimum tape length is note.    Because perpendicular is the shortest distance. (See another method from handout, page 23) 3)   To draw a line parallel to the chain line through a given point:    Let XY be the chain line and Q is the point through which parallel line to be drawn.    Through Q, perpendicular QP is drawn on XY. QP is measured.    Selecting point R on the chain line, perpendicular RS is drawn on XY.    RS is measured. RS must be equal to QP.    Joining points Q and S, QS is parallel to XY. 4)   To run a parallel to a given inaccessible line through a given point: P A1 A1 Y X   A2 A X Y P R S       Let AB be the given inaccessible line and C be the given point through which parallel is to be drawn.    Select point E so that A, C, E remain in the same line.    Select point F and join E, F.    Through C, draw line CG so that CG is parallel to AF.      Through G, draw line GD so that GD is parallel to BF, cutting BE in D.      Then CD  will be the required line. Obstacles in Chaining: (Definition- handout pg-25) a.   Obstacle in ranging This type of obstacle, in which the ends are not intervisible, is quite common except in flat countries. Two cases:    Case 1: Both ends visible from intermediate points on the line.    Case 2: Both ends invisible from intermediate points on the line. Case 1: Method of reciprocal ranging may be used. Case 2: Both ends invisible from intermediate point   A B E F G D C A B D C D C B 1   90 degree 90 degree
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