Obstacle in Chaining

 

Sometimes happens that a survey line passes through some object such as a pond, the building, a river, a hedge etc. which prevents the direct measurement of that part of the line which the object intersects. The interfering object in such a case is called on the obstacle.

The various obstacle may be classed as:

1. When Chaining is free, but Vision is obstructed.

2. When Chaining is Obstructed, but Vision is Free.

3. When Chaining and Vision Both are obstructed.

It is necessary to overcome obstacles so that chaining may be continued in a straight line. Special methods are, therefore, employed in measuring distances across the obstacles.

1. When Chaining is free, but Vision is obstructed.

Such a the problem arise when rising ground or hill intervening, the jungle is interrupted etc. Hence the end stations are not inter-visible. There are two cases

Case-I

In this method difficulty got over by reciprocal ranging. The reciprocal ranging may also, be used in ranging a line across a hollow.

Case-II

The end stations are not visible from intermediate points when a jungle area comes across the chain line. In this case, the obstacle may be crossed over using a random line.

Procedure

1.     Let the AB be the line whose length is required.  From A run a line ‘AB1’ called the random line, in any conventional direction but as nearly towards ‘B’  as can be the jungle and continue it until the point ‘B’ is visible from ‘B1’. Chain the line to ‘B1’ where ‘ BB1’ is perpendicular to ‘AB1’ and measure ‘AB1’ 

                                    AB = √(AB1² + BB1²) 

1.       If any length ‘AC1’ is measured along ‘AB1’ at point ‘C’ is located on the line ‘AB’ by measuring a perpendicular distance ‘CC1’.

                                        (CC1 / BB1) = (AC1 /AB1)

                             CC1  = (AC1 /AB1) x BB1

1.      In this corner a sufficient number of points can be located. The line is cleared and the distance measured.

2. When Chaining is Obstructed, but Vision is Free.

Such a the problem arise when a pound or a river tank, plantation comes across the chain line. The typical obstacle of this type is a sheet of water, the width of which in the direction of measurement exceeds the length of the chain or tape. The problem consists of finding the distance between convenient points on the chain line on either side of the obstacle.

Case-I

In which it is possible to chain around the obstacle e.g. a pound. A bend in the river etc.

Part-1

 Suppose AB is a chain line. Two-point ‘C’ and ‘D’ are selected on it on the opposite bank of the pound.    Equal perpendicular ‘CE’ and ‘DF’  erected at ‘C’  and ‘D’.  Distance ‘EF’ is measured (i.e. CD= EF)

Part-2 The the pond may also be crossed by forming a triangle as shown. Then

                            CD = √(ED² + CE²) 

Part-3 Find by optical square or a cross-staff a point C such that ∠PRQ is right angle (Fig. 4) Measure PR and RQ. Then

                                   CD = √(PR² + RQ²) 

Case-II

Sometimes it is not possible to go around the obstruction {i.e. obstacle is a river}

Part-1 Imagine a small river comes across the chain line 

 Suppose ‘AB’ is the chain line. Two points ‘C’ and ‘D’ selected on this line on the opposite bank of the river.At ‘C’ a perpendicular ‘CE’ is erected and bisected at ‘F’.A perpendicular is set out at ‘E’ and a point ‘G’ are in the same straight line. From triangle ‘DCE’ and ‘GEF’ the distance ‘GE’ is measured and ‘CD’ is obtained indirectly 

Part-2 Imagine a large river comes across the chain line.

1    Let ‘AB’ be the chain line. Point ‘C’, ‘D’ and ‘E’ are selected on this line such that ‘D’ and ‘E’ are on the opposite banks of the river. The perpendicular ‘DE’ and ‘CG’ are erected on the chain line in such a way that ‘E’, ‘F’ and ‘G’ are on the same straight line. Line ‘FH’ take parallel to “CD’. From triangle ‘DEF’ and ‘HEG’.

(ED / DF) = (FH / HG)

ED  = (FH / HG) x  DF

ED  = (FH / (CG - DF)) x  DF

3. When Chaining and Vision Both are obstructed.

 Such a problem arise when a building comes across the chain line. The problem, in this case, consists both in prolonging the line beyond the obstacle and finding the distance across it.

 Suppose ‘AB is the chain line. Two-point ‘C’ and ‘D’ are selected on it at on the side of the building.

1.      Equal perpendicular ‘CC1’ and ‘DD1’ are erected, the line ‘C1D1’ is extended until the building is crossed. On the extended line, two points ‘E1’ and ‘F1’ are selected. Then perpendiculars ‘EE1’ and ‘FF1’ are so erected that.

                                        EE1 = FF1 = =DD1 = CC1

Thus the points ‘C’, ‘D’, ‘E’ AND ‘F’ will lie on the same straight line ‘AB’

    DE =   D1E1