Unit 4 :- Chain Surveying (Obstacle in Chaining)

 Obstacle in Chaining

Sometimes obstacle in chaining occurs while doing chain surveying. Obstacle in chaining are of three kinds:

(a) When chaining is free but vision is obstructed.

(b) When chaining is obstructed but vision is free.

(c) When chaining an vision are both obstructed.

 

(a) When chaining is free but vision is obstructed :-

Such a program arises when a rising ground or a jungle are present. Here the end station are not intervisible. There may be of two cases:

i) 1st Case

The end station may be visible from some intermediate point on the rising ground. In this case, reciprocal ranging is done and the chaining is done be the stepping method.

 

ii) 2nd Case (Random Ranging)

The end station are not visible from intermediate point when a jungle are comes across chainline.

 

Let AB be the actual chain line which can’t be ranged and extended because of interruption by a jungle. Le t the chain line be extended upto R. A point ‘P’ is selected on the chain line and random line and perpendicular are projected from them. The perpendicular at ‘C’ meets the chainline at ‘ C’ ’.

Theoritically the perpendicular at ‘D’ and ‘E’ will meet the chainline at ‘D’’ and ‘E’’. Now the distance PC, PD, PE and CC’ are measured.

From Triangle PDD' and PCC'
\[
\frac{DD'}{PD} = \frac{CC'}{PC}
\]

\[
DD'^2 = \frac{CC'^2}{PC \cdot PD}
\]
 Similarly, from Triangle PEE' and PCC'
\[
\frac{EE'}{PE} = \frac{CC'}{PC}
\]

\[
EE'^2 = \frac{CC'^2}{PC \cdot PE}
\]
 

These calculated distances are measured along the perpendicular at 'D' and 'E'.

'D'' and 'E'' should lie on the chainline AB.

The distance of PD is calculated by the formula,
\[
PD'^2 = \sqrt{PD^2 + DD'^2}
\]

 (b) When chaining is obstructed but vision is free.

Such Problem arises when a pond or a river comes accross the chainline. The situation may be tackled in a following way.

(i) Case I (When a pond interrupted a chainline and it is possible to go around the obstruction)

 

when a pond interrupted a chainine it is possible to go around the obstruction. From figure 'a', suppose 'AB' is a chainline. Two points 'C' and 'D' are selected on the opposite banks of the pond. Equal perpendicular 'CE' and 'DF' are errected at 'C' and 'D'. The distance 'EF' is measured then, 

    CD = EF

From figure 'b' , the pond may also be crossed by forming a traingle. A point 'C' is selected on the chain line and the perpendicular 'CE' set out as 'C'. and a line 'ED' is suitably taken. The distance of 'CE' and 'ED' are measured. then,

\[CD = \sqrt{ED^2 - CE^2}
\]

 (ii) When it is not possible to go around (Like River)

(1) When a small river comes accross the chainline.

 

Suppose AB is a chainline. Two points 'C' and 'D' are selected on this line on opposite banks of river. At 'C' a perpendicular CE is errected and bisected at 'F'. A perpendicular 'C' setout at 'E' and a point G is so selected on it that D,F,G are in same line.

Now,

From Triangle, DCF and FEG 

From axiom AAA these two triangles are similar.

\[
\frac{CD}{EG} = \frac{CF}{FE}
\]

\[
CD = \frac{EG \cdot CF}{FE}
\]

(2) When a large river comes accross the chainline.

Suppose 'AB' is a chainline. 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 'DF' and 'CG' are errected on the chainline in such a way that 'EF' and G are on the same line. The line 'FH' is taken parallel to 'DC'.

From Traingle DEF and HFG

 \[\frac{ED}{FH} = \frac{DF}{HG}\]

\[ED = \frac{DF \cdot FH}{HG}\]

 \[ED = \frac{DF \cdot FH}{CG-CH}\]

 (C) CASE III :- When chaining and vision obstructed.

Suppose AB is a chainline. Two points C and D are selected on it at one side of the building. Equal perpendicular CC' and DD' are errected. The line C'D' is extended until the building is crossed. In the extended line, two points E'F' are selected then perpendicular E'E and F'F are so errected that 

EE' = FF' = CC' = DD'

thus the points C, D, E and F will lie on the same straight line AB. Here Distance D'E' is measured which is equal to DE.

i.e. DE = D'E'