3 Main Types of Obstacles in Chaining of a Line | Land Survey | Surveying

The three main obstacles in chaining of a line are of the following types: 1. Chaining Free, Vision Obstructed 2. Chaining Obstructed, Vision Free 3. Chaining and Vision Both Obstructed.

It sometimes happens that a survey line passes through some object such as a pond, a 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 obstacle.

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.

Type # 1. Chaining Free, Vision Obstructed:

In this type of obstacles, the ends of the lines are not intervisible e.g. rising ground, hill or jungle intervening.

Here two cases may arise:

(i) Both ends may be visible from any intermediate point lying on the line such as in the case of a hill. The obstacle of this kind may easily be crossed over by reciprocal ranging and length measured by stepping method of chaining.

(ii) Both ends may not be visible from any intermediate point such as in the case of a jungle. The obstacle of this kind may be crossed over by “Random line method”. In fig. 3.20, let AB be the line whose length is required. From A, run a line AB’ called a random line, in the approximate convenient direction of AB and continue it until point B is visible from B’ Chain the line to B’ where BB’ is perpendicular to AB’ and measure BB’.

If any other length AC’ is measured along AB’, a point C can be located on the line AB by measuring the perpendicular distance = 

 

Similarly a number of points can be located on the true line. The line is then cleared and chained.

Type # 2. Chaining Obstructed, Vision Free:

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 in finding the distance between convenient points on the chain line on either side of obstacle.

Two cases may arise:

(a) When the obstacle can be chained around, e.g. a pond, a thorny hedge etc.

(b) When the obstacle cannot be chained around e.g. a river.

Case: (i) The distance between two points A and B on either side of the pond may be determined by any of the following methods convenient at site:

(a) Set out equal perpendiculars AC and BD [Fig. 3.21 (a)]. Measure CD which is equal to AB.

(b) Erect perpendicular AC [Fig .3.21(b)] of such a length that CB clears the obstacle and measure AC and CB.

 

(c) Find by optical square or a cross-staff a point C such that ∠ACB is right angle [Fig. 3.21(c)] Measure AC and BC.

(d) Mark a point C so that CA and CB clear the obstacle [Fig. 3.21. (d)]. Range E in line with AC so that CE = AC. Then range D in the line with BC so that CD = BC. The triangles CAB and CED are congruent. Therefore DE = AB.

Case (ii): Any one of the following methods may be employed to find the width of the river along the direction of the chain line:

(a) Select two points A and B on the chain line on opposite banks of the river. [Fig. 3.22 (a)]. From A and C, erect perpendicular or parallel lines AD and CE, such that E, D and B are in line. Measure AC, AD and CE. If a line DF is drawn parallel to AC, meeting CE in F, the triangles ABD and FDE are similar.

(b) Select two points A and B on the chain line on either side of the river [Fig. 3.22. (b)]. Set a perpendicular AC and mark its midpoint D. From C, erect CE perpendicular to AC such that E, D and B are in the same range and measure CE. Then triangles ABD and CED are congruent. Therefore AB = CE.

(c) Select two points A and B as before [Fig. 3.22 (c)]. Erect a perpendicular AC and using an optical square at C, find D on the chain line so that ∠BCD is a right angle. Measure AC and AD. Triangles ABC and ACD are similar.

(d) Fix two points A and B as before [Fig. 3.22 (d)].

Erect a perpendicular AC of such a length that triangle ABC is well conditioned. Measure AC and the angle ACB with prismatic compass or box-sextant or with any other angle measuring instrument.

(e) If a survey line crosses the river obliquely, then the following method is used to find the width of the river:

Select two points A and B as before [Fig. 3.22. (e)].

At A, set out a line AC in a convenient direction so that C is the foot of the perpendicular from B on AC. Produce CA to D and measure AD = AC. At D, erect a perpendicular DE, E being a point on the chain line. Then triangles ABC and AED are congruent. Therefore AB = AE (the oblique width of the river.

Type # 3. Chaining and Vision Both Obstructed:

A building is a typical example of this class of obstacles. The problem in this case consists both in prolonging the line beyond the obstacle and finding the distance across it.

Any one of the following methods may be employed:

(a) Select two points A and B on the chain line [Fig. 3.23 (a)]. At A and B, erect equal perpendiculars AC and BD. Join CD and produce it past the obstacle. Select two points E and F on it. At E and F, set out perpendiculars EG and FH, each equal in length to AC. The points G and H then lie on the chain line and BG = DE.

The direction and length of perpendiculars must be set out with great accuracy. The check can be made by measuring diagonals of the rectangles. For the same rectangle, diagonals should be equal. Here AD should be equal to BC, and EH equal to FG.

(b) Choose two points A and B on the chain line [Fig. 3.23. (b)]. With AB as base, construct an equilateral triangle ABC by swinging equal arcs with a tape. Produce AC to D and take a point E of DA. Again contract an equilateral triangle DEF with DE as the base.

Produce the line DF to G such that DG = DA. ADG then forms an equilateral triangle and G is a point on the chain line. Determine the second point K on the chain line by forming an equilateral triangle GHK on GH as the base. The line joining KG determines the direction of the chain line past the obstacle, and the obstructed length BK = AG-AB-GK = DA – AB – GK.

Example:

There is an obstacle in the form of a pond on the main chain line AB. Two points C and D were taken on the opposite sides of the pond. On left of CD, a line CE was laid out 100 m in length and a second line CF, 80 m long was laid out on the right of CD such that E, D and F are in the same straight line. ED and DF were measured and found to be 60 m and 56 m respectively. Find out the obstructed length CD.

Solution:

In Fig. 3.24, CD is the obstructed length of the pond on the chain line AB. CE and CF are known to be 100 and 80 m respectively and EF = 60 + 56= 116 m.