This article throws light upon the six major problems in ranging and chaining of a line.

#### Problem # 1. To erect, with a tape, a perpendicular to a chain line from a point on it:

It can be done by forming a right angled triangle by taking three sides of a triangle in the proportion of 3, 4 and 5. Let AB be the chain line and C a point on it at which it is desired to erect a perpendicular (Fig. 3.29). Measure CE = 40 units (chain links or decimetres).

Pin zero end of the tape at C and 80th unit at E, the remaining portion of the tape hanging free. Hold 30th unit and pull the tape until both segments DC and DE are taut. Fix an arrow at D. CD will then be the required perpendicular.

#### Problem # 2. To drop a perpendicular to a given chain line from a point outside it:

There may be two cases:

(i) When the point is accessible, and

(ii) When the point is inaccessible.

(i) When the point is accessible:

Let AB be the chain line and C a given point out side it (Fig. 3.30). With C as centre and any convenient length of the tape as radius, describe an arc EF cutting the chain line at E and F. Fix arrows at E and F Measure EF and bisect it at D. CD will then be the required perpendicular.

(ii) When the point is inaccessible:

In fig. 3.31, let C be the inaccessible point. Choose suitable points D and E on the chain line and set out EG and DF perpendiculars to CD and CE respectively and mark H, the intersection point. Locate the point K in line with H and C or alternatively, from H, drop perpendicular UK on AB. CK is then the required perpendicular.

#### Problem # 3. To run a parallel to a given line through a given point:

There may be two cases:

(i) When the point is accessible, and

(ii) When the point is inaccessible.

(i) When the point is accessible:

Let AB be the given line and C the point through which a parallel is to be run (Fig. 3.32), From C. drop a perpendicular CD on AB and measure it. Select another point E on AB and erect a perpendicular EF equal in length to CD. CF is then the required parallel.

(ii) When the point is inaccessible:

In fig. 3.32, let C be the inaccessible point. Locate the foot D of the perpendicular CD on the line AB by the method explained above in 2 (ii) and “find the obstructed perpendicular distance CD as described in the obstacles in chaining” [3. 10(2)].

Choose another point E on AB and erect a perpendicular EF equal in length to CD.CF is then the required parallel.

#### Problem # 4. To run a parallel to a given inaccessible line through a given point:

In fig. 3.33, let AB be the given inaccessible line and C the given point. Fix a point D in line with AC. Fix another point E. Through C, run a parallel CF to AE. cutting DE at F. Through F, run a parallel FG to EB, intersecting DB at G. Then CG is the required parallel to AB.

#### Problem # 5. To find the Intersection of any two Lines meeting in a lake Refer to fig. 3.34:

Let any two lines AO and BO intersect in the lake at O. On the line AO, take any two points X and Y and on the line BO, take a point P. Join YP and produce it to Y1 making PY1 = PY. Similarly, join XP and produce it to X1 making PX1 = PX. Join X1Y1 and range a rod O1 in the direction of X1Y1 meeting BO at O1. Then Δs XOP and X1O1P are congruent.

... PO = PO1 and XO = X1O1.

#### Problem # 6. To find the height of a tower:

There may be two cases:

(i) When the base is accessible, and

(ii) When the base is inaccessible.

(i) When the base is accessible (Refer to fig, 3.35).

Take two rods AB and CD of unequal lengths and range their tops A and C in line with the top T of the tower. Comparing Δs ACD and ATQ, we find that these are similar.

Take two rods AB and CD of unequal length and their tops A and C in line with the top T of the tower. Comparing ∆s ACD and ATQ, we find that these are similar.

(ii) When the base is inaccessible (Refer to fig. 3.36.):

Take four rods AB, CD, EF and GH; each pair having equal lengths i.e. AB = EF and CD = GH. Fix two rods AB and CD in such a way that their tops A and C are in line with the top T of the tower. Fix the other two rods EF and GH ranging their tops E and G again in line with T. Cut off FD’ = BD and erect a perpendicular D’C’. Join EC. ∆s GTC and GEC’ are evidently similar.