OBSTACLES IN CHAINING

Obstacles in Chaining.

  Cases often occur in the field where the distance between two points is required, but direct chaining from one point to the other is impossible because of some sort of obstacle.

  There are two main cases to be considered: (1) obstacles which obstruct chaining but not ranging, (2) obstacles which obstruct both chaining and ranging, and, of those which come under (1), we may distinguish between (a) obstacles round which chaining is possible, and (6) obstacles round which it is not possible to chain.

  Case 1 (a). In fig. 25.a the line crosses a lake between A and B and the distance AB is required.

  At A and B range out lines AC and BD perpendicular to the chain line and make AC equal to BD in length. Then the line CD can be chained and will be parallel and equal in length to AB.

  Other alternative methods are shown in figs. 25.b, 25.c and 25.d.

  In fig. 25.b, E is the middle point of a chained line AC chosen so that the lines EA and EB clear the obstacle. Produce BE to D and make ED = BE. Then DC is equal and parallel to AB.

  In fig. 25.c, C is the middle point of AE and D the middle point of BE. Then AB = 2 x distance CD.

  In fig. 25.d, a perpendicular BC is dropped on a line AC and AC and BC are measured. Calculate from tan Ө = BC/AC, and AB from AB = AC sec Ө = BC cosec Ө.

  Case 1 (b). Here a wide river prevents chaining round the obstacle. In fig. 26.a, take a point C on line AB produced and erect perpendiculars to AB at C and B. Take point D on the perpendicular from C and line in E so as to be on the perpendicular from B and on the line DA. Measure BC, BE and CD. Then

AB/BE = BC/(CD – BE)    or AB = (BE X BC)/(CD – BE)

  In fig. 26.b, lay out and measure BC perpendicular to BA and mark the middle point E. At C lay out line CD perpendicular to BC and find point D on this perpendicular such that D, E and A are all in a straight line. Measure CD, which is equal to BA.

  In fig. 26.c, set out BC perpendicular to BA, and at C set out line CD at right angles to AC. Choose point D on this line so that D is in line with B and A. Measure BD and BC. Then

AB/BC = BC/BD and so AB = BC²/BD

  In fig. 26.d, choose a line AC which makes an angle of about 30⁰ with AB, and from B drop line BC perpendicular to AC. Measure CB and prolong CB to D so that BD = CB. At D lay out line DE at right angles to CBD and find the point E on this line which is in line with B and A. Measure BE, which is equal to BA.

  Case 2. Obstacles which prevent both chaining and ranging. In fig. 27.a, 27.b, 27.c, a building interferes with the direct ranging as well as with the chaining of the line AB.

  (1). In fig. 27.a, set out lines AC and BD perpendicular at A and B respectively to AB, and make AC = BD. From C set out points E and F on line CD, ahead of D and clear of the obstacle. At E and F set out lines EG and FH, equal in length to CA or DB and at right angles to CDEF. Measure DE. Then GH is a continuation of AB and BG = DE.

  (2). In fig. 27.b, choose a suitable point C, measure AC and BC and lay out points D and E on lines BC and AC, so that

CE/CA = CD/CB = k, say,

Set out points F and G in line with D and E and measure OF and CG. Produce CF and CG to I and H, making

CI/CF = CH/CG = CA/CE = 1/k

Measure EF. Then H and I are on BA produced and

AI = EF X CA/CE = 1/k X EF

  (3). In fig. 27.c, lay out line BE and measure BD and BE. Let BD = k X BE. Lay out line BF and line in point C at the intersection of BF and DA produced. Measure BC and make BF = 1/k x BC.

Measure DA and on line EF put in point G so that EG = 1/k X DA.

G is then a point on BA produced. Similarly, find another point L on BA produced. Measure BA.

Then       AG= BA{(1/k) – 1} and GL is a continuation of BA.