Obstacles in chain survey and remedies

The obstacles in chain surveying are of three types:

1. Chaining free but vision obstructed (e.g. a hill or rising ground).
2. Vision free but chaining obstructed (e.g. a pond or river).
3. Both chaining and vision obstructed (e.g. a building).

1. Chaining free, vision obstructed

Rising ground between two stations. The end stations A and B are not intervisible, but the line can be chained over the hill. The line is ranged by reciprocal ranging:

• Two persons stand at M1 and N1, points near the line such that M1 sees N1 and B, and N1 sees M1 and A.
• The person at M1 directs N1 to a point N2 on the line M1B. Then N2 directs M1 to M2 on the line N2A.
• The process is repeated until M, N, A and B are in one straight line.

       M1   _______ N1
  A ___/    hill     \___ B

2. Vision free, chaining obstructed

(a) Pond or lake on the line. Erect two perpendiculars AC and BD of equal length at points A and B on the line, both on the same side of the line. Then CD is parallel and equal to AB, so measure CD. Alternatively, take a point O off the line and range A-O-C and B-O-D with OC = OA and OD = OB; the triangles OAB and OCD are congruent, so CD = AB.

(b) River. To find the width AB: at A set out a perpendicular AC and mark D, its mid-point. At C set out CE perpendicular to AC (away from the river) and move along it until E, D and B are in a line. Then, by congruent triangles ABD and CED, $AB = CE$.

        B   (far bank)
        :
  ~~~ river ~~~
        :
        A ---------- D ---------- C
                                  |
                                  |
                                  E
   (AD = DC; E, D, B in line; AB = CE)

3. Both chaining and vision obstructed

Building on the line. Take two points A and B on the line before the building.

1. At A and B erect perpendiculars AC and BD of equal length, long enough to clear the building.
2. Join C and D and extend CD past the building. Mark E and F on line CD beyond the building.
3. At E and F erect perpendiculars EG and FH equal to AC. G and H lie on the continuation of AB.
4. The hidden distance $BG = DE$. Measure DE and add it to the chainage of B.

    C ------------ D ------------------ E ------------ F
    |              |                    |              |
 ---A--------------B----[ building ]----G--------------H--->

Alternatively, the line can be prolonged by trigonometry, using lines set out at known angles (for example $30^\circ$ and $40^\circ$) from the end of a perpendicular.