Unit-1:- Trigonometrical Levelling (Base is inaccessible)

Case II - Base of the Object Inaccessible.

I) Instrument station in the same vertical plane as the elevated object.
If the horizontal distance between the instrument and the object cannot be measured due to an obstacle, two instrument stations are used so that they are in the same vertical plane as the elevated object.

Fig: Instrument Axis at the Same Level.
Procedures:
i) Set up the theodolite at (\(P\)) and level it accurately with respect to the altitude bubble.
ii) Direct the telescope towards (\(Q\)) and bisect it accurately, clamp both the plates.
iii) Read the vertical angles \(\alpha_1\) and \(\alpha_2\).
iv) Transit the telescope so that the line of sight is reversed. Mark the second instrument station (\(R\)) on the ground.
Measure the distance (\(R\)) accurately. Repeat steps iv and v) for both face observations.
v) The main value should be adopted with the vertical verniers set to 0 reading and the altitude bubble in the center of its run; take the reading on the staff kept at the nearby benchmark.
vi) Shift the instrument to \(R\) and set up the three bolts there. Measure the vertical angle \(\alpha_2\) to \(O\) with both face observations.
vii) With both face observations and with vertical verniers set to 0 reading and the altitude bubble in the center of its run, take the reading on the staff kept at a nearby benchmark.

In order to calculate \(ZL\) of \(O\), we will consider two cases:
a) When instrument axes \(A\) and \(B\) are at the same level.
b) When they are at a different level but the difference is small.
c) When they are at very different levels.

Case-a
When the instrument axes at (\(A\)) and (\(B\)) are at the same level:
Let:
\(h\) = Height difference between the instrument axes at \(A\) and \(B\)
\(\alpha_1\) = Angle of elevation from \(A\) to the target
\(\alpha_2\) = Angle of elevation from \(B\) to the target
\(S\) = Staff reading on the benchmark (BM), taken from both \(A\) and \(B\)
\(b\) = Horizontal distance between the instrument stations
\(D\) = Horizontal distance between point \(P\) and the target

From \(\Delta AQQ'\):
\[h = D \tan \alpha_1\]          (i)

From \(\Delta BQQ'\):
\[h = (b + D) \tan \alpha_2\]     (ii)

Equating equations (i) and (ii):
\[D \tan \alpha_1 = (b + D) \tan \alpha_2\]

Solving for \(D\):
\[D = \frac{b \tan \alpha_2}{\tan \alpha_1 - \tan \alpha_2}\]

\[D = b \tan \alpha_2 - \tan \alpha_1\]

Substituting the value of \(D\) in eqn(i ), we get:
\[h = D \tan \alpha_1\]

\[h = \frac{b \tan \alpha_2 \tan \alpha_1}{\tan \alpha_1 - \tan \alpha_2}\]

\[h = \frac{b \sin \alpha_2 \sin \alpha_1}{\sin(\alpha_1 - \alpha_2)}\]

Then, \(RL\) of \(Q\) = \(RL\) of BM + \(s\)+\(h\)

Case-b:
When they are at a different level but the difference is small.

 When the instrument axes are at different levels as shown in the figure. If \(s_1\) and \(s_2\) are the corresponding staff readings on the staff kept at the benchmark, the difference in levels of the instrument axes will be either (\(s_2 - s_1\)) or (\(s_1 - s_2\)). If the axis of \(A\) is higher, let \(Q'\) be the projection of \(Q\) on the horizontal line through \(A\) and \(Q''\) be the projection of \(Q\) on the horizontal line through \(B\).

 Then deriving the expression for figure when s2 is greater than s1

From \(\Delta QAQ'\):
\[h1 = D \tan \alpha_1\]          (i)

From \(\Delta BQQ''\):
\[h2 = (b+D) \tan \alpha_2\]          (ii)

Subtracting h2 from h1 we get,

\[h1 - h2 =  {D \tan \alpha_1 - (b+D) \tan \alpha_2} \]

h1-h2 = difference in level of instrument axis= s2-s1 = s