Inaccessible object, stations at different levels

When the base of the object is inaccessible and the two instrument stations A and B are at very different levels but in the same vertical plane as the object P, the single-plane method is used. The stations are in line with P, so the horizontal distance and the RL of P are found from vertical angles only, with a measured base AB.

Field work

1. Set up the theodolite at A (the farther station), level it and measure the height of instrument $h_A$. Measure the vertical angle $\alpha_A$ to P (face left and face right, mean value).
2. Place a vane (height $s$ above the foot) at B and measure the vertical angle $\alpha_{AB}$ from A to the vane. Measure the horizontal distance $D$ = AB.
3. Shift the instrument to B (measure $h_B$) and measure the vertical angle $\alpha_B$ to P.
4. Sight a staff on a BM from the station to obtain the RL of the axis, or use the known RL of B.

                                  P
                            _.--'|
                      _.--'      |
            B  _.--'  aB         |
   A _.--'  aA  - - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Computation

Let $x$ = horizontal distance from A to P, so the distance from B to P is $x - D$.

• RL of axis of B $= RL_B + h_B$.
• RL of axis of A from the sight A to the vane at B (vertical angle $\alpha_{AB}$, positive if elevation):

$$RL_{\text{axis }A} = RL_B + s - D\tan\alpha_{AB}$$

The height of P is the same from both stations:

$$RL_{\text{axis }A} + x\tan\alpha_A = RL_{\text{axis }B} + (x-D)\tan\alpha_B$$ $$\boxed{x = \frac{(RL_{\text{axis }B} - RL_{\text{axis }A}) - D\tan\alpha_B}{\tan\alpha_A - \tan\alpha_B}}$$

• Horizontal distance of the object from B: $x - D$
• RL of P:

$$\boxed{RL_P = RL_{\text{axis }A} + x\tan\alpha_A = RL_{\text{axis }B} + (x-D)\tan\alpha_B}$$

If P is a point on top of an object whose base RL is wanted, subtract the height of the object. When the line of sight is long, add the correction for curvature and refraction, $0.0673\,d^2$ (d in km), to each height. Angles read as zenith angles are changed with $\alpha = 90^\circ - z$.