Stadia principle, tangential method depression

Principle of stadia tacheometry: the telescope has two extra horizontal hairs (stadia hairs), one above and one below the central cross-hair. The rays through these hairs make a fixed small angle at the focus, so the staff intercept $s$ between them is proportional to the distance of the staff from the instrument.

 Diaphragm     Objective              Staff
   a |           |   \                  A  top
     | i         |     \  ...           |
  ---+-----------+-------F--------------+---- axis
     |           |      /               |
   b |           |   /                  B  bottom
     |<-- f --->|<--- D1 ---->|

Horizontal sight, staff vertical

From the similar triangles formed at the external focus $F$ (stadia hair interval $i$, focal length $f$, intercept $s$):

$$\frac{D_1}{f}=\frac{s}{i}\ \Rightarrow\ D_1=\frac{f}{i}\,s$$

Adding the distance of the focus from the instrument axis, $(f+d)$, where $d$ is the distance from the objective to the vertical axis:

$$D=\frac{f}{i}\,s+(f+d)=Ks+C$$

$K=f/i$ is the multiplying constant (usually 100) and $C=f+d$ is the additive constant (about 0.3 m for an external focusing telescope, and zero for an internal focusing or anallatic telescope).

Tangential method, both angles of depression

In the tangential system, the vertical angles to two targets (vanes) on a vertical staff, a known distance $s$ apart, are measured. The staff is below the line of collimation, so both angles are angles of depression.

 O ------------------------------ horizontal
  \ \        alpha2 < alpha1
   \  \_______   upper target B   |
    \  alpha1\__ lower target A   | s
     \                           _|
      <--------- H ------------>  staff station C

Let $\alpha_1$ be the angle of depression to the lower target A and $\alpha_2$ that to the upper target B ($\alpha_1>\alpha_2$). Let $H$ be the horizontal distance, $V$ the vertical depth of the upper target B below the horizontal line of sight through the instrument axis, and $h$ the height of the lower target above the staff station C.

Horizontal distance H

$$\begin{aligned} \text{Depth of A}&=V+s=H\tan\alpha_1\\ \text{Depth of B}&=V=H\tan\alpha_2\\ s&=H(\tan\alpha_1-\tan\alpha_2)\\ H&=\frac{s}{\tan\alpha_1-\tan\alpha_2} \end{aligned}$$

Vertical distance V

$$V=H\tan\alpha_2=\frac{s\tan\alpha_2}{\tan\alpha_1-\tan\alpha_2}$$

Reduced level of the staff station

The lower target A is at depth $V+s=H\tan\alpha_1$ below the axis, and the staff station C is $h$ below A:

$$\text{R.L. of C}=\text{R.L. of axis}-H\tan\alpha_1-h$$

or, using the upper target, $\text{R.L. of C}=\text{R.L. of axis}-H\tan\alpha_2-(s+h)$, where R.L. of axis $=$ R.L. of the instrument station $+$ height of instrument.