Anallatic tacheometer distance XY, gradient

Anallatic lens: $C=0$, $K=100$. The intercepts $s$ are given and the central hair readings are used as $h$. The height of instrument cancels in the difference of levels.

Staff held vertical, line of sight inclined at angle $\theta$ (additive constant zero, which is the case for an anallatic lens or $C=0$):

$$D=Ks\cos^2\theta,\qquad V=\tfrac12Ks\sin2\theta$$

where $s$ is the staff intercept, $\theta$ the vertical angle and $K=100$, $C=0$.

Step 1: Horizontal distance and vertical component of each sight

Point X (bearing 38°15', angle of depression $11^\circ 30'$):

$$\begin{aligned} s&=2.230\ \text{m}\\ D&=100\times2.230\times\cos^2 11^\circ 30'=214.136\ \text{m}\\ V&=\tfrac12\times100\times2.230\times\sin 2(11^\circ 30')=-43.567\ \text{m} \end{aligned}$$

Point Y (bearing 68°25', angle of elevation $8^\circ 15'$):

$$\begin{aligned} s&=2.080\ \text{m}\\ D&=100\times2.080\times\cos^2 8^\circ 15'=203.717\ \text{m}\\ V&=\tfrac12\times100\times2.080\times\sin 2(8^\circ 15')=29.538\ \text{m} \end{aligned}$$

Step 2: Difference in level

The height of instrument is the same for both sights, so it cancels (RL of point $=$ RL of axis $+V-h$):

$$\begin{aligned} \Delta H&=(V_{Y}-h_{Y})-(V_{X}-h_{X})\\ &=(+29.538-1.503)-(-43.567-1.980)=+73.581\ \text{m} \end{aligned}$$

Step 3: Horizontal distance XY

The angle at P between the two lines is the difference of the bearings, $30^\circ 10'$ (38°15' and 68°25').

$$\begin{aligned} XY&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{214.136^2+203.717^2-2\times214.136\times203.717\cos 30^\circ 10'}=109.200\ \text{m} \end{aligned}$$

Step 4: Gradient

$$\text{Gradient}=\frac{\Delta H}{XY}=\frac{73.581}{109.200}=0.67382$$

The line from X to Y is rising: 1 in 1.5, i.e. 67.38 %, an angle of $33^\circ 58'23''$ with the horizontal.

Answer: XY = 109.20 m; difference in level = +73.581 m; gradient rising 1 in 1.5 (67.38 %, 33°58').