Anallatic tacheometer AB gradient from P

Anallatic lens: $C=0$, $K=100$. The intercept $s$ is given directly, and the height of instrument is not needed because only the difference in level between A and B is required.

With the staff held vertical and the line of sight inclined at angle $\theta$ (additive constant zero):

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

where $s$ is the staff intercept and $\theta$ the vertical angle.

Step 1: Horizontal distance and vertical component of each sight

Point A (bearing 345°, angle of elevation $15^\circ 10'10''$):

$$\begin{aligned} s&=1.370\ \text{m}\\ D&=100\times1.370\times\cos^2 15^\circ 10'10''=127.619\ \text{m}\\ V&=\tfrac12\times100\times1.370\times\sin 2(15^\circ 10'10'')=34.600\ \text{m} \end{aligned}$$

Point B (bearing 250°, angle of elevation $9^\circ 50'40''$):

$$\begin{aligned} s&=2.425\ \text{m}\\ D&=100\times2.425\times\cos^2 9^\circ 50'40''=235.411\ \text{m}\\ V&=\tfrac12\times100\times2.425\times\sin 2(9^\circ 50'40'')=40.851\ \text{m} \end{aligned}$$

Step 2: Difference in level

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

$$\begin{aligned} \Delta H&=(V_{B}-h_{B})-(V_{A}-h_{A})\\ &=(+40.851-1.835)-(+34.600-1.435)=+5.850\ \text{m} \end{aligned}$$

Step 3: Horizontal distance AB

The angle at P between the two lines is the difference of the bearings: $345^\circ00'-250^\circ00'=95^\circ 00'$.

$$\begin{aligned} AB&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{127.619^2+235.411^2-2\times127.619\times235.411\cos 95^\circ 00'}=277.384\ \text{m} \end{aligned}$$

Step 4: Gradient

$$\text{Gradient}=\frac{\Delta H}{AB}=\frac{5.850}{277.384}=0.02109$$

The line from A to B is rising: 1 in 47.4, i.e. 2.11 %, an angle of $1^\circ 12'30''$ with the horizontal.

Answer: AB = 277.38 m; difference in level = +5.850 m; gradient rising 1 in 47.4 (2.11 %, 1°12'30").