Anallatic tacheometer distance XY, gradient

Anallatic lens: $C=0$, $K=100$. The intercept $s$ is given, the axial hair reading $h$ is the staff reading at the central hair, and the height of instrument cancels in the difference of levels.

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 X (bearing 40°35', angle of depression $10^\circ 20'$):

$$\begin{aligned} s&=2.250\ \text{m}\\ D&=100\times2.250\times\cos^2 10^\circ 20'=217.761\ \text{m}\\ V&=\tfrac12\times100\times2.250\times\sin 2(10^\circ 20')=-39.705\ \text{m} \end{aligned}$$

Point Y (bearing 70°10', angle of elevation $7^\circ 30'$):

$$\begin{aligned} s&=2.050\ \text{m}\\ D&=100\times2.050\times\cos^2 7^\circ 30'=201.507\ \text{m}\\ V&=\tfrac12\times100\times2.050\times\sin 2(7^\circ 30')=26.529\ \text{m} \end{aligned}$$

Step 2: Difference in level

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

$$\begin{aligned} \Delta H&=(V_{Y}-h_{Y})-(V_{X}-h_{X})\\ &=(+26.529-1.500)-(-39.705-1.987)=+66.721\ \text{m} \end{aligned}$$

Step 3: Horizontal distance XY

The angle at P between the two lines is the difference of the bearings: $70^\circ10'-40^\circ35'=29^\circ 35'$.

$$\begin{aligned} XY&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{217.761^2+201.507^2-2\times217.761\times201.507\cos 29^\circ 35'}=108.189\ \text{m} \end{aligned}$$

Step 4: Gradient

$$\text{Gradient}=\frac{\Delta H}{XY}=\frac{66.721}{108.189}=0.61671$$

The line from X to Y is rising: 1 in 1.6, i.e. 61.67 %, an angle of $31^\circ 39'45''$ with the horizontal.

Answer: XY = 108.19 m; difference in level = +66.721 m; gradient rising 1 in 1.6 (61.67 %, 31°40').