Anallatic tacheometer A to B distance, gradient

Anallatic lens: $C=0$, $K=100$, staff vertical.

Reading of the data: no bearings or height of instrument are given, so the horizontal distance from A to B is the distance from the instrument at A to the staff at B. The middle hair readings are used as printed (1.153 m for BM and 1.105 m for B). Since the height of instrument is not given, the gradient is taken from the instrument axis at A (RL of axis) to the staff point B.

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

Step 1: Sight on the benchmark (angle of depression $6^\circ30'$)

Point BM (angle of depression $6^\circ 30'$):

$$\begin{aligned} s&=2.060-1.100=0.960\ \text{m}\\ D&=100\times0.960\times\cos^2 6^\circ 30'=94.770\ \text{m}\\ V&=\tfrac12\times100\times0.960\times\sin 2(6^\circ 30')=-10.798\ \text{m} \end{aligned}$$

Step 2: RL of the instrument axis

$$\text{RL}_{BM}=\text{RL}_{axis}+V-h\ \Rightarrow\ \text{RL}_{axis}=970.000-(-10.798)+1.153=981.951\ \text{m}$$

Step 3: Sight on B (angle of elevation $10^\circ$)

Point B (angle of elevation $10^\circ 00'$):

$$\begin{aligned} s&=1.188-0.982=0.206\ \text{m}\\ D&=100\times0.206\times\cos^2 10^\circ 00'=19.979\ \text{m}\\ V&=\tfrac12\times100\times0.206\times\sin 2(10^\circ 00')=3.523\ \text{m} \end{aligned}$$ $$\text{RL}_B=981.951+3.523-1.105=984.369\ \text{m}$$

Step 4: Distance and gradient

Horizontal distance from A to B:

$$D_{AB}=19.979\ \text{m}$$ $$\text{Gradient}=\frac{\text{RL}_B-\text{RL}_{axis}}{D_{AB}}=\frac{+2.418}{19.979}=0.12102$$

Answer: horizontal distance AB = 19.979 m; RL of B = 984.369 m; gradient rising 1 in 8.3 (12.10 %) from the instrument level at A to B.

Note: the middle hair reading for BM (1.153) is not the mean of the outer hairs (1.580). If 1.580 m were the correct reading, the RL of the axis would be 982.378 m and the RL of B 984.796 m, but the distance and the gradient above would be unchanged.