Anallatic tacheometer distance CD and gradient

Anallatic lens: additive constant $C=0$, $K=100$. With the staff vertical and the line of sight inclined at $\theta$:

$$D=Ks\cos^2\theta,\qquad V=\tfrac12Ks\sin2\theta,\qquad \text{RL}=\text{RL}_{inst}+\text{HI}+V-h$$

Step 1: Distances AC and BD and vertical components

Point C (bearing 330°20', angle of elevation $12^\circ 12'$):

$$\begin{aligned} s&=2.456-1.255=1.201\ \text{m}\\ D&=100\times1.201\times\cos^2 12^\circ 12'=114.737\ \text{m}\\ V&=\tfrac12\times100\times1.201\times\sin 2(12^\circ 12')=24.807\ \text{m} \end{aligned}$$

Point D (bearing 20°36', angle of elevation $10^\circ 36'$):

$$\begin{aligned} s&=2.470-1.300=1.170\ \text{m}\\ D&=100\times1.170\times\cos^2 10^\circ 36'=113.041\ \text{m}\\ V&=\tfrac12\times100\times1.170\times\sin 2(10^\circ 36')=21.155\ \text{m} \end{aligned}$$

Step 2: Coordinates of C and D

Coordinates of the instrument stations: A (218.3 N, 164.7 E) and B (518.2 N, 207.6 E).

$$\begin{aligned} N_C&=218.3+114.737\cos330^\circ20'=317.997,\quad E_C=164.7+114.737\sin330^\circ20'=107.911\\ N_D&=518.2+113.041\cos20^\circ36'=624.013,\quad E_D=207.6+113.041\sin20^\circ36'=247.373 \end{aligned}$$

(a) Distance CD

$$CD=\sqrt{(N_D-N_C)^2+(E_D-E_C)^2}=\sqrt{(+306.016)^2+(+139.462)^2}=336.297\ \text{m}$$

(Bearing of CD = 24°30'.)

(b) Reduced levels of C and D

$$\begin{aligned} \text{RL}_C&=432.550+1.58+24.807-1.86=457.077\ \text{m}\\ \text{RL}_D&=425.500+1.50+21.155-1.885=446.270\ \text{m} \end{aligned}$$

(c) Gradient from C to D

$$\text{Gradient}=\frac{\text{RL}_D-\text{RL}_C}{CD}=\frac{-10.807}{336.297}=-0.03213$$

The ground falls from C to D at 1 in 31.1 (3.21 %).

Answer: CD = 336.297 m; RL of C = 457.077 m, RL of D = 446.270 m; gradient 1 in 31.1 falling from C to D.