Tacheometric RL of B, C, gradient BC

Anallatic lens: $C=0$, $K=100$. The zenith angle $z$ is converted to the vertical angle $\theta=90^\circ-z$: for B, $90^\circ-71^\circ30'=+18^\circ30'$ (elevation); for C, $90^\circ-96^\circ00'=-6^\circ00'$ (depression). The axial hair reading is the middle reading.

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 B (bearing 18°00', angle of elevation $18^\circ 30'$):

$$\begin{aligned} s&=2.345-1.295=1.050\ \text{m}\\ D&=100\times1.050\times\cos^2 18^\circ 30'=94.428\ \text{m}\\ V&=\tfrac12\times100\times1.050\times\sin 2(18^\circ 30')=31.595\ \text{m} \end{aligned}$$

Point C (bearing 127°00', angle of depression $6^\circ 00'$):

$$\begin{aligned} s&=2.570-1.010=1.560\ \text{m}\\ D&=100\times1.560\times\cos^2 6^\circ 00'=154.296\ \text{m}\\ V&=\tfrac12\times100\times1.560\times\sin 2(6^\circ 00')=-16.217\ \text{m} \end{aligned}$$

Step 2: Reduced levels

RL of instrument axis = RL of station + HI = 1086.550 + 1.385 = 1087.935 m.

$$\text{RL of staff point}=\text{RL of axis}+V-h$$ $$\begin{aligned} \text{RL}_{B}&=1087.935+31.595-1.820=1117.710\ \text{m}\\ \text{RL}_{C}&=1087.935-16.217-1.790=1069.928\ \text{m} \end{aligned}$$

Step 3: Horizontal distance BC

The angle at A between the two lines is the difference of the bearings: $127^\circ00'-18^\circ00'=109^\circ 00'$.

$$\begin{aligned} BC&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{94.428^2+154.296^2-2\times94.428\times154.296\cos 109^\circ 00'}=205.453\ \text{m} \end{aligned}$$

Step 4: Gradient

Difference in level: $\text{RL}_C-\text{RL}_B=1069.928-1117.710=-47.782$ m.

$$\text{Gradient}=\frac{\Delta H}{BC}=\frac{47.782}{205.453}=0.23257$$

The line from B to C is falling: 1 in 4.3, i.e. 23.26 %, an angle of $13^\circ 05'33''$ with the horizontal.

Step 5: Bearing of BC

Taking the instrument station as the origin:

$$\begin{aligned} B&=(+89.807,\ +29.180),\qquad C=(-92.857,\ +123.226)\\ \tan\theta&=\frac{\Delta E}{\Delta N}=\frac{+94.046}{-182.664}\ \Rightarrow\ \theta_{BC}=152^\circ 45'29'' \end{aligned}$$

Bearing of BC = 152°45' (S 27°15' E).

Answer: BC = 205.45 m; RL of B = 1117.710 m, RL of C = 1069.928 m; difference in level = -47.782 m; gradient falling 1 in 4.3 (23.26 %, 13°06'); bearing of BC = 152°45'.