Tacheometry booking, plotting, gradient A-B

Tacheometry

Tacheometry (tachymetry) is a method of surveying in which horizontal distances and differences in elevation are determined from optical measurements with a tacheometer (a theodolite with stadia hairs) and a graduated staff, without chaining or levelling. It is quick and is used for contouring, topographic surveys and traversing over rough, steep or obstructed ground.

Booking

Observations are booked in a standard tabular field book, one line for every staff point:

The columns for $s$, $D$, $V$ and RL are filled in during computation. A sketch of the area, with the station names and the position of details, is drawn on the facing page.

Plotting

1. Choose a scale and plot the instrument stations (from the control traverse or by coordinates).
2. From each station, mark each staff point by its bearing (with a protractor) and its horizontal distance $D$ at the scale (polar method), or compute its coordinates and plot on the grid.
3. Write the RL beside each point, join details (roads, buildings, streams) using the sketch, interpolate between spot levels and draw contours.
4. Check with the points common to two stations, and finish with the title, scale and north arrow.

Numerical: gradient between A and B

Anallatic lens, so $C=0$ and $K=100$ (taken as 100, not stated).

Staff held vertical, line of sight inclined at angle $\theta$ (additive constant zero, which is the case for an anallatic lens or $C=0$):

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

where $s$ is the staff intercept, $\theta$ the vertical angle and $K=100$, $C=0$.

Step 1: Horizontal distance and vertical component of each sight

Point A (bearing 30°30', angle of elevation $6^\circ 30'$):

$$\begin{aligned} s&=2.355-1.115=1.240\ \text{m}\\ D&=100\times1.240\times\cos^2 6^\circ 30'=122.411\ \text{m}\\ V&=\tfrac12\times100\times1.240\times\sin 2(6^\circ 30')=13.947\ \text{m} \end{aligned}$$

Point B (bearing 75°30', angle of elevation $9^\circ 15'$):

$$\begin{aligned} s&=2.750-1.250=1.500\ \text{m}\\ D&=100\times1.500\times\cos^2 9^\circ 15'=146.124\ \text{m}\\ V&=\tfrac12\times100\times1.500\times\sin 2(9^\circ 15')=23.798\ \text{m} \end{aligned}$$

Step 2: Reduced levels

RL of instrument axis = RL of station + HI = 1275.000 + 1.550 = 1276.550 m.

$$\text{RL of staff point}=\text{RL of axis}+V-h$$ $$\begin{aligned} \text{RL}_{A}&=1276.550+13.947-1.735=1288.762\ \text{m}\\ \text{RL}_{B}&=1276.550+23.798-2.000=1298.348\ \text{m} \end{aligned}$$

Step 3: Horizontal distance AB

The angle at P between the two lines is the difference of the bearings, $45^\circ 00'$ (30°30' and 75°30').

$$\begin{aligned} AB&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{122.411^2+146.124^2-2\times122.411\times146.124\cos 45^\circ 00'}=105.073\ \text{m} \end{aligned}$$

Step 4: Gradient

$$\text{Gradient}=\frac{\Delta H}{AB}=\frac{9.586}{105.073}=0.09123$$

The line from A to B is rising: 1 in 11.0, i.e. 9.12 %, an angle of $5^\circ 12'46''$ with the horizontal.

Answer: AB = 105.07 m; RL of A = 1288.762 m, RL of B = 1298.348 m; difference in level = +9.586 m; gradient rising 1 in 11.0 (9.12 %, 5°13').