Tacheometry

Anallatic lens: $C=0$, $K=100$. The intercept $s$ is given directly, and the height of instrument is not needed because only the difference in level between A and B is required.

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 345°, angle of elevation $15^\circ 10'10''$):

$$\begin{aligned} s&=1.370\ \text{m}\\ D&=100\times1.370\times\cos^2 15^\circ 10'10''=127.619\ \text{m}\\ V&=\tfrac12\times100\times1.370\times\sin 2(15^\circ 10'10'')=34.600\ \text{m} \end{aligned}$$

Point B (bearing 250°, angle of elevation $9^\circ 50'40''$):

$$\begin{aligned} s&=2.425\ \text{m}\\ D&=100\times2.425\times\cos^2 9^\circ 50'40''=235.411\ \text{m}\\ V&=\tfrac12\times100\times2.425\times\sin 2(9^\circ 50'40'')=40.851\ \text{m} \end{aligned}$$

Step 2: Difference in level

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

$$\begin{aligned} \Delta H&=(V_{B}-h_{B})-(V_{A}-h_{A})\\ &=(+40.851-1.835)-(+34.600-1.435)=+5.850\ \text{m} \end{aligned}$$

Step 3: Horizontal distance AB

The angle at P between the two lines is the difference of the bearings, $95^\circ 00'$ (345°00' and 250°00').

$$\begin{aligned} AB&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{127.619^2+235.411^2-2\times127.619\times235.411\cos 95^\circ 00'}=277.384\ \text{m} \end{aligned}$$

Step 4: Gradient

$$\text{Gradient}=\frac{\Delta H}{AB}=\frac{5.850}{277.384}=0.02109$$

The line from A to B is rising: 1 in 47.4, i.e. 2.11 %, an angle of $1^\circ 12'30''$ with the horizontal.

Answer: AB = 277.38 m; difference in level = +5.850 m; gradient rising 1 in 47.4 (2.11 %, 1°12').

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

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 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, $29^\circ 35'$ (40°35' and 70°10').

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

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').

Anallatic lens: $C=0$, $K=100$. The intercepts $s$ are given and the central hair readings are used as $h$. The height of instrument cancels in the difference of levels.

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 X (bearing 38°15', angle of depression $11^\circ 30'$):

$$\begin{aligned} s&=2.230\ \text{m}\\ D&=100\times2.230\times\cos^2 11^\circ 30'=214.136\ \text{m}\\ V&=\tfrac12\times100\times2.230\times\sin 2(11^\circ 30')=-43.567\ \text{m} \end{aligned}$$

Point Y (bearing 68°25', angle of elevation $8^\circ 15'$):

$$\begin{aligned} s&=2.080\ \text{m}\\ D&=100\times2.080\times\cos^2 8^\circ 15'=203.717\ \text{m}\\ V&=\tfrac12\times100\times2.080\times\sin 2(8^\circ 15')=29.538\ \text{m} \end{aligned}$$

Step 2: Difference in level

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

$$\begin{aligned} \Delta H&=(V_{Y}-h_{Y})-(V_{X}-h_{X})\\ &=(+29.538-1.503)-(-43.567-1.980)=+73.581\ \text{m} \end{aligned}$$

Step 3: Horizontal distance XY

The angle at P between the two lines is the difference of the bearings, $30^\circ 10'$ (38°15' and 68°25').

$$\begin{aligned} XY&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{214.136^2+203.717^2-2\times214.136\times203.717\cos 30^\circ 10'}=109.200\ \text{m} \end{aligned}$$

Step 4: Gradient

$$\text{Gradient}=\frac{\Delta H}{XY}=\frac{73.581}{109.200}=0.67382$$

The line from X to Y is rising: 1 in 1.5, i.e. 67.38 %, an angle of $33^\circ 58'23''$ with the horizontal.

Answer: XY = 109.20 m; difference in level = +73.581 m; gradient rising 1 in 1.5 (67.38 %, 33°58').

A subtense bar is a horizontal bar of fixed length $l$ (usually 2 m) with targets at its ends, mounted on a tripod and set perpendicular to the line of sight by sighting a small telescope on the instrument. The small horizontal angle $\beta$ subtended by the targets at the theodolite is measured (to seconds, by repetition), and the distance is computed. It is used for measuring distances in difficult ground (hilly, water, thick vegetation) without taping.

 Theodolite                       Subtense bar
     O  <------------ D ---------->  A |
       \  beta                       | l (2 m)
         \_____________________________ B

Horizontal line of sight

From the isosceles triangle $OAB$:

$$\frac{l}{2}=D\tan\frac{\beta}{2}\ \Rightarrow\ D=\frac{l}{2}\cot\frac{\beta}{2}\ \approx\ \frac{l}{\beta}\ (\beta\ \text{in radians})$$

Line of sight inclined upward by angle $\alpha$ (derivation)

The bar is normal to the inclined line of sight, so the inclined distance is

$$S=\frac{l}{2}\cot\frac{\beta}{2}$$

Horizontal and vertical components:

$$\begin{aligned} D&=S\cos\alpha=\frac{l}{2}\cot\frac{\beta}{2}\cos\alpha\\ V&=S\sin\alpha=\frac{l}{2}\cot\frac{\beta}{2}\sin\alpha \end{aligned}$$

If the bar is not exactly normal to the line of sight but is out by angle $\delta$, the effective length is $l\cos\delta$ and the distance becomes $D=\frac{l\cos\delta}{2}\cot\frac{\beta}{2}$.

The RL of the bar station is RL of instrument axis $+V-h$, where $h$ is the height of the bar's sighting target above the station.

Reading of the data: only one RL is given (RL of X = 1197.339 m), so the staff point sighted from Q is taken to be the same point X as sighted from P (the common-point layout). With $K=100$ and $C=0$ and the staff vertical:

$$D=Ks\cos^2\theta,\qquad V=\tfrac12Ks\sin2\theta,\qquad \theta=90^\circ-z$$

Vertical angles: from P, $90^\circ-80^\circ10'=+9^\circ50'$ (elevation); from Q, $90^\circ-95^\circ20'=-5^\circ20'$ (depression).

Step 1: Distances and vertical components

Point X from P (bearing 70°20', angle of elevation $9^\circ 50'$):

$$\begin{aligned} s&=2.900-2.230=0.670\ \text{m}\\ D&=100\times0.670\times\cos^2 9^\circ 50'=65.046\ \text{m}\\ V&=\tfrac12\times100\times0.670\times\sin 2(9^\circ 50')=11.274\ \text{m} \end{aligned}$$

Point X from Q (bearing 335°00', angle of depression $5^\circ 20'$):

$$\begin{aligned} s&=1.440-1.000=0.440\ \text{m}\\ D&=100\times0.440\times\cos^2 5^\circ 20'=43.620\ \text{m}\\ V&=\tfrac12\times100\times0.440\times\sin 2(5^\circ 20')=-4.072\ \text{m} \end{aligned}$$

Step 2: RL of P and Q

$$\text{RL}_X=\text{RL of axis}+V-h\ \Rightarrow\ \text{RL of axis}=\text{RL}_X-V+h$$ $$\begin{aligned} \text{Axis at P}&=1197.339-11.274+2.565=1188.630\ \text{m}\\ \text{RL}_P&=1188.630-1.46=1187.170\ \text{m}\\ \text{Axis at Q}&=1197.339-(-4.072)+1.220=1202.631\ \text{m}\\ \text{RL}_Q&=1202.631-1.38=1201.251\ \text{m} \end{aligned}$$

Step 3: Horizontal distance and bearing of PQ

Take P as origin. X is at $D_P=65.046$ m on bearing $70^\circ20'$, and Q is at $D_Q=43.620$ m from X in the direction opposite to the bearing $335^\circ00'$ of the line Q to X:

$$\begin{aligned} X&=(21.891,\ 61.251)\ \text{(N, E)}\\ Q&=X-D_Q(\cos335^\circ,\ \sin335^\circ)=(21.891-+39.533,\ 61.251-(-18.435))=(-17.642,\ +79.686)\\ PQ&=\sqrt{-17.642^2++79.686^2}=81.616\ \text{m},\qquad \theta_{PQ}=\tan^{-1}\frac{+79.686}{-17.642}=102^\circ 29'01'' \end{aligned}$$

Step 4: Gradient from P to Q

$$\text{Gradient}=\frac{\text{RL}_Q-\text{RL}_P}{PQ}=\frac{+14.081}{81.616}=0.17253$$

Answer: RL of P = 1187.170 m, RL of Q = 1201.251 m, PQ = 81.616 m (bearing 102°29'). The gradient from P to Q is rising 1 in 5.8 (17.25 %).

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

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 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, $109^\circ 00'$ (18°00' and 127°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

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

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'.

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

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 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, $109^\circ 00'$ (18°00' and 127°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

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

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'.

Staff is vertical, $k = 100$, $c = 0$, angles are zenith angles $z$, so the vertical angle is $\theta = 90^\circ - z$.

$$H = k\,s\sin^2 z, \qquad V = \frac{k\,s}{2}\sin 2\theta$$

From station A

• $s = 2.025 - 1.025 = 1.000$ m, $z = 75^\circ 35' 25''$, $\theta = +14°24'35"$ (elevation)
• $H_A = 100\times 1.000\times\sin^2 z = 93.807$ m
• $V_A = 50\times 1.000\times\sin 2\theta = +24.103$ m
• RL of P = RL of A + HI + V − r, so $RL_A = 1234.560 - 24.103 + 1.525 - 1.42 = 1210.562$ m

From station B

• $s = 2.125 - 0.925 = 1.200$ m, $z = 96^\circ 52' 45''$, $\theta = -6°52'45"$ (depression, so $V$ is negative)
• $H_B = 100\times 1.200\times\sin^2 z = 118.278$ m
• $V_B = 50\times 1.200\times\sin 2\theta = -14.270$ m
• $RL_B = 1234.560 - (-14.270) + 1.525 - 1.48 = 1248.875$ m

Horizontal distance AB

Bearing of AP = 60°, bearing of BP = 345°, so the bearing of PA = 240° and of PB = 165°. The angle at P is $240^\circ - 165^\circ = 75^\circ$.

            P
           /|\
    H_A   / | \  H_B
         /  75°\
        A ------ B

$$AB = \sqrt{H_A^2 + H_B^2 - 2H_AH_B\cos 75^\circ} = \sqrt{93.807^2 + 118.278^2 - 2(93.807)(118.278)\cos 75^\circ} = 130.561\ \text{m}$$

Gradient

Difference in RL = $RL_B - RL_A = 1248.875 - 1210.562 = 38.312$ m

$$\text{Gradient} = \frac{38.312}{130.561} = 0.2934 = 1\text{ in }3.41$$

Answer: the ground rises from A to B at a gradient of 1 in 3.41 (about 29.3 %); AB = 130.56 m and B is 38.31 m higher than A.

Tacheometric surveying is a rapid method of surveying in which the horizontal distance and the difference in elevation of a point are found from optical measurements on a staff with a tacheometer, with no taping or levelling.

Methods of tacheometry

1. Stadia method: fixed hair (hairs fixed, staff intercept varies) and movable hair (hairs moved to fixed marks on the staff).
2. Tangential method: vertical angles to two targets on the staff.
3. Subtense bar method: the horizontal angle subtended by a bar of fixed length.
4. Special methods: use of the Beaman stadia arc or an EDM/total station.

Tangential method

In the tangential method the stadia hairs are not used. The telescope is pointed at two targets (vanes) on a staff, a known distance $s$ apart, and the two vertical angles are measured. The distance is calculated from the tangent of these angles. It is used when the hairs are not fitted or when a large distance requires a more precise measurement than the stadia method.

                   _ B  (upper target)
               _ -  |
        theta1 -    | s
    O  ----------   A  (lower target)  
        theta2 \_   |
 axis ----------\__ |       horizontal
                  D

Let $D$ be the horizontal distance of the staff from the instrument axis, and $V$ the vertical height of the lower target above (or below) the axis.

Case 1: both angles are angles of elevation

Angle to the upper target is $\theta_1$ and to the lower target $\theta_2$ ($\theta_1>\theta_2$):

$$\begin{aligned} V+s&=D\tan\theta_1,\qquad V=D\tan\theta_2\\ s&=D(\tan\theta_1-\tan\theta_2)\\ D&=\frac{s}{\tan\theta_1-\tan\theta_2},\qquad V=\frac{s\tan\theta_2}{\tan\theta_1-\tan\theta_2} \end{aligned}$$

Reduced level of the staff station $B$, with the lower target at height $h$ above $B$:

$$\text{RL}_B=\text{RL of axis}+V-h$$

Case 2: both angles are angles of depression

Angle to the lower target is $\alpha_1$ and to the upper target $\alpha_2$ ($\alpha_1>\alpha_2$). The lower target is at depth $D\tan\alpha_1$ below the axis and the upper at $D\tan\alpha_2$:

$$\begin{aligned} s&=D\tan\alpha_1-D\tan\alpha_2=D(\tan\alpha_1-\tan\alpha_2)\\ D&=\frac{s}{\tan\alpha_1-\tan\alpha_2},\qquad V=D\tan\alpha_2\ (\text{depth of the upper target})\\ \text{RL}_B&=\text{RL of axis}-V-s-h \end{aligned}$$

where $h$ is the height of the lower target above $B$ (the lower target is $s$ below the upper one, so the staff station is $V+s+h$ below the axis).

Case 3: one angle of elevation $\theta$ (upper) and one of depression $\alpha$ (lower)

$$s=D(\tan\theta+\tan\alpha),\qquad D=\frac{s}{\tan\theta+\tan\alpha}$$

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

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

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

$$\text{Gradient}=\frac{-10.807}{336.297}=0.03213$$

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.

In three-wire stadia tacheometry, the top, middle (axial) and bottom hairs are read on a vertical staff. The method gives the distance from the intercept and a check on the readings (middle reading $\approx$ mean of top and bottom).

Field procedure

1. Reconnaissance and control: establish control stations (traverse) over the area; know the RL of a benchmark or station.
2. Set up the tacheometer on a station, centre and level it. Measure the height of the instrument (HI) with a tape.
3. Orient: with the horizontal circle at zero (or a known bearing) sight a known reference station or magnetic north.
4. Back-sight to a benchmark or known point with the staff to find the RL of the instrument axis.
5. Detail points: the staff-man holds the staff vertical at each ground feature (corners, break lines, spot heights). For each point:
   • bisect the staff and read the horizontal circle (bearing),
   • read the top, middle and bottom hair (staff readings),
   • read the vertical angle, with the middle hair at the same height as the HI where possible (this simplifies the calculation) or at any convenient reading,
   • book the readings with a point description.
6. Check that the mean of top and bottom hair equals the middle reading (within 0.005 m), and close on a known point at the end of the day.
7. Computation: $s=$ top $-$ bottom, $D=Ks\cos^2\theta$, $V=\frac{1}{2}Ks\sin2\theta$, RL $=$ RL of axis $\pm V-h$.
8. Plotting: each point is plotted by its bearing and horizontal distance (protractor and scale) or by coordinates, with its RL, and contours are interpolated for the topographic map.

Field book (typical)

In the tangential method the stadia hairs are not used. The telescope is pointed at two targets (vanes) on a staff, a known distance $s$ apart, and the two vertical angles are measured. The distance is calculated from the tangent of these angles. It is used when the hairs are not fitted or when a large distance requires a more precise measurement than the stadia method.

                   _ B  (upper target)
               _ -  |
        theta1 -    | s
    O  ----------   A  (lower target)  
        theta2 \_   |
 axis ----------\__ |       horizontal
                  D

Let $D$ be the horizontal distance of the staff from the instrument axis, and $V$ the vertical height of the lower target above (or below) the axis.

Horizontal distance and vertical distance

For both angles of elevation $\theta_1$ (upper target) and $\theta_2$ (lower target), vanes $s$ apart:

$$D=\frac{s}{\tan\theta_1-\tan\theta_2},\qquad V=D\tan\theta_2$$ $$\text{RL of staff station}=\text{RL of axis}+V-h$$

where $h$ is the height of the lower vane above the ground at the staff station. For both angles of depression $\alpha_1>\alpha_2$:

$$D=\frac{s}{\tan\alpha_1-\tan\alpha_2},\qquad \text{RL}=\text{RL of axis}-D\tan\alpha_2-s-h$$

The method needs only the vertical circle of the theodolite, and works well even at long distances, but needs careful angle measurement (since the error in $D$ depends on $d\theta/\ (\tan\theta_1-\tan\theta_2)$).

Principle: in optical distance measurement the distance to a point is found from the instrument alone, by measuring a small angle subtended at the instrument by a known length (or a length subtended by a known angle), instead of taping the distance. It rests on the geometry of a thin isosceles triangle:

$$D=\frac{s}{2}\cot\frac{\beta}{2}\approx\frac{s}{\beta}$$

where $s$ is the base (known length) and $\beta$ the parallactic angle at the instrument.

 Instrument                       Staff / bar
     O  <-------------- D -------------->  A
      \  beta                              | s
       \_________________________________  B

Depending on which quantity is fixed:

1. Stadia method: the angle $\beta$ is fixed by two stadia hairs in the telescope, and the intercept $s$ on the staff varies. $D=Ks+C$.
2. Subtense method: the length $s$ is fixed (a subtense bar, 2 m long) and the angle $\beta$ is measured by the theodolite. $D=\frac{s}{2}\cot\frac{\beta}{2}$.
3. Tangential method: the vertical angles to two targets a known distance apart on a staff are measured. $D=s/(\tan\theta_1-\tan\theta_2)$.

The horizontal distance and the difference in elevation are then obtained from the distance and the vertical angle, so that detail surveys and contouring can be done faster than by chaining, especially in rough ground.

Reading of the data: the subtense bar at B gives the horizontal distance AB. The tacheometer at R ($K=100$, $C=0$, staff vertical) gives the distance RB and the heights. For A only the axial reading (1.100 m) is recorded, so RA is not read from stadia hairs but found from the triangle RAB (RA, RB and AB known, bearings known). The instrument is at the same height for both sights, so only readings and vertical components are needed for the difference in level.

Step 1: Distance AB from the subtense bar (angle at A)

The bar is out of normal by $4^\circ$, so its effective length is $l\cos4^\circ=3\cos4^\circ=2.9927$ m.

$$AB=\frac{l'}{2}\cot\frac{\beta}{2}=\frac{2.9927}{2}\cot\frac{0^\circ50'20''}{2}=204.396\ \text{m}$$

Step 2: Distance RB and vertical component from R

Zenith angle $83^\circ45'$ gives $\theta=+6^\circ15'$.

Point B (bearing 210°00', angle of elevation $6^\circ 15'$):

$$\begin{aligned} s&=1.850-0.650=1.200\ \text{m}\\ D&=100\times1.200\times\cos^2 6^\circ 15'=118.578\ \text{m}\\ V&=\tfrac12\times100\times1.200\times\sin 2(6^\circ 15')=12.986\ \text{m} \end{aligned}$$

Step 3: Distance RA from the triangle RAB

Bearings from R: RA $=315^\circ$ and RB $=210^\circ$, so angle ARB $=105^\circ$. By the cosine rule:

$$AB^2=RA^2+RB^2-2\,RA\,RB\cos105^\circ$$ $$RA^2-2(118.578)\cos105^\circ\,RA+(118.578^2-204.396^2)=0\ \Rightarrow\ RA=138.599\ \text{m}$$

(the other root is negative).

Step 4: Vertical component for A

Zenith angle $98^\circ45'$ gives $\theta=-8^\circ45'$ (depression):

$$V_A=RA\tan\theta=138.599\times\tan(-8^\circ45')=-21.332\ \text{m}$$

Step 5: Difference in level and gradient

$$\begin{aligned} \text{RL}_A&=\text{RL}_{axis}+V_A-1.100=\text{RL}_{axis}+(-21.332)-1.100\\ \text{RL}_B&=\text{RL}_{axis}+V_B-1.250=\text{RL}_{axis}++12.986-1.250\\ \Delta H&=\text{RL}_B-\text{RL}_A=(+12.986-1.250)-(-21.332-1.100)=+34.169\ \text{m} \end{aligned}$$ $$\text{Gradient}=\frac{\Delta H}{AB}=\frac{+34.169}{204.396}=0.16717$$

Answer: B is 34.169 m higher than A; AB = 204.396 m; the gradient from A to B is rising 1 in 6.0 (16.72 %).

Principle of stadia tacheometry: a telescope with two additional horizontal hairs (stadia hairs) above and below the central cross-hair. Rays through the hairs make a fixed small angle at the focus, so the staff intercept $s$ between them is proportional to the distance of the staff from the instrument.

 Diaphragm     Objective              Staff
   a |           |   \                  A  top
     | i         |     \  ...           |
  ---+-----------+-------F--------------+---- axis
     |           |      /               |
   b |           |   /                  B  bottom
     |<-- f --->|<--- D1 ---->|

Horizontal sight, staff vertical

From the similar triangles formed at the external focus $F$ (stadia hair interval $i$, focal length $f$, intercept $s$):

$$\frac{D_1}{f}=\frac{s}{i}\ \Rightarrow\ D_1=\frac{f}{i}\,s$$

Adding the distance of the focus from the instrument axis, $(f+d)$, where $d$ is the distance from the objective to the vertical axis:

$$D=\frac{f}{i}\,s+(f+d)=Ks+C$$

$K=f/i$ is the multiplying constant (usually 100) and $C=f+d$ the additive constant (about 0.3 m for an external focusing telescope, zero for an internal focusing/anallatic telescope).

Inclined sight, staff vertical

Let the line of sight be inclined by $\theta$ and the vertical staff intercept $AB=s$. The intercept normal to the line of sight is $s'=s\cos\theta$ (hair interval small). The inclined distance from the axis is $L=Ks\cos\theta+C$. Then

$$\begin{aligned} D&=L\cos\theta=Ks\cos^2\theta+C\cos\theta\\ V&=L\sin\theta=\tfrac12Ks\sin2\theta+C\sin\theta \end{aligned}$$

Reduced level

For a staff station with axial hair reading $h$:

$$\text{RL of staff station}=\text{RL of instrument axis}\pm V-h$$

(+ for angle of elevation, − for depression), where RL of axis $=$ RL of the station $+$ height of instrument.

The principle is used for contouring, detail survey and traversing in hilly ground.

Stadia interval factor and additive constant

• Stadia interval factor (multiplying constant) $K$: the ratio $f/i$ of the focal length of the objective to the stadia hair interval. It is the number by which the staff intercept is multiplied to get the distance, usually 100.
• Additive constant $C$: the distance $(f+d)$ from the instrument's vertical axis to the objective's focus, in metres. About 0.3 m for an external focusing telescope, nearly zero for an internal focusing one, and exactly zero for an anallatic lens.

Distance: $D=Ks+C$.

Determination of the constants (field method)

1. On level ground, set up the instrument on a line and measure out several distances from it, say 50, 100, 150 and 200 m, with a tape. Mark with pegs.
2. Hold the staff vertical at each peg and read the stadia hairs. Compute the intercepts $s_1,s_2,\dots$ with the line of sight horizontal.
3. Write $D=Ks+C$ for two pegs and solve:

$$K=\frac{D_2-D_1}{s_2-s_1},\qquad C=D_1-Ks_1$$

4. Calculate $K$ and $C$ for several pairs and take the mean values, or fit by least squares from all readings.

Example: $D_1=50$ m, $s_1=0.497$ m; $D_2=150$ m, $s_2=1.497$ m give $K=100/1.000=100.0$, $C=50-49.7=0.30$ m.

Reading of the data: the staff is held at right angles (normal) to the line of sight at both points, and no instrument constants are given, so $K=100$ and $C=0$ are taken. No heights of instrument are given, so only the difference in level between P and Q is found; the height of instrument cancels.

Staff normal to the line of sight (anallatic lens or $C=0$):

$$D=Ks\cos\theta,\qquad V=Ks\sin\theta$$

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 P (bearing 84°36', angle of elevation $3^\circ 30'$):

$$\begin{aligned} s&=2.850-1.350=1.500\ \text{m}\\ D&=(K s)\cos\theta=(100\times1.500)\cos 3^\circ 30'=149.720\ \text{m}\\ V&=(K s)\sin\theta=(100\times1.500)\sin 3^\circ 30'=9.157\ \text{m} \end{aligned}$$

Point Q (bearing 142°24', angle of elevation $2^\circ 45'$):

$$\begin{aligned} s&=3.765-1.955=1.810\ \text{m}\\ D&=(K s)\cos\theta=(100\times1.810)\cos 2^\circ 45'=180.792\ \text{m}\\ V&=(K s)\sin\theta=(100\times1.810)\sin 2^\circ 45'=8.684\ \text{m} \end{aligned}$$

Step 2: Difference in level

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

$$\begin{aligned} \Delta H&=(V_{Q}-h_{Q})-(V_{P}-h_{P})\\ &=(+8.684-2.860)-(+9.157-2.100)=-1.233\ \text{m} \end{aligned}$$

Step 3: Horizontal distance PQ

The angle at A between the two lines is the difference of the bearings, $57^\circ 48'$ (84°36' and 142°24').

$$\begin{aligned} PQ&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{149.720^2+180.792^2-2\times149.720\times180.792\cos 57^\circ 48'}=162.030\ \text{m} \end{aligned}$$

Step 4: Gradient

$$\text{Gradient}=\frac{\Delta H}{PQ}=\frac{1.233}{162.030}=0.00761$$

The line from P to Q is falling: 1 in 131.4, i.e. 0.76 %, an angle of $0^\circ 26'10''$ with the horizontal.

Answer: PQ = 162.03 m; difference in level = -1.233 m; gradient falling 1 in 131.4 (0.76 %, 0°26').

If the staff is instead taken as vertical, the corresponding result is $PQ=161.81$ m and gradient $1$ in $132.0$ (difference in level $-1.226$ m), which agrees closely because the vertical angles are small.