Trigonometric Leveling

The observation recorded against "C" in the table is taken as the sight from B to A (instrument station C is a misprint for B; the target height 1.65 m is the vane at A).

Station A is higher than B (the sight from B to A is an elevation), so the farther station is B. Take $D = 118$ m as the horizontal distance AB.

Mean vertical angles

RL of instrument axes

• Axis of B: $RL_{axis\,B} = 1590.000 + 1.47 = 1591.470$ m
• Sight B to A: the vane (1.65 m above the foot) at A is at 1591.470 + 118$\tan$(4°27'00") = 1600.653 m, so the ground at A is 1599.003 m and $RL_{axis\,A} = 1599.003 + 1.45 = 1600.453$ m

Distance and RL of P

Both stations lie in the same vertical plane as P, with A between B and P. Let $x$ = horizontal distance from B to P, so the distance from A to P is $x - D$ with $D = 118$ m.

                                  P
                            _.--'|
                      _.--'      |
            A  _.--'  aA        |
   B _.--'  aB  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of P above axis of B = $x\tan\alpha_B$ and above axis of A = $(x-D)\tan\alpha_A$. The RL of P is the same from both stations:

$$\begin{aligned} RL_{axis\,B} + x\tan\alpha_B &= RL_{axis\,A} + (x-D)\tan\alpha_A \\ x &= \frac{(RL_{axis\,A} - RL_{axis\,B}) - D\tan\alpha_A}{\tan\alpha_B - \tan\alpha_A} \end{aligned}$$

With $\alpha_B = 20°08'30"$ ($\tan = 0.36677$), $\alpha_A = 30°39'00"$ ($\tan = 0.59258$), axis RL of B = 1591.470 m and of A = 1600.453 m:

$$x = \frac{(1600.453 - 1591.470) - 118\times 0.59258}{0.36677 - 0.59258} = 269.884\ \text{m}$$ $$RL_{P} = 1591.470 + 269.884\times 0.36677 = 1690.456\ \text{m}$$

Check from A: $1600.453 + (269.884 - 118)\times 0.59258 = 1690.456$ m (same).

Answer: RL of the top of the hill P = 1690.456 m.

Station A is higher than B (the sight from B to A is an elevation), so the farther station is B. Take $D = 112$ m as the horizontal distance AB.

Mean vertical angles

RL of instrument axes

• Axis of B: $RL_{axis\,B} = 1290.000 + 1.42 = 1291.420$ m
• Sight B to A: the vane (1.75 m above the foot) at A is at 1291.420 + 112$\tan$(4°27'00") = 1300.136 m, so the ground at A is 1298.386 m and $RL_{axis\,A} = 1298.386 + 1.47 = 1299.856$ m

Distance and RL of P

Both stations lie in the same vertical plane as P, with A between B and P. Let $x$ = horizontal distance from B to P, so the distance from A to P is $x - D$ with $D = 112$ m.

                                  P
                            _.--'|
                      _.--'      |
            A  _.--'  aA        |
   B _.--'  aB  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of P above axis of B = $x\tan\alpha_B$ and above axis of A = $(x-D)\tan\alpha_A$. The RL of P is the same from both stations:

$$\begin{aligned} RL_{axis\,B} + x\tan\alpha_B &= RL_{axis\,A} + (x-D)\tan\alpha_A \\ x &= \frac{(RL_{axis\,A} - RL_{axis\,B}) - D\tan\alpha_A}{\tan\alpha_B - \tan\alpha_A} \end{aligned}$$

With $\alpha_B = 20°08'30"$ ($\tan = 0.36677$), $\alpha_A = 30°39'00"$ ($\tan = 0.59258$), axis RL of B = 1291.420 m and of A = 1299.856 m:

$$x = \frac{(1299.856 - 1291.420) - 112\times 0.59258}{0.36677 - 0.59258} = 256.560\ \text{m}$$ $$RL_{P} = 1291.420 + 256.560\times 0.36677 = 1385.519\ \text{m}$$

Check from A: $1299.856 + (256.560 - 112)\times 0.59258 = 1385.519$ m (same).

Answer: RL of the top of the hill P = 1385.519 m.

The observation recorded against "C" is taken as the sight from B to A (C is a misprint for B).

Station A is higher than B (the sight from B to A is an elevation), so the farther station is B. Take $D = 112$ m as the horizontal distance AB.

Mean vertical angles

RL of instrument axes

• Axis of B: $RL_{axis\,B} = 1280.000 + 1.42 = 1281.420$ m
• Sight B to A: the vane (1.75 m above the foot) at A is at 1281.420 + 112$\tan$(4°27'00") = 1290.136 m, so the ground at A is 1288.386 m and $RL_{axis\,A} = 1288.386 + 1.47 = 1289.856$ m

Distance and RL of P

Both stations lie in the same vertical plane as P, with A between B and P. Let $x$ = horizontal distance from B to P, so the distance from A to P is $x - D$ with $D = 112$ m.

                                  P
                            _.--'|
                      _.--'      |
            A  _.--'  aA        |
   B _.--'  aB  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of P above axis of B = $x\tan\alpha_B$ and above axis of A = $(x-D)\tan\alpha_A$. The RL of P is the same from both stations:

$$\begin{aligned} RL_{axis\,B} + x\tan\alpha_B &= RL_{axis\,A} + (x-D)\tan\alpha_A \\ x &= \frac{(RL_{axis\,A} - RL_{axis\,B}) - D\tan\alpha_A}{\tan\alpha_B - \tan\alpha_A} \end{aligned}$$

With $\alpha_B = 20°08'30"$ ($\tan = 0.36677$), $\alpha_A = 30°39'00"$ ($\tan = 0.59258$), axis RL of B = 1281.420 m and of A = 1289.856 m:

$$x = \frac{(1289.856 - 1281.420) - 112\times 0.59258}{0.36677 - 0.59258} = 256.560\ \text{m}$$ $$RL_{P} = 1281.420 + 256.560\times 0.36677 = 1375.519\ \text{m}$$

Check from A: $1289.856 + (256.560 - 112)\times 0.59258 = 1375.519$ m (same).

Answer: RL of the top of the hill P = 1375.519 m.

Mean vertical angles

RL of instrument axes

• Axis of R = RL of R + HI = 1280.000 + 1.42 = 1281.420 m
• Sight R to P: elevation 4°27'. The vane (1.75 m above the foot of P) is at 1281.420 + 112 tan(4°27') = 1290.136 m, so the ground at P = 1288.386 m and axis of P = 1288.386 + 1.47 = 1289.856 m.

Elevation angles to Q: from R = 20°08'30", from P = 30°39'00". Take D = 112 m as the horizontal distance RP.

Distance and RL of Q

Both stations lie in the same vertical plane as Q, with P between R and Q. Let $x$ = horizontal distance from R to Q, so the distance from P to Q is $x - D$ with $D = 112$ m.

                                  Q
                            _.--'|
                      _.--'      |
            P  _.--'  aP        |
   R _.--'  aR  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of Q above axis of R = $x\tan\alpha_R$ and above axis of P = $(x-D)\tan\alpha_P$. The RL of Q is the same from both stations:

$$\begin{aligned} RL_{axis\,R} + x\tan\alpha_R &= RL_{axis\,P} + (x-D)\tan\alpha_P \\ x &= \frac{(RL_{axis\,P} - RL_{axis\,R}) - D\tan\alpha_P}{\tan\alpha_R - \tan\alpha_P} \end{aligned}$$

With $\alpha_R = 20°08'30"$ ($\tan = 0.36677$), $\alpha_P = 30°39'00"$ ($\tan = 0.59258$), axis RL of R = 1281.420 m and of P = 1289.856 m:

$$x = \frac{(1289.856 - 1281.420) - 112\times 0.59258}{0.36677 - 0.59258} = 256.560\ \text{m}$$ $$RL_{Q} = 1281.420 + 256.560\times 0.36677 = 1375.519\ \text{m}$$

Check from P: $1289.856 + (256.560 - 112)\times 0.59258 = 1375.519$ m (same).

Answer: RL of the top of the hill Q = 1375.519 m.

When the base of the object (tower, chimney, hill top) is inaccessible and the two instrument stations are not in the same vertical plane as the object, the horizontal distance to the object is found by the triangulation (base line) method and the height by trigonometric levelling.

Field work

1. Select two stations X and Y a convenient distance $b$ apart (base line) so that P is visible from both and the triangle XYP is well conditioned (angles at X and Y between 30° and 120°).
2. Measure the base line XY and level the instrument at X. Take a back-sight on a BM of known RL to get the RL of the instrument axis; do the same at Y.
3. At X measure the horizontal angle $\alpha$ = $\angle$YXP (face left and right) and the vertical angle $\theta_X$ to P. At Y measure $\beta$ = $\angle$XYP and the vertical angle $\theta_Y$.

                 P (top)
                /  \
               /    \
          XP  /      \ YP
             /        \
            /α        β\
           X ---- b ---- Y
         (plan view)

Computation

In the plan triangle XYP, $\angle$XPY $= 180^\circ - (\alpha + \beta)$. By the sine rule:

$$\begin{aligned} XP &= \frac{b\sin\beta}{\sin(\alpha+\beta)} \\ YP &= \frac{b\sin\alpha}{\sin(\alpha+\beta)} \end{aligned}$$

These are horizontal distances. The height of P above each axis:

$$h_X = XP\tan\theta_X, \qquad h_Y = YP\tan\theta_Y$$

(for zenith angle $z$, use $\theta = 90^\circ - z$.)

RL of P from each station:

$$\begin{aligned} RL_P &= RL_{BM} + BS_X + h_X \quad (\text{from X}) \\ RL_P &= RL_{BM} + BS_Y + h_Y \quad (\text{from Y}) \end{aligned}$$

where $BS$ is the back-sight on the BM (axis RL = RL of BM + BS). The mean of the two values is taken as the RL of P. If the two stations are on very different levels, the axis RLs are found by BM sights at each station, as above. If the angle of elevation is given from the station peg, add the height of instrument $h_i$ to the RL of the peg.

The two values should agree closely; a large difference shows an error in the angles or the base line.

The face right reading of A to P is printed as 304°36', which would not agree with the face left reading (65°18'); it is taken as 294°36' (a misprint).

Mean vertical angles

A to B is a depression ($\alpha_{AB} = -12°48'00"$ below horizontal), so A is higher than B and the farther station is B.

RL of instrument axes

• Axis of B = 1280.000 + 1.47 = 1281.470 m
• Sight A to B: the vane at B is at (axis A) $- 120\tan(12°48'00")$, so ground B + 1.5 m = axis A $+ 120\tan(-12°48'00")$. Axis A = 1280.000 + 1.5 − 120 tan(-12°48'00") = 1308.763 m (the height of instrument of A is already inside the axis level)

Distance and RL of P

Both stations lie in the same vertical plane as P, with A between B and P. Let $x$ = horizontal distance from B to P, so the distance from A to P is $x - D$ with $D = 120$ m.

                                  P
                            _.--'|
                      _.--'      |
            A  _.--'  aA        |
   B _.--'  aB  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of P above axis of B = $x\tan\alpha_B$ and above axis of A = $(x-D)\tan\alpha_A$. The RL of P is the same from both stations:

$$\begin{aligned} RL_{axis\,B} + x\tan\alpha_B &= RL_{axis\,A} + (x-D)\tan\alpha_A \\ x &= \frac{(RL_{axis\,A} - RL_{axis\,B}) - D\tan\alpha_A}{\tan\alpha_B - \tan\alpha_A} \end{aligned}$$

With $\alpha_B = 20°04'00"$ ($\tan = 0.36529$), $\alpha_A = 24°39'00"$ ($\tan = 0.45889$), axis RL of B = 1281.470 m and of A = 1308.763 m:

$$x = \frac{(1308.763 - 1281.470) - 120\times 0.45889}{0.36529 - 0.45889} = 296.717\ \text{m}$$ $$RL_{P} = 1281.470 + 296.717\times 0.36529 = 1389.857\ \text{m}$$

Check from A: $1308.763 + (296.717 - 120)\times 0.45889 = 1389.857$ m (same).

Answer: RL of the hill top P = 1389.857 m.

Zenith angles are converted to elevation angles ($\alpha = 90^\circ - z$). The sight from A to B (z = 105°30') is a depression, so A is above B.

RL of instrument axes

• Axis of B = RL of B + HI = 120.000 + 1.35 = 121.350 m
• Sight A to B: angle -15°30' (depression). The vane at B (1.5 m above the foot) is at 120.000 + 1.5 = 121.500 m, so the axis of A = 121.500 − 150 tan(-15°30') = 163.099 m.

Elevation angles to P: from B = 20°15'00", from A = 30°45'00". Take D = 150 m as the horizontal distance BA.

Distance and RL of P

Both stations lie in the same vertical plane as P, with A between B and P. Let $x$ = horizontal distance from B to P, so the distance from A to P is $x - D$ with $D = 150$ m.

                                  P
                            _.--'|
                      _.--'      |
            A  _.--'  aA        |
   B _.--'  aB  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of P above axis of B = $x\tan\alpha_B$ and above axis of A = $(x-D)\tan\alpha_A$. The RL of P is the same from both stations:

$$\begin{aligned} RL_{axis\,B} + x\tan\alpha_B &= RL_{axis\,A} + (x-D)\tan\alpha_A \\ x &= \frac{(RL_{axis\,A} - RL_{axis\,B}) - D\tan\alpha_A}{\tan\alpha_B - \tan\alpha_A} \end{aligned}$$

With $\alpha_B = 20°15'00"$ ($\tan = 0.36892$), $\alpha_A = 30°45'00"$ ($\tan = 0.59494$), axis RL of B = 121.350 m and of A = 163.099 m:

$$x = \frac{(163.099 - 121.350) - 150\times 0.59494}{0.36892 - 0.59494} = 210.125\ \text{m}$$ $$RL_{P} = 121.350 + 210.125\times 0.36892 = 198.869\ \text{m}$$

Check from A: $163.099 + (210.125 - 150)\times 0.59494 = 198.869$ m (same).

The RL found is for the target on top of the station. As the target is 3.5 m above the ground:

$$RL_{\text{ground at P}} = 198.869 - 3.5 = 195.369\ \text{m}$$

Answer: RL of the target at P = 198.869 m; RL of the hill station (ground) = 195.369 m.

Zenith angles are converted to elevation angles ($\alpha = 90^\circ - z$). The sight from A to B (z = 105°30') is a depression, so A is above B.

RL of instrument axes

• Axis of B = RL of B + HI = 1500.000 + 1.48 = 1501.480 m
• Sight A to B: angle -15°30' (depression). The vane at B (1.75 m above the foot) is at 1500.000 + 1.75 = 1501.750 m, so the axis of A = 1501.750 − 150 tan(-15°30') = 1543.349 m.

Elevation angles to P: from B = 20°15'00", from A = 30°45'00". Take D = 150 m as the horizontal distance BA.

Distance and RL of P

Both stations lie in the same vertical plane as P, with A between B and P. Let $x$ = horizontal distance from B to P, so the distance from A to P is $x - D$ with $D = 150$ m.

                                  P
                            _.--'|
                      _.--'      |
            A  _.--'  aA        |
   B _.--'  aB  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of P above axis of B = $x\tan\alpha_B$ and above axis of A = $(x-D)\tan\alpha_A$. The RL of P is the same from both stations:

$$\begin{aligned} RL_{axis\,B} + x\tan\alpha_B &= RL_{axis\,A} + (x-D)\tan\alpha_A \\ x &= \frac{(RL_{axis\,A} - RL_{axis\,B}) - D\tan\alpha_A}{\tan\alpha_B - \tan\alpha_A} \end{aligned}$$

With $\alpha_B = 20°15'00"$ ($\tan = 0.36892$), $\alpha_A = 30°45'00"$ ($\tan = 0.59494$), axis RL of B = 1501.480 m and of A = 1543.349 m:

$$x = \frac{(1543.349 - 1501.480) - 150\times 0.59494}{0.36892 - 0.59494} = 209.594\ \text{m}$$ $$RL_{P} = 1501.480 + 209.594\times 0.36892 = 1578.803\ \text{m}$$

Check from A: $1543.349 + (209.594 - 150)\times 0.59494 = 1578.803$ m (same).

The target is 2.5 m above the ground at P, so

$$RL_{\text{ground at P}} = 1578.803 - 2.5 = 1576.303\ \text{m}$$

Answer: RL of the target = 1578.803 m; RL of the hill station (ground) = 1576.303 m.

The face right reading of B to P is printed as 239°00', which does not match the face left reading (64°52'); it is taken as 295°00' (a misprint).

Mean vertical angles

A to B is a depression, so A is higher than B.

RL of instrument axes

• Axis of B = RL of B + HI = 1280.500 + 1.42 = 1281.920 m
• Sight A to B: angle -6°15' (depression). The vane at B (2 m above the foot) is at 1280.500 + 2 = 1282.500 m, so the axis of A = 1282.500 − 125 tan(-6°15') = 1296.190 m.

Elevation angles to P: from B = 25°04'00", from A = 27°39'00". Take D = 125 m as the horizontal distance BA.

Distance and RL of P

Both stations lie in the same vertical plane as P, with A between B and P. Let $x$ = horizontal distance from B to P, so the distance from A to P is $x - D$ with $D = 125$ m.

                                  P
                            _.--'|
                      _.--'      |
            A  _.--'  aA        |
   B _.--'  aB  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of P above axis of B = $x\tan\alpha_B$ and above axis of A = $(x-D)\tan\alpha_A$. The RL of P is the same from both stations:

$$\begin{aligned} RL_{axis\,B} + x\tan\alpha_B &= RL_{axis\,A} + (x-D)\tan\alpha_A \\ x &= \frac{(RL_{axis\,A} - RL_{axis\,B}) - D\tan\alpha_A}{\tan\alpha_B - \tan\alpha_A} \end{aligned}$$

With $\alpha_B = 25°04'00"$ ($\tan = 0.46772$), $\alpha_A = 27°39'00"$ ($\tan = 0.52390$), axis RL of B = 1281.920 m and of A = 1296.190 m:

$$x = \frac{(1296.190 - 1281.920) - 125\times 0.52390}{0.46772 - 0.52390} = 911.767\ \text{m}$$ $$RL_{P} = 1281.920 + 911.767\times 0.46772 = 1708.376\ \text{m}$$

Check from A: $1296.190 + (911.767 - 125)\times 0.52390 = 1708.376$ m (same).

Answer: RL of the hill station P = 1708.376 m.

Data are read as: staff reading on BM from A = 1.25 m (horizontal sight, 90°), vane height at B = 1.75 m, horizontal distance AB = 58 m.

RL of instrument axes

• Axis of A = RL of BM + staff reading = 1280.000 + 1.25 = 1281.250 m
• Sight A to B: elevation angle $= 90^\circ - 78^\circ00' = 12°00'$. Vane at B = 1281.250 + 58 tan(12°00') = 1293.578 m. Ground at B = 1293.578 − 1.75 = 1291.828 m.
• Axis of B = 1291.828 + 1.48 = 1293.308 m

Elevation angles to P: from A = $90^\circ - 64^\circ30' = 25°30'$, from B = $90^\circ - 74^\circ15' = 15°45'$. The check below shows which station is farther from P.

Distance and RL of P

Both stations lie in the same vertical plane as P, with A between B and P. Let $x$ = horizontal distance from B to P, so the distance from A to P is $x - D$ with $D = 58$ m.

                                  P
                            _.--'|
                      _.--'      |
            A  _.--'  aA        |
   B _.--'  aB  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of P above axis of B = $x\tan\alpha_B$ and above axis of A = $(x-D)\tan\alpha_A$. The RL of P is the same from both stations:

$$\begin{aligned} RL_{axis\,B} + x\tan\alpha_B &= RL_{axis\,A} + (x-D)\tan\alpha_A \\ x &= \frac{(RL_{axis\,A} - RL_{axis\,B}) - D\tan\alpha_A}{\tan\alpha_B - \tan\alpha_A} \end{aligned}$$

With $\alpha_B = 15°45'00"$ ($\tan = 0.28203$), $\alpha_A = 25°30'00"$ ($\tan = 0.47698$), axis RL of B = 1293.308 m and of A = 1281.250 m:

$$x = \frac{(1281.250 - 1293.308) - 58\times 0.47698}{0.28203 - 0.47698} = 203.763\ \text{m}$$ $$RL_{P} = 1293.308 + 203.763\times 0.28203 = 1350.775\ \text{m}$$

Check from A: $1281.250 + (203.763 - 58)\times 0.47698 = 1350.775$ m (same).

Answer: RL of the top of the hill P = 1350.775 m.

Let O be the instrument station and $D$ the horizontal distance from O to the tower. The staff is held at the plinth (foot) of the tower; the readings 5.00 m and 2.00 m are heights above the plinth. The instrument axis is at $1260.00 + 1.42 = 1261.42$ m.

                    * top (45° elevation)
                  / |
                /   |
              /     |  tower
   axis O --/---- * 5.00 m   (2°30')
              \--- * 2.00 m   (1°15')
                    |_ plinth
   |<------ D ----->|

Elevation angles

• To reading 5.00 m: $\alpha_1 = 90^\circ - 87^\circ30' = 2^\circ30'$
• To reading 2.00 m: $\alpha_2 = 90^\circ - 88^\circ45' = 1^\circ15'$
• To top: $\alpha_3 = 90^\circ - 45^\circ = 45^\circ$

Horizontal distance

Both lines of sight meet the vertical line of the tower, so

$$D\tan\alpha_1 - D\tan\alpha_2 = 5.00 - 2.00$$ $$D = \frac{3.00}{\tan 2^\circ30' - \tan 1^\circ15'} = \frac{3.00}{0.04366 - 0.02182} = 137.357\ \text{m}$$

RL of the plinth and the top

• Plinth: $RL = 1261.42 + 137.357\tan 2^\circ30' - 5.00 = 1262.417$ m
• Top: $RL = 1261.42 + 137.357\tan 45^\circ = 1398.777$ m

Clear height

$$H = 1398.777 - 1262.417 = 136.360\ \text{m}$$

Answer: clear height of the tower = 136.36 m.

Data are read as: staff reading on BM from A = 1.00 m (horizontal sight), staff (vane) reading at B = 1.75 m, horizontal distance AB = 53 m.

RL of instrument axes

• Axis of A = 1260.000 + 1.00 = 1261.000 m
• Sight A to B: elevation $= 90^\circ - 80^\circ = 10°00'$. Reading point at B = 1261.000 + 53 tan(10°00') = 1270.345 m. Ground at B = 1268.595 m.
• Axis of B = 1268.595 + 1.42 = 1270.015 m

Elevation angles to P: from A = $90^\circ - 62^\circ30' = 27°30'$, from B = $90^\circ - 71^\circ15' = 18°45'$. The check below shows which station is farther from P.

Distance and RL of P

Both stations lie in the same vertical plane as P, with A between B and P. Let $x$ = horizontal distance from B to P, so the distance from A to P is $x - D$ with $D = 53$ m.

                                  P
                            _.--'|
                      _.--'      |
            A  _.--'  aA        |
   B _.--'  aB  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of P above axis of B = $x\tan\alpha_B$ and above axis of A = $(x-D)\tan\alpha_A$. The RL of P is the same from both stations:

$$\begin{aligned} RL_{axis\,B} + x\tan\alpha_B &= RL_{axis\,A} + (x-D)\tan\alpha_A \\ x &= \frac{(RL_{axis\,A} - RL_{axis\,B}) - D\tan\alpha_A}{\tan\alpha_B - \tan\alpha_A} \end{aligned}$$

With $\alpha_B = 18°45'00"$ ($\tan = 0.33945$), $\alpha_A = 27°30'00"$ ($\tan = 0.52057$), axis RL of B = 1270.015 m and of A = 1261.000 m:

$$x = \frac{(1261.000 - 1270.015) - 53\times 0.52057}{0.33945 - 0.52057} = 202.114\ \text{m}$$ $$RL_{P} = 1270.015 + 202.114\times 0.33945 = 1338.624\ \text{m}$$

Check from A: $1261.000 + (202.114 - 53)\times 0.52057 = 1338.624$ m (same).

Answer: RL of the top of the hill P = 1338.624 m.

Assumptions: stadia constants $k = 100$, $c = 0$, staff vertical. The height of instrument at B is not given, so the axis of B is taken at the level of the peg B (the RL of T would change by the actual height of instrument at B). Distances are horizontal.

Step 1: RL of the axis of A (from the BM)

• $s = 2.955 - 2.732 = 0.223$ m, $\theta = -6^\circ34'40''$
• $V = \frac{ks}{2}\sin 2\theta = 50\times 0.223\times\sin(-13^\circ09'20'') = -2.538$ m
• Axis of A $= RL_{BM} + r - V = 1200.000 + 2.843 - (-2.538) = 1205.381$ m

Step 2: distance AB and RL of axis at B

• $s = 1.614 - 1.423 = 0.191$ m, $\theta = +10^\circ21'50''$
• $D_{AB} = k s\cos^2\theta = 18.482$ m
• $V = 50 s\sin2\theta = 3.380$ m
• Ground at B $= 1205.381 + 3.380 - 1.518 = 1207.243$ m, so the axis of B is 1207.243 m (peg level).

Step 3: RL of the top T (A, B and T in one vertical plane, B between A and T)

Elevation angles of T: $\alpha_A = 22^\circ26'40''$, $\alpha_B = 27^\circ32'40''$. Let $x$ = horizontal distance A to T.

$$x = \frac{(axis_B - axis_A) - D\tan\alpha_B}{\tan\alpha_A - \tan\alpha_B} = \frac{(1207.243 - 1205.381) - 18.482\times 0.52155}{0.41308 - 0.52155} = 71.696\ \text{m}$$ $$RL_T = 1205.381 + 71.696\times 0.41308 = 1234.997\ \text{m}$$

Check from B: $1207.243 + (71.696 - 18.482)\times 0.52155 = 1234.997$ m.

Answer: RL of the top of the temple = 1234.997 m.

Reciprocal trigonometric levelling

When the two stations A and B are far apart or on opposite sides of a wide obstacle (river, valley) and the effects of curvature and refraction cannot be neglected, the vertical angles are measured from both stations, at nearly the same time, so that the errors cancel. From A the angle of elevation $\alpha$ to B is measured, and from B the angle of depression $\beta$ to A is measured, with the same instrument height (or a vane at the same height as the instrument axis).

        B  .
        |\  `-.  beta
        |  \    `-.
        |   \       `-. A... 
   h    |    \  chord
        |     \
        A ------`--------
          alpha   D

Expression for elevation difference

Let $D$ = horizontal distance, $\phi = D/R$ the angle between the two level lines at A and B (R = radius of earth), and $\varepsilon$ the angular effect of refraction, which is assumed equal at both ends because the sights are taken at the same time. If $\theta$ is the true inclination of the chord AB at A, the observed angles are

$$\alpha = \theta + \varepsilon, \qquad \beta = (\theta + \phi) - \varepsilon$$

Adding, refraction cancels:

$$\alpha + \beta = 2\theta + \phi \quad\Rightarrow\quad \theta = \frac{\alpha+\beta}{2} - \frac{\phi}{2}$$

The true difference in elevation is

$$\boxed{h = D\tan\left(\frac{\alpha + \beta}{2} - \frac{D}{2R}\right) \approx D\tan\frac{\alpha+\beta}{2}}$$

If the heights of instrument at A and B are $h_A$ and $h_B$, and the height of signals are made equal to them, no further correction is needed. Otherwise, the difference in height of instrument is added. Then $RL_B = RL_A + h$.

For short lines the term $D/2R$ is negligible, and the formula becomes $h = D\tan\frac{\alpha+\beta}{2}$, the mean of the two angles. The principal advantage is that the effects of refraction and curvature are eliminated by the mean of the two reciprocal observations.

Assumption: the back sights are taken on the BM (RL 1000.00 m), so the axis RL of an instrument = 1000.00 + back sight. The zenith angles are used as given (elevation $= 90^\circ - z$).

Plan triangle XYP

$\angle XPY = 180^\circ - (45^\circ45' + 60^\circ30') = 73°45'$

              P
             / \
        XP  /   \ YP
           /     \
          X ----- Y
            200 m

$$XP = \frac{200\sin 60^\circ30'}{\sin 73°45'} = 181.315\ \text{m}, \qquad YP = \frac{200\sin 45^\circ45'}{\sin 73°45'} = 149.222\ \text{m}$$

Axis levels

• Axis of X = 1000.000 + 1.8 = 1001.800 m
• Axis of Y = 1000.000 + 0.8 = 1000.800 m

RL of P

• Elevation from X $= 90^\circ - 30^\circ30' = 59^\circ30'$: $h_X = 181.315\tan 59^\circ30' = 307.811$ m, so $RL_P = 1001.800 + 307.811 = 1309.611$ m
• Elevation from Y $= 90^\circ - 29^\circ20' = 60^\circ40'$: $h_Y = 149.222\tan 60^\circ40' = 265.548$ m, so $RL_P = 1000.800 + 265.548 = 1266.348$ m

The two values differ by 43.263 m, so the mean is taken (the given data are not perfectly consistent).

Answer: RL of the top of the tower P = 1287.98 m (mean of 1309.61 m and 1266.35 m).

The vane is held at the plinth of the tower and sighted with the transit theodolite. Readings of 5 m and 2 m are heights of the vane above the plinth. RL of instrument axis = 1200.00 m. Let $D$ be the horizontal distance from the instrument to the tower.

                    * top (22°15')
                  / |
   axis ----------/-* 5 m  (2°15')
                    \* 2 m  (1°30')
                    |_ plinth
   |<------ D ----->|

Angles of elevation

• $\alpha_1 = 90^\circ - 87^\circ45' = 2^\circ15'$ (vane at 5 m)
• $\alpha_2 = 90^\circ - 88^\circ30' = 1^\circ30'$ (vane at 2 m)
• $\alpha_3 = 90^\circ - 67^\circ45' = 22^\circ15'$ (top)

Horizontal distance

The two vane positions are 3 m apart on the same vertical line:

$$D(\tan\alpha_1 - \tan\alpha_2) = 5 - 2$$ $$D = \frac{3}{\tan 2^\circ15' - \tan 1^\circ30'} = \frac{3}{0.03929 - 0.02619} = 228.934\ \text{m}$$

RL of plinth and top

• Plinth: $1200.000 + D\tan\alpha_1 - 5 = 1200.000 + 8.995 - 5 = 1203.995$ m
• Top: $1200.000 + D\tan\alpha_3 = 1200.000 + 93.660 = 1293.660$ m

Clear height

$$H = 1293.660 - 1203.995 = 89.665\ \text{m}$$

Answer: clear height of the flood light tower = 89.66 m.

Take 120 m as the horizontal distance PR. R is the lower station (the sight from R to the staff at P is an angle of elevation), so R is the farther station from the tower and P lies between R and Q.

RL of instrument axes

• Axis of R = RL of R + HI = 112.780 + 1.64 = 114.420 m
• Sight R to 2 m above the foot of the staff at P: elevation $15^\circ11'$. Point sighted = 114.420 + 120 tan 15°11' = 146.986 m, so the ground at P = 146.986 − 2.00 = 144.986 m.
• Axis of P = 144.986 + 1.87 = 146.856 m

Elevation of Q: from P $= 90^\circ - 51^\circ39' = 38°21'$, from R $= 90^\circ - 68^\circ42' = 21°18'$.

Distance and RL of Q

Both stations lie in the same vertical plane as Q, with P between R and Q. Let $x$ = horizontal distance from R to Q, so the distance from P to Q is $x - D$ with $D = 120$ m.

                                  Q
                            _.--'|
                      _.--'      |
            P  _.--'  aP        |
   R _.--'  aR  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of Q above axis of R = $x\tan\alpha_R$ and above axis of P = $(x-D)\tan\alpha_P$. The RL of Q is the same from both stations:

$$\begin{aligned} RL_{axis\,R} + x\tan\alpha_R &= RL_{axis\,P} + (x-D)\tan\alpha_P \\ x &= \frac{(RL_{axis\,P} - RL_{axis\,R}) - D\tan\alpha_P}{\tan\alpha_R - \tan\alpha_P} \end{aligned}$$

With $\alpha_R = 21°18'00"$ ($\tan = 0.38988$), $\alpha_P = 38°21'00"$ ($\tan = 0.79117$), axis RL of R = 114.420 m and of P = 146.856 m:

$$x = \frac{(146.856 - 114.420) - 120\times 0.79117}{0.38988 - 0.79117} = 155.761\ \text{m}$$ $$RL_{Q} = 114.420 + 155.761\times 0.38988 = 175.149\ \text{m}$$

Check from P: $146.856 + (155.761 - 120)\times 0.79117 = 175.149$ m (same).

Answer: RL of the top of the tower Q = 175.149 m.

The face right reading of A to P is printed as 304°36', which would not agree with the face left reading (65°18'); it is taken as 294°36' (a misprint).

Mean vertical angles

A to B is a depression ($\alpha_{AB} = -12°48'00"$ below horizontal), so A is higher than B and the farther station is B.

RL of instrument axes

• Axis of B = 1280.000 + 1.47 = 1281.470 m
• Sight A to B: the vane at B is at (axis A) $- 120\tan(12°48'00")$, so ground B + 2.5 m = axis A $+ 120\tan(-12°48'00")$. Axis A = 1280.000 + 2.5 − 120 tan(-12°48'00") = 1309.763 m (the height of instrument of A is already inside the axis level)

Distance and RL of P

Both stations lie in the same vertical plane as P, with A between B and P. Let $x$ = horizontal distance from B to P, so the distance from A to P is $x - D$ with $D = 120$ m.

                                  P
                            _.--'|
                      _.--'      |
            A  _.--'  aA        |
   B _.--'  aB  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of P above axis of B = $x\tan\alpha_B$ and above axis of A = $(x-D)\tan\alpha_A$. The RL of P is the same from both stations:

$$\begin{aligned} RL_{axis\,B} + x\tan\alpha_B &= RL_{axis\,A} + (x-D)\tan\alpha_A \\ x &= \frac{(RL_{axis\,A} - RL_{axis\,B}) - D\tan\alpha_A}{\tan\alpha_B - \tan\alpha_A} \end{aligned}$$

With $\alpha_B = 20°04'00"$ ($\tan = 0.36529$), $\alpha_A = 24°39'00"$ ($\tan = 0.45889$), axis RL of B = 1281.470 m and of A = 1309.763 m:

$$x = \frac{(1309.763 - 1281.470) - 120\times 0.45889}{0.36529 - 0.45889} = 286.034\ \text{m}$$ $$RL_{P} = 1281.470 + 286.034\times 0.36529 = 1385.955\ \text{m}$$

Check from A: $1309.763 + (286.034 - 120)\times 0.45889 = 1385.955$ m (same).

Answer: RL of the hill top P = 1385.955 m.

When the base of the object is inaccessible and the two instrument stations A and B are at very different levels but in the same vertical plane as the object P, the double-plane (two station) method is used. The stations are in line with P, so the horizontal distance and the RL of P are found from vertical angles only, with a measured base AB.

Field work

1. Set up the theodolite at A (the farther station), level it and measure the height of instrument $h_A$. Measure the vertical angle $\alpha_A$ to P (face left and face right, mean value).
2. Place a vane (height $s$ above the foot) at B and measure the vertical angle $\alpha_{AB}$ from A to the vane. Measure the horizontal distance $D$ = AB.
3. Shift the instrument to B (measure $h_B$) and measure the vertical angle $\alpha_B$ to P.
4. Sight a staff on a BM from the station to obtain the RL of the axis, or use the known RL of B.

                                  P
                            _.--'|
                      _.--'      |
            B  _.--'  aB         |
   A _.--'  aA  - - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Computation

Let $x$ = horizontal distance from A to P, so the distance from B to P is $x - D$.

• RL of axis of B $= RL_B + h_B$.
• RL of axis of A from the sight A to the vane at B (vertical angle $\alpha_{AB}$, positive if elevation):

$$RL_{\text{axis }A} = RL_B + s - D\tan\alpha_{AB}$$

The height of P is the same from both stations:

$$RL_{\text{axis }A} + x\tan\alpha_A = RL_{\text{axis }B} + (x-D)\tan\alpha_B$$ $$\boxed{x = \frac{(RL_{\text{axis }B} - RL_{\text{axis }A}) - D\tan\alpha_B}{\tan\alpha_A - \tan\alpha_B}}$$

• Horizontal distance of the object from B: $x - D$
• RL of P:

$$\boxed{RL_P = RL_{\text{axis }A} + x\tan\alpha_A = RL_{\text{axis }B} + (x-D)\tan\alpha_B}$$

If P is a point on top of an object whose base RL is wanted, subtract the height of the object. When the line of sight is long, add the correction for curvature and refraction, $0.0673\,d^2$ (d in km), to each height. Angles read as zenith angles are changed with $\alpha = 90^\circ - z$.

Let B (RL 1280.00 m, lower station) be the farther station and A the nearer one: the sight from B to A is an angle of elevation, so A is higher. D = 112 m is taken as the horizontal distance AB. Curvature and refraction correction: $C = 0.0673\,d^2$ m with $d$ in km (added to the height of a sighted point).

(a) Without correction

• Axis of B $= 1280.000 + 1.42 = 1281.420$ m
• Sight B to the vane at A (1.5 m above foot): $\alpha = 4^\circ15'$. Vane level $= 1281.420 + 112\tan 4^\circ15' = 1289.743$ m, ground A $= 1288.243$ m
• Axis of A $= 1288.243 + 1.47 = 1289.713$ m

Let $x$ = horizontal distance from B to the temple.

$$x = \frac{(1289.713 - 1281.420) - 112\tan 30^\circ36'}{\tan 20^\circ12' - \tan 30^\circ36'} = \frac{8.293 - 66.237}{0.36793 - 0.59140} = 259.290\ \text{m}$$ $$RL_P = 1281.420 + 259.290\tan 20^\circ12' = 1376.820\ \text{m}$$

(b) With curvature and refraction

Each calculated height is increased by $0.0673\,(d/1000)^2$. For the sight B to A, $d = 112$ m: $C = 0.0008$ m, so the axis of A becomes 1289.714 m.

The height of P must be the same from both stations:

$$1281.420 + x\tan 20^\circ12' + C(x) = 1289.714 + (x - 112)\tan 30^\circ36' + C(x-112)$$

Solving (the correction terms are small, so a few iterations are enough):

$$x = 259.300\ \text{m}, \qquad C(x) = 0.0045\ \text{m}$$ $$RL_P = 1281.420 + 259.300\tan 20^\circ12' + 0.005 = 1376.828\ \text{m}$$

Answer: RL of the top of the temple = 1376.820 m without correction; 1376.828 m after applying curvature and refraction.

Situations for reciprocal trigonometric levelling

Reciprocal trigonometric levelling is conducted when:

• two stations are far apart or are separated by a wide obstacle such as a river, valley or lake, so that ordinary levelling or a level with a staff cannot be used between them, and
• the effects of curvature and refraction are large and must be eliminated.

Vertical angles are measured from both stations at the same time, using the same type of instrument and the same height of signal, so that the errors of curvature and refraction are equal and opposite and cancel in the mean:

$$h = D\tan\frac{\alpha+\beta}{2}$$

where $\alpha$ is the angle of elevation from the lower station and $\beta$ is the angle of depression from the upper station.

Hill top when the stations and the target are in different vertical planes

If the instrument stations X and Y are not in the same vertical plane as the object P, a base line XY is measured and horizontal angles are observed at X and Y. The horizontal distances XP and YP are found by the sine rule in the plan triangle XYP, and the height of P is found from the vertical angles.

                 P (top)
                /  \
          XP   /    \  YP
              /      \
             /a      b\
            X -- XY=b -- Y

Derivation

Let the base line be $b = XY$, the horizontal angles $\alpha = \angle YXP$ and $\beta = \angle XYP$, so $\angle XPY = 180^\circ - (\alpha + \beta)$.

$$\begin{aligned} XP &= \frac{b\sin\beta}{\sin(\alpha+\beta)} \\ YP &= \frac{b\sin\alpha}{\sin(\alpha+\beta)} \end{aligned}$$

Let $\theta_X$ and $\theta_Y$ be the vertical angles (elevation) of P from X and Y, and $h_X$, $h_Y$ the heights of the instrument axes above their stations. If a BM of known RL is sighted from X and Y with back-sights $BS_X$, $BS_Y$:

$$\begin{aligned} RL_{\text{axis X}} &= RL_{BM} + BS_X, \qquad RL_{\text{axis Y}} = RL_{BM} + BS_Y \\ RL_P &= RL_{\text{axis X}} + XP\tan\theta_X \quad (\text{from X}) \\ RL_P &= RL_{\text{axis Y}} + YP\tan\theta_Y \quad (\text{from Y}) \end{aligned}$$

For zenith angle $z$, $\theta = 90^\circ - z$. The mean of the two values is the RL of P. If the stations are at different levels, the difference of the axis RLs is already included through the BM sights. When the sights are long, the correction for curvature and refraction $0.0673\,d^2$ (d in km) is added to each height.