---
title: "Tacheometric RL of B, C, gradient BC"
subject: "Surveying II"
chapter: 2
key: "semester_4/surveying_ii:anallatic-rl-b-c-gradient-bc-2"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/anallatic-rl-b-c-gradient-bc-2
site: BCT Engineering
---

# Tacheometric RL of B, C, gradient BC

Surveying II · Chapter 2: Tacheometry

- Asked in: 2065 Chaitra (old course), 9 marks
- Video: [Tacheometry reduced levels solved example](https://www.youtube.com/watch?v=ypi0CZHjRzA) — Iwords Shots Empire 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/anallatic-rl-b-c-gradient-bc-2
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2

## Question

The following observations were taken in a tacheometric survey from a station A of R.L. 1086.550, the height of instrument being 1.385 m.

| Instrument Station | Height of Instrument | Staff Station | Bearing | Zenithal | Stadia Reading |
|---|---|---|---|---|---|
| A | 1.385 | B | 18°00' | 71°30' | 1.295, 1.820, 2.345 |
| | | C | 127°00' | 96°00' | 1.010, 1.790, 2.570 |

The instrument is fitted with an anallatic lens and the multiplying constant is 100. Determine the R.L. of B and C and the gradient of the line BC.

## Short answer

Use D=Ks cos^2θ and V=1/2 Ks sin2θ with K=100, C=0. For B: D=94.428 m, V=+31.595 m, RL=1117.710 m. For C: D=154.296 m, V=-16.217 m, RL=1069.928 m. BC=205.453 m, fall 47.782 m, gradient falling 1 in 4.3 (23.26%).

## Answer

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'.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The anallatic lens in this problem means:
   A. C = 0 and K = 100
   B. C = 0.3 m and K = 100
   C. K = 0 and C = 100
   D. The staff must be held normal to the line
   Answer: A. An anallatic lens removes the additive constant, so D = Ks with C = 0. (Key point: Anallatic lens constants)

2. Zenith angles 71°30' (to B) and 96°00' (to C) give:
   A. B depression, C elevation
   B. B elevation of 18°30', C depression of 6°00'
   C. Both elevations
   D. Both depressions
   Answer: B. θ = 90° - z gives +18°30' for B and -6°00' for C. (Key point: Elevation or depression)

3. RL of B is found as:
   A. RL of A + V - h
   B. RL of A + HI + V + h
   C. RL of axis (RL of A + HI) + V - h(middle hair)
   D. RL of axis - V - h
   Answer: C. Axis RL = 1086.550 + 1.385; add V and subtract the middle hair reading. (Key point: RL of staff point)

4. The gradient of BC is:
   A. (RL difference) / (horizontal distance BC)
   B. (RL difference) / (D_B + D_C)
   C. (V_B - V_C) / (D_B - D_C)
   D. (RL difference) / (slope distance Ks)
   Answer: A. Gradient = ΔH / horizontal distance BC, where BC is found by the cosine rule with angle 109° at A. (Key point: Gradient of BC)

## Close variants set in other exams

- [Tacheometric RL of B, C, gradient BC](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanallatic-rl-b-c-gradient-bc) — Surveying II, chapter 2, asked 2×

## Related in Surveying II

- [RL, distance, gradient P to Q, instrument RL 1325.750](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanallatic-rl-distance-gradient-pq-1325) — Surveying II, chapter 2
- [Subtense bar, RL difference, gradient, bearing AB](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Asubtense-bar-c-difference-rl-gradient) — Surveying II, chapter 2