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

# Tacheometric RL of B, C, gradient BC

Surveying II · Chapter 2: Tacheometry

- Asked in: 2074 Bhadra, 8 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
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2

## Question

Following observations were made in a Tacheometric survey a station A of RL 1086.550, the height of instrument being 1.385 m.

| Inst. Station | H.I. | Staff Station | Bearing | Zenithal Angle | Staff 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 constants is 100. Determine the R.L. of B and C and the gradient of line BC, and bearing of BC.

## Short answer

For each sight use D=100s cos^2θ, V=0.5(100s)sin2θ, RL axis=1087.935 m, RL=RL axis+V-middle reading. RL_B=1117.710 m, RL_C=1069.928 m; BC=205.45 m; gradient from B to C falling 1 in 4.3 (23.26%); bearing BC=152°45'.

## Answer

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'.

## Check yourself

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

1. What are the instrument constants for an anallatic lens?
   A. C = 0 and K = 100
   B. C = 0.3 and K = 100
   C. C = 0 and K = 0
   D. C = 100 and K = 0
   Answer: A. An anallatic lens makes the additive constant zero, so only the multiplying constant K = 100 remains. (Key point: Anallatic lens constants)

2. A zenith angle of 96°00' is given for point C. What is the vertical angle used?
   A. 6°00' elevation
   B. 6°00' depression
   C. 96°00' elevation
   D. 186°00'
   Answer: B. Vertical angle = 90° - z = -6°, a depression. (Key point: Zenithal to vertical angle)

3. For a vertical staff and inclined sight with C = 0, which pair gives D and V?
   A. D = Ks cosθ, V = Ks sinθ
   B. D = Ks sin^2θ, V = (1/2)Ks sin2θ
   C. D = Ks cos^2θ, V = (1/2)Ks sin2θ
   D. D = Ks cos^2θ, V = Ks cosθ
   Answer: C. With the staff vertical the normal intercept is s cosθ, giving D = Ks cos^2θ and V = (1/2)Ks sin2θ. (Key point: Distance formulae)

4. How is the RL of a staff point found from the RL of the axis?
   A. RL of axis + V + h
   B. RL of station + V - HI
   C. RL of axis - V - h always
   D. RL of axis + V - h (axial hair reading)
   Answer: D. The point is V above the axis (signed) and the staff axial reading h lies below the line of sight, so subtract h. (Key point: RL of staff point)

## 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-2) — 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
- [Gradient A to B, RL of P = 1234.560](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Agradient-ab-tacheometer-100-0-p) — Surveying II, chapter 2, asked 2×
- [Anallatic tacheometer distance CD and gradient](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanallatic-staff-cd-gradient-abcd) — Surveying II, chapter 2