---
title: "Anallatic tacheometer distance XY, gradient"
subject: "Surveying II"
chapter: 2
key: "semester_4/surveying_ii:anallatic-xy-gradient-p-2081c"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/anallatic-xy-gradient-p-2081c
site: BCT Engineering
---

# Anallatic tacheometer distance XY, gradient

Surveying II · Chapter 2: Tacheometry

- Asked in: 2081 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/anallatic-xy-gradient-p-2081c
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2

## Question

A tacheometer fitted with anallatic lens and having multiplying constant of 100 was setup at station 'P'. The following readings were taken with the staff held vertically. Calculate the distance between XY and the gradient between X and Y.

| S.N | Staff Station | Bearing | Vertical Angle | Staff Intercept (s) | Central hair reading |
|---|---|---|---|---|---|
| 1 | X | 38°15' | -11°30' | 2.23 | 1.980 |
| 2 | Y | 68°25' | +8°15' | 2.08 | 1.503 |

## Short answer

For X: D=214.136 m, V=-43.567 m; for Y: D=203.717 m, V=+29.538 m. Difference in level Y-X=+73.581 m. XY=109.200 m. Gradient rising 1 in 1.5, 67.38%, 33°58'.

## Answer

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').

## Check yourself

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

1. In this problem the intercept s is given directly, so D is calculated by:
   A. D = s cos^2θ
   B. D = Ks cos^2θ with K = 100
   C. D = Ks sin^2θ
   D. D = Ks + C
   Answer: B. For a vertical staff with C = 0, D = Ks cos^2θ. (Key point: Staff intercept given)

2. Why is the height of instrument not needed for the difference in level?
   A. It is zero
   B. It is the same for both sights so it cancels
   C. It is replaced by the bearing
   D. It is included in K
   Answer: B. Both RLs contain RL of axis, which cancels in ΔH = (V_Y - h_Y) - (V_X - h_X). (Key point: Height of instrument)

3. The point X has vertical angle -11°30'. This means V for X is:
   A. Positive
   B. Zero
   C. Negative (line of sight goes down)
   D. Not defined for depression
   Answer: C. A depression gives a negative V, which lowers the RL of X. (Key point: Sign of V)

4. The angle used in the cosine rule at P is:
   A. The sum of the bearings
   B. The difference of the bearings, 30°10'
   C. The difference of vertical angles
   D. 90°
   Answer: B. 68°25' - 38°15' = 30°10'. (Key point: Angle at P)

## Close variants set in other exams

- [Anallatic tacheometer distance XY, gradient](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanallatic-xy-gradient-p) — Surveying II, chapter 2, asked 3×
- [Anallatic tacheometer AB gradient from P](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanallatic-ab-gradient-percent-angle) — Surveying II, chapter 2, asked 3×

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- [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
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- [Subtense bar and tacheometer gradient A-B](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Asubtense-bar-3m-level-difference-gradient) — Surveying II, chapter 2