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

# Anallatic tacheometer distance XY, gradient

Surveying II · Chapter 2: Tacheometry

- Asked in: 2075 Baisakh, 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
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2

## Question

A tacheometer fitted with an anallatic lens and having multiplying constant of 100 was setup at station 'P'. The following readings were taken with the staff held vertically.

| S.N | Staff Station | Bearing | Vertical Angle | Staff intercept | Axial hair reading |
|---|---|---|---|---|---|
| 1 | X | 40°35' | -10°20' | 2.25 | 1.987 |
| 2 | Y | 70°10' | +7°30' | 2.05 | 1.500 |

Calculate the distance XY and the gradient between X and Y.

## Short answer

With C=0: X gives D=217.761 m, V=−39.705 m; Y gives D=201.507 m, V=+26.529 m. Using hair readings, ΔH=(26.529−1.500)−(−39.705−1.987)=+66.721 m. The angle at P is 29°35′, so XY=108.189 m; gradient = 66.721/108.189 = 0.6167, rising 1 in 1.6 (61.67%).

## Answer

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

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 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: $70^\circ10'-40^\circ35'=29^\circ 35'$.

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

#### 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').

## Check yourself

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

1. What additive constant applies to an anallatic lens?
   A. C = 0
   B. C = 0.3 m
   C. C = 100
   D. C = f + d
   Answer: A. An anallatic lens gives C = 0, so D = Ks cos^2(theta). (Key point: Constants)

2. X is sighted at 10°20' depression. Its vertical component V is
   A. Positive
   B. Negative
   C. Zero
   D. Equal to D
   Answer: B. A depression gives a negative V (-39.705 m). (Key point: Sign of V for X)

3. What angle is used in the cosine rule for XY?
   A. 70°10' + 40°35'
   B. 70°10' - 40°35' = 29°35'
   C. 180° - 29°35'
   D. 10°20' + 7°30'
   Answer: B. The angle at P is the difference between the bearings, 29°35'. (Key point: Angle at P)

4. Which expression gives the level difference of Y over X?
   A. (V_Y - 1.500) - (V_X - 1.987)
   B. (V_Y + 1.500) - (V_X + 1.987)
   C. V_Y - V_X + HI
   D. (1.987 - 1.500)
   Answer: A. (26.529 - 1.500) - (-39.705 - 1.987) = +66.721 m; HI cancels. (Key point: Level difference)

5. Delta H = 66.721 m and XY = 108.189 m. The gradient is
   A. Rising 1 in 16
   B. Falling 1 in 1.6
   C. Rising 1 in 1.6 (61.67 %)
   D. Rising 1 in 0.6
   Answer: C. 66.721/108.189 = 0.6167, rising about 1 in 1.6. (Key point: Gradient)

## 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-2081c) — 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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