---
title: "Clear height of flood light tower (stadium)"
subject: "Surveying II"
chapter: 3
key: "semester_4/surveying_ii:flood-light-tower-clear-height-stadium"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/flood-light-tower-clear-height-stadium
site: BCT Engineering
---

# Clear height of flood light tower (stadium)

Surveying II · Chapter 3: Trigonometric Leveling

- Asked in: 2078 Poush, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/flood-light-tower-clear-height-stadium
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-3

## Question

It is required to find the clear height of a flood light tower of a stadium and following informations were recorded.

| Inst Stn | Target Stn | Zenithal Angle | Rod Readings |
|---|---|---|---|
| O | P | 87°30' | 5.00 m |
| | P | 88°45' | 2.00 m |
| | P | 45°00' | Top of flood light tower |

Compute the clear height (from plinth to top of tower) of the tower if RL of the peg (instrument station) is 1260.00 m and height of instrument is 1.42 m.

## Short answer

Using two staff readings 5.00 m and 2.00 m at zenith angles 87°30' and 88°45', horizontal distance D = 137.357 m. Plinth RL = 1262.417 m, top RL = 1398.777 m, clear height = 136.360 m.

## Answer

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.

```text
                    * 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.**

## Check yourself

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

1. The zenith angles are 87°30', 88°45' and 45°00'. What are the elevation angles to the 5.00 m reading, the 2.00 m reading and the top?
   A. 87°30', 88°45', 45°
   B. 2°30', 1°15', 45°
   C. 2°30', 1°15', 135°
   D. 92°30', 91°15', 45°
   Answer: B. Elevation = 90° - zenith, giving 2°30', 1°15' and 45°. (Key point: Elevation angles)

2. Which relation gives the horizontal distance D to the tower?
   A. D tan 2°30' + D tan 1°15' = 5.00 + 2.00
   B. D tan 45° = 5.00 - 2.00
   C. D tan 2°30' - D tan 1°15' = 5.00 - 2.00
   D. D / tan 2°30' = 3.00
   Answer: C. Both sights hit the same vertical line, and the two readings are 3.00 m apart. So D(tan a1 - tan a2) = 3.00. (Key point: Finding the distance D)

3. Why is 5.00 m subtracted in: plinth RL = 1261.42 + D tan 2°30' - 5.00?
   A. Because the instrument axis is 5.00 m above the peg
   B. Because curvature reduces the height by 5.00 m
   C. Because the staff is held at the top of the tower
   D. Because the sighted point is 5.00 m above the plinth, so the plinth is 5.00 m lower
   Answer: D. The sight strikes the 5.00 m mark, which is 5.00 m above the plinth. The plinth RL is that sighted level minus 5.00 m. (Key point: RL of the plinth)

4. Which gives the clear height of the tower?
   A. RL of top - instrument axis RL = 137.36 m
   B. RL of top - RL of plinth = 136.36 m
   C. RL of top - RL of the peg = 138.78 m
   D. D tan 45° - 5.00 = 132.36 m
   Answer: B. Clear height is measured from the plinth, so H = 1398.777 - 1262.417 = 136.36 m. (Key point: Clear height)

## Related in Surveying II

- [Clear height of flood light tower (arena)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aflood-light-tower-clear-height-arena) — Surveying II, chapter 3
- [RL of tower top Q from stations P and R](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atower-top-q-from-p-and) — Surveying II, chapter 3
- [RL of tower top, base inaccessible](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atower-inaccessible-base-rl-two-stations) — Surveying II, chapter 3
- [RL of hill top from two stations (A-B 118 m)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-two-stations-2079j) — Surveying II, chapter 3, asked 4×