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

# Clear height of flood light tower (arena)

Surveying II · Chapter 3: Trigonometric Leveling

- Asked in: 2072 Asoj, 10 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-arena
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-3

## Question

It is required to determine the height (clear) of a Flood light tower in an arena by using a transit theodolite and for this zenith angles observation taken at 5 m and 2 m height on a target vane held on the plinth level of tower were 87°45' and 88°30' respectively. From the same instrument, zenith angle observed at top of the tower was found as 67°45'. If the RL of the instrument axis was 1200.00 m, calculate the clear height (plinth to top) of the tower.

## Short answer

Take zenith angles as elevations 2°15', 1°30', and 22°15'. D = 3/(tan2°15' - tan1°30') = 228.934 m. Plinth RL = 1203.995 m, top RL = 1293.660 m, so clear height = 89.66 m.

## Answer

The vane is held at the plinth of the tower and sighted with the transit theodolite. Readings of 5 m and 2 m are heights of the vane above the plinth. RL of instrument axis = 1200.00 m. Let $D$ be the horizontal distance from the instrument to the tower.

```text
                    * top (22°15')
                  / |
   axis ----------/-* 5 m  (2°15')
                    \* 2 m  (1°30')
                    |_ plinth
   |<------ D ----->|
```

#### Angles of elevation

- $\alpha_1 = 90^\circ - 87^\circ45' = 2^\circ15'$ (vane at 5 m)
- $\alpha_2 = 90^\circ - 88^\circ30' = 1^\circ30'$ (vane at 2 m)
- $\alpha_3 = 90^\circ - 67^\circ45' = 22^\circ15'$ (top)

#### Horizontal distance

The two vane positions are 3 m apart on the same vertical line:

$$
D(\tan\alpha_1 - \tan\alpha_2) = 5 - 2
$$

$$
D = \frac{3}{\tan 2^\circ15' - \tan 1^\circ30'} = \frac{3}{0.03929 - 0.02619} = 228.934\ \text{m}
$$

#### RL of plinth and top

- Plinth: $1200.000 + D\tan\alpha_1 - 5 = 1200.000 + 8.995 - 5 = 1203.995$ m
- Top: $1200.000 + D\tan\alpha_3 = 1200.000 + 93.660 = 1293.660$ m

#### Clear height

$$
H = 1293.660 - 1203.995 = 89.665\ \text{m}
$$

**Answer: clear height of the flood light tower = 89.66 m.**

## Check yourself

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

1. The zenith angle to the top of the tower is 67°45'. What elevation angle is used?
   A. 67°45'
   B. 157°45'
   C. 23°15'
   D. 22°15'
   Answer: D. Elevation = 90° - 67°45' = 22°15'. (Key point: Elevation of the top)

2. Which relation gives the horizontal distance D to the tower?
   A. D(tan 2°15' + tan 1°30') = 7
   B. D(tan 2°15' - tan 1°30') = 5 - 2
   C. D tan 22°15' = 5 - 2
   D. D(tan 2°15' - tan 1°30') = 5 + 2
   Answer: B. The vane positions are 3 m apart on one vertical line, so D(tan a1 - tan a2) = 3. (Key point: Finding D)

3. Why is 5 m subtracted in: plinth RL = 1200 + D tan 2°15' - 5?
   A. The vane is 5 m below the plinth
   B. A curvature correction of 5 m is needed
   C. The vane at 5 m is above the plinth, so the plinth is 5 m below the sighted level
   D. The instrument axis is 5 m above the plinth
   Answer: C. The sight hits a point 5 m above the plinth, so the plinth RL is 5 m lower than the sighted level. (Key point: Plinth RL)

4. Which gives the clear height of the tower?
   A. Top RL - plinth RL = 1293.660 - 1203.995 = 89.665 m
   B. Instrument axis to top: 93.660 m
   C. Plinth to axis: 3.995 m
   D. D - 5 = 223.934 m
   Answer: A. Clear height is plinth to top. 93.660 m is only the height above the instrument axis. (Key point: Clear height)

## Related in Surveying II

- [Clear height of flood light tower (stadium)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aflood-light-tower-clear-height-stadium) — Surveying II, chapter 3

## Related in other subjects

- [Theodolite traverse ABCDA with missing VCR (68°30′10″)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atheodolite-traverse-abcda-hcr-vcr-68-30-10) — Surveying I, chapter 7