---
title: "Horizontal length of 30 m span, 50 m tape"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-50m-span-30m-26c"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-50m-span-30m-26c
site: BCT Engineering
---

# Horizontal length of 30 m span, 50 m tape

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2062 Baisakh, 10 marks
- Video: [Surveying | Corrections in Tape Part 1 of 2](https://www.youtube.com/watch?v=MdHdzGSSz7Q) — the Engineer's Insight
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/tape-50m-span-30m-26c
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A 50 m steel tape weighing 0.68 kg was standardized on the catenary and found to have length 49.996 m at 20°C, tension 5 kg. Calculate the horizontal length of a 30 m span at 26°C.

## Short answer

The standardised catenary length (49.996 m) is corrected for sag (+0.03853 m) to obtain true tape length 50.03453 m. For the 30 m span, add tape-length correction +0.0207 m, temperature correction +0.0022 m, and sag correction -0.0083 m. Horizontal length = 30.0146 m.

## Answer

**Assumptions.** The tape is standardised in catenary (50 m span), so the sag in its standard length is added back. The 30 m span is measured in catenary at the same tension, 5 kg. The coefficient of expansion is not given, so the steel value $\alpha = 12\times 10^{-6}$/$^\circ$C is assumed.

#### Data
$W = 0.68$ kg for 50 m, so $w = 0.0136$ kg/m; $P = 5$ kg; standard length $= 49.996$ m at $20^\circ$C; $T = 26^\circ$C; span $L = 30$ m.

#### 1. Tape length
Sag at standardisation: $\dfrac{0.0136^{2}\times 50^{3}}{24\times 5^{2}} = 0.03853$ m.
True tape length $= 49.996 + 0.03853 = 50.03453$ m.

$$
C_a = 30\times\frac{50.03453 - 50}{50} = +0.0207\ \text{m}
$$

#### 2. Temperature
$$
C_t = 12\times 10^{-6}\times 6\times 30 = +0.0022\ \text{m}
$$

#### 3. Sag for 30 m
$$
C_s = -\frac{0.0136^{2}\times 30^{3}}{24\times 5^{2}} = -0.0083\ \text{m}
$$

#### 4. Horizontal length
$$
D = 30 + 0.0207 + 0.0022 - 0.0083 = 30.0146\ \text{m}
$$

**Answer: horizontal length of the 30 m span $= 30.0146$ m.**

## Check yourself

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

1. The standard sag added uses
   A. 50 m span
   B. 30 m span
   C. 26 m
   D. 20 m
   Answer: A. Standardised in 50 m catenary. (Key point: Tape length basis)

2. Since alpha is missing the answer assumes
   A. zero
   B. 12e-6 per C for steel
   C. 2e-6
   D. 1e-5 per F
   Answer: B. Stated. (Key point: Alpha)

3. The field sag correction uses
   A. 50 m
   B. 0
   C. 30 m span
   D. 26 m
   Answer: C. Span of 30 m. (Key point: Field sag)

4. Net horizontal length is
   A. 29.9
   B. 30.0354
   C. 30.0
   D. 30.0146 m
   Answer: D. 30+0.0207+0.0022-0.0083. (Key point: Answer)

## Related in Surveying I

- [Tape standardized in catenary: 20 m span](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-catenary-span-20m) — Surveying I, chapter 2, asked 2×
- [Tape standardized in catenary: 24 m span](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-catenary-span-24m) — Surveying I, chapter 2, asked 2×
- [Tape standardized in catenary: 23 m span](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-catenary-span-23m) — Surveying I, chapter 2
- [Horizontal length of 16 m, 29.8850 m tape](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-29-8850-16m-fully-supported) — Surveying I, chapter 2
- [Correct length of base from sections](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-50m-horizontal-base-sections) — Surveying I, chapter 2
- [True length: tape found 2.5 cm too long](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-825-plus-750-m) — Surveying I, chapter 2