---
title: "Tape standardized in catenary: 24 m span"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-catenary-span-24m"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-catenary-span-24m
site: BCT Engineering
---

# Tape standardized in catenary: 24 m span

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2078 Bhadra, 5 marks
- Video: [Common catenary numerical](https://www.youtube.com/watch?v=AvLhrQbVfDQ) — MD EDUCATIONAL INSTITUTE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/tape-catenary-span-24m
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A tape of nominal length 30 m is standardized in catenary at 50 N tension and found to be 29.8940 m. If the mass of the tape is 0.015 kg/m, calculate the horizontal length of a span recorded as 24 m. (Note: calculation should be done taking at least 4 decimal places.)

## Short answer

Add sag at standardization (0.009744 m) to 29.8940 m to obtain true tape 29.9037 m, then apply tape length correction -0.0770 m and sag correction -0.0050 m. Horizontal length is 23.9180 m.

## Answer

**Method.** A tape standardised in catenary reads 29.8940 m for its nominal 30 m when hanging at 50 N. To find the true tape length, the sag included in this standard value is added back. Then the 24 m span is corrected for tape length and for its own sag. The field pull is taken as 50 N (the standard pull).

#### Data
$w = 0.015 \times 9.81 = 0.14715$ N/m, $P = 50$ N, $L = 24$ m.

#### 1. Sag at standardisation (30 m)
$$
\frac{w^{2}\,l^{3}}{24\,P^{2}} = \frac{0.14715^{2}\times 30^{3}}{24\times 50^{2}} = 0.009744\ \text{m}
$$
True tape length $l_0 = 29.8940 + 0.009744 = 29.9037$ m.

#### 2. Tape length correction
$$
C_a = 24\times\frac{29.9037 - 30}{30} = -0.0770\ \text{m}
$$

#### 3. Sag correction for the 24 m span
$$
C_s = -\frac{0.14715^{2}\times 24^{3}}{24\times 50^{2}} = -0.0050\ \text{m}
$$

#### 4. Horizontal length
$$
D = 24 - 0.0770 - 0.0050 = 23.9180\ \text{m}
$$

**Answer: horizontal length of the span $= 23.9180$ m.**

## Check yourself

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

1. Tape reads 29.8940 m in catenary. The sag at standardisation is 0.009744 m. The flat length is
   A. 29.9037 m
   B. 29.8843 m
   C. 30.0000 m
   D. 29.8940 m
   Answer: A. Add the sag back: 29.8940 + 0.009744 = 29.9037 m. (Key point: True tape length)

2. C_a for the 24 m span is
   A. -0.0770 m
   B. +0.0770 m
   C. -0.0050 m
   D. -0.77 m
   Answer: A. 24 x (29.9037 - 30)/30 = -0.0770 m; the tape is short. (Key point: Tape length correction)

3. Sag correction for the 24 m span is about
   A. -0.0050 m
   B. +0.0050 m
   C. -0.0290 m
   D. -0.0097 m
   Answer: A. 0.14715^2 x 24^3 / (24 x 2500) = 0.0050 m, subtracted. (Key point: Sag for 24 m span)

4. Horizontal length is
   A. 23.9180 m
   B. 24.0820 m
   C. 23.9230 m
   D. 23.9950 m
   Answer: A. 24 - 0.0770 - 0.0050 = 23.9180 m. (Key point: Final length)

## Close variants set in other exams

- [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×

## Related in Surveying I

- [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 30 m span, 50 m tape](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-50m-span-30m-26c) — 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
- [Horizontal length for 120.0 m recorded](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-29-995-120m) — 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
- [Tape correction for 24.726 m distance](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-24-726-m-catenary) — Surveying I, chapter 2, asked 2×