---
title: "Tape standardized in catenary: 20 m span"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-catenary-span-20m"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-catenary-span-20m
site: BCT Engineering
---

# Tape standardized in catenary: 20 m span

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2081 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-20m
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A nominal length of tape is 30 m is standardized in catenary at 50 N tension and found to be 29.895 m. If the mass of the tape is 0.015 kg/m, calculate the horizontal length of a span recorded as 20 m. (Calculation must be done taking at least 4 digits after decimal value).

## Short answer

The tape’s flat length is 29.895 + sag correction (0.0097 m) = 29.9047 m. For a 20 m span: tape-length correction = −0.0635 m, sag correction = −0.0029 m, giving horizontal distance 19.9336 m.

## Answer

**Reading of the data.** A tape standardised in catenary reads 29.895 m between its end graduations when hanging in catenary at 50 N, so it is too short by 0.105 m as a hanging tape. The sag effect included in this standard must be accounted for to get the true length of the tape, then the 20 m span is corrected for its own sag. The field tension is taken equal to the standard tension, 50 N (not stated).

#### Data
- Mass per metre $= 0.015$ kg/m, so weight per metre $w = 0.015 \times 9.81 = 0.14715$ N/m
- Pull $P = 50$ N, standard catenary length of 30 m tape $= 29.895$ m
- Recorded span $L = 20$ m

#### Step 1: Sag of the 30 m tape at standardisation
$$
C_{s,30} = \frac{w^{2}\,l^{3}}{24\,P^{2}} = \frac{0.14715^{2}\times 30^{3}}{24\times 50^{2}} = 0.009744\ \text{m}
$$

Length of tape fully supported (flat):
$$
l_0 = 29.895 + 0.009744 = 29.9047\ \text{m}
$$

#### Step 2: Correction for tape length (for 20 m)
$$
C_a = 20\times\frac{29.9047 - 30}{30} = -0.0635\ \text{m}
$$

#### Step 3: Sag correction for the 20 m span
$$
C_s = -\frac{w^{2}\,l^{3}}{24\,P^{2}} = -\frac{0.14715^{2}\times 20^{3}}{24\times 50^{2}} = -0.0029\ \text{m}
$$

#### Step 4: Horizontal length
$$
D = 20 - 0.0635 - 0.0029 = 19.9336\ \text{m}
$$

**Answer: horizontal length of the span $= 19.9336$ m.**

## Check yourself

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

1. A 30 m tape reads 29.895 m in catenary. How is its fully supported length found?
   A. Subtract the standardisation sag
   B. Add the standardisation sag (0.009744 m) to 29.895 m
   C. Multiply by 30
   D. Use the 20 m sag
   Answer: B. The catenary standard already includes sag, so adding it back gives 29.9047 m. (Key point: Flat length of tape)

2. What is w for mass 0.015 kg/m?
   A. 0.015 N/m
   B. 0.14715 N/m
   C. 1.4715 N/m
   D. 0.0015 N/m
   Answer: B. w = 0.015 x 9.81 = 0.14715 N/m. (Key point: Weight per metre)

3. Why is C_a = -0.0635 m negative for the 20 m span?
   A. The tape (29.9047 m) is shorter than 30 m
   B. The tape is longer than 30 m
   C. Slope
   D. Temperature
   Answer: A. A short tape measures a value that is too large, so subtract. (Key point: Tape length correction)

4. Horizontal length of the 20 m span is
   A. 20.0664 m
   B. 19.9336 m
   C. 19.9971 m
   D. 19.8706 m
   Answer: B. 20 - 0.0635 - 0.0029 = 19.9336 m. (Key point: Final length)

## Close variants set in other exams

- [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×

## 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
- [Correction per tape length (kg data, 58)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-base-line-correction-per-tape-58) — Surveying I, chapter 2, asked 2×