---
title: "Sag correction for 76.35 m line"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-sag-correction-76-35m"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-sag-correction-76-35m
site: BCT Engineering
---

# Sag correction for 76.35 m line

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2067 Asar, 6 marks
- Video: [Sag Correction | Numerical Problems | Surveying | Chain & Tape Correction |Civil Engineering Diploma](https://www.youtube.com/watch?v=ZIhhCn00ItE) — Polytechnic Class (READBY) 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/tape-sag-correction-76-35m
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A steel tape 30 m long weighs 0.7 kg and is used with supports at the ends only. A line is measured in three segments using a 5 kg pull and the length was recorded as 76.35 m. What is the length of the line corrected for sag?

## Short answer

The line is measured in three equal spans of 25.45 m. Sag correction per span is -0.01496 m; total sag correction -0.0449 m, so corrected length is 76.3051 m.

## Answer

#### Data
- Weight of 30 m tape $W = 0.7$ kg, so $w = 0.7/30 = 0.02333$ kg/m
- Pull $P = 5$ kg; recorded length $= 76.35$ m in three segments, taken as three equal spans of $76.35/3 = 25.45$ m

#### Sag correction per span
$$
C_s = -\frac{w^{2}\,l^{3}}{24\,P^{2}} = -\frac{0.02333^{2}\times 25.45^{3}}{24\times 5^{2}} = -0.01496\ \text{m}
$$

#### For the three spans
$$
C_s = -3\times 0.01496 = -0.0449\ \text{m}
$$

#### Corrected length
$$
L = 76.35 - 0.0449 = 76.3051\ \text{m}
$$

**Answer: length of the line corrected for sag $= 76.3051$ m.**

(If the three segments are instead read as two full 30 m tape lengths plus 16.35 m, the sag is $2\times\dfrac{0.7^{2}\times 30}{24\times 5^{2}} + \dfrac{0.02333^{2}\times 16.35^{3}}{24\times 5^{2}} = 0.0490 + 0.0040 = 0.0530$ m, and the corrected length is $76.297$ m.)

## Check yourself

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

1. What span length is used?
   A. 76.35/3 = 25.45 m
   B. 30 m
   C. 76.35 m
   D. 25 m
   Answer: A. Three equal segments. (Key point: Span length)

2. Sag correction per span is
   A. -w l/(24P)
   B. -w^2 l^3/(24 P^2)
   C. +w^2 l^3/24P^2
   D. -w^2 l/(8P^2)
   Answer: B. Standard formula, subtracted. (Key point: Formula)

3. w equals
   A. 0.7 kg/m
   B. 0.7 x 30
   C. 0.7/30 = 0.02333 kg/m
   D. 5/30
   Answer: C. Total weight divided by length. (Key point: Weight per metre)

4. The corrected length is
   A. 76.3949 m
   B. 76.35 m
   C. 76.00 m
   D. 76.3051 m
   Answer: D. 76.35 - 0.0449. (Key point: Result)

## Related in Surveying I

- [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×
- [True length of 780 m line, tape 0.693 kg](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-30m-780m-line) — Surveying I, chapter 2, asked 2×
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- [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

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