---
title: "Correct length of base from sections"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-50m-horizontal-base-sections"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-50m-horizontal-base-sections
site: BCT Engineering
---

# Correct length of base from sections

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2066 Jestha, 9 marks
- Video: [Surveying Tape Correction Numerical 1](https://www.youtube.com/watch?v=VKh33kzPV38) — Studyclass 247
- 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-horizontal-base-sections
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A steel tape weighing 0.68 kg was standardised on the flat and found to have length 49.996 m at 20°C, tension 5 kg. It was used in catenary at the same tension to measure a horizontal base, at average temperature 26°C, and the readings on the successive sections were 49.105, 49.373, 48.976, 49.817 and 34.353. What was the correct length of the line? Take $\alpha = 12\times10^{-6}$ per °C.

## Short answer

Total measured length 231.624 m. Corrections: standardization -0.0185 m, temperature +0.0167 m, sag per section totalling -0.1604 m. Correct length = 231.462 m.

## Answer

**Assumptions.** The tape (50 m) was standardised on the flat, and is used in catenary at the same tension of 5 kg, so there is no pull correction. Each reading is a separate suspended span. The total measured length is $49.105 + 49.373 + 48.976 + 49.817 + 34.353 = 231.624$ m.

#### Data
$W = 0.68$ kg for 50 m, so $w = 0.0136$ kg/m; $P = 5$ kg; standard length $= 49.996$ m at $T_0 = 20^\circ$C; $T = 26^\circ$C; $\alpha = 12\times 10^{-6}$/$^\circ$C.

#### 1. Tape length
$$
C_a = \frac{231.624\times(49.996 - 50)}{50} = -0.0185\ \text{m}
$$

#### 2. Temperature
$$
C_t = 12\times 10^{-6}\times 6\times 231.624 = +0.0167\ \text{m}
$$

#### 3. Sag (for each section, $C_s = -\dfrac{w^{2}\,l^{3}}{24\,P^{2}}$)

| Section length (m) | Sag correction (m) |
|---|---|
| 49.105 | $-0.03650$ |
| 49.373 | $-0.03710$ |
| 48.976 | $-0.03621$ |
| 49.817 | $-0.03811$ |
| 34.353 | $-0.01250$ |

Total $C_s = -0.1604$ m.

#### 4. Correct length
$$
L = 231.624 - 0.0185 + 0.0167 - 0.1604 = 231.4617\ \text{m}
$$

**Answer: correct length of the line $= 231.462$ m.**

## Check yourself

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

1. Why is there no pull correction?
   A. Used at the same tension as the standard
   B. Pull is ignored always
   C. Tape is rigid
   D. Sag cancels it
   Answer: A. P equals P0 gives zero. (Key point: No pull correction)

2. Sag is computed
   A. Once using the total
   B. separately for each section using its own length
   C. Using 50 m for all
   D. Using average
   Answer: B. Sag varies with l^3. (Key point: Sag per section)

3. Standard length 49.996 m means
   A. 4 mm long
   B. no effect
   C. tape is 4 mm short, correction negative
   D. positive correction
   Answer: C. Short tape over-reads. (Key point: Standardised on flat)

4. Temperature correction is
   A. negative
   B. zero
   C. uses 26 C directly
   D. +12e-6 x 6 x 231.624
   Answer: D. 6 C above standard expands the tape. (Key point: Temperature)

## Related in Surveying I

- [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
- [Correct base line length 1120 m](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-base-line-1120m-19-978) — Surveying I, chapter 2
- [Correct horizontal distance 630 m, tape 8.9 N](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-630m-8-9n-30-005) — Surveying I, chapter 2
- [Correct horizontal distance, 28.00 m measured](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-fully-support-28-00m) — Surveying I, chapter 2
- [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×
- [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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