---
title: "Correct base line length 1120 m"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-base-line-1120m-19-978"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-base-line-1120m-19-978
site: BCT Engineering
---

# Correct base line length 1120 m

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2069 Chaitra, 8 marks
- Video: [steel tap, base line correction, surveying solved problem, survey, civil engineering lecture, gate](https://www.youtube.com/watch?v=GeZgXct3xe8) — Civil Engineering Exam
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/tape-base-line-1120m-19-978
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A 20 m steel tape standardized in catenary at a temperature of 12.5°C and a pull of 100 N was found to be 19.978 m. This tape was used to measure a base line. Throughout the measurement the tape was used in catenary for each tape length. Find the correct length of the baseline if the temperature during measurement was 25°C and pull applied was 150 N. Weight of steel is 0.077 N/cm³. The weight of suspended tape was 7.85 N. Take $E = 2.10\times10^{5}$ N/mm² and $\alpha = 12\times10^{-6}$/°C. The measured base line distance was 1120 m.

## Short answer

True tape length is 19.98314 m after adding sag (0.005135 m) to 19.978 m. Correct 1120 m using tape length (-0.9444 m), temperature (+0.1680 m), pull (+0.0523 m), and sag corrections (-0.1278 m). Correct length: 1119.148 m.

## Answer

**Method.** The tape is standardised in catenary, so its true length is found by adding back the sag at standardisation. The tape is used in catenary (20 m spans), 1120 m $= 56$ tapes.

#### Data
- $W = 7.85$ N for 20 m, $w = 0.3925$ N/m; weight density of steel $= 0.077$ N/cm$^3$
- Area: $A = \dfrac{W}{0.077\times 2000} = \dfrac{7.85}{154} = 0.05097$ cm$^2 = 5.097$ mm$^2$
- $E = 2.1\times 10^{5}$ N/mm$^2$, so $AE = 5.097\times 2.1\times 10^{5} = 1.0705\times 10^{6}$ N
- Standard: 19.978 m, $T_0 = 12.5^\circ$C, $P_0 = 100$ N; field: $T = 25^\circ$C, $P = 150$ N
- $\alpha = 12\times 10^{-6}$/$^\circ$C

#### 1. Tape length
Sag at standardisation: $\dfrac{7.85^{2}\times 20}{24\times 100^{2}} = 0.005135$ m.
True tape length $= 19.978 + 0.005135 = 19.98314$ m.

$$
C_a = 1120\times\frac{19.98314 - 20}{20} = -0.9444\ \text{m}
$$

#### 2. Temperature
$$
C_t = 12\times 10^{-6}\times(25 - 12.5)\times 1120 = +0.1680\ \text{m}
$$

#### 3. Pull
$$
C_p = \frac{(150 - 100)\times 1120}{1.0705\times 10^{6}} = +0.0523\ \text{m}
$$

#### 4. Sag
Per tape: $\dfrac{7.85^{2}\times 20}{24\times 150^{2}} = 0.002282$ m; $C_s = -56\times 0.002282 = -0.1278$ m.

#### 5. Correct length
$$
L = 1120 - 0.9444 + 0.1680 + 0.0523 - 0.1278 = 1119.1481\ \text{m}
$$

**Answer: correct length of the base line $= 1119.148$ m.**

## Check yourself

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

1. How is the cross-section area found?
   A. A = W/(unit weight x length)
   B. A = W x length
   C. A = W/length only
   D. A = unit weight/W
   Answer: A. Volume = W/0.077, divided by length gives area. (Key point: Area from weight)

2. True tape length is
   A. 19.978 m
   B. 19.978 + sag at standardisation = 19.98314 m
   C. 19.978 - 0.005135
   D. 20 m
   Answer: B. Catenary standardisation sag is added back. (Key point: Tape length)

3. The tape-length correction is negative because
   A. the tape is longer
   B. the pull is high
   C. the true tape is shorter than 20 m
   D. the temperature is high
   Answer: C. Short tape reads too much. (Key point: Tape length sign)

4. Which temperature difference is used?
   A. 25 - 20
   B. 25 - 0
   C. 12.5 C below
   D. 25 - 12.5 = 12.5 C
   Answer: D. Standard temperature is 12.5 C. (Key point: Temperature)

## Related in Surveying I

- [Correct length of 1350 m line](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-1350m-19-985) — 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×
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- [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
- [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×