---
title: "True length of 1020 m line, 20 m tape"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-20m-1020m-line"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-20m-1020m-line
site: BCT Engineering
---

# True length of 1020 m line, 20 m tape

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2074 Asoj, 8 marks
- Video: [Find the correct length of the line|correction for chain length|length correction](https://www.youtube.com/watch?v=zrthvZDi3VM) — Hamro Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/tape-20m-1020m-line
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A steel tape was exactly 20 m long at 20°C when supported throughout its length under a pull of 100 N. A line was measured with this tape under a pull of 160 N at a mean temperature of 30°C and found to be 1020 m long. The cross sectional area of tape is 0.03 cm², weight per metre length is 24 gm, coefficient of thermal expansion for steel is $11\times10^{-6}$/°C and modulus of elasticity of steel is $2.1\times10^{6}$ kg/cm². Find the true length of the line if the tape was supported at every 10 m during measurement.

## Short answer

Compute temperature correction (+0.1122 m), pull correction (+0.0990 m), and sag correction for 102 spans of 10 m (-0.0092 m). Add to 1020 m to get true length 1020.202 m.

## Answer

**Assumptions.** Standardised fully supported at $20^\circ$C and 100 N, so no standard sag is added back. $g = 9.81$ m/s$^2$.

#### Data
- $L = 1020$ m, supports every 10 m (102 spans of 10 m)
- $P_0 = 100$ N, $P = 160$ N; $T_0 = 20^\circ$C, $T = 30^\circ$C
- $A = 0.03$ cm$^2$, $E = 2.1\times 10^{6}$ kg/cm$^2$, so $AE = 0.03\times 2.1\times 10^{6}\times 9.81 = 618{,}030$ N
- $\alpha = 11\times 10^{-6}$/$^\circ$C; $w = 24$ g/m $= 0.024\times 9.81 = 0.23544$ N/m

#### 1. Temperature correction
$$
C_t = 11\times 10^{-6}\times(30 - 20)\times 1020 = +0.1122\ \text{m}
$$

#### 2. Pull correction
$$
C_p = \frac{(160 - 100)\times 1020}{618{,}030} = +0.0990\ \text{m}
$$

#### 3. Sag correction
Per 10 m span: $\dfrac{w^{2}\,l^{3}}{24\,P^{2}} = \dfrac{0.23544^{2}\times 10^{3}}{24\times 160^{2}} = 0.0000902$ m.

$$
C_s = -102\times 0.0000902 = -0.0092\ \text{m}
$$

#### 4. True length
$$
L_{true} = 1020 + 0.1122 + 0.0990 - 0.0092 = 1020.2020\ \text{m}
$$

**Answer: true length of the line $= 1020.202$ m.**

## Check yourself

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

1. Why is the pull correction positive?
   A. The field pull (160 N) exceeds the standard pull (100 N), so the tape lengthens
   B. The tape is supported
   C. The temperature is higher
   D. The weight is large
   Answer: A. C_p = (P-P0)L/AE is positive. (Key point: Pull correction)

2. How is AE obtained from E in kg/cm^2?
   A. Divide by 9.81
   B. Multiply A x E by 9.81 to convert kg to N
   C. Multiply by 100
   D. Leave in kg and use with N
   Answer: B. Pulls are in newtons, so kg-force must be converted. (Key point: AE in newtons)

3. How many spans are used for the sag correction?
   A. 51
   B. 102 but each of 20 m
   C. 102 (1020/10)
   D. 1020
   Answer: C. Supports every 10 m give 102 spans. (Key point: Number of spans)

4. Why is no sag added back to the standard length?
   A. Because pull is large
   B. Because sag is negligible by definition
   C. Because it was standardised in catenary
   D. The tape was standardised fully supported, so there is no sag in its length
   Answer: D. Fully supported standardisation contains no sag. (Key point: No standard sag)

## Close variants set in other exams

- [True length of 680 m line, 20 m tape](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-20m-680m-line) — Surveying I, chapter 2, asked 3×

## Related in Surveying I

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