---
title: "Correct length of 1350 m line"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-1350m-19-985"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-1350m-19-985
site: BCT Engineering
---

# Correct length of 1350 m line

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2068 Chaitra, 8 marks
- Video: [Correction Due to Incorrect Tape Length | Taping Corrections](https://www.youtube.com/watch?v=EpKIvtmE3Dc) — Jonas' Class Notes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/tape-1350m-19-985
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A 30 m steel tape standardized in fully supported condition at a temperature of 20°C and pull of 100 N was found to be 19.985 m. This tape was used to measure a line under a pull of 120 N and a mean temperature of 17°C and was found to be 1350 m long. Throughout the measurement, the tape was used in catenary condition. Find the correct length of the line. Take weight of steel as 0.081 N/cm³, the weight of tape as 11.775 N, $E = 2.10\times10^{5}$ N/mm² and $\alpha = 11\times10^{-6}$/°C.

## Short answer

Assuming standard length 29.985 m, apply tape length correction (-0.6750 m), temperature (-0.0446 m), pull (+0.0265 m), and sag correction for 45 spans (-0.5416 m). Correct length: 1348.765 m.

## Answer

**Reading of the data.** The tape is stated as "30 m" but the standard length is printed "19.985 m". This is taken as **29.985 m** (a 30 m tape 0.015 m short). The tape is fully supported at standardisation and used in catenary in 30 m spans, so 1350 m $= 45$ tapes.

#### Data
- $W = 11.775$ N for 30 m, $w = 0.3925$ N/m; density of steel $0.081$ N/cm$^3$
- $A = \dfrac{W}{0.081\times 3000} = \dfrac{11.775}{243} = 0.04846$ cm$^2 = 4.846$ mm$^2$
- $E = 2.1\times 10^{5}$ N/mm$^2$, so $AE = 4.846\times 2.1\times 10^{5} = 1.0176\times 10^{6}$ N
- Standard: 29.985 m at $T_0 = 20^\circ$C and $P_0 = 100$ N; field: $T = 17^\circ$C, $P = 120$ N, $L = 1350$ m
- $\alpha = 11\times 10^{-6}$/$^\circ$C

#### 1. Tape length
$$
C_a = 1350\times\frac{29.985 - 30}{30} = -0.6750\ \text{m}
$$

#### 2. Temperature
$$
C_t = 11\times 10^{-6}\times(17 - 20)\times 1350 = -0.0446\ \text{m}
$$

#### 3. Pull
$$
C_p = \frac{(120 - 100)\times 1350}{1.0176\times 10^{6}} = +0.0265\ \text{m}
$$

#### 4. Sag
Per tape: $\dfrac{11.775^{2}\times 30}{24\times 120^{2}} = 0.012036$ m; $C_s = -45\times 0.012036 = -0.5416$ m.

#### 5. Correct length
$$
L = 1350 - 0.6750 - 0.0446 + 0.0265 - 0.5416 = 1348.7654\ \text{m}
$$

**Answer: correct length of the line $= 1348.765$ m** (with the standard length read as 29.985 m).

## Check yourself

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

1. Why is the temperature correction negative?
   A. Pull is high
   B. Field 17 C is below standard 20 C so the tape contracts
   C. Sag is large
   D. Tape is short
   Answer: B. Cold tape is short, reading too large. (Key point: Temperature sign)

2. Because the tape was standardised fully supported, the sag in the standard is
   A. added back
   B. subtracted
   C. not added back
   D. doubled
   Answer: C. Fully supported length has no sag. (Key point: Standard condition)

3. The sag is calculated for
   A. 1350 spans
   B. 30 spans
   C. 90 spans
   D. 45 spans of 30 m
   Answer: D. 1350/30 = 45. (Key point: Sag count)

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