---
title: "Correction per tape length (kgf data, 61)"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-base-line-correction-per-tape-61"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-base-line-correction-per-tape-61
site: BCT Engineering
---

# Correction per tape length (kgf data, 61)

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2061 Baisakh, 10 marks
- 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-correction-per-tape-61
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

To measure a base line, a steel tape 30 m long supported at two ends, standardised at 15°C with a pull of 100 N (or 10 kgf), was used. Find the correction per tape length (fully support condition), if the temperature at the time of measurement was 20°C and the pull exerted was 160 N (or 16 kgf). Weight of 1 cubic cm of steel is 0.0786 N (or 0.00786 kgf), weight of the tape = 8 N (or 0.8 kgf). $E = 2.1\times10^{5}$ kg/sq.cm. Coefficient of expansion of the tape per 1°C = $7.1\times10^{-7}$.

## Short answer

Fully supported: no sag. Temperature correction = +0.000107 m, pull correction = +0.02526 m. Correction per tape length = +0.0254 m (true length 30.0254 m).

## Answer

**Reading.** "Fully support condition" means the tape rests on the ground throughout the measurement, so there is no sag correction. The correction per tape length is made up of the temperature and pull corrections.

#### Data
- $l = 30$ m; $T_0 = 15^\circ$C, $T = 20^\circ$C; $P_0 = 10$ kgf, $P = 16$ kgf
- $W = 0.8$ kgf; density of steel $= 0.00786$ kgf/cm$^3$; $E = 2.1\times 10^{5}$ kgf/cm$^2$ (as given); $\alpha = 7.1\times 10^{-7}$/$^\circ$C

#### Cross-sectional area
$$
A = \frac{W}{\text{density}\times l} = \frac{0.8}{0.00786\times 3000} = 0.03393\ \text{cm}^2
$$
$AE = 0.03393\times 2.1\times 10^{5} = 7124.7$ kgf.

#### 1. Temperature correction
$$
C_t = 7.1\times 10^{-7}\times(20 - 15)\times 30 = +0.000107\ \text{m}
$$

#### 2. Pull correction
$$
C_p = \frac{(16 - 10)\times 30}{7124.7} = +0.02526\ \text{m}
$$

#### 3. Correction per tape length
$$
C = 0.000107 + 0.02526 = +0.02537\ \text{m}
$$

**Answer: correction per tape length $= +0.0254$ m (the true length of one tape length is 30.0254 m).**

**Note on $E$.** The usual value for steel is $2.1\times 10^{5}$ N/mm$^2$ $= 2.1\times 10^{6}$ kgf/cm$^2$, and the printed unit may be a misprint for N/mm$^2$. With that value, $AE = 71{,}247$ kgf, $C_p = +0.00253$ m and the correction per tape length is $+0.00263$ m.

(If the tape were suspended at its ends in the field, a further sag correction of $-\dfrac{W^{2}l}{24P^{2}} = -0.0031$ m would apply.)

## Check yourself

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

1. Fully supported means
   A. no sag correction
   B. sag only
   C. no pull
   D. no temperature
   Answer: A. Rests on ground. (Key point: Fully supported)

2. A computed from
   A. W x length
   B. W/(unit weight x length)
   C. unit weight/W
   D. W only
   Answer: B. Volume/length. (Key point: Area)

3. Largest correction is
   A. temperature
   B. sag
   C. pull
   D. slope
   Answer: C. +0.02526 vs 0.000107. (Key point: Dominant)

4. Correction per tape
   A. -0.0254
   B. 0.0001
   C. -0.0031
   D. +0.0254 m
   Answer: D. Sum. (Key point: Total)

## Close variants set in other exams

- [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×

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