---
title: "Correct length of 360 m line"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-360m-pull-5kg-supports-20m"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-360m-pull-5kg-supports-20m
site: BCT Engineering
---

# Correct length of 360 m line

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2075 Chaitra, 4 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-360m-pull-5kg-supports-20m
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A steel tape was standardized in catenary at 7 kg pull. A distance of 360 m was measured with this tape under a pull of 5 kg. Assuming that the tape was supported at every 20 m length, determine the correct length of the line if the weight of tape = 10 gm/m and cross sectional area of tape = 0.03 cm². Take $E = 210\times10^{3}$ N/mm².

## Short answer

Correct for pull change from 7 kg to 5 kg (-0.0112 m) and for the extra sag relative to the standard (-0.0118 m). Correct length of line is 359.977 m.

## Answer

**Assumptions.** The tape was standardised in catenary at 7 kg with the same support spacing (20 m) as in the field, so its stated length is correct for that condition. Only the changes (pull and sag) are corrected. $g = 9.81$ m/s$^2$.

#### Data
- $L = 360$ m; field pull $P = 5$ kg; standard pull $P_0 = 7$ kg
- $w = 10$ g/m $= 0.01$ kg/m; $A = 0.03$ cm$^2$
- $E = 210\times 10^{3}$ N/mm$^2 = 2.1\times 10^{7}$ N/cm$^2 = 2.1407\times 10^{6}$ kg/cm$^2$
- Spans: $360/20 = 18$ spans of 20 m

#### 1. Pull correction
$$
C_p = \frac{(P - P_0)\,L}{A\,E} = \frac{(5 - 7)\times 360}{0.03\times 2.1407\times 10^{6}} = -0.0112\ \text{m}
$$

#### 2. Sag correction (difference between the sag at 5 kg and that already in the standard at 7 kg)
Sag at 5 kg: $18\times\dfrac{0.01^{2}\times 20^{3}}{24\times 5^{2}} = 0.02400$ m.
Sag at 7 kg: $18\times\dfrac{0.01^{2}\times 20^{3}}{24\times 7^{2}} = 0.01224$ m.

$$
C_s = -(0.02400 - 0.01224) = -0.0118\ \text{m}
$$

#### 3. Correct length
$$
L_{true} = 360 - 0.0112 - 0.0118 = 359.9770\ \text{m}
$$

(If the sag in standardisation is ignored, so that the full sag of $-0.0240$ m is applied, the length is 359.9648 m.)

**Answer: correct length of the line $= 359.977$ m.**

## Check yourself

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

1. Why is the pull correction negative?
   A. Field pull (5 kg) is less than standard pull (7 kg), so the tape is stretched less and reads long
   B. Field pull is greater than standard
   C. The tape is too hot
   D. Because sag is present
   Answer: A. C_p = (P - P0)L/AE with P < P0 is negative. (Key point: Pull correction sign)

2. How is the sag correction found here?
   A. Only the sag at 5 kg
   B. Sag at 5 kg minus the sag already in the 7 kg standard, applied as negative
   C. Only the sag at 7 kg
   D. The sum of both sags
   Answer: B. Standardisation in catenary at 7 kg already includes that sag, so only the difference is corrected. (Key point: Sag difference)

3. How many 20 m spans are used for the sag calculation?
   A. 20
   B. 36
   C. 18
   D. 9
   Answer: C. 360/20 = 18 spans. (Key point: Number of spans)

4. Per span, the sag correction is proportional to
   A. w l / (24 P)
   B. w^2 l / (8 P)
   C. w l^2 / (2 P^2)
   D. w^2 l^3 / (24 P^2)
   Answer: D. The standard catenary correction is w^2 l^3/24P^2. (Key point: Sag formula)

## Related in Surveying I

- [Tape corrections for 86 m, 10 kg pull](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-20m-86m-pull-10kg) — Surveying I, chapter 2, asked 3×
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