---
title: "True length of 358.650 m line"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-358-650-m-line"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-358-650-m-line
site: BCT Engineering
---

# True length of 358.650 m line

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2081 Baisakh, 6 marks
- Video: [True Length Calculation of chain in surveying when chain short & long | Numerical Solution](https://www.youtube.com/watch?v=sDa_gGmN_HM) — Learn Civil Tech & Tricks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/tape-358-650-m-line
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

Determine the correct length of a line as per given conditions. A 30 m steel tape is of standard length under a pull of 5.5 kg when supported throughout its entire length. The tape weighs 0.05 kg/cm [?], has a cross sectional area of 0.04 cm², and its modulus of elasticity is $2.10\times10^{6}$ kg/cm². The tape was used in the field and the measured distance is 358.650 m. At the time the measurement was made, the constant pull applied was 8 kg with the tape supported only at its endpoints. Assume all full tape lengths except in the last one.

## Short answer

Compute pull correction +0.0107 m and sag correction for 11 full 30 m tapes and a 28.650 m last span, giving -0.5217 m. True length is 358.139 m.

## Answer

**Reading of the doubtful value.** The scanned tape weight "0.05 kg/cm" is taken as **0.05 kg per metre** (a 30 m tape then weighs 1.5 kg, a realistic value for steel tape of 0.04 cm$^2$ section). The other values are as printed.

#### Data
- $l = 30$ m; standard pull $P_0 = 5.5$ kg (tape fully supported); $A = 0.04$ cm$^2$; $E = 2.1\times 10^{6}$ kg/cm$^2$
- Measured distance $L = 358.650$ m; field pull $P = 8$ kg; tape supported at the ends only (catenary)
- $w = 0.05$ kg/m, so $W = 0.05\times 30 = 1.5$ kg
- Number of tapes: 11 full tapes (330 m) + a last part of 28.650 m. No temperature or tape-length error is given, so only pull and sag are applied.

#### 1. Correction for pull
$$
C_p = \frac{(P - P_0)\,L}{A\,E} = \frac{(8 - 5.5)\times 358.650}{0.04\times 2.1\times 10^{6}} = +0.0107\ \text{m}
$$

#### 2. Correction for sag
Full tape: $C_s = -\dfrac{W^{2}\,l}{24\,P^{2}} = -\dfrac{1.5^{2}\times 30}{24\times 8^{2}} = -0.043945$ m per tape.

11 full tapes: $-11\times 0.043945 = -0.48340$ m.

Last length of 28.650 m: $C_s = -\dfrac{w^{2}\,l^{3}}{24\,P^{2}} = -\dfrac{0.05^{2}\times 28.65^{3}}{24\times 8^{2}} = -0.03828$ m.

Total sag correction $= -0.48340 - 0.03828 = -0.5217$ m.

#### 3. True length
$$
L_{true} = 358.650 + 0.0107 - 0.5217 = 358.1390\ \text{m}
$$

**Answer: correct length of the line $= 358.139$ m.**

## Check yourself

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

1. Pull 8 kg exceeds standard 5.5 kg. The pull correction is
   A. Negative
   B. Positive, +0.0107 m
   C. Zero
   D. Negative 0.0107 m
   Answer: B. More pull stretches the tape, so C_p = (P - P0)L/(AE) is added. (Key point: Pull correction sign)

2. Which formula gives sag per full 30 m tape?
   A. -W^2 l / (24 P^2)
   B. +W l / P
   C. -L h / R
   D. -h^2 / 2L
   Answer: A. With W = 1.5 kg, l = 30 m, P = 8 kg gives -0.043945 m per tape. (Key point: Sag per full tape)

3. How is the last 28.650 m length handled?
   A. Ignored
   B. Sag computed separately with w^2 l^3 / (24 P^2) for 28.65 m
   C. Same as full tape
   D. Multiplied by 11
   Answer: B. A partial span uses its own length: -0.03828 m. (Key point: Partial last tape)

4. The final corrected length is
   A. 359.18 m
   B. 358.139 m
   C. 358.650 m
   D. 357.77 m
   Answer: B. 358.650 + 0.0107 - 0.5217 = 358.139 m. (Key point: True length)

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