---
title: "True length of AB: 114.095 m, 12°C"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-suspended-ab-114-095-70n"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-suspended-ab-114-095-70n
site: BCT Engineering
---

# True length of AB: 114.095 m, 12°C

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2079 Bhadra, 6 marks
- Video: [Correction in chain surveying|Calculate true length of the two poins|Diploma In Civil Engineering|](https://www.youtube.com/watch?v=2sg7MC5n6q4) — 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-suspended-ab-114-095-70n
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A 30 m steel tape measured 30.015 m, when standardized fully supported condition under a 70 N pull at a temperature of 20°C and had a cross-sectional area of 0.028 cm². The tape weighed 0.9 kg (9 N). What is the true length of the recorded distance AB for the following conditions? Assume all full tape length except in the last one. Take $\alpha = 1.15\times10^{-5}$/°C, $E = 2.11\times10^{6}$ N/cm².

| Item | Value |
|---|---|
| Recorded distance AB | 114.095 m |
| Average temperature | 12°C |
| Condition of support | Suspended |
| Tension | 100 N |
| Elevation difference / 100 m | 2.5 m |

## Short answer

Using weight 9 N per 30 m tape, corrections: standardization +0.0570 m, temperature -0.0105 m, pull +0.0579 m, sag -0.0356 m, slope -0.0357 m. True length = 114.128 m.

## Answer

**Assumptions.** The tape is standardised fully supported; used suspended in 3 full tapes of 30 m and a last part of 24.095 m. The slope is uniform at 2.5 m per 100 m.

#### Data
- $W = 9$ N per 30 m, so $w = 0.3$ N/m
- $A = 0.028$ cm$^2$, $E = 2.11\times 10^{6}$ N/cm$^2$, so $AE = 59{,}080$ N
- Standard: 30.015 m at $20^\circ$C, $P_0 = 70$ N; field: $T = 12^\circ$C, $P = 100$ N; $\alpha = 1.15\times 10^{-5}$/$^\circ$C
- $L = 114.095$ m

#### 1. Tape length
$$
C_a = \frac{114.095\times 0.015}{30} = +0.0570\ \text{m}
$$

#### 2. Temperature
$$
C_t = 1.15\times 10^{-5}\times(12 - 20)\times 114.095 = -0.0105\ \text{m}
$$

#### 3. Pull
$$
C_p = \frac{(100 - 70)\times 114.095}{59{,}080} = +0.0579\ \text{m}
$$

#### 4. Sag
Full tape: $\dfrac{9^{2}\times 30}{24\times 100^{2}} = 0.010125$ m; three tapes $= 0.030375$ m.
Last part: $\dfrac{0.3^{2}\times 24.095^{3}}{24\times 100^{2}} = 0.005246$ m.

$$
C_s = -(0.030375 + 0.005246) = -0.0356\ \text{m}
$$

#### 5. Slope
$$
C_h = -\frac{0.025^{2}\times 114.095}{2} = -0.0357\ \text{m}
$$

#### 6. True length
$$
L_{true} = 114.095 + 0.0570 - 0.0105 + 0.0579 - 0.0356 - 0.0357 = 114.1282\ \text{m}
$$

**Answer: true horizontal length of AB $= 114.128$ m.**

## Check yourself

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

1. The temperature correction is negative because
   A. 12 C is below 20 C
   B. The pull is high
   C. The tape is long
   D. Sag
   Answer: A. Cold tape is shorter. (Key point: Temperature sign)

2. w = 
   A. 9 N/m
   B. 9 N/30 m = 0.3 N/m
   C. 0.9 N/m
   D. 0.03 N/m
   Answer: B. Weight divided by length. (Key point: Weight per metre)

3. Pull correction uses AE =
   A. 0.028 x 2.11 x 10^5
   B. 2.11 x 10^6
   C. 0.028 x 2.11 x 10^6 = 59,080 N
   D. 0.028 N
   Answer: C. A in cm^2 times E in N/cm^2. (Key point: Pull)

4. The slope correction is
   A. +0.0357
   B. -0.357
   C. -0.0036
   D. -0.0357 m
   Answer: D. 0.025^2 x 114.095/2. (Key point: Slope)

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