---
title: "True length of suspended distance AB (66)"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-suspended-ab-114-095"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-suspended-ab-114-095
site: BCT Engineering
---

# True length of suspended distance AB (66)

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2066 Bhadra, 9 marks
- Video: [Surveying | Corrections in Tape Part 1 of 2](https://www.youtube.com/watch?v=MdHdzGSSz7Q) — the Engineer's Insight
- 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
- 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 under a 70 N tension at a temperature of 20°C. The density of tape material is $7.75\times10^{3}$ kg/m³ and it had a cross-sectional area of 0.028 cm². What is the true length of the recorded suspended distance AB for the following condition? Recorded distance = 114.095 m. Mean temperature = 32°C, tension applied = 100 N, elevation difference per 100 m = 2.5 m, $\alpha = 1.15\times10^{-5}$/°C, $E = 2.10\times10^{5}$ N/mm². Assume all full tape length except in the last one.

## Short answer

Corrections: standardization +0.0570 m, temperature +0.0157 m, pull +0.0058 m, sag -0.0179 m, slope -0.0357 m. True horizontal length = 114.120 m.

## Answer

**Assumptions.** The tape is standardised fully supported and used suspended (catenary) in 3 full tapes of 30 m plus a last part of 24.095 m. The ground slopes uniformly at 2.5 m per 100 m. $g = 9.81$ m/s$^2$.

#### Data
- $A = 0.028$ cm$^2 = 2.8$ mm$^2$; density $= 7750$ kg/m$^3$; $w = 7750\times 2.8\times 10^{-6}\times 9.81 = 0.21288$ N/m ($W = 6.386$ N per 30 m)
- $E = 2.1\times 10^{5}$ N/mm$^2$, so $AE = 2.8\times 2.1\times 10^{5} = 588{,}000$ N
- Standard: 30.015 m at $20^\circ$C, $P_0 = 70$ N; field: $T = 32^\circ$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
$$
C_t = 1.15\times 10^{-5}\times(32 - 20)\times 114.095 = +0.0157\ \text{m}
$$

#### 3. Pull
$$
C_p = \frac{(100 - 70)\times 114.095}{588{,}000} = +0.0058\ \text{m}
$$

#### 4. Sag
Full tape: $\dfrac{w^{2}\,l^{3}}{24\,P^{2}} = \dfrac{0.21288^{2}\times 30^{3}}{24\times 100^{2}} = 0.005098$ m; three tapes $= 0.015294$ m.
Last part: $\dfrac{0.21288^{2}\times 24.095^{3}}{24\times 100^{2}} = 0.002641$ m.

$$
C_s = -(0.015294 + 0.002641) = -0.0179\ \text{m}
$$

#### 5. Slope
For each tape length $l$, the height difference is $h = 0.025\,l$, so summing $h^{2}/2l$ over all lengths gives:
$$
C_h = -\sum\frac{h^{2}}{2l} = -\frac{0.025^{2}\times 114.095}{2} = -0.0357\ \text{m}
$$

#### 6. True length
$$
L_{true} = 114.095 + 0.0570 + 0.0157 + 0.0058 - 0.0179 - 0.0357 = 114.1200\ \text{m}
$$

**Answer: true horizontal length of AB $= 114.120$ m.**

## Check yourself

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

1. How is the sag handled for the incomplete last tape?
   A. Sag computed separately using 24.095 m
   B. Ignored
   C. Uses 30 m
   D. Uses 114.095 m
   Answer: A. Spans are 3 x 30 m and a 24.095 m last part. (Key point: Last part)

2. Slope correction for the uniform 2.5 m per 100 m gradient is
   A. +0.025 L
   B. -(0.025^2 x L)/2
   C. -0.025 L
   D. -0.025^2 L
   Answer: B. Sum of h^2/2l with h/l = 0.025. (Key point: Slope correction)

3. w is found from
   A. density x g
   B. area/density
   C. density x area x g
   D. density/area
   Answer: C. Mass per metre times g. (Key point: Unit weight)

4. Which corrections are positive?
   A. Sag and slope
   B. Slope and tape length
   C. Sag only
   D. Tape length, temperature, pull
   Answer: D. Sag and slope are negative; other three are positive. (Key point: Signs)

## Related in Surveying I

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