---
title: "Correct horizontal distance 630 m, tape 8.9 N"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-630m-8-9n-30-005"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-630m-8-9n-30-005
site: BCT Engineering
---

# Correct horizontal distance 630 m, tape 8.9 N

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2070 Chaitra (old course), 8 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-630m-8-9n-30-005
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A 30 m tape weighing 8.9 N and having a cross sectional area of 2.58 mm² was standardized and found to be 30.005 m at 20°C with 52 N tension at fully supported condition. This tape was used for measuring the distances at constant temperature of 31.2°C and pull applied 110 N. The tape was supported at 0 and 30 m end. The observed distance was 630 m. Calculate the correct horizontal distance between points. Take coefficient of linear expansion of tape $\alpha = 12\times10^{-6}$/°C and Young's modulus of elasticity of tape material $E = 12\times10^{11}$ N/m².

## Short answer

Apply tape length correction (+0.1050 m), temperature (+0.0847 m), pull (+0.0118 m), and sag correction for 21 spans of 30 m (-0.1718 m). Correct horizontal distance is 630.030 m.

## Answer

**Reading.** The modulus is used as given, $E = 12\times 10^{11}$ N/m$^2$. Standardised fully supported (flat), so no standard sag is added back. Field supports are at the 0 and 30 m ends only, so each tape length is a 30 m catenary span.

#### Data
- $W = 8.9$ N for 30 m; $A = 2.58$ mm$^2 = 2.58\times 10^{-6}$ m$^2$; $AE = 2.58\times 10^{-6}\times 12\times 10^{11} = 3.096\times 10^{6}$ N
- Standard: 30.005 m at $20^\circ$C, $P_0 = 52$ N
- Field: $T = 31.2^\circ$C, $P = 110$ N, $L = 630$ m (21 tapes of 30 m)
- $\alpha = 12\times 10^{-6}$/$^\circ$C

#### 1. Tape length
$$
C_a = \frac{630\times 0.005}{30} = +0.1050\ \text{m}
$$

#### 2. Temperature
$$
C_t = 12\times 10^{-6}\times(31.2 - 20)\times 630 = +0.0847\ \text{m}
$$

#### 3. Pull
$$
C_p = \frac{(110 - 52)\times 630}{3.096\times 10^{6}} = +0.0118\ \text{m}
$$

#### 4. Sag
Per tape: $\dfrac{W^{2}\,l}{24\,P^{2}} = \dfrac{8.9^{2}\times 30}{24\times 110^{2}} = 0.008183$ m.

$$
C_s = -21\times 0.008183 = -0.1718\ \text{m}
$$

#### 5. Correct horizontal distance
$$
D = 630 + 0.1050 + 0.0847 + 0.0118 - 0.1718 = 630.0296\ \text{m}
$$

**Answer: correct horizontal distance $= 630.030$ m.**

## Check yourself

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

1. What sag formula is used per tape of 30 m?
   A. W^2 l / (24 P^2)
   B. W l / (24 P)
   C. W^2 l^3/(24 P^2)
   D. W^2/(8P)
   Answer: A. With total weight W = w l, it becomes W^2 l/24P^2. (Key point: Sag formula with W)

2. The 30.005 m standard length gives
   A. -0.105 m
   B. +0.105 m over 630 m (630 x 0.005/30)
   C. +0.005 m
   D. 0
   Answer: B. Tape is long, so the measured length is increased. (Key point: Tape length correction)

3. Which pull is used in the sag term?
   A. Standard pull, 52 N
   B. The difference 58 N
   C. Field pull, 110 N
   D. Average 81 N
   Answer: C. Sag depends on the pull in use. (Key point: Pull in field)

4. Number of 30 m tapes in 630 m is
   A. 20
   B. 63
   C. 30
   D. 21
   Answer: D. 630/30 = 21. (Key point: Number of tapes)

## Related in Surveying I

- [Correct horizontal distance, 28.00 m measured](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-fully-support-28-00m) — Surveying I, chapter 2
- [Tape standardized in catenary: 20 m span](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-catenary-span-20m) — Surveying I, chapter 2, asked 2×
- [Horizontal length of 30 m span, 50 m tape](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-50m-span-30m-26c) — Surveying I, chapter 2
- [True length: tape found 2.5 cm too long](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-825-plus-750-m) — Surveying I, chapter 2
- [Horizontal length of 16 m, 29.8850 m tape](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-29-8850-16m-fully-supported) — Surveying I, chapter 2
- [Tape standardized in catenary: 24 m span](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-catenary-span-24m) — Surveying I, chapter 2, asked 2×

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