---
title: "Tape corrections for 86 m, 10 kg pull"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-20m-86m-pull-10kg"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-20m-86m-pull-10kg
site: BCT Engineering
---

# Tape corrections for 86 m, 10 kg pull

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2076 Asoj, 6 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-20m-86m-pull-10kg
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A 20 m steel tape was standardized in catenary conditions under a pull of 10 kg and found to be 20.006 m. This tape was used to measure a distance of 86 m in catenary conditions at a pull of 10 kg. Supports were provided at every 10 m. The weight of the tape was 30 gm/m. Apply necessary tape corrections for the measured length of line.

## Short answer

Add standard sag 0.0030 m to get true tape 20.009 m, apply tape length correction +0.0387 m, subtract field sag -0.0031 m. Corrected length is 86.036 m.

## Answer

**Method.** The tape was standardised in catenary, so the sag in the standard length is added back to find its true length. Then the sag for the supports at every 10 m is applied.

#### Data
- $w = 30$ g/m $= 0.03$ kg/m; $P = 10$ kg (same as standard); $L = 86$ m
- Standard: 20 m tape $= 20.006$ m in catenary (20 m span)
- Field spans: 8 spans of 10 m + 1 span of 6 m

#### 1. Tape length
Sag at standardisation: $\dfrac{0.03^{2}\times 20^{3}}{24\times 10^{2}} = 0.0030$ m.
True tape length $= 20.006 + 0.0030 = 20.009$ m.

$$
C_a = \frac{86\,(20.009 - 20)}{20} = +0.0387\ \text{m}
$$

#### 2. Sag correction
10 m span: $\dfrac{0.03^{2}\times 10^{3}}{24\times 10^{2}} = 0.000375$ m; for 8 spans $= 0.003000$ m.
6 m span: $\dfrac{0.03^{2}\times 6^{3}}{24\times 10^{2}} = 0.000081$ m.

$$
C_s = -(0.003000 + 0.000081) = -0.0031\ \text{m}
$$

#### 3. Correct length
$$
L_{true} = 86 + 0.0387 - 0.0031 = 86.0356\ \text{m}
$$

**Answer: corrected length of the line $= 86.0356$ m.**

## Check yourself

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

1. Sag at standardisation (20 m span, P = 10 kg) is
   A. 0.0030 m
   B. 0.0013 m
   C. 0.030 m
   D. 0.00030 m
   Answer: A. 0.03^2 x 20^3 / (24 x 100) = 0.0030 m, so the flat tape is 20.009 m. (Key point: Standard sag)

2. Compared with 15 kg, a pull of 10 kg gives a larger sag because sag
   A. Varies inversely with P^2
   B. Varies directly with P
   C. Is independent of P
   D. Varies with P^3
   Answer: A. Sag correction is proportional to 1/P^2. (Key point: Effect of pull)

3. C_a = 86 (20.009 - 20)/20 equals
   A. +0.0387 m
   B. -0.0387 m
   C. +0.387 m
   D. +0.0045 m
   Answer: A. The tape is longer than 20 m so the correction is positive. (Key point: Tape length correction)

4. Corrected length is
   A. 86.0356 m
   B. 85.9644 m
   C. 86.0418 m
   D. 86.0031 m
   Answer: A. 86 + 0.0387 - 0.0031 = 86.0356 m. (Key point: Result)

## Close variants set in other exams

- [Tape corrections for 86 m, 15 kg pull](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-20m-86m-supports-10m-15kg) — Surveying I, chapter 2, asked 3×
- [Tape corrections for 86 m, supports at 15 m](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-30m-86m-supports-15m) — Surveying I, chapter 2, asked 3×

## Related in Surveying I

- [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×
- [Tape corrections for 66 m in three spans](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-66m-three-equal-spans) — Surveying I, chapter 2, asked 2×
- [Tape correction for 24.726 m distance](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-24-726-m-catenary) — Surveying I, chapter 2, asked 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×
- [Tape standardized in catenary: 23 m span](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-catenary-span-23m) — Surveying I, chapter 2