---
title: "Tape corrections for 86 m, supports at 15 m"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-30m-86m-supports-15m"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-30m-86m-supports-15m
site: BCT Engineering
---

# Tape corrections for 86 m, supports at 15 m

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2076 Chaitra, 4 marks
- Video: [Surveying Tape Correction Numerical 1](https://www.youtube.com/watch?v=VKh33kzPV38) — Studyclass 247
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/tape-30m-86m-supports-15m
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A 30 m steel tape was standardized in catenary conditions under a pull of 15 kg and found to be 30.006 m. This tape was used to measure a distance of 86 m in catenary conditions at a pull of 15 kg. Supports were provided at every 15 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 0.0045 m to 30.006 m to obtain true tape 30.0105 m, apply tape length correction +0.0301 m, subtract field sag -0.0030 m. Corrected length is 86.027 m.

## Answer

**Method.** The tape was standardised in catenary, so first the sag included in its standard length is added back to find its true length. Then the sag for the supports at every 15 m is applied.

#### Data
- $w = 30$ g/m $= 0.03$ kg/m; $P = 15$ kg (same as standard); $L = 86$ m
- Standard: 30 m tape $= 30.006$ m in catenary
- Field spans: 5 spans of 15 m + 1 span of 11 m

#### 1. Tape length
Sag in standardisation (30 m span):
$$
\frac{w^{2}\,l^{3}}{24\,P^{2}} = \frac{0.03^{2}\times 30^{3}}{24\times 15^{2}} = 0.0045\ \text{m}
$$
True tape length $= 30.006 + 0.0045 = 30.0105$ m.

$$
C_a = \frac{86\,(30.0105 - 30)}{30} = +0.0301\ \text{m}
$$

#### 2. Sag correction
15 m span: $\dfrac{0.03^{2}\times 15^{3}}{24\times 15^{2}} = 0.000563$ m; for 5 spans $= 0.002813$ m.
11 m span: $\dfrac{0.03^{2}\times 11^{3}}{24\times 15^{2}} = 0.000222$ m.

$$
C_s = -(0.002813 + 0.000222) = -0.0030\ \text{m}
$$

#### 3. Correct length
$$
L_{true} = 86 + 0.0301 - 0.0030 = 86.0271\ \text{m}
$$

**Answer: corrected length of the line $= 86.0271$ m.**

## Check yourself

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

1. Sag at standardisation of the 30 m tape (w = 0.03 kg/m, P = 15 kg) is
   A. 0.0045 m
   B. 0.045 m
   C. 0.00045 m
   D. 0.45 m
   Answer: A. 0.03^2 x 30^3 / (24 x 225) = 0.0045 m. (Key point: Standard sag)

2. 86 m with supports every 15 m gives
   A. 5 spans of 15 m + 1 span of 11 m
   B. 6 spans of 15 m
   C. 4 spans of 15 m + 26 m
   D. 5 spans of 17.2 m
   Answer: A. 75 m + 11 m = 86 m. (Key point: Field spans)

3. True tape length 30.0105 m gives C_a = 
   A. -0.0301 m
   B. +0.0301 m
   C. +0.0105 m
   D. +0.3 m
   Answer: B. 86 x 0.0105 / 30 = +0.0301 m; the tape is long, so add. (Key point: Tape length correction)

4. The corrected length is
   A. 86.0271 m
   B. 85.9729 m
   C. 86.0331 m
   D. 86.0030 m
   Answer: A. 86 + 0.0301 - 0.0030 = 86.0271 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, 10 kg pull](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-20m-86m-pull-10kg) — Surveying I, chapter 2, asked 3×

## Related in Surveying I

- [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: 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 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