---
title: "Tape corrections for 86 m, 15 kg pull"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-20m-86m-supports-10m-15kg"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-20m-86m-supports-10m-15kg
site: BCT Engineering
---

# Tape corrections for 86 m, 15 kg pull

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2080 Baisakh, 4 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-supports-10m-15kg
- 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 15 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 15 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 back the sag included in standardization (0.00133 m) to get true tape length 20.00733 m, apply tape length correction +0.0315 m and subtract field sag -0.0014 m. Corrected length is 86.030 m.

## Answer

**Method.** A tape standardised in catenary has the sag of the standardisation span built into its stated length. We first add this sag back to get the tape's true length, then apply the field sag for supports every 10 m.

#### Data
- $w = 30$ g/m $= 0.03$ kg/m; pull $P = 15$ kg (same as standard); $L = 86$ m
- Standard: 20 m tape $= 20.006$ m in catenary (one 20 m span)
- Field spans: 8 spans of 10 m + 1 span of 6 m

#### 1. Tape length
Sag in standardisation (20 m span):
$$
\frac{w^{2}\,l^{3}}{24\,P^{2}} = \frac{0.03^{2}\times 20^{3}}{24\times 15^{2}} = 0.00133\ \text{m}
$$
True (flat) tape length $= 20.006 + 0.00133 = 20.00733$ m.

$$
C_a = \frac{86\,(20.00733 - 20)}{20} = +0.0315\ \text{m}
$$

#### 2. Sag correction in the field
Each 10 m span: $\dfrac{0.03^{2}\times 10^{3}}{24\times 15^{2}} = 0.000167$ m; for 8 spans $= 0.001333$ m.
The 6 m span: $\dfrac{0.03^{2}\times 6^{3}}{24\times 15^{2}} = 0.000036$ m.

$$
C_s = -(0.001333 + 0.000036) = -0.0014\ \text{m}
$$

Temperature and pull corrections are zero (same as standard, no data).

#### 3. Correct length
$$
L_{true} = 86 + 0.0315 - 0.0014 = 86.0302\ \text{m}
$$

**Answer: corrected length of the line $= 86.0302$ m.**

## Check yourself

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

1. The tape reads 20.006 m in catenary. What is done first?
   A. Add back standardisation sag (0.00133 m) to find the true flat length
   B. Subtract it
   C. Ignore it
   D. Double it
   Answer: A. The standard value includes sag, so true length = 20.00733 m. (Key point: Standard in catenary)

2. For 86 m with supports every 10 m, the spans are
   A. 9 spans of 10 m
   B. 8 spans of 10 m + 1 span of 6 m
   C. 10 spans of 8.6 m
   D. 4 spans of 20 m
   Answer: B. 8 x 10 = 80, plus 6 m remaining. (Key point: Field spans)

3. Which governs the sag correction per span?
   A. w^2 l^3 / (24 P^2), negative
   B. w l / P positive
   C. h^2 / 2L
   D. L h / R
   Answer: A. Sag grows with the cube of span and falls with pull squared; here total is -0.0014 m. (Key point: Sag formula)

4. Corrected length is
   A. 86.0302 m
   B. 85.9686 m
   C. 86.0329 m
   D. 86.0014 m
   Answer: A. 86 + 0.0315 - 0.0014 = 86.0302 m. (Key point: Final length)

## Close variants set in other exams

- [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×
- [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 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 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 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: 23 m span](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-catenary-span-23m) — Surveying I, chapter 2