---
title: "True length of base line AB (four bays)"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-base-line-four-bays"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-base-line-four-bays
site: BCT Engineering
---

# True length of base line AB (four bays)

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2078 Kartik, 6 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-base-line-four-bays
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

Four bays of base line AB were measured under a tension of 120 N and the data is given below. If the tape was standardised on the flat under a pull of 89 N and a temperature of 20°C, calculate the true length of the line.

| Bay | Length (m) | Difference in level (m) |
|---|---|---|
| 1 | 29.478 | +0.294 |
| 2 | 29.208 | -0.384 |
| 3 | 29.396 | +0.923 |
| 4 | 29.916 | -0.726 |

Field temperature = 31°C; cross-sectional area of tape = 3.24 mm²; density = 7700 kg/m³; coefficient of linear expansion = 0.000001/°C; Young modulus = $15.3\times10^{4}$ MN/m².

## Short answer

For each bay, apply slope correction h^2/2l, sag, pull and temperature corrections, then sum. True horizontal length of the base line is 117.962 m.

## Answer

**Assumptions.** The tape is nominally 30 m long, standardised on the flat (fully supported), so it has no length error. Each bay is measured with the tape in catenary at 120 N. Slope corrections use $h^2/2l$.

#### Data
- Area $A = 3.24$ mm$^2 = 3.24\times 10^{-6}$ m$^2$; density $= 7700$ kg/m$^3$
- Mass per metre $= 7700\times 3.24\times 10^{-6} = 0.024948$ kg/m, so $w = 0.24474$ N/m
- $E = 15.3\times 10^{4}$ MN/m$^2 = 1.53\times 10^{11}$ N/m$^2$, so $AE = 495{,}720$ N
- $P = 120$ N, $P_0 = 89$ N, $T = 31^\circ$C, $T_0 = 20^\circ$C, $\alpha = 1\times 10^{-6}$/$^\circ$C

#### Formulas for each bay of length $l$ with level difference $h$
$$
C_h = -\frac{h^{2}}{2l},\quad C_t = \alpha\,(T-T_0)\,l,\quad C_p = \frac{(P-P_0)\,l}{AE},\quad C_s = -\frac{w^{2}\,l^{3}}{24\,P^{2}}
$$

#### Calculation

| Bay | $l$ (m) | $h$ (m) | $C_h$ | $C_s$ | $C_p$ | $C_t$ | Horizontal length (m) |
|---|---|---|---|---|---|---|---|
| 1 | 29.478 | +0.294 | $-0.00147$ | $-0.00444$ | $+0.00184$ | $+0.00032$ | 29.4743 |
| 2 | 29.208 | $-0.384$ | $-0.00252$ | $-0.00432$ | $+0.00183$ | $+0.00032$ | 29.2033 |
| 3 | 29.396 | +0.923 | $-0.01449$ | $-0.00440$ | $+0.00184$ | $+0.00032$ | 29.3793 |
| 4 | 29.916 | $-0.726$ | $-0.00881$ | $-0.00464$ | $+0.00187$ | $+0.00033$ | 29.9048 |

Sum of measured lengths $= 117.998$ m.

$$
\text{True length} = 29.4743 + 29.2033 + 29.3793 + 29.9048 = 117.9617\ \text{m}
$$

**Answer: true horizontal length of the base line AB $\approx 117.962$ m** (the sum of the bays computed with unrounded values is $117.9616$ m).

## Check yourself

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

1. Which formula is used for the slope correction of each bay?
   A. C_h = -h^2/(2l)
   B. C_h = +h^2/(2l)
   C. C_h = -h/(2l)
   D. C_h = -h^2/(24l)
   Answer: A. The slope correction for a small gradient is h^2/2l, subtracted from the measured length. (Key point: Slope correction formula)

2. Why is the pull correction positive for every bay?
   A. The tape was standardised in catenary
   B. Field pull (120 N) is greater than the standard pull (89 N), so the tape stretches and reads short
   C. The temperature is above standard
   D. The bays are on slopes
   Answer: B. C_p = (P - P0)l/AE is positive when P exceeds P0. (Key point: Pull correction)

3. Why is no tape-length correction applied?
   A. Because the tape is invariable
   B. Because sag cancels it
   C. The tape is standardised on the flat and taken as nominally 30 m, so it has no length error
   D. Because temperature is standard
   Answer: C. The assumption is that the tape is exactly nominal under standard conditions. (Key point: Standardised on the flat)

4. How is the true length of AB obtained?
   A. Sum of measured lengths, 117.998 m
   B. Sum of measured lengths minus the sag only
   C. Average of the four bays times 4 without slope correction
   D. Sum of the corrected horizontal bay lengths, about 117.962 m
   Answer: D. Each bay is corrected for slope, sag, pull and temperature, then added. (Key point: Final result)

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