Question

Difficulty: HardElementary Surveying Instruments and Fieldwork

During a fieldwork measurement along a baseline, a geography student uses a steel tape with a nominal length of 30.00 m30.00\text{ m}. Due to high ambient temperature, the actual length of the tape has expanded to 30.15 m30.15\text{ m}. If the uncorrected recorded distance for the baseline is 450.00 m450.00\text{ m}, what is the true ground distance and the nature of the measurement error?

  1. The true ground distance is 452.25 m452.25\text{ m}, representing a cumulative negative error.Answer
  2. B
    The true ground distance is 447.75 m447.75\text{ m}, representing a cumulative positive error.
  3. C
    The true ground distance is 452.25 m452.25\text{ m}, representing a compensatory positive error.
  4. D
    The true ground distance is 447.75 m447.75\text{ m}, representing a random field error.

Answer

The true ground distance is 452.25 m452.25\text{ m}, representing a cumulative negative error.
The correct answer states that the true ground distance is 452.25 m452.25\text{ m} and represents a cumulative negative error. When a 30.00 m30.00\text{ m} tape expands to 30.15 m30.15\text{ m}, every time the tape is laid out, 30.15 m30.15\text{ m} of ground is covered, but only 30.00 m30.00\text{ m} is added to the tally. Multiplying 450.00 m450.00\text{ m} by the ratio 30.15/30.0030.15 / 30.00 gives a true ground distance of 452.25 m452.25\text{ m}. Because the recorded distance is shorter than the actual ground distance, the error is negative, and because it acts consistently in one direction throughout the survey, it is cumulative.

Step-by-Step Solution

1
Determine the total number of tape applications made during the baseline measurement.
Number of applications N=450.00 m30.00 m=15N = \frac{450.00\text{ m}}{30.00\text{ m}} = 15.
The recorded distance is measured in standard 30.00 m30.00\text{ m} units.
2
Calculate the true ground distance using the actual expanded length of the tape.
True Distance =15×30.15 m=452.25 m= 15 \times 30.15\text{ m} = 452.25\text{ m} (or using Ltrue=Lrecorded×ll=450.00×30.1530.00=452.25 mL_{\text{true}} = L_{\text{recorded}} \times \frac{l'}{l} = 450.00 \times \frac{30.15}{30.00} = 452.25\text{ m}).
Each application of the expanded tape spans 30.15 m30.15\text{ m} of ground, while only 30.00 m30.00\text{ m} is recorded.
3
Classify the nature and sign of the survey error.
The recorded measurement is 2.25 m2.25\text{ m} less than the true distance (450.00 m<452.25 m450.00\text{ m} < 452.25\text{ m}). The error in the recorded distance is negative and accumulates proportionally with line length, making it a cumulative negative error.
When a measuring tape is too long, the recorded distance is always less than the true distance, causing a cumulative negative error.

Key Concept

Tape Length Error Corrections & Cumulative vs. Compensatory Errors
Estimated Time:2m 0s
Rate this question