An offset is laid out 5° from its true direction on the field. If the scale of plotting is 20 m to 1 cm, find the maximum length of the offset so that the displacement of the point on the paper may not exceed 0.5 mm.

Discussion - MCQs on Compass Surveying - Maximum Offset Length

An offset is laid out 5° from its true direction on the field. If the scale of plotting is 20 m to 1 cm, find the maximum length of the offset so that the displacement of the point on the paper may not exceed 0.5 mm.

A. 5 m
B. 9.22 m
C. 11.47 m
D. 13.33 m
Correct Answer: C. 11.47 m

🎯 Understanding Plotting Accuracy

This problem links an error made in the field (an incorrect angle) to its effect on the final drawing (displacement on paper). The key principle is that the longer the offset, the larger the plotting error will be for a given angular mistake. We need to find the maximum length the offset can be before this plotting error becomes unacceptably large.

🔬 Step-by-Step Calculation

The relationship between the field error and the plotting displacement is given by trigonometry.

Step 1: Identify Given Data and Convert Units

It's crucial to work with consistent units.

  • Angular Error (Δ):
  • Maximum Permissible Displacement on Paper: 0.5 mm = 0.05 cm
  • Scale: 1 cm on paper = 20 m in the field. The scale factor (S) is 20.
  • Maximum Offset Length (l): To be determined.

Step 2: The Formula

The displacement on the paper is the scaled-down version of the displacement in the field. The field displacement is `l × sin(Δ)`.

Maximum Displacement on Paper = (Field Displacement) / Scale

Maximum Displacement on Paper = (l × sin(Δ)) / S

Step 3: Rearrange and Solve for 'l'

We need to find 'l', so we rearrange the formula:

l = (Maximum Displacement on Paper × S) / sin(Δ)

Now, substitute the known values:

l = (0.05 cm × 20) / sin(5°)

l = 1 / sin(5°)

Since sin(5°) ≈ 0.08715:

l ≈ 1 / 0.08715

l ≈ 11.47 m

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