If an offset is laid out 5° from its true direction in the field, find the resulting displacement of the plotted point on paper:(a) in a direction parallel to the chain line and (b) in a direction perpendicular to the chain line.

Surveying Offset Error Calculation

Surveying Problem: Offset Angular Error

Problem: If an offset is laid out 5° from its true direction in the field, find the resulting displacement of the plotted point on paper: (a) in a direction parallel to the chain line and (b) in a direction perpendicular to the chain line.

Given Data:

  • Length of the offset (l): 20 m
  • Angular error (α): 5°
  • Scale of plotting: 1 cm = 5 m

Solution:

(a) Displacement Parallel to Chain Line

Ground Displacement = l * sin(α)
= 20 m * sin(5°)
= 20 m * 0.0872   (sin(5°) ≈ 0.0872)
= 1.744 m

Paper Displacement = Ground Disp. / Scale Factor
= 1.744 m / (5 m/cm)
= 0.3488 cm
Displacement Parallel ≈ 0.35 cm

(b) Displacement Perpendicular to Chain Line

Ground Displacement = l * (1 - cos(α))
= 20 m * (1 - cos(5°))
= 20 m * (1 - 0.9962)   (cos(5°) ≈ 0.9962)
= 20 m * 0.0038
= 0.076 m

Paper Displacement = Ground Disp. / Scale Factor
= 0.076 m / (5 m/cm)
= 0.0152 cm
Displacement Perpendicular = 0.0152 cm

Explanation of Principles

This problem demonstrates how a small angular error during fieldwork (laying out an offset) translates into positional errors on the final plotted map or plan. The principles used are:

1. Trigonometry: When an offset of length 'l' is laid out with an angular error 'α' from its intended direction (usually perpendicular to the chain line), the actual endpoint is displaced. We can visualize this as a right-angled triangle (or nearly so for small angles).

  • The displacement parallel to the chain line (along the main survey line) is calculated using the sine function: l * sin(α). This represents the side opposite the angle α in the triangle formed by the intended offset direction, the actual offset, and the parallel displacement component.
  • The displacement perpendicular to the chain line (away from the main survey line, but in the direction the offset *should* have gone) is calculated as the difference between the actual offset length 'l' and its projection onto the intended perpendicular direction (l * cos(α)). Therefore, this displacement is l * (1 - cos(α)).

2. Scale Conversion: Surveying measurements are taken on the ground, but the final output is often a scaled drawing (plan or map). The scale relates distances on the ground to distances on the paper.

  • In this case, the scale is 1 cm = 5 m. This means every 5 meters measured on the ground is represented by 1 centimeter on the paper.
  • To find the displacement *on the paper*, the calculated ground displacements (in meters) must be divided by the scale factor (5 m per cm). This converts the real-world error into its representation on the drawing.

How it was Solved:

  1. Identify the given parameters: offset length (l=20m), angular error (α=5°), and scale (1cm = 5m).
  2. Apply the trigonometric formula l * sin(α) to find the ground displacement parallel to the chain line.
  3. Divide the result from step 2 by the scale factor (5) to get the paper displacement parallel to the chain line.
  4. Apply the trigonometric formula l * (1 - cos(α)) to find the ground displacement perpendicular to the chain line.
  5. Divide the result from step 4 by the scale factor (5) to get the paper displacement perpendicular to the chain line.
  6. Round the final answers to appropriate significant figures.

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