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Levelling methods; errors eliminated by balancing

Short answer

Direct (spirit), trigonometric, barometric and hypsometric levelling. Balancing sights (equal BS and FS distances) eliminates errors from collimation, curvature, refraction and non‑adjustment of the bubble tube, because equal errors cancel in the difference. Example: with 0.002 m/20 m collimation error, 40 m BS and FS give no net error.

Explain the principle methods of levelling. Which types of errors are eliminated by balancing of sight? Illustrate with suitable examples.

Principle methods of levelling

  1. Direct (spirit) levelling: a level and levelling staff are used; the difference in level is found from staff readings with a horizontal line of sight. It is the most accurate and most used method. Types: simple, differential, fly, check, profile, cross-section, reciprocal.
  2. Trigonometric levelling: the difference of elevation is found from the vertical angle (theodolite) and the horizontal distance: h=Dtan⁡αh = D \tan\alpha. It is used in hilly ground and for inaccessible points.
  3. Barometric levelling: based on the variation of atmospheric pressure with altitude. Used for reconnaissance.
  4. Hypsometric (boiling point) levelling: based on the variation of the boiling point of water with altitude.

Errors eliminated by balancing of sights

Balancing of sights means keeping the backsight and foresight distances equal from each instrument position. The following errors are eliminated:

  • Error due to collimation (line of collimation not parallel to the axis of the bubble tube).
  • Error due to curvature of the earth.
  • Error due to atmospheric refraction.
  • Error due to non-adjustment of the bubble tube (the same effect as collimation error).

Each of these errors depends only on the sight distance (collimation in proportion to the distance, curvature and refraction in proportion to its square). Equal sights therefore give equal errors in the BS and FS, which cancel in the difference of the readings.

Example: an instrument has a collimation error of 0.002 m per 20 m. If the BS and FS are both 40 m, each reading is wrong by 0.004 m and the difference is correct. If BS = 20 m and FS = 60 m, the difference is wrong by 0.002×(60−20)/20=0.0040.002 \times (60 - 20)/20 = 0.004 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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