Calculate the curvature correction (in m) for a distance of 500 m between the instrument and the staff.
Correct Answer: A. 0.0196 m
📚 Detailed Explanation: Curvature Correction at 500 m = 0.0196 m
Why A (0.0196 m) is correct: The curvature correction formula is Cc = 0.0785 × D² (D in km). For D = 500 m = 0.5 km, the correction magnitude = 0.0196 m.
Formula: Cc = 0.0785 × D² (D in km)
D = 500 m = 0.5 km
Cc = 0.0785 × (0.5)²
Cc = 0.0785 × 0.25
Cc = 0.019625 m ≈ 0.0196 m (magnitude)
Note: The correction is applied as negative (subtracting from observed reading),
but the magnitude 0.0196 m is always positive.
Corrections at Different Distances
| Distance (km) | Curvature Cc (m) | Refraction Cr (m) | Combined (m) |
|---|---|---|---|
| 0.5 | 0.0196 | 0.0028 | 0.0168 |
| 1.0 | 0.0785 | 0.0112 | 0.0673 |
| 2.0 | 0.3140 | 0.0448 | 0.2692 |
- Cc = 0.0785 × D² (D in km) → at 0.5 km: 0.0196 m.
- Verification: Cr = Cc/7 = 0.0196/7 = 0.0028 m (matches Q40 result ✓).
- Combined = Cc − Cr = 0.0196 − 0.0028 = 0.0168 m.