The combined correction for curvature of Earth and refraction in levelling for a distance of 6 km would be __________ m (rounded off to two decimal places). Assume the radius of earth is 6370 km.
This section calculates the combined correction required in leveling due to Earth's curvature and atmospheric refraction over a specific distance.
For surveying calculations, the combined correction ($C$) in meters is typically estimated using the distance ($d$) in kilometers and the Earth's radius ($R$) in kilometers with the formula:
$ C = \frac{7}{8} \times \frac{d^2}{2R} \times 1000 $
Substitute the known values into the formula:
$ C = \frac{7}{8} \times \frac{(6)^2}{2 \times 6370} \times 1000 $
Calculate the square of the distance:
$ 6^2 = 36 $
Calculate the denominator term:
$ 2 \times 6370 = 12740 $
Now, compute the correction value:
$ C = \frac{7}{8} \times \frac{36}{12740} \times 1000 $
$ C = \frac{252}{101920} \times 1000 $
$ C = \frac{252000}{101920} $
$ C \approx 2.4725 $
Rounded to two decimal places, the combined correction is 2.47 m.
| Staff readings at station | ||
|---|---|---|
| Instrument set very near to station | A (m) | B (m) |
| A | 1.275 | 2.005 |
| B | 1.040 | 1.660 |