A seventeenth century navigator would have been pleased to estimate his longitud

Question

A seventeenth century navigator would have been pleased to estimate his longitude with an error of 1 degree. At a WGS84 LLH (Lat: 42 degree, Long: 83 degree, Geodetic Height: 100 m), what would have been the resultant position error in meters? First, try the simple spherical model of the earth with radius of 6371 km. Next, use the WGS 84 ellipsoidal model (Table 4.1). Note that 6371 km is the approximate radius of a sphere whose volume equals that of the WGS 84 ellipsoid. Parameter Value Ellipsoid Semi-major axis (a) 6378137.0m Reciprocal flattening (1/f) 298.257223563 Earth's angular velocity (omega_E) 7292115.0 times 10^11 rad/sec Earth's gravitational constant (GM) 398600418 times 10^8 m^3/s^2 Speed of light in a vacuum (c) 2.99792458 times 10^8 m/s

Explanation / Answer

1 Degree Error meand = 60 Minutes = 60 Seconds

1 Seconds = 30 Meters

60 Seconds = 60 x 30meters = 1800 Meters

1 Minutes = 1800 Meters

60 Minutes = 60 x 1800 Meters

1 Degree Error = 10,8000 Meters (108km)

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