Module 3 : Photogrammetry
  Lecture 21 : Flight planning
Example 2:
The following data is given for flight planning:
  • Format 18 x 18 cm
  • Focal length = 21 cm
  • Scale = 1/20,000
  • Longitudinal overlap = 60%
  • Lateral overlap = 20%
  • East-west terrain length = 100 km
  • North-south terrain width = 50 km
  • Flight direction: East to west
  • Aircraft velocity = 296 km
  • Permissible image movement = 0.02 mm
  • Wind velocity = 10 m /s from SSE direction
Calculate the following
  • Exposure interval
  • maximum exposure time
  • Ground speed of aircraft
  • Drift angle
Solution
  • Airbase = 0.4 x 18 cm at photo scale. On ground
  • Exposure interval
 
  • Image movement = 0.02 mm on photo. On ground,
  • Time taken to cover this distance = maximum exposure time (shutter speed)
  • Wind speed = 10 m/s = 36km/hr; flying speed = 296 km/hr. The ground speed of the aircraft and the angle of drift can be found out by vector operations.
    Effective ground speed = (2962 + 362 - 2x296x36xcos67.5) 1/2 = 284.176 km/hr.
  • Drift angle ( θ ) is given as
 
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