Discussion
For a conventional airplane in straight, level, constant velocity flight condition, which of the following condition(s) is/are possible on Euler angles \((\phi, \theta, \psi)\), angle of attack \((\alpha)\) and the sideslip angle \((\beta)\)?
Answer: A
The possible set is \(\phi = 0, \theta = 2^{o}, \psi = 0, \alpha = 2^{o}, \beta = 0\).
In straight, level, constant-velocity flight the wings are level (\(\phi = 0\)), the flight path is horizontal so the pitch angle equals the angle of attack (\(\gamma = \theta - \alpha = 0\)), and a conventional airplane flies with no sideslip (\(\beta = 0\)). The heading \(\psi\) can be any constant value. The first set meets every condition.
A 5-degree bank with zero sideslip tilts the lift vector and turns the airplane, and \(\theta = 0\) with \(\alpha = 2^{o}\) would mean a descent, so the second set is neither straight nor level. The third and fourth sets have a 5-degree sideslip with the wings level, which leaves an unbalanced side force, and the fourth also has \(\theta - \alpha = 3^{o}\), a climb. The official key accepts only the first set.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development