Discussion
An aircraft has a design weight of \( 640 \) kN and a chosen wing loading of \( 5.8 \) kN/m\(^2\). The wing area required is
Answer: A
Correct answer: 110.3 m\(^2\)
Wing area is the design weight divided by the chosen wing loading:
\[ S = \frac{W}{W/S} = \frac{640}{5.8} = 110.3\ \mathrm{m}^2 \]
The distractor \( 3712 \) multiplies instead of dividing, which is the error to guard against - a quick sanity check is that a wing loading in kilonewtons per square metre divided into a weight in kilonewtons must leave square metres. This area, with the installed thrust from the other coordinate, is the principal output of the sizing exercise.
No comment is present. Be the first to comment.
Loading…
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