Discussion
Flow angularity of \( 0.5 \) degrees resolves what fraction of a lift of \( 1200 \) N into the drag channel?
Answer: D
The component is \( 1200 \sin 0.5^\circ = 1200 \times 0.00873 = 10.5 \) N. Against a true drag of perhaps 90 N that is a twelve per cent error from half a degree of misalignment, which shows why upright-and-inverted testing is worth the extra runs.
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