A control run has a gearing such that \( 1 \ \mathrm{N\,m} \) of hinge moment gives \( 0.25 \ \mathrm{N} \) at the column. A hinge moment of \( 320 \ \mathrm{N\,m} \) therefore needs
Answer: A
Applying the gearing directly:
\[ F = 320 \times 0.25 = 80 \ \mathrm{N} \]
so 80 newtons, a firm but manageable pull. 1280 N divides by the gearing instead of multiplying, and would be far beyond any pilot. 320 N ignores the gearing altogether. 40 N halves the correct answer. Gearing is the designer's cheapest lever on stick force, limited by the column travel the cockpit can accommodate.
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