Machine Theory – Maximum Efficiency of a Screw Jack\nFor a square-thread screw jack with friction angle φ and lead angle λ, what is the maximum theoretical efficiency and the condition for it?

Introduction / Context:
A screw jack's efficiency depends on its lead angle λ and the friction angle φ. Designers often seek the condition that maximizes efficiency, subject to self-locking and strength constraints.

Given Data / Assumptions:

• Square-thread screw; standard friction model with friction angle φ.
• Lead angle λ is the helix angle measured on the mean radius.
• General efficiency formula: η = tan λ / tan(λ + φ).

Concept / Approach:
To find the maximum, differentiate η with respect to λ and set the derivative to zero. Using a trigonometric identity via the derivative of ln η, the optimality condition emerges as 2λ + φ = 90°. Substituting this into the efficiency formula yields a compact expression in terms of φ alone.

Step-by-Step Solution:
Start with η = tan λ / tan(λ + φ). Set d(ln η)/dλ = 0 ⇒ sec^2λ/tanλ − sec^2(λ+φ)/tan(λ+φ) = 0. This simplifies to sin(2λ) = sin(2λ + 2φ) ⇒ 2λ + φ = 90° (non-trivial solution). At 2λ + φ = 90°, tan(λ + φ) = tan(90° − λ) = cot λ, so η_max = tan^2 λ. Using λ = 45° − φ/2, tan λ = tan(45° − φ/2) ⇒ η_max simplifies to (1 − sin φ)/(1 + sin φ).

Verification / Alternative check:
Deriving η_max directly from tangent half-angle identities also leads to η_max = (1 − sin φ)/(1 + sin φ).

Why Other Options Are Wrong:
(b) Inverts the correct ratio. (c) Gives the general η expression, not the maximized value or correct condition. (d) Uses an incorrect condition and value. (e) Ignores dependence on λ and gives an incorrect constant value.

Common Pitfalls:
Confusing lead angle λ with friction angle φ. Assuming maximum efficiency occurs at maximum λ without proof.

Final Answer:
η_max = (1 − sin φ) / (1 + sin φ), attained when 2λ + φ = 90°