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328 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

13-8 (a) The smallest pinion tooth count that will run with itself is found from Eq. (13-10)

NP ≥ 2k

3 sin2 φ

(1 +

√1 + 3 sin2 φ

)

≥ 2(1)

3 sin2 20°

(1 +

√1 + 3 sin2 20°

)

≥ 12.32 → 13 teeth Ans.

(b) The smallest pinion that will mesh with a gear ratio of mG = 2.5, from Eq. (13-11) is

NP ≥ 2(1)

[1 + 2(2.5)] sin2 20°

{2.5 +

√2.52 + [1 + 2(2.5)] sin2 20°

}

≥ 14.64 → 15 pinion teeth Ans.

The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-12) is

NG ≤ N 2P sin2 φ − 4k2

4k − 2NP sin2 φ

≤ 152 sin2 20° − 4(1)2

4(1) − 2(15) sin2 20°

≤ 45.49 → 45 teeth Ans.

(c) The smallest pinion that will mesh with a rack, from Eq. (13-13)

NP ≥ 2k

sin2 φ= 2(1)

sin2 20°

≥ 17.097 → 18 teeth Ans.

13-9 φn = 20°, ψ = 30°, φt = tan−1 (tan 20°/cos 30°) = 22.80°

(a) The smallest pinion tooth count that will run itself is found from Eq. (13-21)

NP ≥ 2k cos ψ

3 sin2 φt

(1 +

√1 + 3 sin2 φt

)

≥ 2(1) cos 30°

3 sin2 22.80°

(1 +

√1 + 3 sin2 22.80°

)

≥ 8.48 → 9 teeth Ans.

(b) The smallest pinion that will mesh with a gear ratio of m = 2.5, from Eq. (13-22) is

NP ≥ 2(1) cos 30°

[1 + 2(2.5)] sin2 22.80°

{2.5 +

√2.52 + [1 + 2(2.5)] sin2 22.80°

}

≥ 9.95 → 10 teeth Ans.

The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-23) is

NG ≤ 102 sin2 22.80° − 4(1) cos2 30°

4(1) cos2 30° − 2(20) sin2 22.80°

≤ 26.08 → 26 teeth Ans.

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