For a planetary gear set with sun teeth 36 and ring teeth 84, two technicians debate the ratio when the sun is held and ring is input with carrier as output? The sun 36 teeth and ring 84 teeth. Who is correct?

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Multiple Choice

For a planetary gear set with sun teeth 36 and ring teeth 84, two technicians debate the ratio when the sun is held and ring is input with carrier as output? The sun 36 teeth and ring 84 teeth. Who is correct?

Explanation:
When a planetary gear set has the sun gear held stationary and the ring gear is the input with the carrier as the output, the carrier speed is fixed relative to the ring speed by the tooth counts: ω_C = (N_R / (N_S + N_R)) × ω_R. This comes from the no-slip constraint at the two meshing interfaces (sun–planet and ring–planet) and the fact the planets are carried by the carrier. With N_S = 36 and N_R = 84, the total is 120, so ω_C = 84/120 × ω_R = 0.7 × ω_R. That means the carrier turns at 70% of the ring’s speed, and the ring turns 120/84 ≈ 1.4286 times for every carrier revolution. So the correct relation is that the carrier speed is 0.7 times the ring input speed (ring-to-carrier ratio about 1.43:1). If Technician A stated that specific ratio, that is the correct claim.

When a planetary gear set has the sun gear held stationary and the ring gear is the input with the carrier as the output, the carrier speed is fixed relative to the ring speed by the tooth counts: ω_C = (N_R / (N_S + N_R)) × ω_R. This comes from the no-slip constraint at the two meshing interfaces (sun–planet and ring–planet) and the fact the planets are carried by the carrier.

With N_S = 36 and N_R = 84, the total is 120, so ω_C = 84/120 × ω_R = 0.7 × ω_R. That means the carrier turns at 70% of the ring’s speed, and the ring turns 120/84 ≈ 1.4286 times for every carrier revolution.

So the correct relation is that the carrier speed is 0.7 times the ring input speed (ring-to-carrier ratio about 1.43:1). If Technician A stated that specific ratio, that is the correct claim.

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