Compound planet epicyclic train numerical

Geometry check (same module $m$)

Centre distance of S-P1: $\dfrac{m(20 + 40)}{2} = 30m$.

Centre distance of P2-R (internal mesh): $\dfrac{m(90 - 30)}{2} = 30m$.

Both are equal, so the compound planet P1-P2 can mesh with both S and R at the same arm radius.

        R (ring, 90T, internal)
       /  \
   P2(30)=P1(40)  <- carried on arm
        \
         S (sun, 20T, fixed)

Tabular method (clockwise positive, arm speed $y = +90$ rpm)

With the arm fixed and S turning $+x$:

• P1 (external mesh with S): $-\dfrac{20}{40}x = -0.50x$.
• P2 turns with P1: $-0.50x$.
• R (internal mesh with P2, same sense as P2): $-0.50x \times \dfrac{30}{90} = -0.1667x$.

Sun fixed

$$y + x = 0 \Rightarrow x = -y = -90\ \text{rpm}$$

Ring gear:

$$N_R = y - 0.1667x = 90 - 0.1667(-90) = 105\ \text{rpm}$$

Compound planet:

$$N_P = y - 0.5x = 90 - 0.5(-90) = 135\ \text{rpm}$$

The ring gear turns in the same direction as the arm, faster than the arm by the ratio $N_R/N_{arm} = 1.1667$. The planet spins about its own axis relative to the arm at $N_P - y = 45$ rpm in the same sense.

Answer: ring gear $= 105$ rpm clockwise; compound planet $= 135$ rpm clockwise.