Question 5 of 7: Two-stage planetary gear train with ring brakes

Question 5: Two-stage planetary gear train with ring brakes (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Stage 1: sun $2$ ($N_2=18$), planet $3$, ring $4$ ($N_4=60$), carrier $c_1$. Stage 2: sun $5$ ($N_5=32$), planet $6$, ring $7$ ($N_7=86$), carrier $c_2$. The input shaft is the common carrier of both stages ($\omega_{c1}=\omega_{c2}=\omega_{in}=1800$ rpm), and sun 2 is rigidly joined to ring 7 on a floating sleeve ($\omega_2=\omega_7$); the output is sun 5. Brake $C$ holds ring 4; brake $D$ holds ring 7 — and therefore sun 2 with it.

Find. Planet teeth $N_3,N_6$; the maximum equally-spaced planet count in each stage; and the output speed with brake $C$ alone, and with brake $D$ alone.

[Figure not reproduced: Skeleton of the two-stage train traced from the section drawing: the input shaft is the common carrier of both stages, sun 2 and ring 7 are rigidly joined on a floating sleeve, and the output is sun 5. Brakes $C,D$ ground ring 4 / ring 7. See the official exam paper.]

(i) Planet tooth numbers

Approach. Coaxial (concentric) meshing requires the ring diameter to equal the sun plus two planets: $N_{\text{ring}}=N_{\text{sun}}+2N_{\text{planet}}$ (all gears share the pitch).

$$N_3=\frac{N_4-N_2}{2}=\frac{60-18}{2}=\boxed{21},\qquad N_6=\frac{N_7-N_5}{2}=\frac{86-32}{2}=\boxed{27}.$$

(ii) Maximum number of equally-spaced planets

Two conditions must hold together: the assembly (equal-spacing) condition requires $(N_{\text{sun}}+N_{\text{ring}})/n_p$ to be an integer, and the adjacent-planet clearance requires $\sin(\pi/n_p)>(N_{\text{planet}}+2)/(N_{\text{sun}}+N_{\text{planet}})$ (full-depth addendum).

1. Stage 1. $N_2+N_4=78$; divisors $\{1,2,3,6,13,\dots\}$. Clearance: $\sin(\pi/n_p)>(21+2)/(18+21)=0.590\Rightarrow n_p\le4$. The largest divisor of 78 within that limit is $n_p=3$ ⇒ 3 planets.
2. Stage 2. $N_5+N_7=118=2\times59$; divisors $\{1,2,59,118\}$. Clearance: $\sin(\pi/n_p)>(27+2)/(32+27)=0.492\Rightarrow n_p\le6$. The largest divisor of 118 within that limit is $n_p=2$ ⇒ 2 planets.

(iii)–(iv) Output speeds

Approach. Each stage obeys the planetary relation $N_s\omega_s+N_r\omega_r=(N_s+N_r)\omega_c$. Both carriers are on the input shaft, so $\omega_{c1}=\omega_{c2}=1800$ rpm, and the single floating sleeve gives one coupling unknown $Z\equiv\omega_2=\omega_7$:

$$\underbrace{18\,Z+60\,\omega_4=78(1800)}_{\text{stage 1}},\qquad \underbrace{32\,\omega_5+86\,Z=118(1800)}_{\text{stage 2}}.$$

1. Brake $C$ (ring 4 held, $\omega_4=0$). Stage 1 becomes a fixed-ring, carrier-driven, sun-output stage: $$Z=\frac{78}{18}(1800)=\boxed{7800\text{ rpm}}\qquad(=\omega_2=\omega_7).$$ That ring speed drives stage 2, whose carrier also turns at 1800 rpm: $$32\,\omega_5=118(1800)-86(7800)=-458\,400\;\Rightarrow\;\boxed{\omega_{out}=-14\,325\text{ rpm}}$$ i.e. 14 325 rpm reversed (CW) relative to the input — a 7.96:1 step-up.
2. Brake $D$ (ring 7 held, $\omega_7=0$). Sun 2 shares the floating sleeve with ring 7, so holding $D$ grounds sun 2 as well and $Z=0$. Stage 2 is then carrier-driven with its ring fixed: $$32\,\omega_5=118(1800)\;\Rightarrow\;\boxed{\omega_{out}=+6637.5\text{ rpm}}$$ CCW, the same sense as the input — a 3.69:1 step-up. Stage 1 transmits nothing in this mode: with its sun held and its carrier driven, ring 4 simply free-wheels at $\omega_4=78(1800)/60=2340$ rpm.