Question 3 of 6: Two-stage planetary gear train — design for 1500→39 rpm

Question 3: Two-stage planetary gear train — design for 1500→39 rpm (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. $\omega_i=1500$ rpm (CCW viewed from the output end); target $\omega_o=39$ rpm, i.e. a reduction of $1500/39=38.46\!:\!1$; one common module $m$ and one pressure angle (take $20^\circ$ full depth).

Find. the general expression $\omega_o(\omega_i,N_k)$, and tooth numbers $N_1,N_2,N_{2'},N_3,N_4$ satisfying constraints (i)–(iv).

Reading the schematic

The exam calls the unit a “2-stage PGT” and the drawing bears that out. Three features settle it, all measurable on the printed figure:

• The arm labelled H runs from the main axis out to a horizontal shaft that carries gear 2 at one station and gear 2′ at another, with a journal-bearing symbol where that shaft crosses the arm — that is a rotating carrier with a compound planet, not a fixed bearing.
• The radii obey the planetary coaxial relations: the arm sits at $r_1+r_2$ from the axis, and each ring tooth at the arm radius plus its own planet radius, i.e. $r_3=r_1+2r_2$. On fixed-axis shafts there is no reason for those identities to hold.
• Gears 3 and 4 are drawn with their teeth pointing inward (toward the axis) on bodies that wrap around the train — the standard symbol for internal ring gears. Ring 3’s body is flanged to the casing wall (ground hatching), so $\omega_3=0$; ring 4’s body is continuous with the output shaft.

So: sun 1 in, ring 3 fixed, compound planet 2–2′ on free carrier H, ring 4 out — a two-stage (three-central-member, 3K) epicyclic. Stage 1 is the meshes $1\!-\!2\!-\!3$; stage 2 is the mesh $2'\!-\!4$.

General velocity expression

Use the Willis (relative-to-carrier) equation: viewed from the arm, an epicyclic becomes an ordinary train.

Stage 1 — from sun 1 to ring 3 through planet 2 (one external mesh, then one internal mesh):

$$\frac{\omega_3-\omega_H}{\omega_1-\omega_H}=\left(-\frac{N_1}{N_2}\right)\left(+\frac{N_2}{N_3}\right)=-\frac{N_1}{N_3}.$$

Ring 3 is held, $\omega_3=0$, which gives the carrier speed

$$\omega_H=\omega_1\,\frac{N_1}{N_1+N_3},\qquad \omega_1-\omega_H=\omega_1\,\frac{N_3}{N_1+N_3}.$$

Planet speed, from the same first-stage mesh, $\dfrac{\omega_2-\omega_H}{\omega_1-\omega_H}=-\dfrac{N_1}{N_2}$.

Stage 2 — the compound planet turns as one body ($\omega_{2'}=\omega_2$) and drives ring 4 through an internal mesh:

$$\frac{\omega_4-\omega_H}{\omega_{2'}-\omega_H}=+\frac{N_{2'}}{N_4}.$$

Substituting the previous two lines into this one and collecting $\omega_1$:

$$\boxed{\;\frac{\omega_o}{\omega_i}=\frac{\omega_4}{\omega_1}=\frac{N_1}{N_1+N_3}\left(1-\frac{N_3\,N_{2'}}{N_2\,N_4}\right)\;}$$

The bracket is the signature of the two-stage design: the output is the small difference between the carrier motion and what the second ring takes back from it, which is how such a compact train reaches a 38:1 reduction with only three meshes.

Constraint (i): the geometric (coaxial) constraint

With one module $m$, pitch radii are $r=\tfrac{m}{2}N$. The first stage is coaxial only if the sun, planet and ring radii close on the same centre:

$$r_3=r_1+2r_2\;\Longrightarrow\; \boxed{N_3=N_1+2N_2}\qquad\text{(the first-stage geometric constraint).}$$

The second stage must put planet 2′ on the same planet shaft, i.e. at the same carrier radius $r_1+r_2$, so

$$r_4=r_1+r_2+r_{2'}\;\Longrightarrow\; N_4=N_1+N_2+N_{2'}.$$

Choosing the teeth

Searching integer sets that satisfy both coaxial conditions and hit $39$ rpm gives a clean solution:

$$N_1=26,\quad N_2=25,\quad N_{2'}=29,\quad N_3=76,\quad N_4=80.$$

• Geometric (i): $N_1+2N_2=26+50=76=N_3$ $\checkmark$; and $N_1+N_2+N_{2'}=26+25+29=80=N_4$ $\checkmark$.
• Ratio (ii): $\dfrac{\omega_o}{\omega_i}=\dfrac{26}{102}\left(1-\dfrac{76\times29}{25\times80}\right)=0.254902\,(1-1.102)=-0.026$, so $\omega_o=1500(-0.026)=-39.00$ rpm — an error of $0.0\%$, well inside $\pm3\%$ $\checkmark$. The minus sign means the output turns opposite to the input: input CCW from the output end $\Rightarrow$ output CW from that end.
• Non-integer ratios (iii): the three mating pairs are $N_2/N_1=25/26=0.962$, $N_3/N_2=76/25=3.040$ and $N_4/N_{2'}=80/29=2.759$ — none an integer, and $\gcd(26,25)=\gcd(25,76)=\gcd(29,80)=1$, so every pair is a full hunting-tooth set and wear is spread over all teeth $\checkmark$.
• Interference (iv): the only external mesh is $1\!-\!2$, with $N_1=26$ and $N_2=25$, both comfortably above the $18$-tooth undercut minimum for $20^\circ$ full-depth teeth, so no too-few-teeth interference. For the two internal meshes the relevant limit is tip interference between ring and planet, avoided when the tooth difference is large enough: $N_3-N_2=51$ and $N_4-N_{2'}=51$, far above the usual minimum of about $10$–$12$. No gear exceeds $80$ teeth, so no too-many-teeth interference $\checkmark$.

Intermediate speeds, useful for bearing and shaft sizing:

$$\omega_H=1500\frac{26}{102}=382.4\ \text{rpm (same sense as the input)},\qquad \omega_2=\omega_{2'}=\omega_H-\frac{26}{25}(\omega_1-\omega_H)=-780.0\ \text{rpm}.$$

$$\boxed{\;N_1{=}26,\ N_2{=}25,\ N_{2'}{=}29,\ N_3{=}76,\ N_4{=}80\ \Rightarrow\ \omega_o=-39.0\ \text{rpm (reversed)},\ \text{error }0.0\%\;}$$

As a design note the train also assembles with equally spaced planets: $(N_1+N_3)/n_p=102/n_p$ is an integer for $n_p=3$, and the adjacency check $\sin(\pi/3)=0.866>(N_{2'}+2)/(N_1+N_2)=31/51=0.608$ is satisfied, so three compound planets can be fitted — sharing the load three ways at the 38:1 reduction the rolling mill needs.