Simple and Planetary gear trains

Types of gear trains

For a reverted train, the centre distances must be equal: $(z_A + z_B) = (z_C + z_D)$ for the same module (or equal modules times teeth). Here the train is not reverted because $20 + 60 \ne 18 + 54$.

Formula

$$\text{Train value} = \frac{N_{out}}{N_{in}} = \frac{\text{product of teeth on drivers}}{\text{product of teeth on followers}} = \frac{z_A \times z_C}{z_B \times z_D}$$

Calculation

Drivers: A (20) and C (18); followers: B (60) and D (54).

$$\frac{N_D}{N_A} = \frac{20 \times 18}{60 \times 54} = \frac{360}{3240} = 0.1111$$ $$N_D = 1440 \times 0.1111 = 160\ \text{rpm}$$

Intermediate shaft speed: $N_B = N_C = 1440 \times \dfrac{20}{60} = 480$ rpm.

Direction

Each external mesh reverses the sense of rotation. There are two external meshes (A-B and C-D), so the output rotates in the same direction as the motor shaft.

   motor --> A(20) o--o B(60)=C(18) o--o D(54) --> output
   1440 rpm            480 rpm            160 rpm
   (cw)    (ccw)            (ccw)   (cw)

Answer: output speed $= 160$ rpm in the same direction as the motor; train value $= 0.1111$ (speed reduction 9:1).

Tabular method

Step 1: Fix the arm and give the sun gear +1 revolution (anticlockwise taken positive). Step 2: Find the revolutions of the other gears from the teeth ratios. Step 3: Give $+x$ revolutions to the whole train; add the columns to get the actual motions. Step 4: Use the fixed-gear condition to find $x$, then use the arm speed.

Check geometry: $z_{annulus} = z_{sun} + 2z_{planet} = 20 + 60 = 80$, consistent.

Clockwise is taken as positive here, so the arm speed is $y = +100$ rpm.

(The planet turns opposite to the sun in an external mesh. The annulus turns the same way as the planet in the internal mesh, so its sign is opposite to the sun.)

(a) Annulus fixed

$$y - 0.25x = 0 \Rightarrow x = 4y = 400\ \text{rpm}$$ $$N_{sun} = y + x = 100 + 400 = 500\ \text{rpm}$$ $$N_{planet} = y - 0.6667x = 100 - 0.6667(400) = -166.67\ \text{rpm}$$

The sun turns at 500 rpm clockwise (same direction as the arm, speed ratio 5:1 up). The planet turns at 166.67 rpm anticlockwise (negative means opposite to the arm).

(b) Sun fixed

$$y + x = 0 \Rightarrow x = -y = -100\ \text{rpm}$$ $$N_{annulus} = y - 0.25x = 100 - 0.25(-100) = 125\ \text{rpm}$$ $$N_{planet} = y - 0.6667x = 100 - 0.6667(-100) = 166.67\ \text{rpm}$$

The annulus turns at 125 rpm clockwise and the planet at 166.67 rpm clockwise.

Answer: (a) sun $= 500$ rpm cw, planet $= 166.67$ rpm ccw; (b) annulus $= 125$ rpm cw, planet $= 166.67$ rpm cw.

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.

Condition for equal spacing

Let the sun have $z_s$ teeth and the ring $z_r$ teeth. Fix the ring and place the first planet in mesh with both. Now rotate the arm to bring the next planet position, a fraction $1/N$ of a revolution away (angle $360^\circ/N$). The sun turns through

$$\theta_s = \frac{360^\circ}{N}\left(1 + \frac{z_r}{z_s}\right)$$

(arm rotation $1/N$, and the sun speed relative to the ring-fixed arm is $1 + z_r/z_s$ times the arm speed).

The second planet will mesh correctly only if the sun has turned through a whole number of tooth pitches, that is, $\theta_s$ equals $k$ teeth $= k \times \dfrac{360^\circ}{z_s}$:

$$\frac{z_s + z_r}{N} = k \quad \text{(integer)}$$

So the sum of the teeth on the sun and ring must be divisible by the number of planets. Also, for the geometry: $z_r = z_s + 2z_p$ (equal module).

Check: $z_s = 24$, $z_r = 72$

Geometry: $z_p = (72 - 24)/2 = 24$ teeth. The planet teeth given (24) agree.

$$z_s + z_r = 24 + 72 = 96$$

Also the planets must not touch each other: for $N = 6$, the centre distance $m(24+24)/2 = 24m$ and the chord between neighbouring planet centres is $2 \times 24m\sin 30^\circ = 24m$, which is equal to the planet pitch diameter $24m$, so they just touch. Hence $N = 6$ is not practical, while $N = 3$ and $N = 4$ are fine.

Applications

1. Automatic transmission of vehicles (speed change by locking elements).
2. Differential of automobiles (bevel epicyclic train).
3. Reduction gearbox of wind turbines, hoists and cranes.
4. Turbine speed reducers; watch mechanisms and epicyclic screwdrivers/hand drills.