Epicyclic gear train tabular method

Tabular method

Step 1: Fix the arm and give the sun gear $+x$ revolutions (clockwise is taken as positive in this problem). Step 2: Find the revolutions of the other gears from the teeth ratios. Step 3: Give $+y$ revolutions to the whole train (arm included); add the rows to get the actual motions. Step 4: Use the fixed-gear condition and the arm speed to find $x$ and $y$.

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

With clockwise positive, 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's.)

(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; a 5:1 speed increase). 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.