M029 · Epicyclic transmission

Planetary Gears & Bevel Differential

In a planetary gear set, why does holding a different part change how the others turn? Because sun, ring and carrier share one speed relation, and holding one decides how the other two drive each other. Seen from the carrier, the planet axles stand still and the sun and ring turn opposite ways like an ordinary gear pair. Hold the ring and the carrier follows the sun, more slowly. Hold the carrier and the ring turns backward. Hold the sun and the ring drives the carrier.

Explore planetary gears and a bevel differential. Change the held member or axle speeds and see how motion, torque and power are shared.

Drive speed $30\,\mathrm{rpm}$

Final output speed $43.5\,\mathrm{rpm}$

Left axle speed $16.5\,\mathrm{rpm}$

External output load torque $-0.5\,\mathrm{N\,m}$

Build a gearbox

Turn one shaft. Follow every output.

Drive speed

n_{\mathrm{in}}

30\,\mathrm{rpm}

Applied drive torque

T_{\mathrm{in}}

1\,\mathrm{N\,m}

Drive starting phase

\phi_0

0.55\,\mathrm{rad}

Ring drive speed

n_r

0\,\mathrm{rpm}

Left axle / case speed

n_L/n_c

0.55

All parallel shaft coordinates point toward the front; counterclockwise viewed from the front is positive. Lengths use one uniform millimetre scale, with module two millimetres. White markers belong to the gear material. Playback follows prescribed steady speeds, one model second per real second.

Torques are external torques applied to the gear network. Positive power enters it. Held parts and fixed-axis bearings supply the required support torque. A second drive may supply or absorb power. Speeds alone do not determine a dynamic load; the applied drive torque supplies the extra quasistatic condition.

The builder has up to three serial stages with keyed intermediate shafts. A mixed planetary stage holds its ring; a single planetary set also supports held sun, held carrier and two inputs. The bevel differential terminates the train. Its left axle speed is imposed by an external road or servo condition; both side torques are equal in this ideal open differential.

Spur tooth flanks use sampled twenty-degree involutes. Bevel teeth use schematic radial flanks on intersecting pitch cones, with a face band equal to twenty-seven percent of cone distance. Tooth roots, shafts, bearings and supports are schematic. The model omits inertia, clutch transients, backlash, friction, tooth forces, elastic deformation, strength, heat and limited-slip action. The integer design search uses the displayed finite catalogue and does not certify manufacture.

Model assumptions

Rigid, lossless, quasistatic gearing at imposed speeds. Spur flanks are sampled involutes; bevel flanks and supports are schematic. No inertia, friction, strength or clutch transients are computed.

Start with the essentials

Focus question: Why can two axles turn at different speeds and still carry equal torque?
One-sentence intuition: Subtract the carrier motion first. Relative to a differential case, the two equal side gears turn equally in opposite directions.

Core mathematical model

One mesh and a compound train

N_a n_a+N_b n_b=0,\qquad \frac{n_o}{n_{\mathrm{in}}}=\prod_j r_j

The two default pairs have tooth ratios one-third and one-third. Two reversals give a positive one-ninth speed ratio. The intermediate gears share a rigid keyed shaft.

Planetary geometry and motion

N_r=N_s+2N_p,\qquad N_s(n_s-n_c)+N_r(n_r-n_c)=0

The ring surrounds three equally spaced planets. Subtracting carrier speed fixes their axes in the moving frame. Two shaft speeds determine the third.

Bevel differential

n_L+n_R=2n_c,\qquad T_L=T_R=-\frac{T_c}{2}

Both side gears have the same tooth count. Torque signs refer to external loads applied to the mechanism. A positive case drive balances two negative axle loads, even when one axle stops.

Power is shared by speed

P_j=T_j\frac{2\pi n_j}{60},\qquad \sum_jP_j=0,\qquad \frac{P_L}{-P_c}=\frac{n_L}{2n_c}

The last fraction assumes a driven case, equal ideal side torques and nonzero case power. It is the fraction of input power delivered at the left axle when both axles are outputs. A reversed axle may instead supply power.

Intersecting reference cones

r_s=\frac{mN_s}{2},\quad r_p=\frac{mN_p}{2},\quad \tan\delta_s=\frac{N_s}{N_p},\quad \delta_s+\delta_p=\frac{\pi}{2}

The two bevel axes intersect at the shared cone apex. The visible radial teeth are schematic; these reference cones supply the ideal rolling relation. The module is two millimetres.

Common difficulties

Equal torque is not equal power

Typical misconceptionAn open differential always divides power in half.

Better mental modelThe two ideal side torques are equal. Power also includes speed: an axle held still carries torque but takes no power, so the moving axle can take all the case power.

A keyed shaft differs from an idler

Typical misconceptionAny intermediate gear cancels from the ratio.

Better mental modelA single idler’s teeth cancel across its two meshes. Two different gears keyed to one shaft introduce two different tooth counts; their product changes the overall reduction.

The plotted loads do not accelerate the gears

Typical misconceptionA larger applied torque must make the animation speed up.

Better mental modelThis model prescribes steady speeds and solves compatible quasistatic loads. A speed transient requires shaft inertia, motor and load dynamics, which are outside this model.

Run the experiment

01
Watch the untouched default
Watch the two axle gears, then choose Ring held to follow a planet spinning and orbiting.
What to observe: A held ring gives reduction. A held carrier reverses the ring. Equal sun and ring speeds rotate the set as one.
02
Open the differential
Choose the two-axle preset. Set left-axle speed to zero, then equal to the case speed. Watch the blue and green side gears.
What to observe: With the left held, the right runs at twice case speed. At equal axle speeds the spider gears stop spinning relative to the case, though the whole case keeps rotating.
03
Solve a tooth-count task
Open Design a transmission, choose a target and apply one integer solution. Inspect its stages and power ports.
What to observe: The solution uses the actual builder configuration. The left-power task also imposes a left-axle speed. An impossible target reports no exact solution in this catalogue.