What are the three fundamental rules for assembling a planetary gearset?

A true planetary gear set must satisfy three geometric rules: 1. Coaxiality (Distance Sun-Planet = Distance Ring-Planet), requiring Z_r = Z_s + 2 Z_p assuming identical standard modules. 2. Planet Equi-spacing Phase Check, requiring that (Z_s + Z_r) / N_p equals an exact integer, allowing equally spaced planets to mesh simultaneously with the sun and ring gears. 3. Adjacent Planet Tip Clearance, establishing a maximum number of planet gears based on physical diameter interference (c = distance between neighboring planet tip circles > 0).

How does the Willis equation predict epicyclic gear ratios?

The Willis kinematic equation (ω_s - ω_c) / (ω_r - ω_c) = - (Z_r / Z_s) provides the fundamental relationship between the sun speed (ω_s), carrier speed (ω_c), ring speed (ω_r), and tooth counts. By setting one member's speed to zero (the fixed grounding member), the equation simplifies uniquely for all 6 possible active inversion architectures (where a rotating input drives a single rotating output).

What is a Standard Planetary Reducer vs a Star or Solar Reducer?

The 3 common reduction modes are: 1. Standard Reducer (Ring fixed): Sun drives Carrier (Output) – Highest ratio (u = 1 + Zr/Zs), same rotation direction. 2. Solar Reducer (Sun fixed): Ring drives Carrier (Output) – Moderate ratio (u = 1 + Zs/Zr), same rotation direction. 3. Star Reducer (Carrier fixed): Sun drives Ring (Output) – Known as a parallel star set, acts as a simple reverse idler (u = -Zr/Zs), reverses rotation direction.

How is epicyclic gear train mechanical efficiency calculated?

Unlike parallel-shaft gears, planetary efficiency must account for power path splitting. Using AGMA/Dudley's approach, the base stationary-carrier basic efficiency (η_0, typically ~0.98-0.99 for two sequential spur/helical meshes) is transformed into the true epicyclic structural efficiency: η = 1 - (1 - η_0²) · |(ω_s - ω_c) / ω_in|. Standard Reducers typically exhibit extremely high real-world mechanical efficiencies (99%+), as the carrier rotates most of the power without forcing it through full sliding gear meshes.

How does steady-state torque distribute within a planetary gearbox?

Ignoring frictional losses for idealized static equilibrium, the sum of external torques acting on the complete assembly must equal zero: T_s + T_r + T_c = 0. Therefore, the holding torque required on the fixed structural housing (e.g., the Ring gear in a standard reducer) always equals the difference between the high output torque and the low input torque. By identifying the input component mathematically, T_out = T_in × Ratio × Efficiency.

Machine Design

AGMA 6123 / ISO 6336 Base

PRO Suite

Planetary / Epicyclic Gear Ratio & Torque Calculation PRO Suite

Calculate speed ratios, torque distribution, and mechanical efficiency for single-stage and compound epicyclic gear trains across all 6 Willis inversion modes (standard reducer, solar, star) incorporating AGMA design standards and exact kinematic mesh generation.

Input Parameters

Authoritative Presets (AGMA / ISO)

Standard planetary stage operating in high reduction mode (Ring fixed, Sun input, Carrier output). Characteristic of modern 8-speed automatic step-gear transmissions.

Unit System

Kinematic Operating Mode (Willis)

Gear Tooth Geometry (Stage 1)

Sun Teeth (Z_s):24

Planet Teeth (Z_p):18

Ring Teeth (Z_r):60

Number of Planets (N_p):3

Module (m) [mm]

Pressure Angle (α)

Operating Speed & Torque

Input Shaft Speed:3500 RPM

Input Driving Torque:250 N·m

Mesh Base Efficiency (η₀):98.5%

Load Sharing (K_γ):1.05

Compound 2-Stage Train

Epicyclic Gear Ratio & Torque Synthesis

Input Parameters

Driving Member	3500RPM
Input Torque	250N·m
Sun Teeth (Z_s)	24
Planet Teeth (Z_p)	18

Calculated Results

Speed Ratio (i)	3.5:1 KEY
Output Speed	1,000RPM
Delivered Torque	856.39N·m
Epicyclic Efficiency	97.87%

AGMA 6123-B06 / ISO 6336 Epicyclic Train Executive Audit

Standard Reducer (Ring Fixed)

Geometric Assembly Feasible

Overall Speed Ratio (i)

3.5:1

Direct (Same Dir)

Output Shaft Speed

1,000RPM

From 3,500 RPM In

Delivered Output Torque

856.4N·m

Reaction: 606.4 N·m

Epicyclic Efficiency (η)

97.87%

Loss: 1.95 kW

Stage 1 Geometric Assembly & Packaging ValidationModule m = 2.50 mm

1. Coaxiality Match

MATCH (Δ=0)

Z_r (60) = Z_s (24) + 2·Z_p (2×18 = 60)

2. Symmetrical Spacing

VALID (rem=0)

(24 + 60) mod 3 = 0 (3 planets @ 120°)

3. Planet Tip Clearance

c = 40.9 mm

Adjacent chord 90.9 mm - Tip dia 50.0 mm

Member Rotational Speeds (Stage 1)

Sun Gear Speed (n_s):3,500 RPM

Carrier Orbit Speed (n_c):1,000 RPM

Ring Gear Speed (n_r):0 RPM

Planet Pin Spin Relative (n_p,rel):-3,333.3 RPM

Planet Absolute Speed:-2,333.3 RPM

Torque Equilibrium & Mesh Forces

Input Torque (T_in):250 N·m

Delivered Torque (T_out):856.4 N·m

Reaction on Fixed Ring Gear (Annulus):606.4 N·m

Per-Planet Tangential Mesh (F_t,p):2,916.7 N

Planet Pin Bearing Load (F_pin = 2·F_t):5,833.3 N

Epicyclic Gear Mesh Cross-Section (Stage 1)

Pitch Circles

Carrier Spider

Clearances

Rotations

Sun: Z_s = 24

Planet (3×): Z_p = 18

Ring: Z_r = 60

Carrier: a_sp = 52.5mm

Kinematic & Statics Breakdown

Static Torque Equilibrium Matrix

Input (Sun Gear)250 N·m

Output (Planet Carrier)856.4 N·m

Reaction (Fixed Housing)606.4 N·m

Static balance: T_in — T_out + T_reaction = 0

Planet Gear Dynamic Loads

Tangential Load (F_t)2,917 N

Radial Separation (F_r)1,062 N

Total Pin Reaction (2·F_t)5,833 N

Epicyclic Power Distribution

Mechanical Input Power:91.63 kW

Delivered Output Power:89.68 kW

Meshing Dissipation (Heat):- 1.949 kW

Verification Calculation Log

1*Step 1: Verify Coaxiality & Center Distance Match (Spur Gears)** $$ Z_r = Z_s + 2 \cdot Z_p $$ $$ Z_r = 24 + 2 \cdot 18 = 60 \quad \text{vs actual } Z_r = 60 $$ $$ \text{MATCH: } Z_r = 60 \quad (\Delta = 0) $$ *Note:* Coaxiality condition is perfectly satisfied. Pitch centers are concentric.
- •*Step 2: Verify Equi-Spacing & Planet Assembly Condition** $$ (Z_s + Z_r) \pmod{N_p} = 0 $$ $$ (24 + 60) \pmod{3} = 84 \pmod{3} = 0 $$ $$ \text{VALID: } \frac{84}{3} = 28 \in \mathbb{Z} $$ *Note:* Planets can be installed at exact 120.0° intervals.
- •*Step 3: Calculate Adjacent Planet Tip-to-Tip Clearance** $$ c_{\text{tip}} = 2 \cdot a_{sp} \cdot \sin\left(\frac{\pi}{N_p}\right) - m \cdot (Z_p + 2) $$ $$ c_{\text{tip}} = 2 \cdot 52.50 \cdot \sin\left(\frac{\pi}{3}\right) - 2.5 \cdot (18 + 2) = 90.93 - 50.00 $$ $$ c_{\text{tip}} = 40.93 \text{ mm} $$ *Note:* Clearance of 40.93 mm prevents planet tip clashing during orbital rotation.
- •*Step 4: Compute Kinematic Velocity Ratio (Standard Planetary Reducer (Ring Fixed, Sun In, Carrier Out))** $$ i = \frac{\omega_{\text{in}}}{\omega_{\text{out}}} $$ $$ i = 3.5000 $$ $$ i = 3.5000 : 1 \quad (\text{Reduction factor } = 3.5000) $$ *Note:* Output rotates in SAME direction as input. Speed ratio i = 3.5000.
- •*Step 5: Determine Absolute Member Speeds & Planet Pin Spin** $$ n_{\text{out}} = \frac{n_{\text{in}}}{i}, \quad n_{p,\text{rel}} = -(n_s - n_c) \frac{Z_s}{Z_p} $$ $$ n_{\text{out}} = \frac{3500}{3.5000} = 1000.0 \text{ RPM}, \quad n_{p,\text{rel}} = -(3500 - 1000.0) \frac{24}{18} $$ $$ n_{\text{out}} = 1000.0 \text{ RPM}, \quad n_{p,\text{spin}} = 3333.3 \text{ RPM} $$ *Note:* Sun: 3500 RPM | Carrier: 1000.0 RPM | Ring: 0 RPM | Planet Spin: 3333.3 RPM.
- •*Step 6: Evaluate Epicyclic Meshing Efficiency (AGMA / Dudley)** $$ \eta = 1 - (1 - \eta_0^2) \cdot \left| \frac{\omega_s - \omega_c}{\omega_{\text{in}}} \right| $$ $$ \eta = 1 - (1 - 0.985^2) \cdot \left| \frac{3500 - 1000.0}{3500} \right| $$ $$ \eta = 97.87\% \quad (P_{\text{loss}} = 1.949 \text{ kW}) $$ *Note:* Input power: 91.63 kW | Output power: 89.68 kW.
- •*Step 7: Calculate Output Torque & Housing Reaction Torque** $$ T_{\text{out}} = T_{\text{in}} \cdot |i| \cdot \eta, \quad T_{\text{reaction}} = |T_{\text{out}} - T_{\text{in}}| $$ $$ T_{\text{out}} = 250 \cdot 3.5000 \cdot 0.9787 = 856.4 \text{ N}\cdot\text{m} $$ $$ T_{\text{out}} = 856.4 \text{ N}\cdot\text{m}, \quad T_{\text{reaction}} = 606.4 \text{ N}\cdot\text{m} $$ *Note:* The fixed Ring Gear (Annulus) must be anchored to react 606.4 N·m into the frame.
- •*Step 8: Calculate Per-Planet Tangential Load & Carrier Pin Reaction** $$ F_{t,p} = \frac{2 \cdot T_{\text{sun}}}{d_{s} \cdot N_p} \cdot K_{\gamma}, \quad F_{\text{pin}} = 2 \cdot F_{t,p} $$ $$ F_{t,p} = \frac{2 \cdot 0.0600}{(60.0 \cdot 10^{-3}) \cdot 3} \cdot 1.05 $$ $$ F_{t,p} = 2916.7 \text{ N}, \quad F_{\text{pin}} = 5833.3 \text{ N} $$ *Note:* Separating radial force per mesh F_r = 1061.6 N (pressure angle 20°).

Mathematical Models & Equations

Coaxiality & Center Distance Match Condition

Z_r = Z_s + 2 \cdot Z_p

Planet Assembly & Equi-Spacing Rule

\frac{Z_s + Z_r}{N_p} = K \quad (K \in \mathbb{Z}) \iff (Z_s + Z_r) \pmod{N_p} = 0

Adjacent Planet Tip-to-Tip Clearance

c_{\text{tip}} = 2 \cdot a_{sp} \cdot \sin\left(\frac{\pi}{N_p}\right) - m \cdot (Z_p + 2) \ge 0.5m

Willis Relative Speed Equation (Fundamental Epicyclic Ratio)

\frac{\omega_s - \omega_c}{\omega_r - \omega_c} = -\frac{Z_r}{Z_s} = -R_0

Standard Planetary Reducer (Ring Fixed)

i = \frac{\omega_{\text{in}}}{\omega_{\text{out}}} = 1 + \frac{Z_r}{Z_s} = 1 + R_0

Solar Reducer (Sun Fixed)

i = \frac{\omega_{\text{in}}}{\omega_{\text{out}}} = 1 + \frac{Z_s}{Z_r} = \frac{1 + R_0}{R_0}

Star Gear Reducer / Inverter (Carrier Fixed)

i = \frac{\omega_{\text{in}}}{\omega_{\text{out}}} = -\frac{Z_r}{Z_s} = -R_0

Static Torque Equilibrium & Reaction Torque

T_{\text{in}} + T_{\text{out}} + T_{\text{reaction}} = 0 \implies T_{\text{reaction}} = T_{\text{in}} (i \cdot \eta - 1)

Per-Planet Tangential Mesh Force & Load Sharing

F_{t,p} = \frac{2 \cdot T_{\text{sun}}}{d_{\text{pitch,sun}} \cdot N_p} \cdot K_{\gamma}

Epicyclic Meshing Power Efficiency

\eta = 1 - (1 - \eta_0^2) \cdot \left| \frac{\omega_s - \omega_c}{\omega_{\text{in}}} \right|

Engineering DisclaimerEngineering calculations are provided for preliminary design and educational/reference purposes only. Users must verify all results according to applicable engineering standards, supplier data, manufacturing requirements, and professional engineering judgment.