Usually we are used to describe the power of the electronic device with a load capacity possessed, but for servo motor system, the main part is the output torque which could guide the relative concepts.  From the notion of the motor driving circuit design, the power is focused especially.  For the purpose we must clarify the relationship among the power, the torque, and the speed as following:

1. Power ( P _W ) = Force ( F ) x Linear Velocity ( V )
2. Force ( F ) = Torque ( T _Nm ) / Radius ( r )
3. Linear Velocity ( V ) = 2 π r x Angular Velocity ( n _RPS ) = ( 2 π r x n _RPM ) / 60 = ( π r x n _RPM ) / 30

• Guide formula 2 and 3 into formula 1, and then we can get:
• Power ( P _W ) = ( T _Nm / r ) x ( π r x n _RPM / 30 ) = ( π / 30 ) x T _Nm x n _RPM

We can transfer the unit of Power from ( P _W ) to ( P _KW ), and then get the following:

• Power ( P _KW ) = { ( π / 30 ) x T _Nm x n _RPM } / 1000
• Power ( P _KW ) = ( 3.1415926 / 30000 ) x T _Nm x n _RPM
• Power ( P _KW ) = ( T _Nm x n _RPM ) / 9549.297 ≒ ( T _Nm x n _RPM ) / 9550

Through this finalized fomula:  Power ( P _KW ) = ( T _Nm x n _RPM ) / 9549.297 ≒ ( T _Nm x n _RPM ) / 9550  we can make the choice for selecting the proper planetary gearbox.  For example:

1. Select the servo motor with proper power capacity
   ⇒  example: 100W / defined output speed 3000 RPM
2. Checking the required working speed (RPM) and Torque (Nm) for applying machine
   ⇒ example: 10 RPM / 70 Nm
3. Reduction ratio of gearbox
   ⇒ sevo motor output speed 3000 RPM / gearbox output speed 10 RPM = 1:300
4. The defined output torque of servo motor
   ⇒ (100W / 1000 ) x 9550 = ( T _Nm x 3000 RPM ), T _Nm = ( 0.1 KW x 9550 ) / 3000 RPM = 0.3183 Nm
5. Result: 
   ⇒ through reduction ratio (1:300) of gearbox, the output speed is 10 RPM, and the output torque = 0.3183 Nm x 300= 95.49 Nm, and then, according to the output torque we can choose the proper gearbox in applying.