The inertia converted to the output of the motor J = ?

The inertia converted to the output of the motor J = J load + J minus = 0.021 7 + 0.000 9 = 0.022 6kg·m2, where J minus is the load moment of inertia of the reducer itself, in kg·m2. Choose two ßis22/3000 servo motors, the moment of inertia of the motor is J electricity = 0.005 3kg·m2. The moment of inertia of the dual drives of the two motors is J double electricity = 1.5 J electricity = 1.5 × 0.005 3 = 0.007 95 kg·m2. 

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