What does a motor with 8 N·m at 1,500 rpm deliver after a 10:1 reduction?
Define the ratio up front
Mechify uses i = n₁ / n₂. A value greater than 1 is a reduction; a value between 0 and 1 increases the speed. The output speed is n₂ = n₁ / i. This explicit definition avoids confusion when catalogues express the ratio in a different order.
Torque with efficiency
The ideal output torque is T₁ · i. For a driving motor with efficiency η, we use T₂ = T₁ · i · η. With 1,500 rpm, 8 N·m, i = 10 and η = 0.92, you get 150 rpm and 73.6 N·m. The input delivers about 1.257 kW; the output about 1.156 kW. The difference is lost in this simple model.
Check more than the ratio
Look at the nominal and peak torque, backlash, torsional stiffness and the allowed radial and axial shaft load. A pulley mounted far outside the bearing can exceed the allowable shaft load even while the torque fits fine. Also assess lubrication, mounting position and heat dissipation for the actual cycle.
Assumptions and applicability limits
This model assumes a fixed ratio, constant efficiency and energy flow from input to output. Real efficiency varies with load and speed. Back-driving, self-locking, starting friction and dynamic peaks are not calculated.