www.jbj.co.uk/couplings.html #DriveLineHarmony Series ‘E’ Technical Characteristics SIT Universal Joints 2 Example » Power: 0,65 kW » -1 rpm: 230 min » With working angle α = 10, Factor F = 1,00 ( 0,65 kW : 1,00 = 0,65 kW) we get point P and Torque M = 27 Nm corresponding to joint size D = 25/26 mm T (type 04E, 1EB ) » With working angle α = 30, Factor F = 0,45 (0,65 kW : 0,45 = 1,44 kW) we get point P1 and Torque M = 60 Nm corresponding to joint size D = 32 mm T (type 1E, 3EB). Consider that: ( ( ( ( Working Angle “a” 5° 10° 15° 20° 25° 30° 35° 40° 45° Correction Factor “F” 1,25 1,00 0,80 0,65 0,55 0,45 0,38 0,30 0,25 Torque M in (Nm) T Joint size - Outside diameter “D” Speed [rpm] M = 9.550 x T M = 7.020 x T Power (kW) -1 rpm (min ) Power (CV) -1 rpm (min ) (Nm) (Nm) kW
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