n13_helical_gears page 1 of 3 MOTOR X Wa Z Wt TOP VIEW Wr Wr Wa Wt Y Y X MOTOR Z J K DRIVER LJ = 2 inch LK = 3 inch SIDE VIEW FRONT VIEW d = 4 inch 18 HP = 20° pressure angle r = 2 inch 250 rpm desire 108 rotations 250 rev min 250 rpm = min 60 sec 2 rad = 26.18 rad/sec rev P=T sec 550 ft .lbf T = P / = 18 HP 26.18 rad HP. sec T = Wt r Wt = T / r = 2269 lbf Wr = Wt tan = 825.8 lbf = 30° helix angle 12 in = 4538 in.lbf ft Wa = Wt tan = 1310 lbf n13_helical_gears MOTOR page 2 of 3 FXJ FXK X Z Wa FZJ FZK Wt TOP VIEW Wr Wr Wa Y X MOTOR FYJ FXJ Wt FYK FXK J Y Z K DRIVER LJ = 2 inch LK = 3 inch SIDE VIEW FRONT VIEW TOP VIEW M Y about J Wt L J FZK L J L K 0 CCW positive FZK = +907.6 lbf TOP VIEW M Y about K CCW positive Wt L K FZJ L J L K 0 FZJ = +1361.4 lbf FRONT VIEW M Z about J CCW positive Wr L J Wa r FYK L J L K 0 FYK = -193.7 lbf FRONT VIEW M Z about K Wr L K Wa r FYJ L J L K 0 CCW positive FYJ = +1019.5 lbf n13_helical_gears page 3 of 3 FJ _ RADIAL FYJ FZJ = 1701 lbf 2 2 FK _ RADIAL FYK FZK = 928 lbf 2 2 assume worst case - Wa acts on bearing J ball bearing selection Fr = 1701 lbf P = X V Fr + Y Fa V=1 try desire L10 = 100 for 100x106 rev Fa = 1310 lbf inner race rotating Eq. 11.22a Fig. 11-24 X=1 L10 = (C / P)3 Y=0 P = Fr = 1701 lbf Eq. 11.20a CDESIRED = P (L10)1/3 = 1701 lbf (100)1/3 = 7895 lbf try 6309 bearing Fa / C0 = 0.1955 C = 9150 lbf C0 = 6700 lbf interpolate e = 0.3493 Fa / (V Fr) = 0.7701 > e X = 0.56 P = X V Fr + Y Fa = 2620 lbf Fig. 11-23 Fig. 11-24 interpolate Y = 1.273 Fig. 11-24 Eq. 11.22a CDESIRED = P (L10)1/3 = 2620 lbf (100)1/3 = 12,161 lbf try 6311 bearing Fa / C0 = 0.131 C = 12,900 lbf interpolate e = 0.314 Fa / (V Fr) = 0.7701 > e X = 0.56 P = X V Fr + Y Fa = 2788 lbf C0 = 10,000 lbf Fig. 11-23 Fig. 11-24 interpolate Y = 1.401 Fig. 11-24 Eq. 11.22a CDESIRED = P (L10)1/3 = 2788 lbf (100)1/3 = 12,940 lbf probably OK 6311 bearing C = 12,900 lbf very close to CDESIRED
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