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A 15-tooth spur pinion has a module of 3 mm and runs at a speed of 1600 rev/min. The driven gear has 60 teeth. Find the speed of the driven gear, the circular pitch, and the theoretical center-to-center distance.

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Answer:

Following are the solution to this question:


N_p= 1600 \ rpm\\\\N_G = 400 \ (rev)/(min)\\\\ p = 9.42 \ mm\\\\C = 112.5 \ mm\\

Step-by-step explanation:


t= \text{number of tooth in pinion}= 15 \\\\T= \text{number of tooth on Gear}= 60\\\\m= module = 3 \ mm\\\\N_p= 1600 \ rpm\\\\G= gear ratio= (I)/(t) = (60)/(15) =4 =(N_p)/(N_G) \\\\N_G =(1600)/(4) = 400 \ rpm \\\\Circular\ pitch = (\pi D)/(T) = \pi m= \pi (3) = 9.42 mm \\\\\\text{Center to center distance} = (m)/(2)(T+t)\\\\


=(3)/(2) (15+60)\\\\=(3)/(2) * 75\\\\=3 * 37.5\\\\= 112.5 \ mm

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