The specifications of the generator is as follows:
Efficiency- 70% Rated Torque- 0.56Nm Rated current- 14.3A Voltage- 24V Rotor shaft dimension- 1cm(diameter)
The gear train is designed such that a spool/winch will be attached tot he last gear of the gear train. A weight block will be suspended through that spool/winch and the gear box must be designed in such a manner that it will provide a rotation of 2900 rpm but the rate of drop of the weight must be 1meter/min.
Diameter of shaft= 1cm Circumference of shaft= Πd = (22/7) *1 cm Circumference is equal to one revolution Generator RPM= 2900 Total length covered by generator shaft per minute = 2900*22/7 = 63800/7 cm Total length required at output of motor 1meter/min =100 cm Required gear ratio= (63800/7)/100 =638/7 Output torque is= gear ratio* input torque* efficiency // This is the torque acting on the gear attached with the spool/winch = (638/7)* 0.56* 0.7 =35.728 Nm The gear teeth to achieve the drop rate of 1meter/min is: 1st layer= 7 and 25 2nd layer= 5 and 29 3rd layer= 5 and 22
I am a CS graduate so this is not my field of expertise. It happened so that during this quarantine i am trying to indulge myself into some DIY mechanical projects. Please guide me in case i am wrong somewhere.