# Can we calculate the torque acting on the last gear determine teeth of the gears of the gear train in the below addressed manner?

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.

• Please make a sketch to clarify. You have your torque and the radius. That's enough for getting the force. But what do you need it for. Do you need the tooth breaking force? Do you have a specification? Aug 13 '20 at 16:26
• HINT: The tangential force ($F_T$), tangential acceleration ($a_T$) and tangential velocity ($v_T$) of each of the mating gears in a gear train ideally remain constant. The torque $T$ transmitted on a gear is directly to its radius $R$ (or diameter $D$) i.e. $T\propto R$ for any two mating gears $$\frac{T_1}{R_1}=\frac{T_2}{R_2}\iff T_1\frac{v_T}{R_1}=T_2\frac{v_T}{R_2}\iff T_1\omega_1=T_2\omega_2$$ Aug 16 '20 at 14:03