# The torque I need to lift the device and correct gear ratio for the device to be as fast as possible

The device have to climb a pipe vertically. I have a DC motor with power rating 21.2 w, output speed 13360 rpm and maximum output torque 154.4 gcm. My device is 1 kg and I have wheel with 0.012 m radius. I did calculations and I found that my gear box should be 1:72 but I am not sure about that. I need the device to go upwards as fast as possible. With gear ratio 1:72 is enough for the DC motor to lift the device? And what I have to do to make it go faster? Make the gear ratio smaller or bigger?

• Can you please share specifications of the DC motor? – user8055 Mar 7 at 16:28
• Yes of course! It is a DC motor. Supply voltage 4.5 -> 15 V. Power rating 21.1 W. Output speed 13360 rpm shaft diameter 3.18 mm. Maximum output torque 154.4 gcm. Current rating 2.85 A. – Christos LA Mar 7 at 18:35

lifting a 1kg weight requires 9.8 WATT, we need to add to that the power needed to accelerate your mass from zero to the final speed.

So we assume an acceleration of.

a =1.8m/s

The power needed $$\ P= 9.8 + 1.8*1kg =11.8N.s =11.8W$$

Say your system efficiency is 80% then you need 11.8W* 1.25 = 14.75 W which is less than your DC motor output.

As for the gear.

you need a torque of $$14.75* (1.2cm/2*100cm/m) = 0.0885Nm \ <$$

your DC motor torque of $$0.1513Nm = 154.4*9.8/100cm/m*1000g/kg$$

This means no gear needed, your motor with the 1.2cm wheel directly attached to its shaft will lift the load.

As for the question of benefit of increasing gear ratio, you need to have the motor's torque and performance ratio diagrams, but adding gear will add to friction.

• For the torque i need you devide my radius devide by 2, maybe shouldn't be tge 2 there right?? Becausenit us already radius ( maybe you confused i gave the diameter before) – Christos LA Mar 7 at 22:52
• Yes, radius. If I wrongly assumed 1.2 as diamer, then you multiply demand torque by 2. – kamran Mar 8 at 0:03