How Does Gear Ratio Affect Motor Torque and Load Capacity?

[nomisme]

HI,
please help me... I am thirsty for any insights to my problem.

I have a motor:

Output: 60W
VOLT:12V DC
AMP: 7.5A
RPM:1800
TORQUE: 3.5kgcm

I have a motor reducer:
for converting rotation movement of motor to up and down motion(for lifting heavy load)
look like this http://www.yhcdsb.com/Eweb/uploadfile/20120405235504775.jpg
It has a speed reduction ratio of 12:1.

Based on theory, if I have a 3.5kgcm input torque to the motor reducer,I will get 3.5kgcm x 12= 42kgcm from the motor reducer. It means that I will have no difficulties in lifting loads less than 30kg without damaging the motor, right?

The problem is I tried both AC and DC motors. Both of them have difficulties lifting the 30kg loads and generate high heat which cause the motors to slowly die out.

I have checked the circuitry and no problem was found on the power supply.

I am really confused about the gear ratio stuff... I get my speed reduced by a factor of 12. Does it mean that my torque is multiplied by a factor of 12 too? It seems not to be the case. Does the efficiency of the motor reducer places an important part on it? how significant is it?

 

[Baluncore]

My first idea was that the weight of the central screw must be lifted in addition to the load.

The device you link to is a screw jack driven by a worm gear.

The screw is raised by a nut that is rotated by the input worm shaft.
The 12:1 ratio would be the ratio of the input worm to one turn of the internal geared nut.
So 12 turns of the input shaft will lift the screw by one pitch only.

A 12:1 ratio worm is quite a low ratio and so will have quite high friction.
Worm gears are not in rolling contact like spur gears.
If not lubricated correctly, any worm gear will have significant friction.

It appears that the screw has a simple thread, not a ball screw. A ball screw has balls that roll between the nut and screw. A simple thread has friction proportional to the load on the jack.

The internal nut carries the load as it rotates against the housing and should have some form of low friction thrust bearing.

So there are three sources of friction.
1. The worm driving the nut.
2. The nut bearing on the housing.
3. The nut against the screw thread.

But after all that analysis of the mechanical inefficiencies, the problem appears to be the motor specification of 1800 RPM.
A screw jack is a very heavy slow device. You may be driving it too fast, or the jack may be lubricated with thick grease, not light oil. Is the 3.5 kg.cm torque specified at a zero RPM stall, or at 1800 RPM ?

If the 1800 RPM is the “no load” DC motor maximum speed then there will be zero torque remaining at that speed. The RPM should drop to about 900 RPM for maximum power. That is 15 turns per second which will lift about 1.25 screw pitches per second. At 900 RPM I would expect the torque to be half of the stall torque specification because at half speed a DC motor will draw half the stall current.

Remember that power is torque * RPM so there is no available power from a DC motor when stalled or running at maximum RPM.

That is the quick summary of my thoughts on the subject. If you are lucky there will now be a rush of posts telling me I'm wrong. I don't know your level of understanding so some feedback would be good, maybe more questions, observations, or product details would help.

 

[Lok]

Torque force is very dependent on shaft diameter.
From what that picture shows there might be a gear there that if not of 2cm Diameter will reduce or amplify your torque. Which is good only 50% of times :P
the 42kgcm= 42kg of Force on a 2cm shaft diameter.
yet 42kgcm= 21Kg of Force on a 4cm shaft diameter.

And of course as said by Baluncore friction is always in play.

 

[Baluncore]

@Lok. I think you will find that the torque of 3.5 kg cm is independent of the shaft diameter that it drives.

It would be necessary to know the screw thread pitch in order to do the lifting calculation.

 

[Lok]

@Lok. I think you will find that the torque of 3.5 kg cm is independent of the shaft diameter that it drives.

It would be necessary to know the screw thread pitch in order to do the lifting calculation.

I have seen the mistake after the post as I usually do. And replaced the "Torque" with "Torque Force" thus implying that i meant the Force generated by said torque with a specific shaft diameter in mind.