Gears create more torque? Prove it

[Luchekv]

Hi guys,

I was sizing up a servo motor the other day and started thinking. While we all know gear ratios will effectively "convert" speed into torque. I couldn't help wondering about the initial starting conditions.. The problem below...

Lets say I have a motor with a max load of 1Kg (holding). The motor has a driving gear with a 1cm radius. Therefore:

Required torque = 1Kg_F x 1cm = 1Kg_F.cm = 0.098 N.m of torque

Now let's say we also have another gear with twice the radius. To lift the same 1Kg weight we now require 0.196 N.m of torque.

Required torque = 1Kg_F x 2cm = 2Kg_F.cm = 0.196 N.m of torque

So the question is this:

Lets pretend that motor has a 1cm radius driving gear...that is connected to the larger 2cm radius gear.

If my motor cannot lift(move) a weight of 1Kg through its driving gear (requiring T > 0.098 N.m)

How can it be expected to start turning an even larger gear that will require twice the torque to get it moving?

 

[andrewkirk]

The problem statement is a bit muddled.
Motors generate torque, not force, and certainly not mass, so the 1kg figure doesn't fit.

First describe the torque the motor generates, calling it T.
Then if the motor interacts at radius R1 with another cog with radius R2 that is driving a linear shaft at radius R3, the force on the linear shaft will be T x (R2 / R1) / R3.

It looks like you have stated R1 and R2 but not R3, and it is not clear what T is either.

 

[Luchekv]

Apologies I should have included a diagram, please see below:

This is what I was referring to with the calculation: 1KgF x 1cm = 1KgF.cm = 0.098 N.m of torque
As for the motor, by "Max Load (holding)" I meant the max torque output is 0.098 N.m

As for the gear coupling, see below:

A motor with T = 0.098 N.m drives Gear1(1cm) which impact onto Gear2 (2cm)

Hope that clarifies the question :)

EDIT: I did confuse my units N.m is definitely not force haha

 

[andrewkirk]

In that case we have R2 = R3 = 2cm and R1 = 1cm so plugging those into the formula above we see that the upward force on the weight is 9.8N which is what is required to keep the weight from falling.

 

[A.T.]

As for the gear coupling, see below:

That is not how you build a gear box, to lift heavier weights.

 

[Luchekv]

That is not you build a gear box, to lift heavier weights.

and that's probably why I'm confused - All the YouTube videos 'introducing' the concept of gears display the configuration like this. So how ARE they coupled to achieve greater lifting strength?

It just didn't make sense to me how simply having a bigger gear next to the first one would amplify the torque. When we can clearly see here that a larger radius requires more torque to move the same 1Kg weight.

Required torque = 1KgF x 1cm = 1KgF.cm = 0.098 N.m of torque
Required torque = 1KgF x 2cm = 2KgF.cm = 0.196 N.m of torque

The only thing I can see happening with the configuration in my last post is it wouldn't be able to hold the weight anymore. Would you agree?

 

[andrewkirk]

and that's probably why I'm confused

instead of wrapping the cord around the outside of the large cog, wrap it around a disc that is locked concentrically to the big cog but has a smaller radius. That way R3 << R2 and the lifting force can be much greater.

 

[Dale]

Using a bigger gear and then another locked concentrically to it just seems redundant?

Think of it like a lever. You get more force at points near the fulcrum. You always need the concentric bits to change torque.

 

[Luchekv]

Hmm ok..but let's say:
Load = 1Kg
R1=1cm
R2=2cm
R3=1cm

Aside from R3 now turning slower due to R1 & R2 sizing...would you still not only be generating 0.098 N.m at R3? Doesn't that in the end just boil down to a 1:1?

Scenario 2:
Load = 1Kg
R1=1cm
R2=2cm
R3=1.5cm
R3: 1KgF x 1.5cm = 1.5KgF.cm = 0.147 N.m of torque which is more than what the motor can output

 

[A.T.]

So how ARE they coupled to achieve greater lifting strength?

Around 4:00 min:

 

[256bits]

That is not how you build a gear box, to lift heavier weights.

Could be that the cord is winding around a larger drum, with the small gear engaging a rim gear on the drum.
The cord suffers less bending stress on the larger diameter, even though no mechanical advantage is produced.

 

[Luchekv]

This doesn't seem right..

In the video it states that to balance the gears you would apply a second equal weight to the blue gear.

The blue gear has 3x the rim force. Meaning for the gears to balance in this case, the red gear would need a weight of 3Kg, right? So again, your motor has to do 3x the work to lift the same weight.

This however does make sense:

So in effect, you build up a gear train with different size gears to try maximize that 'lever' length?

 

[A.T.]

This doesn't seem right..

In the video it states that to balance the gears you would apply a second equal weight to the blue gear.

The blue gear has 3x the rim force. ...

No, that would violate Newton's 3rd Law. The rim forces between meshed gears are equal, but opposite.

 

[FactChecker]

In the video it states that to balance the gears you would apply a second equal weight to the blue gear.]

The diagram is balanced with no additional weight. Any torque multiplier that the weight on the larger gear has is exactly canceled by the torque multiplier that the smaller gear also has on the larger gear. The result is exactly as though the second weight was attached directly to the smaller gear. So the torque on the smaller gear is exactly 0.

 

[andrewkirk]

Scenario 2:
Load = 1Kg
R1=1cm
R2=2cm
R3=1.5cm
R3: 1KgF x 1.5cm = 1.5KgF.cm = 0.147 N.m of torque which is more than what the motor can output

It's not clear what calculation was done here because the working was not shown.

If the load has mass 1kg then 9.8N force is needed to lift it.
If the torque required from the motor to lift the mass is T then, using the formula I gave above, we have:

T x (R2 / R1) / R3 = 9.8 Nm.

Substituting R2 = 0.02m, R1 = 0.01m, R3 = 0.015m and solving for T we get

T = 9.8 Nm x 0.015 m / (0.02 m / 0.01 m) = 0.0735 Nm

which is easily manageable.