If two gears are connected by a shaft do they have the same torque?

[Stevengouws]

TL;DR Summary

1 axle. 2 gears on the same axle fixed in place. Does the larger gear with 48 teeth have the same amount of (force/strength) as the smaller gear of 12 teeth. IE: if I attached a chain to the 12 tooth gear and it could lift a max load of 10 kg, would the 48 tooth gear be able to lift 10 kg as well. Please speak plainly I am not a mechanical engineer and do not understand technical lingo/jargon.
thank you in advance for your time. Highly appreciated.

1 axle. 2 gears on the same axle fixed in place. Does the larger gear with 48 teeth have the same amount of (force/strength) as the smaller gear of 12 teeth. IE: if I attached a chain to the 12 tooth gear and it could lift a max load of 10 kg, would the 48 tooth gear be able to lift 10 kg as well. Please speak plainly I am not a mechanical engineer and do not understand technical lingo/jargon.
thank you in advance for your time. Highly appreciated.

 

[Baluncore]

Welcome to PF.

If one gear is driven, the torque will be transmitted through the shaft to the second gear, so the two gears will have the same torque.

But torque is the force (measured perpendicular to a radius arm), multiplied by the length of the arm. In this case using a chain, that arm length will be the diameter of the gears.

For identical module gears, the 4:1 tooth count ratio will give a force ratio of 1:4.

 

[Stevengouws]

Thanks for the answer. The axle is driven separately. The two gears are attached to it. All I need to know is, if the 12 tooth gear can lift 10 kg will the 48 tooth gear be able to lift 10 kg as well please?

 

[Stevengouws]

Welcome to PF.

If one gear is driven, the torque will be transmitted through the shaft to the second gear, so the two gears will have the same torque.

But torque is the force (measured perpendicular to a radius arm), multiplied by the length of the arm. In this case using a chain, that arm length will be the diameter of the gears.

For identical module gears, the 4:1 tooth count ratio will give a force ratio of 1:4.

Thanks for the answer. The axle is driven separately. The two gears are attached to it. All I need to know is, if the 12 tooth gear can lift 10 kg will the 48 tooth gear be able to lift 10 kg as well please?

 

[Baluncore]

If the chain meshes with the gear teeth, and the 12 tooth gear will lift 10 kg, then for the same torque, the 48 tooth gear would only lift 2.5 kg.

 

[Stevengouws]

If the chain meshes with the gear teeth, and the 12 tooth gear will lift 10 kg, then for the same torque, the 48 tooth gear would only lift 2.5 kg.

Thanks for the help.

 

[Tom.G]

Simple answer to title: Yes.
Simple answer to 'lift same weight': No

Next step answer: The larger gear will lift a weight inversely proportional to the radius of the two gears. In this case 10kg/4 = 2.5kg. But it will lift the 2.5kg 4 times as fast as the smaller gear would.

The gory details:
That 'Turning Force' you asked about is called Torque. It is defined as the product of Force multiplied by the Distance over which it acts. The Distance is measured from the center of rotation to where the force is applied.

In your example of a 12 tooth gear and a 48 tooth gear, the radius (from the center to the teeth) differ by a factor of 4.

Let's try it with some numbers. Maybe the smaller gear is 100mm in diameter, and the larger gear has a 400mm diameter. That makes the radius of them 50mm., and 200mm.

If you put a turning force on the small gear by pushing on a tooth with 10kg, you are applying a Torque of 10 x 50 = 500kg.mm.

So the rotational Torque is 500kg.mm. on the shaft.

For the larger 400mm. gear the torque is the same, that is 500kg.mm., but you are working with a distance (radius) of 200mm. now, not 50mm. So the force available at the teeth of the larger gear is 500kg.mm./200mm = 2.5kg.

Cheers,
Tom

oops, I see @Baluncore answered while I was typing this (too) long version

 

[Stevengouws]

Simple answer to title: Yes.
Simple answer to 'lift same weight': No

Next step answer: The larger gear will lift a weight inversely proportional to the radius of the two gears. In this case 10kg/4 = 2.5kg. But it will lift the 2.5kg 4 times as fast as the smaller gear would.

The gory details:
That 'Turning Force' you asked about is called Torque. It is defined as the product of Force multiplied by the Distance over which it acts. The Distance is measured from the center of rotation to where the force is applied.

In your example of a 12 tooth gear and a 48 tooth gear, the radius (from the center to the teeth) differ by a factor of 4.

Let's try it with some numbers. Maybe the smaller gear is 100mm in diameter, and the larger gear has a 400mm diameter. That makes the radius of them 50mm., and 200mm.

If you put a turning force on the small gear by pushing on a tooth with 10kg, you are applying a Torque of 10 x 50 = 500kg.mm.

So the rotational Torque is 500kg.mm. on the shaft.

For the larger 400mm. gear the torque is the same, that is 500kg.mm., but you are working with a distance (radius) of 200mm. now, not 50mm. So the force available at the teeth of the larger gear is 500kg.mm./200mm = 2.5kg.

Cheers,
Tom

oops, I see @Baluncore answered while I was typing this (too) long version

Thank you, that was a fantastic explanation, just what I needed and more, cheers!

 

[Filip Larsen]

Not sure if its relevant for your problem or not, but remember that in a static situation (e.g. when the axle rotates with a constant, including zero, angular speed) the net sum of all torques are zero, so in your case (where I understand you have a drive and two gears) the drive torque has to counter the torque on both gears.

 

[Stevengouws]

Not sure if its relevant for your problem or not, but remember that in a static situation (e.g. when the axle rotates with a constant, including zero, angular speed) the net sum of all torques are zero, so in your case (where I understand you have a drive and two gears) the drive torque has to counter the torque on both gears.

Thanks Filip, it isn't relevant to my problem, but I did learn something, so thank you.