Why is the gear ratio reversed in a planetary gear system?

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SUMMARY

The discussion centers on the confusion regarding the gear ratio in planetary gear systems compared to simple gear systems. It establishes that in a lossless system, the torque ratio is inversely proportional to the gear teeth ratio, expressed as T_input/T_output = N_input/N_output. However, in planetary systems, the relationship appears reversed due to the configuration of gears, specifically the internal ring gear. Participants confirm that the Wikipedia article on epicyclic gearing contains inaccuracies, particularly regarding the torque relationships and angular velocities.

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Trying2Learn
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What is the relation between torque to teeth in an epicyclic gearing system?
Hello,

I am confused about something. I will take the time to work it out, but right now, I am using it to remember. And I am now confused.

Go here:
https://www.smlease.com/entries/mechanism/gear-train-gear-ratio-torque-and-speed-calculation/

Scroll down to the section on: Gear Ratio and Torque

In a loss-less system, I get that that ratio of output to input torque is inverse to the output to input number of gear teeth:
To/Ti = Ni/No

Now go hear for a planetary gear system:
https://en.wikipedia.org/wiki/Epicyclic_gearing

Scroll down about one third to: Torque Ratios of Standard Epicyclic Gearing

(Granted, there are three gears here and in an planetary style, not simple).

But why (say, for the first one), is this ratio reversed.

Tr/Ts = Nr/Ns

Am I missing something, or is there an error in one of these`? I get the simple gear system (and I am confident of working it out), but for the planetary one, is there an inverse relation because the teeth are INSIDE on the ring gear? Can someone point me to where (assuming the wiki page is correct) it is worked out?
 
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When gears are in normal external mesh, the angular velocity changes sign.
Running a pinion in an internal ring gear does not reverse the sign of the angular velocity.
The absolute ratio remains the same, it is not the reciprocal.

Trying2Learn said:
Can someone point me to where (assuming the wiki page is correct) it is worked out?
In the wikipedia epicyclic article, N is angular velocity, z is tooth count, and T is torque.
For the same power the product of shaft torque and shaft angular velocity should be constant.
With a fixed carrier, the planets are idlers, so Tr * Nr = Ts * Ns;
∴ Tr = Ts * Ns / Nr;
Which suggests the wikipedia article got it wrong.
My guess is that someone incorrectly thought N was the toothcount.
 
Baluncore!

THANK YOU VERY MUCH

May I ask you to repeat this, but just by affirming? (I am shocked that I found this error.)

Set aside your comment on the sign and reciprocity--perhaps my wording was off, and that was not my issue, anyway.

May I ask you to confirm this.

In a gearing system, where T is the torque and (in my question) N is the tooth count

T_input/T_output = N_input/N_output

Also, in the same wikipedia article, the very next line for the torque of the ring to the carrier: did they incorrectly get that one wrong, too?

Can you suggest a source that explains the input/output for the torque/teeth, simply.
(If possible, an online source?, but I will gladly accept anything)
 
Trying2Learn said:
In a gearing system, where T is the torque and (in my question) N is the tooth count

T_input/T_output = N_input/N_output
For tooth count N, the relative angular velocities are Vi = 1 / Ni; and Vo = 1 / No.
Power = Torque * Angular velocity.
Conservation of energy requires; Ti * Vi = To * Vo.
∴ Ti * No = To * Ni;
∴ Ti / To = Ni / No;
If there are only two shafts, one input and one output, you are correct.

Wikipedia gets things wrong. It is consistent in that article.

I will see what references I can find.
 
Baluncore said:
For tooth count N, the relative angular velocities are Vi = 1 / Ni; and Vo = 1 / No.
Power = Torque * Angular velocity.
Conservation of energy requires; Ti * Vi = To * Vo.
∴ Ti * No = To * Ni;
∴ Ti / To = Ni / No;
If there are only two shafts, one input and one output, you are correct.

Wikipedia gets things wrong. It is consistent in that article.

I will see what references I can find.
This is great. I am really excited I am on the right track.

Please do not forget me and please provide a reference if you can.
 
A thought experiment may help.
  • Draw an imaginary line between the centers of the two gears
  • When the Input gear turns, some number of teeth will cross that line
  • Since the teeth of the gears are meshed, the same number of teeth of the other, Output, will also cross that line
  • If the Output gear is bigger, then it will turn only a partial revolution while the Input gear turns turns a full revolution
    • Now you can see that if you swap the Input and Output, putting the big gear on the input
    • The smaller gear will turn faster than the big gear
Here is a link with lots of diagrams and explanation:
https://www.wikihow.com/Determine-Gear-Ratio

This one has formulas and a sketch:
https://www.engineeringtoolbox.com/gear-output-torque-speed-horsepower-d_1691.html

(above found with:
https://www.google.com/search?&q=gearbox+calculations)

Just remember that if the Speed goes UP by some factor, say 3.5, then the Torque goes DOWN by the same factor.

Code: 
    IN               Out
Speed Torque     Speed Torque
200   100        700    28.6     speed UP by 3.5, torque DOWN
200   100         57   350       speed DOWN by 3.5, torque UP

Cheers,
Tom
 
Last edited:
Baluncore said:
Hidden behind the wikipedia epicyclic page is a talk page with another set of formulas.
https://en.wikipedia.org/wiki/Talk:Epicyclic_gearing#Formulas_for_calculating_gear_ratios

Thank you Baluncore. I will look.
Tom.G said:
A thought experiment may help.
  • Draw an imaginary line between the centers of the two gears
  • When the Input gear turns, some number of teeth will cross that line
  • Since the teeth of the gears are meshed, the same number of teeth of the other, Output, will also cross that line
  • If the Output gear is bigger, then it will turn only a partial revolution while the Input gear turns turns a full revolution
    • Now you can see that if you swap the Input and Output, putting the big gear on the input
    • The smaller gear will turn faster than the big gear
Here is a link with lots of diagrams and explanation:
https://www.wikihow.com/Determine-Gear-Ratio

This one has formulas and a sketch:
https://www.engineeringtoolbox.com/gear-output-torque-speed-horsepower-d_1691.html

(above found with:
https://www.google.com/search?&q=gearbox+calculations)

Just remember that if the Speed goes UP by some factor, say 3.5, then the Torque goes DOWN by the same factor.

Code: 
    IN               Out
Speed Torque     Speed Torque
200   100        700    28.6     speed UP by 3.5, torque DOWN
200   100         57   350       speed DOWN by 3.5, torque UP

Cheers,
Tom

Tom.G said:
A thought experiment may help.
  • Draw an imaginary line between the centers of the two gears
  • When the Input gear turns, some number of teeth will cross that line
  • Since the teeth of the gears are meshed, the same number of teeth of the other, Output, will also cross that line
  • If the Output gear is bigger, then it will turn only a partial revolution while the Input gear turns turns a full revolution
    • Now you can see that if you swap the Input and Output, putting the big gear on the input
    • The smaller gear will turn faster than the big gear
Here is a link with lots of diagrams and explanation:
https://www.wikihow.com/Determine-Gear-Ratio

This one has formulas and a sketch:
https://www.engineeringtoolbox.com/gear-output-torque-speed-horsepower-d_1691.html

(above found with:
https://www.google.com/search?&q=gearbox+calculations)

Just remember that if the Speed goes UP by some factor, say 3.5, then the Torque goes DOWN by the same factor.

Code: 
    IN               Out
Speed Torque     Speed Torque
200   100        700    28.6     speed UP by 3.5, torque DOWN
200   100         57   350       speed DOWN by 3.5, torque UP

Cheers,
Tom
Thank you, Tom. This helps a lot
 

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