# Talk:Epicyclic gearing

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## Discussion

The mathematical exegesis of this article is extremely poor. After introducing the concepts of Sun (s), Planet (p) and Annulus (a), the reader is then presented with two equations involving a 4th factor c (which I assume in the gear ratio, but this is never explained) and two quantities, N and w (omega). From the following text, it emerges that N refer s to the number of teeth, but it's not at all clear what w refers to. This is unfortunately typical of math articles on Wikipedia, in which the equations are pasted into an article may indeed be correct and canonical, but are not explained for the casual reader and often do not seem to be fully understood by the article editors.98.248.125.108 (talk) 22:48, 17 July 2013 (UTC)

Why does a planetary geartrain have multiple planets? Is this about power transfer or stability or something?

See below.

Isn't the sun gear in the picture gray? --snoyes 22:36, 19 Nov 2003 (UTC)

yes, only the drive shaft of the sun gear is in yellow. This drive shaft goes through the (green) planet gear carrier and extends on the oter side into the sun gear. Maybe the OP should also make the sun gear yellow. MH 213.236.117.2 08:21, 24 May 2004 (UTC)
Kinda depends on your monitor, I guess. On the computer I created them on, it looks slightly yellow, but I checked it on a different screen and it's much darker and muddier. The same problem occurred with the images on Differential. I'll be uploading brightened-up versions soon - thanks for pointing it out! -- Wapcaplet 23:42, 19 Nov 2003 (UTC)
It seems that my version of galeon (or galeon in general) doesn't display pngs all that well. In both konqueror and mozilla the sun gear is clearly yellow, while in galeon the whole picture is significantly darker. (And the picture is viewed from a different angle - very strange). What browsers were you using? --snoyes 00:49, 20 Nov 2003 (UTC)
Hey, another Linux user! :-) The angle is something I changed in the last revision. I thought it showed the red arrows a little better. It may have been a browser issue - in Firebird under Linux on a newer 19" monitor, the colors were brighter; in Mozilla on Windows on an older 15" monitor, they were much darker. Could just be differences in gamma correction between the two monitors. In your case, maybe Galeon is still using a cached version of the image? That'd explain why one of them still shows the other angle. I've noticed that different software often uses different gamma correction on images - especially with things like Gimp and Photoshop, which are tweaked to get images to look closer to print output.
Anyhow, let me know what you think of the newer colors. I've used the same materials and similar lighting for the illustrations for Differential and Pin tumbler lock; I hope to make similar illustrations in this vein, and would like to find good distinct colors. -- Wapcaplet 04:47, 20 Nov 2003 (UTC)
Yip, you pretty much can't help avoiding linux users on wikipedia. Which is a good thing, since suggestions such as support for proprietary stuff like flash animations get struck down immediately. (A couple of days ago on the Village pump). Anyway, I cleared my cache and the pictures now look identical and the colours clear and distinct in galeon, konqueror, firebird, etc. --snoyes 05:35, 20 Nov 2003 (UTC)

Isn't the maximum gear ration always less than one? If so, it would be nice to note it in the descriptions.

Not necessarily. With the given scheme, sun gear as input and carrier as output, I think yes, but you could just as easily use the carrier as input and sun as output to get a ratio greater than 1. You could also hold the carrier fixed, and use the outer ring as output. But you're right in noting that a discussion of how the ratios work would be nice... -- Wapcaplet 16:50, 2 Jun 2004 (UTC)

In the section under discussion, it is written that N(s)*omega(s)+N(r)*omega(r)=(N(s)+N(r))*omega(c). This is consistent with equations found at http://www.roymech.co.uk/Useful_Tables/Drive/Epi_cyclic_gears.html and www.md.kth.se/~fredrikr/AM2D/gearReport.pdf. The latter references "Vedmar Lars, Maskinelement, Lunds Tekniska Högskola, 2002" for the equation.

On the other hand, the equation given in the gear ratio section, with some simplification (notably substituting N(r) for 2N(p)+N(s)), may be written as Omega(r)*N(r)+Omega(s)*N(s)=(N(p)+N(c))*2*omega(c).

The equations need to be made consistent; based on references, I believe that the one in this section is more accurate, but I would prefer some kind of consensus before changing the other.

The derivation for the equation given in the section under discussion states that omega(r)/omega(s)=N(s)/N(r), when it ought to be -N(s)/N(r). With that negative sign, I can derive the equation as given with the method described. I would like to write that derivation more clearly in the article, but seem to be unable to edit a flagged section. Tomvreomfodj (talk) 18:56, 13 December 2010 (UTC)

## Other uses for Epicyclic gearing than just gear ratios?

It seems that many combinations of power flow are available from the gearset. If power input is to the outer ring gear, wouldn't the "planetary" carrier rotate in the ring gear direction and the central "sun" gear in the opposite direction? Useful for driving grinding wheels in opposite directions?

• Yes, if the carrier is fixed. One of the gear units (outer ring, planetary carrier, sun gear) must be held fixed for any useful output to result. But yeah, this is sometimes used in simple transmissions to give two forward gears plus a reverse gear. Rather nifty! -- Wapcaplet 22:10, 22 Jul 2004 (UTC)

Further, what about its use as a "differential"? The whole article is written on the assumption that it is a way of achieving a few fixed ratios by holding one part stationary. In the Hybrid Synergy Drive, which is referenced, it's used as a 3-way power-split device, and all parts are usually in motion. --KJBracey 08:37, 18 November 2005 (UTC)

Who makes a two speed planetary gearbox? —Preceding unsigned comment added by 67.201.136.122 (talk) 20:31, 5 October 2008 (UTC)

All due respect, but "must be held fixed for any useful output" is not correct. In the optics industry, double side polishers make use of epicyclic gears. Optical windows are placed within the planets, or "carriers". The Ring and Sun gears can be controlled independently, and the carriers rotate/revolve based upon both the ring and sun gear speeds. "Plates" sandwiching the carriers from the top and bottom provide the friction necessary for polishing to occur. I'm currently seeking a general solution on how to determine the carrier revolution velocity based upon the speed of the Ring and Sun gears. From the article,
(2+n)ωa+nωs-2(1+n)ωc = 0
appears to be the best relation I've found so far but no derivation or citation is provided.PHiZiX —Preceding undated comment added 01:14, 29 December 2009 (UTC).
Norton gives the general equation for epicyclic angular velocity to be: ±(product of number of teeth on driver gears)/(product of number of teeth of driven gears) = (ωL - ωarm)/(ωF - ωarm); where ωL is the angular velocity of the last gear and ωF is the angular velocity of the first gear. Note that this equation is limited because the first and last gear must be pivoted to the ground (i.e. not orbiting). Hope that helps. Wizard191 (talk) 13:40, 29 December 2009 (UTC)

## My math is rusty

OK, it's about time we got some examples of how the gear ratios work in this thing. Preferably a version that is easy to understand for the non-mathematician (namely, me). I found this site that attempts an explanation, but it leaves me somewhat baffled. Ideally, a thorough exposition of how the ratios are calculated would be nice, but a simple formula would suffice for now. Three cases:

• Annulus (outer ring) fixed
• Planetary carrier fixed
• Sun gear fixed

In each case, what is the ratio between the remaining gearsets, in simple terms of how many teeth each gear has? For the second case, it's fairly easy, since the sun and annulus simply turn in opposite directions, in the ratio sun/annulus = -teethannulus/teethsun (this is our reverse gear, as mentioned above). For the others, there's some angular velocity of the carrier involved, which taxes my brain. Assistance would be welcome. -- Wapcaplet 00:41, 23 Jul 2004 (UTC)

Nevermind. I found a much better explanation, which I will try to work into the article.

## Sun and planet gear article

Hello, I just added a stub article on the above & got a message that it seems to be the same as Epicyclic gearing, looking at the page it seems to be the case that it is either the same thing or epicyclic gearing is a more advanced form of the original sun and planet gear. I am not an engineer & have some problems understanding technical articles (only created the s&pg as I'm working on the article of the engineer who invented it - William Murdoch) so I'm not too sure about the differences & the terminology. Does anyone know if the terminology 'sun and planet gear' is outdated or whether epicyclic gearing is the common modern (or American) phrasing for this? If so it might be better if I add a link from s&pg to epicyclic gearing as the s&pg article (while a stub) deals with the original invention. Otherwise it might be better to merge the articles & put in a re-direct. Can you please let me know what you think. AllanHainey 12:35, 6 September 2005 (UTC)

I'm a Mechanical Engineer. They arent the same thing at all. They do both contain sun and planet gears, as this just refers to one gear orbiting another. This article is about using them to make a gear train which converts shaft torques, the other is using them to convert piston force to torque.
On the Internet, everyone is a Mechanical Engineer —Preceding unsigned comment added by 71.37.50.137 (talk) 07:01, 3 March 2008 (UTC)

## Why planetary geartrains have multiple planets

Planetary gear connected to pistons

Someone posted this question above: "Why does a planetary geartrain have multiple planets? Is this about power transfer or stability or something?"

The reason for having multiple planets is that you can connect more than one shaft to the planetary gear. That's part of the ingenuity of planetary gears: you can input power from multiple shafts and output power in one, or more, shafts, as necessary.

Take, for instance, the planetary gear connected to two pistons on the right. Here a drive shaft inputs power into the sun gear, which in turn rotates the two planets, timed so the pistons move up and down one right after the other. (Note that the pistons are not actually pictured, but are connected to the "moving crankshafts.")

Of coarse stability is the first and main reason.

Multiple planets also increase torque capacity; I believe that this is noted in the article.66.235.27.68 (talk) 16:09, 13 December 2010 (UTC)
As 66.235.27.68 (may I call you .68?) correctly puts it, it's usually about torque. The simple ability to increase torque capability by adding more planets is often useful. Gear trains are somewhat inefficient in terms of their torque / weight capacity, because they usually have a great many teeth (costing money and weight), but only use very few of them at a time. An epicyclic geartrain that can easily use three or four times as many saves here.
Another reason, probably even more important, is to reduce the radial loading on the bearings. As this load is largely balanced in the epicyclic gearbox, the bearings can be much lighter and simpler.
The claim above that epicyclic gearboxes can be used with multiple crankshafts is simply bogus. Unless the pistons driving those cranks are happy to rotate with the planet carrier, that multi-crank arrangement is no longer an epicyclic gearbox. If you look at Commons:Category:Bristol_Centaurus, all those many gears driving the sleeve valves are not epicyclic planet gears, as their "planet carrier" is fixed to the crankcase, not rotating. However the propeller reduction gearbox (the large bevels) is an epicyclic gear. Andy Dingley (talk) 17:40, 13 December 2010 (UTC)

## Animated gifs

Animated gifs, showing each locking configuration would be very helpful - its very difficult to visualise this kind of thing.

would this be any good? just the pic in the article now put together --Astrokey44 13:47, 26 May 2006 (UTC)
I like it. It would be even better with an intermediate frame; a little easier to follow. ike9898 22:45, 12 June 2006 (UTC)
Is it possible to do this with photoshop? When I try to do tweening the in between frame only changes the opacity of the two not the position --Astrokey44 12:33, 22 June 2006 (UTC)

THE ARTICLE NEEDS AN ANIMATION!!!

This one is not smooth enough. You don't need to animate very much time, but you do you need to see the individual gear teeth move. Helvitica Bold 21:36, 11 May 2011 (UTC)

# Question

What is method to calculate the maximum planet gears can involved in a Epicyclic gear system?

To me, this article feels incomplete. I would expect to see a discussion of the advantages/disadvantages of planetary gears versus other gearing arrangements (i.e. when would you want to use planetary gears? when would you not?) For example, advantages include high power density, large reduction in a small volume, multiple kinematic combinations, pure torsional reactions, coaxial shafting. Disadvantages include high bearing loads, inaccessibility, design complexity. I have studied planetary gears in graduate school and could add such a section if others think it would add value... This would be my first attempt at contributing to wikipedia so I wanted to test the waters before jumping in... Kiracofe8 02:51, 6 February 2007 (UTC)

## terminology

(First, I am harmless).

In the page on epicyclic gears, one section calls the annulus the ring gear. Either is correct but at least stick to one.

Malcolmcochran@hotmail.com —Preceding unsigned comment added by 86.137.178.200 (talk) 11:58, 9 September 2009 (UTC)

I've converted them all to annulus. Thanks for the heads up. Wizard191 (talk) 15:48, 9 September 2009 (UTC)

## Image

Although not essential, the image seems a bit weird, ie the carrier design is not really standard, see the image at http://www.mvwautotechniek.nl/Motor/Transmissie/automatisch.htm M. van Wijk 14:01, 22 March 2016

There is no "standard design" for planet carriers. They're all correct. Do not be tempted to invent another of your "fixes". Andy Dingley (talk) 17:43, 13 December 2010 (UTC)

## Switching gears by use of pinions

I think that Regular Planetary Gearsets use basic pinions for switching gears, and not oil, see http://mysite.du.edu/~jcalvert/tech/planet.htm

Is this correct ? I also wonder whether other setups than the double reduction, single overdrive can be used for regular planetary gearsets (that way, ie 3 overdrives or 3 reductions can be put in place which would be more useful for specific tasks such as for wind turbines) 91.182.226.145 (talk) 14:27, 13 December 2010 (UTC)

It's impossible to tell what you're babbling about, what a Regular Planetary Gearset (which must be important because It's All Capitalised) is, or what the perfectly innocent website has to do with hydraulics. It's just talking about lubricating the gearbox on assembly, not anything about switching ratios. If you're interested (and you can't afford the old Graupner epicyclic stack) take a look here http://www.active-robots.com/products/motorsandwheels/tamiya-gearbox-kits.shtml where they sell the whole range of cheapo and excellent Tamiya gearboxes, which are great components for home robotics. Andy Dingley (talk) 17:48, 13 December 2010 (UTC)
I agree with Wizard191's comment that this article is about the epicyclic gears, and not the gearbox in general, but epicyclic gearbox redirect to this page, so this article would normally also need to have this info. A solution could be to remove the redirect and place the new info from the old edit to the new article. At present, the 2 different systems are not yet mentioned anywhere. As for the capitals, these were used in the original articles, so that's not something I put in, rather I reverted some things such as Ring to Annulus, hence downplaying the popular text a bit.
I also assume that Watt sun-and-planet Gearset is the type of gearset which I refer to when stating "regular" (again based on the reference article). In addition, as I make up from http://mysite.du.edu/~jcalvert/tech/planet.htm , there are even more epicyclic trains/gearsets, including Tamiya, and Graupner (as you mentioned), Simpson Gearset and Ravigneaux Gearset (see http://www.mvwautotechniek.nl/Motor/Transmissie/automatisch.htm) All need to be mentioned at the new article.

91.182.79.197 (talk) 13:40, 14 December 2010 (UTC)

The use of band brakes in an epicyclic gearbox is commonplace enough that it belongs here, but not the use (other than a brief mention and link) for automatic transmissions in cars. That has its own article. If you care, you can go and read about them over there.
Like most "vanilla" terms, the "Regular Planetary Gearset" is not a term in any sort of common use. It was only created for that article, to give a distinguishing name from the compound sets.
You added a links page to wind turbine early today. One of the PDFs in that has a good explanation of the role of flexible pins to support planet gears, and that's worth adding too. Andy Dingley (talk) 13:41, 14 December 2010 (UTC)
I feel that epicyclic gearbox is correctly redirecting here, because epicyclic gearboxes are commonly used outside of transmissions. I use them all of the time at my workplace and they are found on all sorts of industrial machinery. Moreover, there is already a paragraph about how epicyclic gearing is used in auto transmissions, so I think that info should be placed over in the automatic transmission article, if anywhere. Wizard191 (talk) 17:51, 14 December 2010 (UTC)

## Article should be called planetary gears, not epicyclic gears.

This article should be called planetary gearing because planetary gear-trains are just a specific instance of epicyclic gear-trains and there are many other configurations of epicyclic gear trains, for example a differential gear-train. — Preceding unsigned comment added by Cdhickam (talkcontribs) 23:46, 29 March 2011 (UTC)

## Protection

Is it possible to have the section on formulas protected, som that edits like this/vandalism will be prevented? Keanu (talk) 08:43, 6 May 2011 (UTC)

# Formulas for calculating gear ratios

I've deleted the formulas section, because a lot of these formulas are incorrect (at least 7b and 8b are, and probably more). This should be checked upon! I don't have time to do this right now but I think no formulas is better than wrong formulas so I deleted them all. Here they are for future notice: (I will see if I can get back on this, but I'm leaving on a vacation in a few hours)

Notations of planetary gearing elements for the next table [1]
Name Number of teeth Speed
${\displaystyle {\color {blue}Driving}\,}$ ${\displaystyle {\color {blue}z}\,}$ ${\displaystyle {\color {blue}n}\,}$
${\displaystyle {\color {magenta}Auxiliary\ driving}\,}$ ${\displaystyle {\color {magenta}z}\,}$ ${\displaystyle {\color {magenta}n}\,}$
${\displaystyle {\color {red}Driven}\,}$ ${\displaystyle {\color {red}z}\,}$ ${\displaystyle {\color {red}n}\,}$
${\displaystyle Fixed\,}$ ${\displaystyle z\,}$ ${\displaystyle n\,}$
${\displaystyle {\color {green}Planetary}\,}$ ${\displaystyle {\color {green}z}\,}$ ${\displaystyle {\color {green}n}\,}$
${\displaystyle {\color {cyan}Planetary}\,}$ ${\displaystyle {\color {cyan}z}\,}$ ${\displaystyle {\color {cyan}n}\,}$

Sketch and output speed of planetary gearings
Sketch Output speed Sketch Output speed Sketch Output speed Sketch Output speed
${\displaystyle {\color {red}n}={\color {blue}n}(1+{\frac {z}{\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}(1-{\frac {z}{\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}(0+{\frac {z}{\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}(\cos \beta +{\frac {z}{\color {red}z}})}$
${\displaystyle {\color {red}n}={\color {blue}n}(1+{\frac {{\color {green}z}z}{\color {cyan}z\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}(1+{\frac {{\color {green}z}z}{\color {cyan}z\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\frac {1}{(1+{\dfrac {{\color {green}z}z}{\color {cyan}z\color {blue}z}})}}}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\frac {1}{(1+{\dfrac {{\color {green}z}z}{\color {cyan}z\color {blue}z}})}}}$
${\displaystyle {\color {red}n}={\color {blue}n}(1+{\frac {z}{\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}(1+{\frac {z}{\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\dfrac {1}{1+{\dfrac {z}{\color {blue}z}}}}}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\dfrac {1}{1+{\dfrac {z}{\color {blue}z}}}}}$
${\displaystyle {\color {red}n}={\color {blue}n}(1-{\frac {{\color {cyan}z}z}{{\color {green}z}{\color {red}z}}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}(1-{\frac {z\color {green}z}{\color {cyan}z\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}(1-{\frac {z\color {green}z}{\color {cyan}z\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\dfrac {1}{1+{\dfrac {z}{\color {blue}z}}}}}$
${\displaystyle {\color {red}n}={\color {blue}n}{\dfrac {1}{1-{\dfrac {{\color {cyan}z}z}{{\color {green}z}{\color {blue}z}}}}}}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\dfrac {1}{1-{\dfrac {{\color {cyan}z}z}{{\color {green}z}{\color {blue}z}}}}}}$ ${\displaystyle {\color {red}n}={\color {blue}n}(1-{\frac {z\color {green}z}{\color {cyan}z\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}(1-{\frac {{\color {green}z}z}{\color {cyan}z\color {red}z}})}$
${\displaystyle {\color {red}n}={\color {blue}n}\left[1-\left({\frac {\color {magenta}n}{\color {blue}n}}-1\right){\frac {\color {magenta}z}{\color {red}z}}\right]}$ ${\displaystyle {\color {red}n}={\color {blue}n}\left[1-\left({\frac {\color {magenta}n}{\color {blue}n}}-1\right){\frac {\color {magenta}z}{\color {red}z}}\right]}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\frac {1+{\dfrac {\color {magenta}n\color {magenta}z}{\color {blue}n\color {blue}z}}}{1+{\dfrac {\color {magenta}z}{\color {blue}z}}}}}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\frac {1+{\dfrac {\color {magenta}n\color {magenta}z}{\color {blue}n\color {blue}z}}}{1+{\dfrac {\color {magenta}z}{\color {blue}z}}}}}$
Specify which formulas are incorrect so they can be corrected. 7b and 8b, which ones are these? Dont delete something and then "leave for a vacation". I have reverted your deletion. Keanu (talk) 10:53, 15 July 2011 (UTC)

7b and 8b probably refer to the image the formula belongs to. move your mouse over it and you see the name an the bottom of your screen. Deleting wrong information would be better than showing it is my opinion. Better not no post anything which is not veryfied at all. I found mistakes in other formulas too. Formula for Gear 13 contains a blue Z. However the arm is blue and has no teeth at all. Formula 12a and 12b are wrong, If all gears have the same number of teeth it would still rotate, but the formula results in devide by 0 Formula 9 If the planets are equal and the rings are equal both rings will rotate at the same speed. Since one is locked, the other will not rotate. the formula does however not result in 0. — Preceding unsigned comment added by 62.194.118.254 (talk) 13:16, 5 February 2012 (UTC)

I've fixed formula 8a, 8b, 9, 10a, 10b, 13 although the graphics for 10a and 10b need to be properly tweaked (the blue gears should be removed). 7b, as the original deleter posited, is not wrong. If there are other wrong formulas, I haven't found them, and as far as I can tell, the whole section is now correct. 10a and 13, and 9 and 10b are the same gear formations, and are redundant, although both are correct.97.124.82.32 (talk) 17:57, 24 September 2012 (UTC)

Thanks. I think we will need some good references before we can use these formulas in the article. Keanu (talk) 15:45, 26 September 2012 (UTC)
If 10a and 10b are wrong, and 10a and 13 are the same, along with 10b and 9, why don't we delete 10a and 10b? -- :- ) Don 16:07, 26 September 2012 (UTC)
Do you have a reference to a trusted source that may confirm that the formulas are correct? I would like to check for myself before adding them to the article. Keanu (talk) 13:06, 1 October 2012 (UTC)
No, I don't. Not into ME much, but I might be, the math is much more interesting that EE. Forget my previous question, I now realize it was stupid. But, are not 1, 2 and 3 identical? I will put a reference book on my list, but finding all this in one reference is not likely. If I might make a suggestion to reduce some confusion to me and perhaps others, the drawings (which are very nice) are never going to do well in B&W, so maybe we could put the grey levels at the same point by darkening the green and eliminate two blues by changing the light blue to maybe orange? The shading could be a light grey. Or at least make the coupling gear dark blue and light blue? -- :- ) Don 14:46, 1 October 2012 (UTC)
Yes, I do have a good source, and being a backyard tinkerer I personally built each system, as well as several other configurations I hopefully will make a graphic for to replace this one. Here is the website I obtained the formulas from, http://www.roymech.co.uk/Useful_Tables/Drive/Epi_cyclic_gears.html I agree that formulas 10a and 10b should be deleted, and posit that 3 and 12b should be deleted as well, since both are redundant as well. 3 being a special and easily solved case of 4a, and 12b being the same as 12a. Before I mend the graphic for that, I'd like to be sure others agree with my justification. — Preceding unsigned comment added by 97.124.82.32 (talk) 06:00, 3 October 2012 (UTC)
Notations of planetary gearing elements for the next table [2]
Name Number of teeth Speed
${\displaystyle {\color {blue}Driving}\,}$ ${\displaystyle {\color {blue}z}\,}$ ${\displaystyle {\color {blue}n}\,}$
${\displaystyle {\color {magenta}Auxiliary\ driving}\,}$ ${\displaystyle {\color {magenta}z}\,}$ ${\displaystyle {\color {magenta}n}\,}$
${\displaystyle {\color {red}Driven}\,}$ ${\displaystyle {\color {red}z}\,}$ ${\displaystyle {\color {red}n}\,}$
${\displaystyle Fixed\,}$ ${\displaystyle z\,}$ ${\displaystyle n\,}$
${\displaystyle {\color {green}Planetary}\,}$ ${\displaystyle {\color {green}z}\,}$ ${\displaystyle {\color {green}n}\,}$
${\displaystyle {\color {cyan}Planetary}\,}$ ${\displaystyle {\color {cyan}z}\,}$ ${\displaystyle {\color {cyan}n}\,}$

Sketch and output speed of planetary gearings
Sketch Output speed Sketch Output speed Sketch Output speed Sketch Output speed
${\displaystyle {\color {red}n}={\color {blue}n}(1+{\frac {z}{\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}(1-{\frac {z}{\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}(1-{\frac {{\color {green}z}z}{\color {cyan}z\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}(\cos \beta +{\frac {z}{\color {red}z}})}$
${\displaystyle {\color {red}n}={\color {blue}n}(1+{\frac {{\color {green}z}z}{\color {cyan}z\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}(1+{\frac {{\color {green}z}z}{\color {cyan}z\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\frac {1}{(1+{\dfrac {{\color {green}z}z}{\color {cyan}z\color {blue}z}})}}}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\frac {1}{(1+{\dfrac {{\color {green}z}z}{\color {cyan}z\color {blue}z}})}}}$
${\displaystyle {\color {red}n}={\color {blue}n}(1+{\frac {z}{\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}(1+{\frac {z}{\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\dfrac {1}{1+{\dfrac {z}{\color {blue}z}}}}}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\dfrac {1}{1+{\dfrac {z}{\color {blue}z}}}}}$
${\displaystyle {\color {red}n}={\color {blue}n}(1-{\frac {{\color {cyan}z}z}{{\color {green}z}{\color {red}z}}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\dfrac {1}{1-{\dfrac {{\color {cyan}z}z}{{\color {green}z}{\color {blue}z}}}}}}$ ${\displaystyle {\color {red}n}={\color {blue}n}(1-{\frac {z\color {green}z}{\color {cyan}z\color {red}z}})}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\dfrac {1}{1+{\dfrac {z}{\color {blue}z}}}}}$
${\displaystyle {\color {red}n}={\color {blue}n}\left[1-\left({\frac {\color {magenta}n}{\color {blue}n}}-1\right){\frac {\color {magenta}z}{\color {red}z}}\right]}$ ${\displaystyle {\color {red}n}={\color {blue}n}\left[1-\left({\frac {\color {magenta}n}{\color {blue}n}}-1\right){\frac {\color {magenta}z}{\color {red}z}}\right]}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\frac {1+{\dfrac {\color {magenta}n\color {magenta}z}{\color {blue}n\color {blue}z}}}{1+{\dfrac {\color {magenta}z}{\color {blue}z}}}}}$ ${\displaystyle {\color {red}n}={\color {blue}n}{\frac {1+{\dfrac {\color {magenta}n\color {magenta}z}{\color {blue}n\color {blue}z}}}{1+{\dfrac {\color {magenta}z}{\color {blue}z}}}}}$

Here's my updated table, I haven't arranged it, but I have deleted 4 redundant formulas and graphics. ADAzriel (talk) 02:04, 8 October 2012 (UTC)

## Intuitive explanations?

It would be nice if it gave an intuitive explanation for the basic operation of increased rotation. With the sun gear held stationary, and carrier rotated, the annulus will spin faster. My intuitive explanation is that at any moment each planetary gear is like a tiny lever, with the fulcrum resting on a sun gear tooth, the effort at the center of the planetary gear, and the load at an annulus tooth. Clearly, this configuration increases motion of the load.

Also, the gear ratio section has an example with the carrier held stationary. It shows the ratio of the sun gear to a planetary gear, but this seems irrelevant; just the number of teeth on the sun and annulus matter. The planetary gears essentially "move" the sun gear closer to the annulus so it's as if its teeth were touching the annulus directly (except moving in the opposite direction). For every tooth of the sun gear that moves past a line from the center outward, one tooth of the planetary gear moves past the same point where this line intersects the annulus. I see that it eventually shows this with math. I'd like more "intuitive" explanations like this. 72.48.75.131 (talk) 19:55, 19 December 2011 (UTC)

The comments above on formulas 9, 12a, 12b is at best incomplete. Under the conditions stated the equations degenerate and kinematic relations are no more determined, i.e. one component becomes loose. The formula in 13, however is indeed misprinted, the blue z is to be red. Snoloven (talk) 20:32, 28 March 2012 (UTC)

## History

Hold on, the wikipedia article on the Antikythera mechanism says that the device was definitively found not to contain any planetary gear mechanisms, and this was realized over a decade ago. Why does this page contradict the other wikipedia page? I think we ought to take down the reference to the Antikythera mechanism on this page. — Preceding unsigned comment added by 50.131.61.86 (talk) 11:46, 19 September 2013 (UTC)

It doesn't contain any planetary gear systems (I suppose, I'm not 100% on the definition), but does contain an epicyclic set of gears. Picture two gears mounted face to face, but on different axes (that is, there is an offset between the axes). As the gears aren't meshed, they're free of each other. Now put a pin on one that engages in a slot on the other such that the slot is positioned radially and is large enough to contain the full travel of the pin (that length is equal to the distance between the gear axes) through an entire revolution. Now, when one gear is driven at and even angular velocity, the other will speed up and slow down relative to the even velocity because its point of being driven moves radially across the gear. The speed of the variable gear averages the same as the driver. The Antikythera Mechanism used such a pair of epicyclic gears to model the ellipticity of the moon's orbit, and it even mounted the two gear on a gear that rotated once every 9 years of simulated time to model the elliptical precession of the moon. Of course, the Greeks didn't see it as ellipticity, that had to wait for Kepler. But they weren't as much into why as they were into making the model track with reality. It is possible, but not certain, that the mechanism may have used other pin/slot epicycles to model the retrograde orbits of the other planets. SkoreKeep (talk) 07:10, 17 June 2014 (UTC)

## Intuitive explanation!

At the end of par. Gear Ratio, I've added an argument that I think is simple, intuitive and precise, establishing the general formula that relates the angular velocities of sun, carrier and annulus (plus two standard applications). This might cause other explanations superfluous, but I made no further changes.KeesDoe (talk) 21:57, 27 March 2015 (UTC)

Your intuitive explanation is correct but it duplicates the material provided in the following section. Your "imagine a you are riding on the carrier" is the exactly the principle of the "fixed carrier train ratio." You proceed to obtain the same equation included in the relationship of speed averaging of a differential, all of which are shown in the following section. Prof McCarthy (talk) 02:21, 31 March 2015 (UTC)
Dear prof McCarthy, yes, I have been duplicating. The point is I find both arguments in the article incomprehensible and my aim was to add one that would be simple as well as precise. I hope most readers will find the existing exposition, like you, illuminating. (In view of this talk, I have doubts.) Anyway - thank you!KeesDoe (talk) 19:58, 1 April 2015 (UTC)
• ^ Pattantyús Gépész- és Villamosmérnökök Kézikönyve 3. tom. Műszaki Könyvkiadó, Budapest, 1961. p.632.
• ^ Pattantyús Gépész- és Villamosmérnökök Kézikönyve 3. tom. Műszaki Könyvkiadó, Budapest, 1961. p.632.