# Sun and Planet gear ratio formula

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Small reduction gearbox "sun" of 9 teeth , 3 "planets " of 18 teeth each and outer static ring of 44 teeth. Output is connected to the anualar that houses the axles of the planets. Output is co-axial with the sun axis. Output is approx 5:2 reduction or is it exactly 5:2 ? Does anyone know where to find the general case formula for gear ratio ?

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appology: ratio is approx 5:1 not 5:2 I think i've answered myself - drawing out on paper Probably ratio 44:9 and not 45:9 (5:1)

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Hi Nigel:

The formula is

ratio = 1 + R/S

Where R is the number of teeth of the ring gear and S is the number of teeth of the sun gear. A 44-toothed ring gear won't work though. The pitch diameters of spur gears are directly proportionate to their numbers of teeth, so you can use them to figure gear relationships interchangeably. So with a sun gear of nine teeth and enough space on each side to roll 18-toothed planet gears takes

18 + 9 + 18 = 45 teeth

for your internal gear. This makes your ratio 1 + 45/9 = 6 ...and it *is* exact.

Hope this helps!

Don

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Mm, yes i recounted the ring teeth and 45 not 44 but I don't see where the 1 of 1+R/s comes from. Gear ratio = number of turns of input to number of turns of output ?

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As I recall, the ratios differ between which gear(s) is stationary and which is driven. So, it could have the above the ratio of 1:6 at one instance, but the stated 5:2 in another. I forget which is which, so I'm not being very helpful here. Dan

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Nigel:

What you have to take into account when dealing with planetary gears is the fact that the planet gears have a moving center, not fixed, so they behave differently than stationary gears would.

Let's take a look at your gear train. If the sun gear (9T) revolves exactly once around, it pays out nine teeth, which drive nine teeth of the planet gear (18T) and causes it to revolve one-half a revolution and take up nine teeth in the internal gear for a ratio of 9/45ths of a revolution, right?

Wrong.

Let's say the planet gear starts exactly in the 12 o'clock position. When the sun gear has revolved once (360 degrees), the tooth at 12 o'clock will have returned to 12 o'clock, but the planet gear has still not contacted all nine teeth. It is in a different place now. Had it remained stationary, nine teeth would have driven nine teeth and it would have rotated half way around its axis. Instead, it ended up ahead of the first tooth because it's rolling inside another gear. The sun gear has to roll an additional amount (lost angle) before the original 12 o'clock tooth catches up with the planet gear and is tangent to it again. This happens with each revolution.

If you do the math, it turns out that by the time the planet gear has passed all the way around, the sum of all the lost angles is equal to exactly one revolution, so an "extra" revolution is required to get a full rotation of the planet--hence the "1+" in the equation. That's why your answer comes out to 6:1 instead of 5:1.

Make sense now?

Don

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I make it 44 / (9X2) = 22/9 reduction precisely.

Brian W (General case for outer r>Small reduction gearbox "sun" of 9 teeth , 3 "planets " of 18 teeth each

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Correction to my OP 45 teeth on Ring ,not 44 and I forgot I only measured half a turn originally and gear ratio is /= 5:1.

I've had another go at turning by hand and by sight it is 5:1 reduction. I could properly index and measure but any error is not going to make it 6:1 unless I'm very much mistaken. Looking at the problem on paper S turns 9,P turns 9 so the axis of any of the P gears move through arc of 360 degrees x 9/45. The external R ring is static and output taken from the anular connecting the axles of the P cogs

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Nigel, My Machinery's Handbook has 24 different planetary gear formulas for these mechanisms based on what type and what is stationary. No wonder we are all confused. For yours: ring fixed, sun is driver, planet annular is follower (output). Ratio=1+ring/sun = 1+45/9 = 6 Ratio of 1:6 That's what the book says. Dan

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I dug out my 1944 version of Machinery's Handbook from the loft and in Planetary/ Epicyclic it has the 1 + C/B formula. So how many gearboxes made to that (wrong) formula ? When i first looked at the problem I thought it was more complicated because of this 'secondary drift' but it is not involved

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If this newgroup had propagated just a little bit faster than quasi snail mail, reading this note would have prevented my embarassingly wrong answer - wherein I provided a badly posed answer to a badly posed question!

Thanks anyway

Brian W

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Nigel:

The formula is not wrong. The Machinery's Handbook is over a hundred years old. I use those formulas all the time and have for eighteen years. Either you miscounted the teeth (which seems unlikely), or you're not describing the mechanism correctly.

Look; It's not rocket science, nor is it magic. Use your head! Gears are simple.

If you turn the sun gear until its point of contact with the planet comes back around into contact with it, you will have turned the planet by nine gear teeth. In doing so it has walked nine teeth on the outer gear. For the planet this is one-fifth of the way around (9 out of 45 teeth), but for the sun gear it took one revolution PLUS one-fifth of the way around, or 1.2 revolutions. To get the planet all the way around requires five of these cycles, so for the sun gear

5 x 1.2 = 6 revolutions

This is high-school math here, kids.

Don

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I removed the sun and shaft so I could turn by hand and see the operation. I marked one tooth of one planet and then 2 turns of the output shaft produced 5 whole turns on this planet. Then the 2 to 1 ratio between sun and planets but I had ignored the rotational positional change of each of these planet gears.

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Ta Da! Mechanical things are logical. This matches the Machinery's Handbook, too. Finally, truth in this thread. :)

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Yep, agreed 1+ 45/9 = 6

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