Faulhaber gearheads?

[GaryHinshaw]

Sorry, TLDR.  I think you're in good hands with Max.  But if someone wants to state the problem succinctly, I'd be happy to have a look.

-gfh

[peteski]

This seems like a @GaryHinshaw problem solving issue.

"Planetary", and "astrophysicist"?  I see what you did there! 

[mmagliaro]

Alrighty Gary... planet pun or not,  I appreciate you jumping in.
Given a planetary gear set, where the outer ring is held fixed, the input is the rotating sun gear,
and the output is rotating carrier that has the planetaries on it, how is the gear reduction ratio
computed?

This drawing is close (although the planetaries show only 14 teeth because the simulator I used to draw it wouldn't
let me specify 16, but they should have 16)




When I analyzed this, my reasoning went like this:
1. For the carrier to go around once, there must be 41 tooth movements (imagine just
one of the planetaries rolling over the ring gear as the carrier makes one full rotation)

2. There are 12 teeth on the input Sun gear.   For that gear to advance 41 teeth requires
41/12 = 3.41 rotations

Therefore, I concluded that the ratio would be 3.41:1 because the input would go around 3.41 times
and the carrier would go around once.

But somehow, this is not correct.  I am off by exactly 1.  The correct answer is 4.41:1, so says a few
different planetary gear ratio web sites I looked at.

So this is where you come in, Gary.  What am I missing?  Why is it 4.41:1?

[nickelplate759]

From what I can see, the formula depends on which part of the system is fixed.

For a fixed ring, the ratio of sun rotation to carrier (that holds the planets) rotation is (teeth-of-sun + teeth-of-ring) / teeth-of-planet
(12+41) / 16 = 3.94
If the ring actually has 42 teeth, the ratio is exactly 4.
[doh!] this was wrong!  .(12+41) / 16 = 3.31 [/doh!]

If the carrier is fixed, then the planets are just idler gears, and the ratio is (ring/sun) - 41/12=3.42
If the ring actually has 42 teeth, the ratio is 3.5

If the sun is fixed, then the ratio is (ring/planet) : 41/16 = 2.565
If the ring actually has 42 teeth, the ratio is 2.625

[davefoxx]

Wow.  The calculations and math here have my head spinning faster than a planetary gear. 

DFF

[mmagliaro]

From what I can see, the formula depends on which part of the system is fixed.

For a fixed ring, the ratio of sun rotation to carrier (that holds the planets) rotation is (teeth-of-sun + teeth-of-ring) / teeth-of-planet
(12+41) / 16 = 3.94
If the ring actually has 42 teeth, the ratio is exactly 4.

If the carrier is fixed, then the planets are just idler gears, and the ratio is (ring/sun) - 41/12=3.42
If the ring actually has 42 teeth, the ratio is 3.5

If the sun is fixed, then the ratio is (ring/planet) : 41/16 = 2.565
If the ring actually has 42 teeth, the ratio is 2.625

Every place I look this up says something different.  It's ridiculous that this is so hard.

https://sciencing.com/calculate-planetary-gear-ratio-6002241.html
says, for fixed ring, carrier as output (which is what we have),
"divide the number of teeth on the planetary gears (the driven gear) by the number of teeth on the sun gear (the driving gear)"
And it says the number of of teeth on the planetary gears is sun + ring, which would be 41+12 = 53 in our case.
So it would be 53 / sun teeth or 53 / 12 = 4.416666  (4.42)

[mmagliaro]

Okay, HERE, FINALLY.

GREAT presentation, step by step, with mathematical derivation of how it works out that for a fixed ring gear (like in our gearheads),
the ratio is

1 + Zr/Zs 
Zr = teeth in ring
Zs = teeth in sun

The number of planetary gears, and planetary teeth do not matter.
So in this case, it is 1 + 41 / 12 = 4.41666 : 1    (4.42:1)

[woodone]

OK, after a bit of thinking I took a photo of the gears-down loaded to the computer-enlarged them so I could see them to get an accurate tooth count.  The pinion gear (sun gear) has 12 teeth. The planetary gears have 16 teeth (there are three used) the ring gear has 45. We are using the planetary gear carrier out put has the final drive.
So 16 dived by 12 = 1.333  and dividing the ring gear (45 teeth) by the planetary gear (16 teeth) you get 2.8125.
How do we work with these two numbers?  If you add the two you end up with 4.1455.
Divide the 2.8125 by the 1.333 you end up with 1.72
Multiply  you get  3.749
Do not think 1.72 is correct, so that means we add or multiply ?

[GaryHinshaw]

Thanks Max.  I am going to analyze this without looking at anyone else's answer, then you can tell me if I'm right. 

This system is just one of compound circular motion where the gears are just wheels that move synchronously without slipping; all that matters are the radii of each wheel.  The tooth count of a given gear is a (quantized) proxy for the radius, so we have sun gear with radius r_s=12, a set of planetary gears with radii r_p=16, and a ring gear with radius r_r=41.   I'll start off assuming the sun gear spins with angular velocity omega_s in radians/sec (more on that unit later, but it is proportional to revolutions per minute), then the speed of a point on the perimeter of the sun gear moves with speed

v_s = r_s*omega_s = 12*omega_s

Note that omega_s is the input angular velocity.

The perimeter of the planetary gear must move with the same speed v_s (the gear teeth enforce that).  Here we need to account for both the spin of the planetary gear about it own axis, omega_p, and the translational motion of that gear's axis.  The speed of the planetary gear at the contact point of the sun gear is given by

v_p = v_s = - r_p*omega_p + (r_s+r_p)*omega_c

where the first term arises from the spin of the planetary gear (proportional to omega_p), and the second term is the translational motion of the gear's axis (proportional to omega_c, the carrier's angular velocity).  Note that (r_s+r_p) is the radius at which the axis of the planetary gear is located, and the minus sign in the first term accounts for the vector component of the spin motion.  (This is easiest to picture at a point when the gears are stacked vertically.)

The gear ratio of interest is:

R = omega_s/omega_c

But we still need one more fact to constrain omega_c and omega_p: namely that the ring gear is fixed.  This comes from requiring that the outermost point on the planetary gear have zero speed:

v_p(outer) = 0 = +r_p*omega_p + (r_s+r_p)*omega_c

We now have two equations and two unknowns (omega_p and omega_c).  We do a bit of algebra to solve for omega_c in terms of things we know and find

R = omega_s/omega_c = 2*(r_s+r_p)/r_s

With r_s=12 and r_p=16, I get

R = 4.67

How did I do? 

-gfh

Caveat: it should be the case the r_c = r_s + 2*r_p, but that gives 44, not 41, so my math does not quite account for the fact that the effective radius of each ring is a bit smaller than the gear count would imply.  So I would expect my answer to be too large by a factor of ~44/41 = 1.07, so actually ~4.35.

[GaryHinshaw]

Ah, the answer was posted.  Given r_r = r_s + 2*r_p, my formula reduces to:

R = 1 + r_r/r_s <--> 1 + Z_r/Z_s

so we agree when expressed that way.  Phew. 

[mmagliaro]

Ah, the answer was posted.  Given r_r = r_s + 2*r_p, my formula reduces to:

R = 1 + r_r/r_s <--> 1 + Z_r/Z_s

so we agree when expressed that way.  Phew. 

After all the reading and watching, I think you drove to the core of understanding how to derive this by thinking about angular velocity, and the instantaneous velocity at the teeth.  Watching the YouTube video I linked made that light bulb go on for me.
But you have to work strictly in terms of the tooth count because relying on the radius is what introduced some error in your numbers.  Even after you scaled it for the 44 vs 41 tooth discrepancy, you got 4.35:1 when the correct answer is 4.42.

[mmagliaro]

OK, after a bit of thinking I took a photo of the gears-down loaded to the computer-enlarged them so I could see them to get an accurate tooth count.  The pinion gear (sun gear) has 12 teeth. The planetary gears have 16 teeth (there are three used) the ring gear has 45. We are using the planetary gear carrier out put has the final drive.
So 16 dived by 12 = 1.333  and dividing the ring gear (45 teeth) by the planetary gear (16 teeth) you get 2.8125.
How do we work with these two numbers?  If you add the two you end up with 4.1455.
Divide the 2.8125 by the 1.333 you end up with 1.72
Multiply  you get  3.749
Do not think 1.72 is correct, so that means we add or multiply ?

The 45 and 12 are all that matter.  The ratio is really 1 + 45/12 = 4.75:1
When we thought it was a 41 tooth ring, it was 1+ 41/12 = 4.42

You are trying to do what I did: look at it as just a couple of standard gear ratios.  But that is not correct.
I think I can explain it in more "human" terms this way:

When the sun gear turns one tooth, the planetary turns one tooth, true.  That is a conventional gear ratio.
But when that happens,  the carrier MOVES.  It doesn't just move one tooth like a gear.  The carrier is not a gear.  The carrier motion is really the center of one of the planetaries moving around a circle somewhere in between the sun and ring gears.  And the radius of that circle is determined by the radii of the various gears.  That's why the YouTube video I linked, and Gary, analyze the whole thing in terms of radius and angular velocity.  That's the only way to do it right, because not everything moving in the system is just a gear.  You've got that carrier in there.  You have to convert everything to velocities at the outside edges of circles.

[woodone]

Wow, did not know that I would get a math education here on the Railwire. Just shows how much you can learn when you do Model Railroading!

[Hiroe]

I do have a lot going on. However I can get some gearheads shipped out next week. I'll send PM

Tom, were there any 2020C012 gearhead motors left? (20x20mm, 59:1 gearing, 12v)
I could use several for my HOn3 boxcabs.

Thanks!

[DKS]