Faulhaber gearheads?

[peteski]

I have two or three small motors with gear reduction. I was able to remove one geared section. The way I am figuring the gear ratio I come up with 4:1. This seems way too slow? Here is the gear tooth count- I must be doing something wrong.

While I have never calculated the ratio, the answer should be on the Interwebs.
https://woodgears.ca/gear/planetary.html
https://www.physicsforums.com/threads/question-about-calculation-of-planetary-gear-ratios.927470/

Well, maybe not - it is not that clear cut.  I'll look some more . . .

This looks promising.
https://sciencing.com/calculate-planetary-gear-ratio-6002241.html

[mmagliaro]

I have two or three small motors with gear reduction. I was able to remove one geared section. The way I am figuring the gear ratio I come up with 4:1. This seems way too slow? Here is the gear tooth count- I must be doing something wrong.
The sun gear is 12 teeth (motor driven) the planetary gears (3 of them) each with 16 teeth, and the ring gear (stationary} 41 teeth.
I tried this into an 4-4-0 loco and the top speed is very slow 10-15 MPH guessing on that.
The biggest problem is I do not have any of the motor or gear specks.
The loco can creep so slow you can take a cat nap waiting for it to creep to the next tie. Very smooth and lots of toque.

How did you compute the ratio?
This is what I get:

Focus on just one 16-tooth planetary.  It runs against the 41 tooth ring
and must go all the way around for the output shaft to rotate once.

To run over 41 teeth, the 16-tooth planetary has to rotate 41/16 = 2.56 times.

Now, the 12-tooth sun gear has to rotate 16/12 1.3333 times to turn the planetary once.

So I get 1.333 x 2.56 = 3.41:1 ratio (not 4:1)

As for your top speed, what 4-4-0 is this?  A Model Power?  That was a pretty nicely geared loco as it was.
It wouldn't surprise me if it's top speed was only 60 mph, so at 3.41, you get about 17.5 mph.

If you actually measure the top speed of it, it would help know if we're crazy or not.

[woodone]

Pete- I have looked at several ways to figure this out- still confused.
Max , I came up with my figure much like you did. 16 divided by 12. = 1.33. Then 41 dived by 16. = 2.56 but I added the two.
Came up with 3.89. Close enough to 4;1 for me..

[mmagliaro]

I don't understand adding the two ratios.  It's really just like a chain of spur gears.

If you had a 20 tooth gear meshed to a 40 tooth gear, then you get a 2:1 reduction.  If that 40 tooth gear was then
meshed to a 100 tooth gear, that would give you an additional 2.5:1, and the total reduction for the 3 gears in a row would be 2 x 2.5 = 5:1, not 4.5.

You can check it like this, much more simply: 
If the 20-tooth gear rotates 5 times, that's 100 tooth movements..   It doesn't even matter what the intermediate gear does.
100 tooth movements will turn the final gear one time, because it is a 100 tooth gear, so the ratio is 5:1 because the first
gear will go around 5 times for each rotation of the last gear.

In the planetary,  we need to move the planetary assembly over 41 teeth of the ring in order to rotate the output shaft
one time because the output shaft is connected to the planetary assembly. 
If the sun gear has 12 teeth, how many times does it rotate to go 41 teeth?
41/12 = 3.41

You don't even have to consider how many teeth the planetaries have.

[woodone]

That’s why I did not come up with the right Ratio.
But at any rate the output speed is too slow, I will have to see what is in the worm drive.

[mmagliaro]

That’s why I did not come up with the right Ratio.
But at any rate the output speed is too slow, I will have to see what is in the worm drive.

What 4-4-0 is this?  The Model Power?  I know they were geared slow, which was good.  The MR review shows them at a top speed of 75 mph at 12v (they say "80" in the text, but the speed graph they published with the review clearly shows that it is 75).
So 75 / 3.41 = 22 mph.   That is slow for the top speed of a 4-4-0.  Personally, I would like that.  I'm in the camp that believes that we should never be running our locomotives anywhere near their real-life top speed because they look ridiculous.  For most of us, our layouts are massively compressed representations of real life, so an engine running at 50 mph through our scenery looks like it's going 100, at least to me

[woodone]

Not sure who manufactured it.
And before this go much farther, it is an HO n3 model.
Yes, I agree with you about speed, 90% are running twice has fast has they should be.
I will take a look at the model to see if I can ID it.

[peteski]

From https://sciencing.com/calculate-planetary-gear-ratio-6002241.html

First Steps
To make calculating planetary gear ratios as simple as possible, note the number of teeth on the sun and ring gears. Next, add the two numbers together: The sum of the two gears’ teeth equals the number of teeth on the planetary gears connected to the carrier. If the sun gear has 12 teeth and the ring gear has 41, the planetary gear has 53 teeth. The next steps depend on the state of the planetary gears connected to the carrier, although all use the same formula. Calculate gear ratio by dividing the number of teeth on the driven gear by the number of teeth on the driving gear.

Carrier as Output
If the carrier is acting as the output in the planetary gear system, being rotated by the sun gear while the ring gear stays still, divide the number of teeth on the planetary gears (the driven gear) by the number of teeth on the sun gear (the driving gear).  53 / 16 = 3.31.

That is close, but not the same as what Max came out with.

[mmagliaro]

Pete, I saw that calculation page too.  But this is the part I don't understand:
"Note the number of teeth on the sun and ring gears. Next, add the two numbers together:
The sum of the two gears’ teeth equals the number of teeth on the planetary gears connected to the carrier."

But it DOESN'T.  There are 12 teeth on the sun in this case, and 16 on the planetaries, and 41 on the ring.
12 + 41 = 53
There are 3 planetaries  = 3 x 16 = 48 teeth, which does not equal 53.

Then they say, "divide the number of teeth on the planetary gears (the driven gear) by the number of teeth on the sun gear (the driving gear).   You did this as 53/16, but that's a mistake.  The sun gear has 12 teeth, not 16.
So it would be 53 / 12 = 4.41

I'm still looking for a page that actually explains how the ratio is what it is and doesn't just give me formulas.

They go on to compute 53 / 16

[mmagliaro]

I took a Faulhaber 4:1 gearhead and ran it through the formula Peteski found:
Faulhaber 4:1

3 planetaries, 12 teeth each
Case/Ring  36 teeth

Sun gear (pinion) also has 12 teeth
Using the formula from that website Peteski found...
Sun + Ring = 12 + 36 = 48  (although this still does not actually equal the number of planetary teeth.  That would be 36, so
I have no idea what they are talking about with that)

Then divide the number of teeth on the planetary gears (which they calculate as 48 in this case)
by the  teeth on the sun gear 12).  48/12 = 4.

Which is exactly right because this is a 4:1 gearhead.

I just need somebody to explain how adding the sun teeth + the ring teeth equals "the number of teeth on the planetary gears connected to the carrier", because it never does, even when the formula works.

[mmagliaro]

I found this animated simulator for planetary gears:

http://www.thecatalystis.com/gears/

It is very cool, but it will not let me configure it to have 12 teeth on the sun, 16 on the planets and 41 on the ring.
The closest I can get with 12 sun and 41 ring is a planet gear of 14 teeth, and then it gives  4.42:1

That is almost exactly what the formula on that other page computed (when I ran through it in the previous posts, and corrected
the sun gear to be 12, I got 4.41:1

Is that 41 teeth really right?  That seems very weird to have an odd number of teeth on the ring gear.

========================
Another reference:
https://khkgears.net/new/gear_knowledge/gear_technical_reference/gear_systems.html

It is a very intense, detailed page, but if you dig down to Table 17.1, it boils down the ratio to:
Zc/Za + 1  or   (number or ring teeth) / (number of sun teeth )  + 1
for the case where the planetary carrier is the output (which it is for our gearheads).

That would be consistent with my observation that the number of teeth on the planetary gears is irrelevant (notice
how the equation only includes the teeth on the sun and ring gears, and the planetary teeth are not part of the
equation).
41/12 + 1 = 4.41:1
Interestingly, it is my value of 3.41 + 1.

I just don't see the logic of where that +1 comes from.  I do not doubt it.  There is something else physically going on
that I just don't see.

[peteski]

I wonder if Jerry miscounted the teeth?

[woodone]

Pete, that is possible. These teach are small and being an internal tooth are hard to count. I marked the first tooth and counted the teeth twice but I still could have mis counted. But even  if I did not count the ring gear correctly (missed or added a tooth) would not the ratio be very close?
Max, do not understand the planetary gear count. Why would you add the planetary gear teeth and add the number of planetary gears used?
 We really do not need three planetary gears, do we? Two would also work?
I am headed up stairs now to recount the gears.
Something to think about. I worked on many torque hubs and you had to time the planetary gears to the ring gear when you put them back together, if you did not do this you, would break the ring gear case. 

[mmagliaro]

Pete, that is possible. These teach are small and being an internal tooth are hard to count. I marked the first tooth and counted the teeth twice but I still could have mis counted. But even  if I did not count the ring gear correctly (missed or added a tooth) would not the ratio be very close?
Max, do not understand the planetary gear count. Why would you add the planetary gear teeth and add the number of planetary gears used?
 We really do not need three planetary gears, do we? Two would also work?
I am headed up stairs now to recount the gears.
Something to think about. I worked on many torque hubs and you had to time the planetary gears to the ring gear when you put them back together, if you did not do this you, would break the ring gear case.

Yes, if you miscounted, the ratio would still be close.  For example, note how in the simulator I cited, I had to
run it at 12 sun, 41 ring, planet 14 (not 16), and it gives a ratio of 4.42:1, which is very close to the formula solution of 4.41:1

And yes, you don't necessarily need 3 planetary gears, and the output ratio would still be the same.  The choice of the number of planetaries is a function of designing for how much torque you want the gearhead to be able to handle and the strength of the individual gears.  The load is distributed over 3 teeth if you have 3 planetaries, only 2 if you have 2.  Of course, how many planetaries you can have is a function of the physical size of the gears.  They have to all fit in there.

As for "timing" the planetaries to the ring gear, I read something about that somewhere while I've been investigating this.  It only matters for certain ratios between the sun/planet/ring.  Imagine dropping all three planetaries onto the carrier, inside the ring gear, and now having to insert the sun gear into the center of them.  It could be possible that there is no way to get the sun gear in there so that it meshes with all 3 of the planetary gears.  One or more planetaries might have to be lifted out and rotated one tooth one way or the other.  In the gearheads we are using for Faulhaber and Maxon, I have never seen any such thing.  The planetaries just drop onto the carrier inside the ring gear, and the pinion (the sun gear) always fits.  There is no "wrong way" to put them in.

[CRL]

This seems like a @GaryHinshaw problem solving issue.