Faulhaber gearheads?

[CRL]

Using the pure radius of the gears does introduce a small error since the gear intermeshing creates an overlap that reduces the effective radius of each gear. If the interface between the gears was a flat surface, the pure calculation would be accurate. It seems logical that the effective radius of one gear could be determined by reducing the radius by the depth of the gear teeth, but I have no idea if that is correct. Never took Physics or Calculus in HS.

[svedblen]

Is it really necessary to mess with the radius of each gear? Why not just think like this:

• When a planetary moves one tooth the sun rotates one tooth. This gives the [ring teeth]/[sun teeth] contribution as discussed earlier.
• But when the planetary moves one tooth it also traverses a short distance of its circular path around the sun, meaning the planetary to sun "contact point" has moved. This also adds to rotation of the sun, i.e. the sun does not only rotate because the planetary rotates around its own center but also beacuse the planetary center has moved.
• This extra sun rotation is 1/[ring teeth] of a revolution, for each tooth passed by the planetary, and repeats itself [ring teeth] times, and always comes out as exactly one extra revolution for the sun. Regardless of what the actual radii of the gears happen to be.
• Thus the additional "1" term, and the complete formula is  [ring teeth]/[sun teeth] + 1. And with 41 ring teeth and 12 sun teeth it becmes 41/12 + 1 = 4.42 if you round the result and 4.41 if you truncate.

Please correct me if I am wrong and just has shown my ignorance 
 

[CRL]

But teacher, you didn’t say we were going to have Word Problems on the test. 

[mmagliaro]

I do believe svedblen has hit upon a good physical explanation of what's going on and where the extra "1" comes from.
I wish we could have an animated simulation of this, narrated!

[GaryHinshaw]

@svedblen 's analysis is sound.  I think it boils down to the fact that for every cycle of the planetary gear around the ring, it spins one less time than it would if it were rolling along a straight pinion gear with the same number of teeth.

Personally, I find it easier to think in terms of angular and linear velocities, but that is probably just because of my physics training.  That said, I have come up with a puzzle about the number of teeth in this system which seems inconsistent to me.  To make a long story short, I don't see how this system can work with Z_p=16 teeth in the planetary gear.  If the system really has a tooth count of 12/16/41, then the teeth cannot all be equally spaced.  What do we know about the tooth spacing in this system? 

I'll put my math in fine print so as to take up less ink.

When expressed in terms of radii, the formula I worked out in my first post was

R = omega_s/omega_c = 2*(r_s+r_p)/r_s = 1 + r_r/r_s

The last equality follows because we have a constraint r_r = r_s + 2*r_p.  Expressed in terms of tooth counts, the latter formula translates to

R = 1 + Z_r/Z_s = 1 + 41/12 = 4.41666...

per the video Max posted.  QED

However, I could also translate the first formula into tooth counts as well:

R = 2*(1+Z_p/Z_s) = 2*(1+16/12) = 4.666...

which is the number I originally posted.  Why the difference?  My analysis assumes that all the gears have the same linear tooth spacing at their midpoints (then I can replace a radius ratio with a count ratio).   In that case, I can translate the radius constraint into a count constrain:

r_s + 2*r_p = r_r  -->  Z_s + 2*Z_p = Z_r

but

12 + 2*16 = 44 ≠ 41

So the tooth spacing cannot be the same for all 3 gears in the system.  You might argue that r_p cannot be as large as allowed by the geometry because there needs to be some mechanical tolerance in the system.  Ok, then we make an inequality:

r_p ≤ (r_r-r_s)/2  -->  Z_p ≤ (Z_r-Z_s)/2 = (41-12)/2 = 14.5

So now the problem is even worse.  If the teeth are all equally spaced, the planetary gear can have no more than 14 teeth for this system to work.

I'm guessing that the tooth spacing of the planetary gear is finer to account for the fact that the sun gear is so small.  (With small gears, the linear tooth spacing changes quite a bit from base to tip.)  But now we're in to detailed implementation questions.  If someone knows actual gear radii, as well as counts, we could work it out.   

[DKS]

My uneducated $0.02 to toss into the pot... I see another problem with the tooth count. If the outer gear indeed has 41 teeth, and there are three planetary gears, then they can't all be "indexed" with the same starting point, and hence evenly spaced, because 41/3 = 12.666, whereas the sun gear has a tooth count evenly divisible by 3. It seems to me there would be a problem physically assembling the gears. Shouldn't the outer gear have 42 teeth (or some other count with a modulus of 3) for it to work, or does it not matter? Do I make any sense, or am I just being a blithering idiot?

[GaryHinshaw]

That's a great point David.  To maintain the equilateral symmetry of the carrier, the outer ring needs to have a multiple of 3 teeth.  To make matters worse, 41 is a prime number, so maybe one of the worst possible choices.  A ratio of 42/15/12 would work out nicely though.  42 and 15 are both divisible by 3, and:

R = 1 + 42/12 = 4.5
R = 2*(1+15/12) = 4.5

That said, I don't think equilateral symmetry is a requirement of a planetary system, as long as the planetary pivots are equidistant from the sun pivot.  But surely it must be good for balance.

[DKS]

Just as an experiment, I took an empirical approach rather than mathematical: I drew the gears in a drawing program that allowed me control tooth count on the fly. I could not "assemble" the gears when the outer gear had 41 teeth; the planetary gears (which I made 15 teeth) had to be unevenly spaced (i.e. not arranged exactly 120 degrees apart) in order to come close. But everything dropped together perfectly when I made the outer gear 42 teeth. It makes no sense to me to design a gear system that forces the planetary gears to be oddly positioned relative to one another.

FWIW. YMMV. Etc.

[Missaberoad]

Reading this discussion got me thinking "I bet Matthias Wandel has something about this"
And he does...

https://woodgears.ca/gear/planetary.html

Now I'm not even gonna pretend to be smart enough to know if its at all helpful but figured I would leave it here. 

It always amazes me the level of intelect we have here...

[peteski]

Guys,  On the previous page Jerry corrected himself.  Just like I suspected all along (and asked Jerry for a recount), the ring gear does not have 41 teeth.

He wrote: 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.  Does that help out to make sense out of all this?   

[mmagliaro]

All of you who observed that a ring gear count of 41 is impossible with a sun = 12 and  planetary = 16  are correct.
woodone corrected his count about a page ago, and reposted that the correct ring gear count on his motor is 45, not 41.

So the ratio is 12/45 + 1, or  4.75:1

[DKS]

I confess I did not read every single post here. That said, I think it was a good exercise, nonetheless. And 45 makes perfect sense.

[svedblen]

All of you who observed that a ring gear count of 41 is impossible with a sun = 12 and  planetary = 16  are correct. woodone corrected his count about a page ago, and reposted that the correct ring gear count on his motor is 45, not 41.

Which then contradicts the constraint R = 2*P + S as given in the link that Missaberoad supplied! Will this ever be settled? 

[GaryHinshaw]

This was exactly my point my second analysis post (in the fine print, admittedly).  But at least now we have S+2P < R (44 < 45) rather than S+2P > R (44 > 41).  I think the small difference allows for mechanical tolerance.  It may or may not require slightly different tooth spacing for the planetary gear, but only by a small amount, at most.  I think this makes the actual gear ratio somewhere in the range 4.66 < R < 4.75, and probably closer to the high end.  But I suspect we have saturated this topic by now.

[mmagliaro]

Gary,
Very good observation: the radius formula almost HAS to be S+2P < R  or in real life, is might be too hard to make the mechanism work.  Constructing a zero-tolerance gearhead with perfect gear diameters is not easy.

There are examples where the equation is exactly equal:
In a Faulhaber 10mm 10/1 gearhead, there are 12 sun teeth, 12 pinion teeth and 36 ring teeth.
12 + 2 x 12 = 36.  S + 2P = R, exactly  But then, that's Faulhaber.  There's a reason their gearheads (and Maxon's) are so expensive and so much better than everybody else's.