Planetary Gear Torque Output & Distribution 2

[AxeMan]

I am admittedly a rookie when it comes to planetary gear trains, so bear with me. I am unsure how to determine the strength of a given planetary gear train. I don't understand the dynamics of power distribution in them. Determining the strength of a given gear is one thing, but I don't know how they all share the power transmitting duties. Is there a formula or so to determin the strength? Also, this gear train will be operated manually, so it is difficult to base anything on horsepower ratings because of RPM fluctuation. This unit will need to generate around 600 ft. lb. of torque at extremely slow RPM. Any help is much appreciated.

 

[plasgears]

Work with the basics. Use vectors to determine forces between mating gears. Load sharing is assumed in the planets. Once you have a handle on tooth forces, then you can stress analyze the gears using the Lewis equation for starters.

 

[AxeMan]

Thanks for the reply. I am familiar with the Lewis equation. If the sun & planet gears are exactly the same, which gear(s) am I to assume is/are the weekest in the train and why? I'm working with a 4:1 ratio. I realize the sun gear will have 3 points of contact (3 planet gears) and the planet gears will each have 2 points of contact. I would really like to understand how power is distributed in such a gear train. That's where a formula would help.

 

[EnglishMuffin]

You also might want to take a look at this thread, which was about different methods of ensuring good load sharing in planetaries.
thread406-59082

 

[EnglishMuffin]

And a few more tips :

Assuming you have decided upon the basic arrangement, ratio, numbers of teeth, helical or spur etc, the next thing you need to do is to figure out the torques and then the forces.
When figuring the pitting fatigue strength, the multiple meshes need to be taken into account.
Just as in designing any other gearbox, it is frequently possible to get balanced strength on both teeth of a mesh by adjusting the addenda. On the other hand, one might decide to adjust the addenda to give mimimum slide/roll ratios.
One thing to remember is that the planet gears will experience reverse bending, like any idler gear, and will need to be derated because of that.
The degree to which load sharing needs to be considered depends on what steps you have taken, if any, to ensure it.
Some people use a de-rating factor. It depends on the design.
Sometimes, a good way to picture a sun and ring gear planetary when you are figuring out torques, ratios etc is to imagine that you are viewing the train from a reference frame that is rotating at the same speed as the planet carrier. That way, the train appears as a simple non-planetery, and you can figure things out more easily..
For a low speed such as you appear to have, you really shouldn't have to contend with "dynamics" as you put it.

 

[AxeMan]

Thanks for the reply. I was aware of some of the issues with actually achieving equal load sharing. I hadn't thought about the reverse bending issue. Actually, since this unit will be bi-directional, all the gears will experience reverse bending, although the planet gears will experience twice the effect. I do plan on de-rating the gears to account for all such issues. Forgive me if I am misunderstanding your explanation, but I still need help. To simplify things let's assume that everything is perfectly made and positioned in my gear train. I want to boost torque with my planetary gear train. If the 1 sun & 3 planet gears are the same, as they will be in my 4:1 unit, in order for me to apply the Lewis equation to the proper place I need to know which gear's teeth are going to carry the most load. There are going to be 6 engagement points. Again, assuming perfect construction for simplicity's sake, where (which engagement point) is the greatest load going to be, and how do I calculate what percent of the input load on the sun gear is going to be exerted on the teeth at that engagement point?

 

[EnglishMuffin]

I'm probably being a little slow on the uptake - but bear with me.

Assuming perfect load sharing, if the input torque on the sun gear is Ti and its radius is Ri, then:

The tangiential loads on the three meshing teeth on this gear will be Ti/(3*Ri).
The three planet gears will each see these same tangiential loads on one of their teeth, and the same load on a diametrically opposed tooth, in the same absolute direction.
The planet bearings will each see twice this load in a tangiential direction.
The internal gear will also see tangiential loads on three teeth of Ti/(3*Ri).
If we agree so far - what exactly are you asking beyond that ?
By the way - I would recommend ANSI/AGMA 6123-A88 - Design manual for Epicyclic Gear Drives.

 

[AxeMan]

You're not slow on the uptake, it's my ignorance on this issue that's causing the communication slowdown. I can't thank you enough for taking the time to offer your help. Anyway, if I understand you correctly, all the teeth that are engaged at any one moment are up against the same tangential load. Okay. Now I know what I'm up against with the gear tooth load. Next I need to figure out how much force the planet carrier can deliver to the output shaft with a given input torque (again, with a perfect system) so I can choose appropriate bearings, etc. How can I calculate the output torque? Do I just multiply input by the ratio (assuming 100% efficientcy)?

 

[EnglishMuffin]

Well, if you add up all those forces on the planet pinions and apply them to the planet carrier, taking into account the radius of the planet shafts, you will find that they give you an increase in torque of 4:1, as you would expect, since that is your speed ratio. If I were rating the gears, I wouldn't consider efficiency. Its normal to ignore it when doing the strength calculations. Do the efficency calculation at the end if you need to - calculate it at each mesh. You will generally see 0.5-3% loss per mesh for internal and external spur and helical. To calculate it see, for example, Chapter 12 of Dudley's Gear Handbook. In general, it is very dependant on lubrication churning and windage, not just tooth friction. But in your case, probably just the tooth friction will be relevant.

 

[diamondjim]

AxeMan,
You might want to consider using a 50 percent
long addendum for the sun, 25 percent long for
the idler and a special pitch diameter for the
internal gear to balance the strengths.
If you submit the number of teeth and dp and
pressure angle, I will be glad to offer several
center distance options and still hold the 4:1
ratio for the system. If your center distance
is fixed for the Sun and Annulus, you can drop
one tooth in the idler and still use some
combination on long addendums on the sun and
idler to arrive at the same center distance.
You might also look at shaving the addendum on
the internal annulus to prevent any involute
interference between the idler and the annulus.
Just a few ideas to balance the system.
My email is J.GEISEY@juno.com

 

[EnglishMuffin]

Although it looks like I'm out of the loop here, I'll just mention one other suggestion, which is to use a multple of three for the number of teeth on the sun (and by implication ring) gears. It simplifies the assembly issues with equispaced planets. If the issue of hunting teeth arises, in my view its a myth, unless you have a cyclic torque which always occurs in the same place - in which case its a very good idea. But if you do what Diamondjim suggests, which looks like a good plan, you will probably get hunting teeth as a bonus. Good luck.

 

[AxeMan]

English Muffin,

Out of the loop?! Says who? You have been most helpful to me. I was planning on using 18 teeth for the sun & planet gears, 54 for the ring. This unit is going to see very little action(2 output revolutions per day would be a lot), so wear is not even an issue. Diamond Jim's idea (thanks for the input Jim!) would be great if useage was going to be much higher, but since it's not I want to keep things simple. Getting back to an earlier post of yours, using the formula Ti/(3*Ri), if input torque is 100 ft-lb and the sun & planet gears have a 1-1/2" PD, then each tooth in the 6 mesh points on all gears will experience a tangential force of 44.44? What unit of measure is this answer in? If I seem a little incoherant, it's just one of those days for me where I feel like the proverbial chicken without a head...

 

[EnglishMuffin]

So you're not gonna use Diamondjims idea huh ?
And he got the star ! (You see, I want to get tipster of the week, and put it on my resume so I can find a job !
Actually, kidding aside, Diamondjim's design is no more complicated or expensive to make, once you have designed it - just a little more design work that's all. But you are probably smart to keep it simple in this case.
Now :
If the input torque is 100 lbf.ft, that's 1200 lbf.in.
That means that, if the pitch radius of the input pinion is 3/4", you will have a force of 533.33 lbf at each of three equispaced teeth on the input pinion. On the planets, each will have a force of 533.33 lbf on one tooth, and 533.33 lbf on a diametrically opposed tooth. So each planet gear exerts a force of 2*533.33 lbf on the planet carrier, at a radius of 1.5 ins. Thats a total output torque of 6*533.33*1.5 lbf.in = 4800 lbf.in, which lo and behold is exactly 4 times the input torque as it should be. Hope that helps. By the way, I have no clue at this point whether those forces are reasonable.

 

[AxeMan]

English Muffin,

Your last post just earned a star from me because it finally pulled the concept of power distribution into focus for me. Thanks again for your help. I must confess that although I liked Diamond Jim's idea, I was not the person who gave his post a star. Someone else must have done it.
Please don't drop out of the loop. You never know what I'll come up against next. ;o)

 

[EnglishMuffin]

Thanks - glad to be of help. I once designed a 4:1 planetary very similar to yours - although it had rather more teeth - I think thirty something on the pinions. It had no addendum mods or any design pyrotechnics of that sort and worked just fine. But you have only eighteen teeth. Thats about the smallest number you can go without undercut - so increasing the addendum has something to be said for it. I did allude to doing something like that in one of my posts too. But it'll work OK, it just won't be optimum. This doesn't sound like a real critical application anyway. Don't take my remarks too seriously by the way - I just have a rather warped sense of humor!