Achieving desired epicyclic ratios 2

[zachalbert]

Greetings. I have a job I'm working on for building a mechanical orrery. I'm a 3d modeler normally tasked with less technical jobs, so I thought I'd ask for some good web or print resources to help with my question:

I've been crash-studying epicyclic trains but a brick wall I keep running up against is a way to kind of work backwards; say I know that I need a 1:8.32 raio, how to achieve this mathematically (other than my guess & check methodology thus far!). Everything I've found online so far has explained more or less how to calculate possible ratios for existing planetary trains, but how to start from scratch?

Any help, or even links or names of books would be really great. Thanks!

 

[Philrock]

I would have thought it would be possible to rearrange the equations to get what you want. If not, set up an Excel model that calculates the ratio for a given setup. Have Excel Solver modify the inputs so the ratio comes out as you want it to.

 

[eromlignod]

If you're building an orrery, I'm guessing that you are talking about a sun gear with a planetary gear revolving about it. Is your ratio the spin of the planet gear to its revolution about the sun gear? If so, then the ratio is the same as the ratio of numbers of gear teeth of the two gears. So, for example, if the sun gear has 100 teeth and the planet gear has 25 teeth, then the ratio is 100:25 or 4:1. So the circumference of the sun gear is four times that of the planet gear and the outer gear rotates one full revolution when the connecting link revolves a quarter of the way around the sun gear.

To calculate tooth-ratios in reverse, knowing the desired motion ratio, use the "method of conjugate fractions" (fractions are the same as integer gear ratios). I was going to simply post a link to the thread where I had explained this method, but it seems to be gone, so I'll explain it again.

Take your ratio and make it a decimal between 0 and 1. If it's greater than one, then you can simply take the reciprocal (1/x). In your case this is

1/8.32 = .12019

Then we start with the conjugate fractions 0/1 and 1/1, which we know the ratio falls between. You can find the simplest fraction between two conjugate fractions by adding the numerators and denominators to create the new conjugate fraction. In our case this would be

0+1 / 1+1 = 1/2 = .500

Which is the simplest fraction between 0 and 1 and is conjugate to both the other fractions. Our ratio falls between 0/1 and 1/2, so we add the numerators and denominators of these

0+1 / 1+2 = 1/3 = .333

Now we're between 0/1 and 1/3, so

0+1 / 1+3 = 1/4 = .250

To continue

0+1 / 1+4 = 1/5 = .200
0+1 / 1+5 = 1/6 = .167
0+1 / 1+6 = 1/7 = .143
0+1 / 1+7 = 1/8 = .125
0+1 / 1+8 = 1/9 = .111

But now our ratio is greater than 1/9, so it falls between 1/9 and the next conjugate up 1/2, so we get

1+1 / 9+2 = 2/11 = .182

now our ratio is between 1/9 and 2/11, so

1+2 / 9+11 = 3/20 = .150

and so on. With each iteration you get closer and closer to your ratio and are guaranteed that you didn't miss any simpler fractions (and thus have the fewest number of gear teeth). This can be done with any ratio and to any accuracy desired.

There are also tables of gear ratios (in the Machinery's Handbook, for example) and tables of gear ratio logarithms which are also handy. But the conjugate method is just as effective.

Of course, one obvious choice, if your ratio is exactly 8.32:1, is 832:100, which reduces to 208:25. But that's a lot of gear teeth.

Hope this helps.

Don
Kansas City

 

[rb1957]

i followed that right up to "and so on" ...

i think you can achieve the same ratio with two steps (of about 2.44 = sqrt(8.32)) ...

 

[eromlignod]

I followed yours right up to "2.44 = sqrt(8.32)", which doesn't make sense (and is untrue).

If you've only got two gears in a simple ratio, which I believe is the case here, then conjugate fractions will get you the simplest ratio (fewest gear teeth) for a given desired error.

Two fractions a/b and c/d are conjugate if

ad - bc = +/-1

and have the property that no fraction between them can have a denominator less than b or d.

All you're doing is continually finding which two conjugate fractions that your particular ratio falls between and then finding the simplest fraction in that gap, then determining which side of the new fraction the ratio falls into and doing it again, etc. As long as you keep adding numerators and denominators to find the next simplest fraction in between, you will continue to approach the ratio with the simplest fraction possible.

Don
Kansas City

 

[rb1957]

sorry typo ... sqrt(8.32) = 2.88 ... my point was you can achieve a gear ratio with more than one set of gears, and if your use the sqrt of the final ratio, isn't that an efficient gearing ?? ...

and my original statement wasn't meant as a dig, just that in the limited space of these posts it's hard to get the full message across (google didn't help much either). i don't work with gear design, but why are conjugate fractions required ... other than to bound the exact solution ... in this case couldn't you start with 832/100 (=208/25) and see how that fits your available space. maybe try 50/6 = 8.3333

 

[zachalbert]

I really appreciate the help. Eromlignod, this is exactly what I'm looking for. And the reason I'm running into trouble is because 832/100 is just too big. I'm really trying to keep the teeth below 300 in order to keep the orrery from being gigantic. I was looking for a method to discover the fewest teeth possible in any given ratio.

In answer to both your questions, it isn't necessary that these ratios be achieved in one step. For instance, mercury's revolution is .25 (roughly) earth's while saturn is nearly 30 times. In order to keep things from getting too huge, I think I may need to have my epicyclic trains set up in more than one stage.

I appreciate you guys point me in the right direction! Any more thoughts or comments are a big help as well.

 

[eromlignod]

Oops. I made a mistake in my example. .12019 (your ratio) is between 1/8 and 1/9, so the rest of the fractions would be

2/17 = .117
3/25 = .120
4/33 = .1212
7/58 = .1207
10/83 = .1205
13/108 = .1204
16/133 = .12030075
19/158 = .120253
22/183 = .120218
25/208 = .12019

Which is the reduction of your 100/832 fraction. You can take your pick of these fractions and know that there are no simpler ones that were missed.

Don
Kansas City

 

[eromlignod]

...oh and don't forget that you can also use unreduced forms of these fractions to get the same ratios. So 2/17 = .117, but so does 4/34, 6/51, 8/68, 10/85, etc. This gives you even more possibilites to match up with what gears are available.

Don
Kansas City

 

[rb1957]

actually that cleared up a confusing point in your post (going back to 1/2) ...

 

[JCJackson]

Remember that actual [internal] gear ratio is 1 - 1/E, where E is the epicycloidal (overall) ratio. So for your 8.32 ratio, his can be achieved with (8 x 10)/(7 x 13). See Machinery's Handbook. Mine's antique 19th edition.

 

[Designmechman]

Hi everyone im pretty new to the forum and just wanted to ask some q's about gear ratios and compound gear trains. I need to acheive an overall ratio of 1500:1 from the energy released by a spiral spring with 20 turns mounted on a 40mm dia shaft. Thickness of spring is 1.5mm with 30mm width. max torque is 10200 Nmm.

What is energy released from spring?
I have come up with gear designs of using four sets of gear trains compound with 100T and 10T which comes to 1/10,000. This is too high, so i went to reduce it to 60T and 14T using 5 sets to get 1500:1. Is this a practical mesh of 60 teeth to 14 teeth i.e 6dia to 1.4 dia all in cm.????

Please help me out and give some solutions, deadline is closing in!!!

 

[GregLocock]

I suggest you start a new thread.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[rb1957]

yeah, something like "how to calculate the energy released from a spiral spring?" ... or is the question more about gear trains ??

whilst i'm no expert (but that never stopped me before !) ... I'd've thought that 5 compound gears was a simple (possibly "agricultural") solution to achieve a 1500:1 ratio ((60/14)^5 = 1500). I wouldn't think it's an issue working in metric units. I wouldn't think it's a problem to find a gear design to mesh a 60T gear with a 14T gear ... tho' putting 14T on a shaft 40mm dia sounds a bit rough (say the mean dia. of the gear is 50mm, this means 1 tooth every 11mm (1/2") ... big teeth, no? and how much load are they transmitting (at the high torque end of the gear train) ... 1 tooth in 11mm says that the 60T gear has dia. = 660mm (no?) ... 10200Nmm of applied torque means a tooth load of 15N (well that's not much !) ... and what's your output torque going to be (10200/1500 = 7Nmm) delivered for 20*1500 = 30,000 revs

 

[Designmechman]

Thanks for that, what i gathered is that its too simple (the gear trains solution). What i would like to know is calculating Epicyclic Ratio of 1:1500. I looked on the net but couldnt even understand how its all done.

1) There is an input shaft horizontal 40mm dia with spiral spring mounted on it. This is to connect to epicyclic gears.
2) There is an output shaft vertical which has a spinning disc on top 15cm dia.
3) Gear ratio of input to output must be 1:1500

Please hwlp me calculate the number of teeth and whatever is necessary like diameters of all gears etc. Remember the spinning disc on top, is 15cm dia, so cant have the gears too big in dia.

Once this is done i will model it all in solidworks.

Work deadline for this is in two days, please help!!!!

 

[GregLocock]

Designmechman - If you want me to help then start a new thread.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.