Planetary gear sets

P

Probability

Full Member
Joined
Jan 26, 2012
Messages
425
An interesting problem I have encountered. I have a automatic transmission and I don't want to dismantle it to manually work out the gear ratios. Why do I want to do this anyway? Well, diagnostics cost money and time is money, so the faster and more accurately a diagnosis can be carried out the better for the customer.

I know that gear ratios can be calculated from (Driven/Driver) counting the number of teeth on each pulley, and I've also understood that the output (rpm) divided by the input (rpm) will calculate the gear ratio, although the math looks different.

Suppose then that I had a planetary gear set and first gear was selected. The number of teeth on the Sun gear and the number of teeth on the planet gear calculate to a ratio of 2.5:1. This information is given. The input shaft speed is 4800 rpm and the output shaft speed is 2800 rpm.

Creating a mathematical model...

(Input speed + output speed/output speed) = (4800 + 2800)/(2800) = 2.7:1

Now the question says it is 2.5:1.

Now I also believe that something else is going on inside that planetary gear set before the output speed is calculated because if I divide (4800/2.5) = 1920 rpm, and if I then divide (4800/1920) = 2.5:1

With this in mind it seems something else is also going on in the planetary gear set before the output speed becomes final. (I don't know what that is yet)

Without getting too technical, Newton said that for every action there is an equal and opposite reaction, so for now if possible I'd like to know how I could "reverse" engineer the gear ratio 2.5:1 to produce the mathematical model to prove it works!
 
Probability said:
An interesting problem I have encountered. I have a automatic transmission and I don't want to dismantle it to manually work out the gear ratios. Why do I want to do this anyway? Well, diagnostics cost money and time is money, so the faster and more accurately a diagnosis can be carried out the better for the customer.

I know that gear ratios can be calculated from (Driven/Driver) counting the number of teeth on each pulley, and I've also understood that the output (rpm) divided by the input (rpm) will calculate the gear ratio, although the math looks different.

Suppose then that I had a planetary gear set and first gear was selected. The number of teeth on the Sun gear and the number of teeth on the planet gear calculate to a ratio of 2.5:1. This information is given. The input shaft speed is 4800 rpm and the output shaft speed is 2800 rpm.

Creating a mathematical model...

(Input speed + output speed/output speed) = (4800 + 2800)/(2800) = 2.7:1

Now the question says it is 2.5:1.

Now I also believe that something else is going on inside that planetary gear set before the output speed is calculated because if I divide (4800/2.5) = 1920 rpm, and if I then divide (4800/1920) = 2.5:1

With this in mind it seems something else is also going on in the planetary gear set before the output speed becomes final. (I don't know what that is yet)

Without getting too technical, Newton said that for every action there is an equal and opposite reaction, so for now if possible I'd like to know how I could "reverse" engineer the gear ratio 2.5:1 to produce the mathematical model to prove it works!
Click to expand...
Do you know the respective radii of the gears at the contact point?
 
Probability said:
An interesting problem I have encountered. I have a automatic transmission and I don't want to dismantle it to manually work out the gear ratios.
Click to expand...
Without those data, in all probability you will not be able to calculate with any more accuracy.
 
I'm not 100% sure at the moment. I wonder if the author in the book has made some mistakes?

I know in normal gear calculations they say (Driven/Driver) = GR, but supposing a planetary gear set should be worked out like this...

(Input/Output) = GR

It seems to work on a good few examples but the odd one seems incorrect. By examples;

(4800/2800) = 1.7:1. The example says it is 2.5:1

(4800/3200) = 1.5:1. The example says it is 1.5:1

(4800/2162) = 2.22:1. The example says it is 2.2:1

Moving onto some Ford and GM transmissions...

(4800/1690) = 2.84:1. The example says exactly the same answer.

(4800/3057) = 1.57:1. The example says exactly the same answer.

(4800/6957) = 0.69:1. The example says exactly the same answer.

(4800/2172) = 2.21:1. The example says exactly the same answer.

Should I conclude that the author has made some mistakes?
 
As I understand it, the ratio of number of teeth should be the same as the ratio of speeds (assuming there isn't something else going on, as you suggested). You don't need to separately know diameters.

But I have no idea where you got "Input speed + output speed/output speed)" from. Also, you mention "the question", "the example", and "the author", but never mentioned where this comes from. What are you referring to? Is it something we could look at? At the least, we need to see the actual examples as given, in their entirety, in order to be sure.
 
Dr.Peterson said:
As I understand it, the ratio of number of teeth should be the same as the ratio of speeds (assuming there isn't something else going on, as you suggested). You don't need to separately know diameters.

But I have no idea where you got "Input speed + output speed/output speed)" from. Also, you mention "the question", "the example", and "the author", but never mentioned where this comes from. What are you referring to? Is it something we could look at? At the least, we need to see the actual examples as given, in their entirety, in order to be sure.
Click to expand...
Thank you for replying. Just to clarify the (Input speed + output speed/Output speed).

Initially I advised that I did not wish to dismantle the transmission to prove data. When a planetary gear set is on your bench you have a Sun gear, followed by some planet gears secured in a carrier. Each gear has teeth and the planet gears are smaller in diameter than the sun gear, which can also change specifications depending upon how complex the planetary gear sets become.

Keeping it simple.

The sun gear has a number of teeth, and the ring gear has a number of teeth. To calculate the ratio between the gears the teeth are counted. Let's say you count 40 teeth on the sun, followed by 76 teeth on the ring, then you have;

(40 + 76)/(40) = 2.9:1

You can also carryout the reciprocal of the ratio as follows;

(40)/(40 + 76) = 0.34:1

All this very much depends on which way the torque is passing through the planetary gear set and what gear is selected. Now that method is very good in a class room or a college workshop to learn, but in the real world workshop time is money and labour must be kept to a minimum, and diagnostics must be accurately carried out, hence finding alternative methods to calculate and prove the correct results required.

Now if you refer to the attachment you'll see the examples. I didn't want to post them or mention the author because he is a very respectable author in our trade, but it does not help when he will not answer his emails or provide a correction list. There can be many reasons why text is not accurate in a book after going to print, it's not necessarily down to the author, but it does not help people who buy his books when he does not respond to inquiries about his text, and he is asking very good money for each book too.

Referring now to the final part of my previous discussion regarding (input/output) = gear ratio. That part is correct but typo problems in the book caused me confusion.

planetary gears.jpg

Sorry the picture looks upside down. Bottom right example shows input 4800 rpm and output 2800 rpm. Clearly that example is incorrect for the ratio presented, i.e. 2.5:1. If you divide 4800 by 2.5 you'll end up with 1920. If you divide 4800 by 1920 you'll get 2.5. Such simple mistakes put the learner off, they have nowhere to turn as the book is the teacher and the learner is the student, and if the author is not willing to communicate with the people who buy his books, then that is very unprofessional. I've spent a lot of good money on his books and only every needed to ask his advise twice in the last fifteen years, and neither time did he reply to my emails. And no it's not that his email is redundant, his last book I bought from Amazon was purchased for £100 and paid in full, and then Amazon contacted me with the excuse the book was not available and then saying in the next breath that the author wanted £130 for the book. The books are good but very overpriced for the pitch of information provided. I've resonantly found ASE NATEF books very much better in my opinion.
 
That's both more and less information than I needed. I don't care about the author or what you think of the publisher's policies. I just want to know the right formulas and numbers. I'd like to see what the author has said about how to calculate gear ratios, and what was said about the formula you gave.

My initial confusion was that you gave both (Driven/Driver), with respect to number of teeth, and (Input speed + output speed/output speed), with respect to speeds. I wanted you to tell me why the latter is true, or at least how you know. (I was also confused because you initially just talked about doing some measurements, and then later you started comparing to some book without having mentioned it.)

I'm not an expert on gears, but my understanding is that the "gear ratio" for a combination of gears (which I think includes a planetary system) refers to the ratio of input : output speeds (or turns), and is calculated from the various numbers of teeth according to how they are connected. What we need to do is to find out whether your formulas are correct, and what the book is talking about, in order to make a comparison. Also, having now looked at information about planetary gears (which I intended but didn't do before), I realize that they can be used in different ways, and we need to know how the gears you refer to are being used.

According to what I found in Wikipedia, or more simply here, with power transferred from sun to planets with the ring stationary, the gear ratio is 1 + R/S = (R+S)/S, which sort of corresponds to your formula when we use teeth rather than speeds (and parenthesize correctly): (input teeth + ring teeth)/(input teeth). Is that what you meant? I don't know how the gears are being used in your examples.

In the picture you show this time, taking the gear ratio as a simple ratio of speeds as I expect (not your formula), and assuming the input speed for third gear that was cut off is 3200, you are right that they are correct (though reverse gear is rounded to the nearest tenth) except for first gear, which ought to be 1.7:1 if the speeds shown are correct.

Anyway, I don't think any of the calculations you show used numbers of teeth, so the only question really is the inconsistency in that one picture, for first gear. It does appear that either the ratio or one of the speeds is not what they intended. I take it the publisher does not post errata, which would be the appropriate thing to do.
 
I'll come back to you on this one as soon as I can. The author has now emailed, probably to keep face as now he'll of seen the mistakes/errors pointed out. No he does not seem to do a errata as you mention. Sometimes I just feel like its a case of the blind leading the blind.
 
You must log in or register to reply here.
Top