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Mechanical Engineering

Planetary Gear Challenge, #55


DrD

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    Mechanics Corner
    A Journal of Applied Mechanics and Mathematics by DrD, #55


                                                             Planetary Gear Challenge


Introduction

    Back in the mid-1990s, I worked as the "in-house consultant" for a small aerospace manufacturing firm in the Chicago suburbs (500 employees). The general rule was simply that any engineer in the company could bring me any problem, and I would try to give them a solution. It was one of the happiest jobs I've ever had because I was constantly being given new problems to work on. Some of the problems were simple, the sort that could be solved in 20 minutes with pencil and paper. Others took many days and lots of computer work, but I had a free hand and I really enjoyed it.

 

For the figure and the remainder of this challenge, please see the PDF file. I am unable to save the figure in a format that will allow me to upload it here. -- DrD

Planetary Gear Challenge.pdf

DoubleCompoundEpicyclic3.png

17 Comments


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It's an interesting challenge and I have had to find my old university notes to try and answer the questions.

Not quite there yet but I will certainly be uploading my solution shortly

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The Kinematic coeff  say X)   , i.e the ratio (wR-wPC) /(wS   - wPC) = (-Ns / N1) * (N2 / N3) * (-N4 / Nr)   where (wR - wPC) is the  angular velocity of the  ring gear relative to wPC , the angular velocity of the planet carrier.  Likewise, wS is the angular velocity of the sun gear and wS - wPC the relative angular velocity ( though the ring gear is not fixed  to the arm unlike the other gears, we are measuring velocities relative to the planet carrier, hence the subtraction).      But since this is stationary , wR is zero. Hence (- wPC / wS - wPC) = X. . Therefore, wS*X - wPCX = -wPC . Therefore, -wS*X = (1-X)*wPC

Therefore, wPC / wS = - X / (1 -  X). 

In our case, the kinematic coeff X = (-16 / 29) * (17 / 35) * (-17/107) = 4624 / 108605 = .042576 (app) . Therefore, in our case, wPC / wS = -.042576 / (1-.042576) = - .04447 which is in the opposite direction. 

In the  3rd problem, if we take clockwise rotations as positive, then wS - wPC = 3500 - (-1575) = 5075

and the relative velocity of the ring gear , wR - wPC = 5075*X = 216.07 (app) . Here, i was confused whether to add the velocity of carrier arm to the relative angular velocity obtained above. But going by the above argument, to be consistent one would have to add back the arm velocity. Hence, one would have to subtract 1575 , since the angular velocity of the planet carrier arm is in the anti-clockwise direction. Therefore, angular velocity of the ring = 216.07 - 1575 = - 1358.925, i.e 1358.925 in the anti-clockwise direction  

 

 

 

 

 

 

 

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Thank you for this response. I'm very pleased to see someone make an attempt.

To begin, you said,

(wR-wPC) /(wS   - wPC) = (-Ns / N1) * (N2 / N3) * (-N4 / Nr)

Why would we believe that the ratio of differences on the left is equal to the triple product of ratios on the right? This is not obvious, and I think it requires justification. As it stands, it is simply pulled out of the blue.

DrD

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20 hours ago, Sundaram Ramchandran said:

I applied the formulae given in standard textbooks on mechanisms. It also  seems intuitive in the sense that if the angular velocity of the arm/planet carrier is zero, then the ratio of angular velocities  of 2 gears in contact would be inversely proportional to the number of teeth / radii. Also,  the angular velocities of  the concentrically attached inner gears would be same as that of the outer gears to which they are attached. Also, i applied the rule of direction (and hence sign) reversal with every gear contact. I admit i was confused about whether to subtract the ang. velocity of the planet carrier from the ring gear since the former is not attached to the latter but i looked at some worked examples and felt that from the point of vector algebra, it is just a change of coordinates.  

 

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(1) Regarding your equation

(wR-wPC) /(wS   - wPC) = (-Ns / N1) * (N2 / N3) * (-N4 / Nr)

you say that you simply applied a formula given in standard textbooks on mechanism. You do not say what standard textbooks, nor do you show a derivation. I could certainly show you a number of "standard textbooks on mechanisms" that do not include this, so I still think your equation requires justification. The best justification would, of course, be to simply show how it is derived. Note the wording of the first question: "... develop the mathematical relation ..." This implies more than simply grabbing an equation from a book.

(2) Regarding the second question, you did obtain results for this item.

(3) Regarding the third question, you did obtain results for this item.

(4) Regarding the fourth question, there does not appear to be any attempt to obtain an answer. Why is this?

DrD

 

 

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The book is : "Theory of Machines and Mechanisms - Dicker, Pennock and Shigley" - 3rd Edition, Oxford University Press, 2003 . Further details ( Chapter 10 - Mechanism Trains, equation 10.4 and solved example 10.2 ). 

I realized that the energy calculations would be considerably more complex involving :

- The kinetic energy (KE)  of the sun gear (with its moment of inertia etc)

- The KE of the Planet Carrier / arm (whose mass / moment of inertia would possibly include those of the planet gears)

The  above would be the input energy and would be distributed between the KE's of the 3 sets of planet gears (only 1 of which is shown in the diagram) and the KE of the ring gear

The KE of the planet gears would involve both translational KE due to the velocity of the center of mass as well as their rotational KE's involving moment of inertia etc. 

The KE of the ring gear is purely rotational. 

I did not attempt this because it is considerably more involved !! 

 

 

 

 

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I do not have this textbook; I have it's "ancestor," the first edition, before Pennock joined the other authors and Shigley was listed as the first author. I will have to see if I can locate a copy of the 3rd edition. Can you copy the relevant pages and send them to me via a PM, perhaps?

(Just as an aside, there is an error in the authors as you have them listed. The name you show as Dicker should actually be Uicker, and he is (or was?) on the faculty of the University of Wisconsin-Madison. This error pops up ever where, and does not seem to have ever been corrected.)

You are correct about the energy calculation being more involved. However, it would not be much of a difficulty if you had worked through the development of the necessary kinematics as opposed to simply following a textbook example.

DrD

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I found a copy of the textbook you are using, at least I think that it is the same. Just to be sure we are on the same page (literally), you are speaking of eq(10.4) on p,.318 and Example 10.2 on p. 319; is that correct?

DrD

 

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An alternative answer would be not to subtract wPC from wR (since the ring gear is not attached to the arm) , in which case, wR / (wS - wPC) =   X. 

That is , (wS - wPC) * X = wR = 0. Since X is non-zero, we would have wS = wPC, That is , when the ring gear is fixed, the angular velocities of the arm / planet carrier and the sun gear are equal and in the same direction. I was somehow not satisfied with this answer. 

Using the  same logic, for question 3, we would get wR /  (3500 - (- 1575)  = wR / 5075 = X = .042576. Therefore wR = 5075*.042576 = 216.0732 in the same direction as S , anti-clockwise

 

 

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It appears to me that, in all the discussion and examples given by your reference, the velocity vectors at the contact points are exactly parallel to each other. In the Challenge problem, this is most certainly not true. How can you assure that the results taken from your reference apply to the Challenge problem when the geometry is significantly different?

DrD

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I guess the essential relationships in terms of angular velocities still hold, irrespective of the shape of the arm / carrier which would  , i suppose , still fulfill its function of transmitting the input angular momentum to the gears attached to it. 

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An argument that begins with the words, "I guess ..." is not very persuasive! As an engineer, aren't you supposed to be able to justify your work, to prove your result?

I've written lots of engineering reports that were required to convince the customer that our product was soundly engineered and adequate for their application. I would not dare to begin with "I guess ..."

I've also reviewed lots of engineering reports written by others to either accept or reject their results. I doubt that I would ever accept a report that includes the phrase "I guess..."

As to the central point, what you have stated as "the essential relationships in terms of angular velocities still hold, irrespective of the shape of the arm / carrier," I don't see any justification for that at all. I would like to be convinced.

DrD

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