Epicyclic Gear Train Homework: Why Does $\omega_3=\omega_6$?

[boboYO]

Homework Statement

http://img8.imageshack.us/img8/2393/geartain.jpg

The Attempt at a Solution

I got the right answer, by assuming that $$\omega_3=\omega_6$$ , but i don't know why this has to be so.
Here's my working:$$\frac{\omega_2-\omega_3}{\omega_4-\omega_3}=-\frac{N_4}{N_2}$$

$$\frac{\omega_6-\omega_7}{\omega_4-\omega_7}=-\frac{N_4}{N_6}$$Then, subbing in $$\omega_4 = 0$$ and $$N_6=N_2$$, equating the above 2 equations, and rearranging, I end up with

$$\frac{\omega_2}{\omega_7}=\left(1+\frac{N_4}{N_2}\right)^2\frac{w_3}{w_6}$$

which gives the right answer, 68.0625, if i let $$\frac{w_3}{w_6}=1$$

but why?

 

[tiny-tim]

Welcome to PF!

Hi boboYO! Welcome to PF!

(have an omega: ω )

Perhaps I'm reading it wrong, but doesn't that crooked link, together with "the carrier for the first stage is rigidly connected to gear 6", mean that ω₃ is defined as being equal to ω₆?

 

[boboYO]

that's probably it, but i still don't really understand it very well:"the carrier for the first stage is rigidly connected to gear 6" --

does this mean that gear 6 is able to rotate about the axle? I assume so because I don't think the mechanism would work otherwise.

now, if 6 was allowed to rotate around the axle, doesn't it mean w_6 could be different to w_3? like for example the earth+sun system, Earth is rotating around the sun at ang. velocity of w_3, while it's also rotating about its axis at w_6?

 

[Dr.D]

boboYo, I think that there is a flaw in your formulation. When you wrote your two equations, you treated omega3 and omega7 as comparable quantities, but they are not. Omega7 is comparable to omega6 which is the output from the first stage of the epicyclic gear train.

From where do you conclude that 68.0625 is the correct value of the train ratio?

 

[boboYO]

i got 68.1 from the (unworked) solutions, but the answers tend to have a lot of mistakes in them so it might not be right.

sorry, I'm not really sure what you mean by comparable quantities?edit: ohhh, i think i can finally visuallise what's happening. it wasn't that w_3= w_6, it was that w_6 should have been where w_3 was.

for some reason, i kept thinking 3 was rotating about its own centre (top dotted line)
its so much clearer now. thanks all.edit2: hmm how come the edit button has disappeared from the first post? i can't mark the thread as solved.

 

[Dr.D]

Planet gear 3 does turn about its own center line. It must roll on the sun gear 2 and the ring gear 4.

You might want to draw an end view for a better look at what happens.

 

[boboYO]

yes, it does, but it's irrelevant isn't it? what i meant before was i thought it _only_ rotated about its centre line.

all that matters is the rotation of the carrier, which is equal to w_6

 

[Dr.D]

It is not irrelevant because it establishes the rate at which w_6 moves. It is pretty hard to call any part of the system irrelevant.

 

[boboYO]

sorry, i didn't mean irrelevant, just that it doesn't need to be worked out to solve the problem.

using

$$\frac{\omega_2-\omega_c}{\omega_4-\omega_c}=-\frac{N_4}{N_2}$$

and

$$\frac{\omega_6-\omega_7}{\omega_4-\omega_7}=-\frac{N_4}{N_6}$$

would be correct, right?

 

[Dr.D]

Those appear to be in the same form, provided you recognize that what you have called w_c is w_6.

I don't know where you got these equations. They are not equations that I use. Have you ever derived them? I would not use them without personally deriving them.

 

[boboYO]

yes, i realize that w_c = w_6.

we call it the method of relative velocities, i'll just copy a paragraph from wikipedia explaining how they are derived:

To derive this, just imagine the arm is locked, and calculate the gear ratio wring / wsun = Nsun / Nring, then unlock the arm. From the arms reference frame the ratio is always Nsun/Nring, but from your frame all the speeds are increased by the angular velocity of the arm. So to write this relative relationship, you arrive at the equation from above.


What equations do you use?

 

[Dr.D]

I always make it a practice to work problems from first principles, so I would not start from a "canned equation" such as the one you gave. The "derivation" you provided was just so many words, but it does not translate easily into a true mathematical derivation. So...

First notice that you have two identical planetary trains, so you only need to analyze the first stage. The overall train ratio will be square of the train ratio for the first stage.

I would write a relation in terms of the pitch radii,
r2 + 2*r3 = r4
just from the geometry.
Now the pitch radius is in direct proportion to the number of teeth for a given diametral pitch or module, so this can be rewritten as
N2 + 2*N3 = N4
From this we find out that N3 = 25, the number of teeth on the planet.

Now back to pitch radii, and write the rolling constraint equations that must be satisfied:
r2*theta2 = - r3*theta3 + (r2+r3)*theta6
r4*theta4 = +r3*theta3 + (r2+r3)*theta6
and the condition, theta4 = 0
This is quickly solved to give
(theta2/theta6) = 2*(r2+r3)/r2 = 2*(1+r3/r2) = 2*(1+N3/N2) = 8.25
(theta2/theta6)^2 = 68.025