Calculating Simpson Compound Planetary Ratio

[jgatlin]

I am trying to calculate the first gear ratio of a compound Simpson gear set. Power is transmitted thru the front planetary ring gear (66 tooth). The output shaft is splined to the rear planetary ring gear (66 tooth) and the front planetary carrier. The sun gear is common (34 Tooth on each end). Also the rear planetary carrier is held with a one-way sprag. Any help would be appreciated.

 

[Matt51]

I would use the standard tabular approach. "Planetary Gear Design" by Oliver Kelley, from Gear Design Nicholas Chironis, shows ratios for both Ravigneaux and Simpson. I can fax the article. Auto Drive Trains Technology, by Duffy & Johanson, page 248, shows power flow for first gear in a Simpson. Two stages of planetary gearing, common sun gear, first stage planetary carrier connects to second stage ring gear which is the output shaft. Input is first stage ring gear.

I have another paper that gives the overall kinematic eqn for this drive as
(1+2alpha)ws = (-alpha^2)(wr1)+{(1+alpha)^2)wc2 where alpha is ring teeth/sun teeth. w is angular velocity, or rpm.
So for this first gear, the overall eqn reduces to
(1+2(Nr/Ns)rpmsun=(-(Nr/Ns)^2)(rpmring1)
solve for the rpmsun. Then output = rpmsun x Ns/Nr.
Example
60 teeth ring, 18 tooth sun. Then I get 2.52 as first gear ratio.
Please someone check for errors.

 

[Matt51]

I was assuming identical tooth counts in both planetaries. If they are different, the overall eqn is
(1+alpha1 x alpha2)ws =
(-alpha1xalpha2)(wr1)+[(1+alpha1)(1+alpha2)]wc2

 

[jgatlin]

There are a few parts of the equation that I do not understand. Please fax me a copy to 954-916-8990.

Both planetary are the same tooth counts. 66 Ring gear and 34 tooth sun gear.

Thank you for the quick reply.

 

[Matt51]

In particular, see page 56, "applying kinematic equations to complex planetaries".

 

[jgatlin]

Thank you for the fax. Some of the text in the equations is very hard to read so I have ordered the book and should receive it in a few days. I am having difficulty in calculating the speed of the sun gear speed out of the front planetary. Power is applied to the front ring gear and the carrier is connected to the output shaft. Therefore the carrier is turning with the ring but at a slower speed. I am going to attempt to upload a drawing for reference.

 

[Matt51]

Hi,
Using the first paper,

(1+2(Nr/Ns)rpmsun=(-(Nr/Ns)^2)(rpmring1)
solving for rpm sun using 34-16-66 teeth, gives
rpm sun = .772 rpm ring, and the inverse is 1.295.

As a check, looking at building a superposition table.

Carrier1 Sun planet1 ring1 carrier2 planet2 ring2
+1 +1 +1 +1 +1 +1 +1
0 a b c -1 0

sum +1 1+a 1+b 1+c 0

a = -1(100/34) = -2.941 = planetary gear ratio
b = (34/16)a = 2.125a =
c = (34/66)a = .5151a = 1.514
ratio of sun to ring1 = (1+a)/(1+c) = -1.941/2.514 = .772

So we have a check.

 

[Matt51]

Sorry, the formatting shifted the second row in the table. c is under ring1. a is uncer sun, be is under planet1, -1 under carrier2.