planetary ratio problem
planetary ratio problem
(OP)
Hi
Please view the image file below.
I am at lose on how to find the number of revolution and the torque of the S shaft when the L shaft is turned 1 revolution at 1 foot-pound.
If possible, please give a step by step with explanation.
Thanks a million.
Please view the image file below.
I am at lose on how to find the number of revolution and the torque of the S shaft when the L shaft is turned 1 revolution at 1 foot-pound.
If possible, please give a step by step with explanation.
Thanks a million.





RE: planetary ratio problem
RE: planetary ratio problem
-handleman, CSWP (The new, easy test)
RE: planetary ratio problem
How can we help when there's no dimensions for sizes on the sketch?
desertfox
RE: planetary ratio problem
I was reading some examples on http:/
but I could not work my problem out.
thx
RE: planetary ratio problem
Yes trying to see if your set up is there, have you tried the links at the bottom of the page on the site you gave
desertfox
RE: planetary ratio problem
Revolution in the output shaft S will be
1 + (Nd/Nx).(Nb/Nd).(Ne/Nb).(Na/Ne)
But I think I am missing the Ny part of it. can some one help please?
thx
RE: planetary ratio problem
it should be
1 + (Nd/Nx).(Nb/Nd).(Ne/Nc).(Na/Ne)
and not
1 + (Nd/Nx).(Nb/Nd).(Ne/Nb).(Na/Ne)
so if we let
Nx = Nc = Nd = Ne = 30
Nb = Na = 90
then
Na = 1 + (1 . 3 . 1 . 3) = 10
i.e. speed increase from 1 to 10 turns.
Now I will go and find out how to figure the torque independent of the speed.
RE: planetary ratio problem
I tried to deduce the concept from
http:/
I will need help understanding it step by step in the link above Example 1.
why P = -NA /NP in the second raw?
I understand that - means opposite direction of rotation to the input shaft. but why NA/NP is it because A has more teeth than P? or is it not the output gear number of teeth / input gear number of teeth?
I give up....
RE: planetary ratio problem
I am no expert in this field either an in answer to your question it is to do with the teeth ratio's on the sun gear panet gear etc but any one of them could be the input or output, I was trying to look for an example like your sketch or even break it down into sub-sections but no luck.
desertfox
RE: planetary ratio problem
The sungear on left train call S1, and the carrier C1 which is linked to the outpuit sungear S2 and the carrier on the 2nd train is fixed.
I think it goes like this:
temporarily uncouple the linkage C1 to S2
1)rotate the ring -1, keeping C2 =0,C1=0
R=-1
C2=0, C1=0
s2=Nr/Ns2
S1=Nr/Ns
3)Now rotate each train +1 including R to make
R=0
S1=1+Nr/Ns1
C1=1
S2=1+Nr/Ns2
C2=1
This part is fairly standard .
Next, to make C1=S2, the train1 is sped up by factor(1+Nr/Ns2)( train2 is not changed).Now reconnect the linkage between C1 and S2 and rotate the linked assenmbly -1 (to make C2=0 and R=-1). which yields
S1=(Nr/Ns1+1)(Nr/Ns2+1)-1
S2=Nr/Ns2
S1/S2=Nr/Ns1+Ns2/Ns1+1
Please check this.
RE: planetary ratio problem
You assumed a common ring gear is hard for me to see.
the representation shows the right train ring gear is part of the left train sun gear but has a different radius i.e number of teeth.
RE: planetary ratio problem
RE: planetary ratio problem
I found that there is one equation that can handle this quite readily ( I'm sure it must be in the literature but is easily developed)I was taught the sometimes tedious method I used above.
Now the fundamental equation connecting the three rotations is simply
R=-(Ns/Nr)S+(1+Ns/Nr)C
R ring speed
S sun gear speed
C carrier speed
since we have 2 interconnected trains we write
R1=-(Ns1/Nr1)S1+(1+Ns1/Nr1)C1
R2=-(Ns2/Nr2)S2+(1+Ns2/Nr2)C2
and the links and constraint
C2=0
S1=R2
C1=S2
Substituting
R1=-(Ns1/Nr1)*(-NS2/Nr2)S2+Ns1/Nr1)S2
R1=Ns1/Nr1)(Ns2/Nr2+1)S2
R1/S2=Ns1/Nr1)(Ns2/Nr2+1)
RE: planetary ratio problem
RE: planetary ratio problem
How do I go about finding the torque of R1 if I has 1 foot-pound?
RE: planetary ratio problem
Samjesse,
Well, thats the easy part.
From the conservation of energy, for any system having an input rotation and an output motion, like yours we have
Power 0ut= efficiency * power in
Power in is S2*torquein
power out is R1*torqueout
R1*torqueout=E*S2*torquein
therefore
torqueout=E*S2/R1*torquein
Use the S2/R1 developed above
E in this case is probably in the 90% range.
Hope this answers you question.
Regards
Zeke
RE: planetary ratio problem
I want to calculate the torque using T = F.d
and to simplify things we will write c1 to mean the radius of left train carrier, s2 to mean the radius of the right train sun p2 to mean the radius of the right train planetary and so on.
RE: planetary ratio problem
R1/S2=Ns1/Nr1)(Ns2/Nr2+1)+1= gear ratio
Keep in mind that R1 and S2,have nothing to do with dimensions; they are the output and input rotational speeds (RPMs) in my convention and therefore R1/S2 is the gear ratio for that setup. So asking for the output torque at Ring1, given the input torque at S2 is
torque at Ring 1=efficiency *input torque at Sungear2/R1/S2.
Your first question: for 1 lb foot torque at Sungear2 yields
torque at Ring 1=efficiency /(R1/S2)
R1/S2 is found by evaluating the first formula above.
Ifnthis isn't clear just put numbers in fot the teeth and first get R1/S2.
RE: planetary ratio problem
what I am after at this stage is how to use the above in order to arrive to the answer which I see that you maybe able to obtain it via other ways as indicated previously using efficiency and power.
If 1 foot-pound is applied at I and using T=Fd, what are the steps in order to calculate the T at the output shaft O?
d = distance = the distance from the axis of rotation to the pitch circle of a gear.
Would the input torque be divided to S2 and C1?
How it will all come together?...
thx
RE: planetary ratio problem
For the sake of this discussion, let's say it is 34.
Now take the 1 pound of torque at the input and divide it by 34, giving 1/34 pound-foot at the output. With efficiency considerations, it would be slightly less.
RE: planetary ratio problem
OK, let me have an attempt at it.
I will need your help/correction to establish some facts.
Shaft I is the input and is turning CW
Torque of shaft I is divided
Ti = Ts2 + Tc1 (1)
Since C2 is not turning "leave the direction out for now"
Fs2 = Fr2 (2)
Given from the drawings
Ts1 = Fr2 . (Dr2 - Ds1) (3)
Fs1 = Ts1 / Ds1 (4)
Since C1 is turning CW and S1 CCW then the force at the R1 is the combination of both as:
Fr1 = Fs1 + Fc1 (5) is this correct?
Finally the output torque
To = Fr1 . Dr1 (6)
If you agree then I can solve for To.
RE: planetary ratio problem
Very unorthodox method even though it looks like a direct solution, but ok if done correctly. Should lead to same result.
I added my comments to your input as follows:
Shaft I is the input and is turning CW
Torque of shaft I is divided
Ti = Ts2 + Tc1 (1) OK
Since C2 is not turning "leave the direction out for now"
Fs2 = Fr2 (2) OK
Given from the drawings
Ts1 = Fr2 . (Dr2 - Ds1) (3) NO Ts1=Tr2=
Fr2*Dr2=Fs1*Ds1
Fs1 = Ts1 / Ds1 (4) OK
Since C1 is turning CW and S1 CCW then the force at the R1 is the combination of both as:
Fr1 = Fs1 + Fc1 (5) is this correct? NO
Fs1=Fr1
Finally the output torque
To = Fr1 . Dr1 (6)OK
If you agree then I can
You also need Tc1 to complete the analysis,
which is
Tc1=Fs1*(Ds1+Dr1)
Keep going, I think you will get there eventually.
RE: planetary ratio problem
I will put the facts and solve as below:
Ti = Ts2 + Tc1 (1)
Fs2 = Fr2 (2)
Ts1 = Tr2 (3)
Fs1 = Ts1 / Ds1 (4)
Fs1 = Fr1 (5)
Tc1 = Fs1 * (Ds1+Dr1) (6)
To = Fr1 * Dr1 (7)
using (3)
Fs1 = Fr2*Dr2/Ds1 (8)
substituting (6) into (1)
Ts2 = Ti - Fs1 * (Ds1+Dr1) (9)
substituting (8) into (9) eliminating Fs1
Fs2*Ds2 = Ti - (Fr2*Dr2/Ds1)*(Ds1+Dr1)
using (2) and solving for Fs2
Fs2*Ds2 = Ti - (Fs2*Dr2/Ds1)*(Ds1+Dr1)
Ti/Fs2 = Ds2 + (Dr2/Ds1)*(Ds1+Dr1)
Fs2 = Ti / (Ds2+(Dr2/Ds1)*(Ds1+Dr1)) (10)
using (2) and substituting (10) into (8)
Fs1 = Ti/(Ds2+(Dr2/Ds1)*(Ds1+Dr1))* Dr2/Ds1 (11)
using (5) and substituting (11) into (7)
To = Fs1 * Dr1
To = Ti/(Ds2+(Dr2/Ds1)*(Ds1+Dr1))*(Dr2/Ds1)*Dr1
are you OK with that?
RE: planetary ratio problem
You must get the same answer.
BTW, your solution, when simplified should be identical to the one I posted on 12/2, 16:59.
Good luck and BTW, don't use this method; in my opinion, it is fraught with pitfalls; use the standard velocity method I outlined earlier.