planetary ratio problem 1

[samjesse]

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.

 

[samjesse]

Please add to the OP that K is fixed.

 

[handleman]

It's that time of year... Please add to the OP some indication that this is not a homework/exam problem.

-handleman, CSWP (The new, easy test)

 

[desertfox]

Hi samjesse

How can we help when there's no dimensions for sizes on the sketch?

desertfox

 

[samjesse]

what I am after is the equation to solve this. BTW it is not a homework/exam request.

I was reading some examples on

http://www.roymech.co.uk/Useful_Tables/Drive/Epi_cyclic_gears.html

but I could not work my problem out.

thx

 

[desertfox]

Hi samjesse

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

 

[samjesse]

Could some one second this please...

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

 

[samjesse]

I think I made a mistake.
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.

 

[samjesse]

I think I got it all wrong.
I tried to deduce the concept from

http://www.roymech.co.uk/Useful_Tables/Drive/Epi_cyclic_gears.html

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....

 

[desertfox]

Hi samjesse

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

 

[zekeman]

Not sure but from your diagram, I assume that you have two epicyclic trains with a moving common ring gear, I'll call R
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.

 

[samjesse]

zekeman:
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.

 

[samjesse]

Here is an updated kinematical representation.

 

[zekeman]

My mistake. I can redo it another way without the manipulations as follows:
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)

 

[samjesse]

How do I go about finding the torque of R1 if S2 has 1 foot-pound?

 

[samjesse]

sorry, let me revise the question.
How do I go about finding the torque of R1 if I has 1 foot-pound?

 

[zekeman]

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

 

[samjesse]

zekeman:
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.

 

[zekeman]

Firstly, I checked my previous answer and left off a 1. It should read
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.

 

[samjesse]

Can T = F . d by used?
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