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.

 

[zekeman]

OK,let's try this. Take the input as a crank 1 foot in length and apply 1 pound of force to it giving 1 pound- foot of torque. Next evaluate the gear ratio formula I gave as R1/S2.
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.

 

[samjesse]

but this does not utilize T = F.D "Torque = Force x Distance.

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.

 

[zekeman]

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.

 

[samjesse]

Thank you.

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?

 

[zekeman]

Looks OK but put some numbers in and check against my formula by substituting number of teeth for the various diameters.
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.