Force Distribution in planetry gear train

Dear Nagaraja

More data (input RPM, gear radius, input torque) are required to analyse a planetary gear box. You must read and digg more, I'll suggest to calculate the RPM of each shaft (input, planetary, output) by block the apropriate parts, calculate the corresponding torques (remember M_input + M_planet +M_output=0). With torques and rpm you will calculate the force on each gear.

lgeo

Hi lgeo

Thnks for your valuable info,and here is the following details

Ratio ; 6:1
Configuartion:planetry type(Sun:input,Carrier:Output,&Ring:fixed,Three Planetry gears)
Input details
Speed N =1000rpm
Torque Minput =88.50lb-in

Diametral pitch=20
Teeth on sun gear(Zs) :20
Teeth on Planet gear(Zp) :40
Teeth on Ring gear(Zr) :100
PCD of Sun(Ds):1"
PCD of Planet(Dp):2"
PCD of Ring(Ds):5"
Now
Torque at Sun (Ms) :88.5lb-in
Speed at Sun (Ns) :1000rpm
Transmitting force on sun(Fs):176lb
And
Torque at Planetry gear(Tp):(40/20)*88.5=177lb-in
Speed at Planetry gear(Np):500rpm
Force at Planetry gear(Fp):177lb

Now i am having 3planetry gears,so total torque transmitted to carrier:3*177=531lb-in.

But,How the following equation works in this

Minput + Mplanet +Moutput=0

Thanks in before

nagaraja

Simply: Power=torque x rpm

so, you know rpm exit = 6 rpm input, then torque exit = 1/6 torque input.

For others, same.

Be happy

Last time I designed a planetary gear set, I seem to remember, that to have a symmetrical spider 120⁰, I had to have the teeth on ring and sun divisible by 3 - you might want to check if you have meshing with your teeth numbers.

However the tooth stress with 3 planets is 1/3 of the torque - not 3 times.

I.e. if your reduction ratio is 5+1, then torque out is ~ 6 times torque in. Or the torque is shared between planets - not 'increased' by number of planets.