Inertia of Planetary Gearbox

If the axis of rotation goes not change = zero precession forces you can calculate it from what you have.

With precession, it depends on the speed of rotation and the plane etc = complex beast.

Why do you need to know this?

The input shaft (the sun) is on the same axis as the output shaft (carrier).

I'm reviewing an existing gearhead design to be used in a servo for a customer who states a maximum inertia requirement for the entire servo in his specification.

I do not know the load mass, or load inertia, at the output. Might be able to work with an output torque requirement. Anyway, I don't think this is needed to calculate inertia for the unit alone.

The description that you make of your gearbox seems to be an epicicloid gearbox not a planetary.

If you have the inertia of all the part they are given for there rotation speed.

for example:

inertia of the input shaft: Jhs Speed of the input shaft: Nhs

inertia of the planet gear: Jms Speed of the planet gear: Nms

inertia of the ouput shaft: Jls Speed of the output shaft: Nls

Equivalent inertia at the output speed: Jeq

Jeq=Jhs.Nhs^2/Nls^2+3.Jms.Nms^2/Nls+Jls

> Jeq=Jhs.Nhs^2/Nls^2+3.Jms.Nms^2/Nls+Jls

Sorry but I'm not getting all of your notation. Looks like "." represents multiplication?

Yes "." represents multiplication.

I'm guessing that there are multiple sun gears and multiple stages. I'm curious why you need to figure this out. The only contribution I can make is that wikipedia has a section on Epicyclic gearing to help with internal rpm and ratios. http://en.wikipedia.org/wiki/Planetary_gear

Thanks, I'll check out Wikipedia.

The configuration is as I stated in my original post - a single stage with one input (sun), three planets, a stationary ring gear (annulus) and and an output shaft (carrier). Ratio is 10:1.

Sorry, there was a mistake in my first answer with , planetary and epicyclic are the same type! But the formula is the same.

Sounds like a dynamics question out of a text book, being that it's probably easier to calculate the inertia of the system emperically, without finding the individual inertias by the same means and then summing them by the process you are in search of.

Generally speaking, you will have the input shaft multiplied by a unity factor (1) as one component.

For the planet gears, each one of them, you'll have to figure out both their rotational speed around their own axis, and that of the input shaft. For the planet gears around their own axis, you will have to use the tooth/diameter ratio of the input shaft compared to the housing. I'm not sure on the math, but im sure it's readily available out there. Either way, to add in the inertia of the planet gears themselves, take the mass separated by the radius to the center of the input shaft.

Add in the intertia of the carrier/output shaft, and vioila.

Sorry I didn't have any equations for the most complex part, but hopefully you can find some.

Here is the possible way to calcualate the inertia at the output, assuming you have all the individual ineritas

Ip = Inertia of planets about center of system (sun) =

[ Ineritia about own axis + (mass * d^2) ] * (ratio to output )^2

Here "d" is distance between two axis of rotation.

Is = Inertia of sun * (ratio to the output)^2

Ic = Inertia of carrier (output). The ratio to the output is 1.

Total = Ip + Is + Ic

***Only questionable thing in this calcualtion is does the additional inertia (m*d^2) go through the ratio if this rotation is about carrier's center where ratio is 1?