An epicyclic gear train (also known as a planetary gearset) consists of two gears mounted so that the center of one gear revolves around the center of the other. A carrier connects the centers of the two gears and rotates to carry one gear, called the planet gear or planet pinion, around the other, called the sun gear or sun wheel. The planet and sun gears mesh so that their pitch circles roll without slip. A point on the pitch circle of the planet gear traces an epicycloid curve. In this simplified case, the sun gear is fixed and the planetary gear(s) roll around the sun gear.
An epicyclic gear train can be assembled so the planet gear rolls on the inside of the pitch circle of a fixed, outer gear ring, or ring gear, sometimes called an annular gear. In this case, the curve traced by a point on the pitch circle of the planet is a hypocycloid.
The combination of epicycle gear trains with a planet engaging both a sun gear and a ring gear is called a planetary gear train. In this case, the ring gear is usually fixed and the sun gear is driven.
Hi,
Just have a question about the method in attempting this two-stage epicyclic gear-box.
We are given the following information about the number of teeth: S1 = 48, P1 = 28, S2 = 96, and P2 = 48
My attempt:
1. Work out the number of teeth for the two ring/annulus gears
From geometry, we can...
I'm looking to calculate the ratio of a planetary gear train. With the ultimate goal of producing a spreadsheet/calculator to show the state of the gearbox (i.e. all shaft speeds) based on applied constraints.
A gear set consists of Ring Gear, Planet Gear, Sun Gear. The sick diagram is the...
All About Epicyclic gearing
This Video illustrates :
- How does Epicyclic gear Work
- some Application of Epicyclic gear
- How to make Epicyclic gear calculation
Homework Statement
The teacher said it is possible to use an epicyclic gearing with no axis fixed but nobody use the gearing like that. I watched this video:
at time 42s it's possible to watch it. I have 2 questions:
1) Is it possible to used the epicycloidal train with no axis fixed ?
2)...
Imagine I am standing but not moving forward/backward, but I am rotating my body about my own axis. Now imagine if I moved in a circular path as I was spinning on my own axis. Is this considered to be precessing (assuming the axis of rotations are always in the z direction)?
My friend...
Homework Statement
Here the gear B is fixed and C and D are on the same drive. If C has 100 teeth, B has 50, D has 50 and E has 200. Calculate the ratio of the gearbox.
Homework Equations
The Attempt at a Solution
My attempt is uploaded as an attachment as putting tables in...
Is the module(M) of the sun,planet and ring gear in a 'planetary gear set' equal to each other always? Is it possible for M to be different for each of the gears in the set?[-Applications of this?]
$$PCD_{Ring}=PCD_{Sun}+2*PCD_{Planet}\Rightarrow N_{Ring}=N_{Sun}+2*N_{Planet}$$
is valid only...
Hi, A quick clarification question. Is epicyclic a correct description for our moons path around the sun. I can only find references to epicycles in a historical context.
Thanks.
Homework Statement
A gear ratio is req from A/D of 1:100 and that it will be necessary to redesign D so that it has 100 teeth. Determine:
a)number of teeth on c
b)number of teeth on planet gears.
This followed on from another question.
An epicyclic gear c is the annulus and is fixed. The...
Hi all, I'm Chris, a first time poster!
Currently stuck on an assignment question for my mechanical engineering course. The question is shown in the picture below. The capital T refers to 'teeth', as in 20T is 20 teeth on that particular gear. Ignore the little arrows with P and S - that is...
Hi,
Cant understand this at all, even modeled the set up with lego but still can't fathom it! I can't see how the arm can rotate with locking something.
Cheers
Homework Statement
http://img8.imageshack.us/img8/2393/geartain.jpg
The Attempt at a Solution
I got the right answer, by assuming that \omega_3=\omega_6 , but i don't know why this has to be so.
Here's my working:
\frac{\omega_2-\omega_3}{\omega_4-\omega_3}=-\frac{N_4}{N_2}...