An epicylic gear is a planetary gear arrangement consisting of one or more planet(epicyclic)
gears (P) meshed and rotating round a central sun gear (S). The planet gears
are also meshed and rotate within an internal ring gear (A). The planet gears
are fixed to a planet carrier-crank arm(L) designed to rotate on the same centre as the sun
gear. Only one planet-carrier /crank arm is used in a single epicyclic gear train. This complicated arrangement (see below)
has a number of modes of operation depending on which members are locked. Epicyclic gears
can be based on spur gears, helical gears, or bevel gears.
Epicyclic gearboxes are generally purchased as complete units from specialist suppliers.
Design Features
|
Epicyclic Gearboxes (Planetary gearboxes)have the following features
Typical Epicyclic Gear Arrangement
Example 1..
|
|
|
| Action | Rotation (CW= +ve) from action....................Nx = number of teeth | |||
| L | A | P | S | |
| Turn whole gear thro 1 rev Clock Wise | 1 | 1 | 1 | 1 |
| Fix arm L and rotate A back CCW 1 rev | 0 | -1 | -NA /NP | +NA / NS |
| Add the two motions above | 1 | 0 | 1 - ( NA / NP) | 1 + ( NA / NS) |
Summary : Rotation of the planet shaft(L) 1 rev CW results in the rotation of the shaft (S) 1 + ( NA / NS) revs (CW). Ratio 1 : ( 1 + ( NA / NS) )
Example 2..
|
|
|
| Action | Rotation (CW= +ve) from action....................Nx = number of teeth | |||
| S | L | P | A | |
| Fix L and rotate Sun 1 rev CW | 1 | 0 | - NS / NP | - NA / NS |
Summary : Rotation of the Sun shaft(S) 1 rev CW results in the rotation of the shaft (L) NS / NA revs (CCW). Ratio 1: ( - NS / NA )
Example 3..
|
|
|
| Action | Rotation (CW= +ve) from action....................Nx = number of teeth | |||
| L | A | B-D | S | |
| Turn whole gear through 1 rev CW | 1 | 1 | 1 | 1 |
| Fix L and rotate A back CCW 1 rev | 0 | -1 | - NA / NB | - ( NA / NB).( N D / N S) |
| Add the two motions above | 1 | 0 | 1 -( NA / NB) | 1 - ( NA . ND)/( N B . N S) |
Summary : Rotation of the Shaft L 1 rev CW results in the rotation of the shaft (S) 1 - ( NA . ND)/( N B . N S) revs (CW).
Ratio 1: [1 - ( NA . ND)/( N B. N S)]
Example 4..
|
|
|
| Action | Rotation (CW= +ve) from action....................Nx = number of teeth | |||
| L | A | B-D | S | |
| Turn whole gear through 1 rev CW | 1 | 1 | 1 | 1 |
| Fix L and rotate A back CCW 1 rev | 0 | -1 | - NA / NB | + ( NA / NB).( N D / N S) |
| Add the two motions above | 1 | 0 | 1 -( NA / NB) | 1 + ( NA . ND)/( N B . N S) |
Summary : Rotation of the shaft L 1 rev CW results in the rotation of the shaft (S) 1 + ( NA . ND)/( N B . N S) revs (CW).
Ratio 1: [1 + ( NA . ND)/( N B . N S)]
Example 5..
|
|
 |
| Action | Rotation (CW= +ve) from action....................Nx = number of teeth | |||
| L | A | B-D | S | |
| Turn whole gear through 1 rev CW | 1 | 1 | 1 | 1 |
| Fix L and rotate A back CCW 1 rev | 0 | -1 | + NA / NB | + ( NA / NB).( N D / N S) |
| Add the two motions above | 1 | 0 | 1 + ( NA / NB) | 1 + ( NA . ND)/( N B . N S) |
Summary : Rotation of the shaft L - 1 rev CW results in the rotation of the shaft (S) 1 + ( NA . ND)/( N B . N S) revs (CW).
Ratio 1: [1 + ( NA . ND)/( N B . N S)]
Example 6..
|
|
 |
>| Action | Rotation (CW= +ve) from action-...................... Nx = number of teeth | |||
| L | A | B-D | S | |
| Turn whole gear through 1 rev CW | 1 | 1 | 1 | 1 |
| Fix L and rotate A back CCW 1 rev | 0 | -1 | + NA / NB | - ( NA / NB).( N D / N S) |
| Add the two motions above | 1 | 0 | 1 + ( NA / NB) | 1 - ( NA . ND)/( N B . N S) |
Summary : Rotation of the shaft L 1 rev CW results in the rotation of the shaft (S) 1 - ( NA . ND)/( N B . N S) revs (CW).
Ratio 1: [1 - ( NA . ND)/( N B . N S)]
Epi-cyclic gear Train Variations
The figure below shows the range of possible epicylclic gear arrangements.. Those in
section I & III are classed as simple arrangements because the planet gears mesh with both sun gears.
Those in sections II & IV are classed as complex trains because the planet gears partially match with each other
and partially mesh with the sun gears.
Calculation of epicyclic Gear Ratio
1) First calculate the ratio of the gears with the planet carrier fixed..
r f = (±) Product of
Driving Gear Teeth /Product of Driven Gear Teeth
Note: when two external gears are in contact there is a sign change (change of direction)
when an internal gear meshes with an external
gear both gears rotate in the same direction and there is no change in direction..
2) Designate and input gear (x) and an output gear (y).
The speed of the input gear relative to the carrier arm =
ω x - ω L
The speed of the output gear relative to the carrier arm =
ω y - ω L
3) The train value with the carrier fixed =
r f = ( ω y - ω L ) / ( ω x - ω L )
This relationship is used to solve the planetary gear train ratios.
Using this method for the examples above
Example 1.
1) r f = (NA /NP ).( - NP /NS )= - NA /NS
2) Select ω A as input speed = 0 and ω S as output speed (unknown) .
3) r f =
( ω S - ω L ) / ( ω A - ω L )
Therefore - r f.ω L = ( ω S - ω L )
Therefore ω S/ω L = 1 - r f
ω S/ω L = 1 + NA /NS
Example 2.
1) r f = (-NS /NP ).( NP /NA )= - NS /NA
2) Select ω S as input speed and ω A as output speed .ω L =0
3) r f =
( ω A - ω L ) / ( ω S - ω L )
Therefore ω A /ω S = r f
ω A/ω S = - NS /NA
Example 3.
1) r f = (NA /NB ).(ND /NS )= ( N A.N D) / (NB .NS )
2) Select ω A as input speed = 0 and ω S as output speed.
(solve for ω S/ ω L)
3) r f =
( ω S - ω L ) / ( ω A - ω L )
Therefore 1 - r f = ω S /ω L
ω S/ω L = 1 - ( N A.N D) / (NB .NS )
Example 4.
1) r f = (NA /NB ).( - ND /NS )= - ( N A.N D) / (NB .NS )
2) Select ω A as input speed = 0 and ω S as output speed.
(solve for ω S/ ω L)
3) r f =
( ω S - ω L ) / ( ω A - ω L )
Therefore 1 - r f = ω S /ω L
ω S/ω L = 1 + ( N A.N D) / (NB .NS )
Example 5.
1) r f = (- NA /NB ).( ND /NS )= - ( N A.N D) / (NB .NS )
2) Select ω A as input speed = 0 and ω S as output speed.
(solve for ω S/ ω L)
3) r f =
( ω S - ω L ) / ( ω A - ω L )
Therefore 1 - r f = ω S /ω L
ω S/ω L = 1 + ( N A.N D) / (NB .NS )
Example 6.
1) r f = (- NA /NB ).(- ND /NS )= ( N A.N D) / (NB .NS )
2) Select ω A as input speed = 0 and ω S as output speed.
(solve for ω S/ ω L)
3) r f =
( ω S - ω L ) / ( ω A - ω L )
Therefore 1 - r f = ω S /ω L
ω S/ω L = 1 - ( N A.N D) / (NB .NS )