Transcript Advanced Gear Analysis

Advanced Gear Analysis
• Epicyclic Gearing
• Tooth Strength Analysis
Epicyclic Gears 1
Epicyclic Gearset
An epicyclic gear set has some gear or
gears whose center revolves about
some point.
Here is a gearset with a stationary ring
gear and three planet gears on a
rotating carrier.
The input is at the Sun, and the output
is at the planet carrier.
The action is epicyclic, because the
centers of the planet gears revolve
about the sun gear while the planet
gears turn.
INPUT
Planet
Sun
CARRIER
Ring
Finding the gear ratio is somewhat
complicated because the planet gears
revolve while they rotate.
Epicyclic Gears 2
Epicyclic Gearset
Let’s rearrange things to make it simpler:
1) Redraw the planet carrier to show arms rotating about the center.
2) Remove two of the arms to show only one of the planet arms.
OUTPUT
INPUT
INPUT
OUTPUT
Epicyclic Gears 3
The Tabular (Superposition) Method
To account for the combined rotation and
revolution of the planet gear, we use a
two step process.
Planet
OUTPUT
First, we lock the whole assembly and
rotate it one turn Counter-Clockwise.
(Even the Ring, which we know is fixed)
INPUT
Sun
We enter this motion into a table using
the convention:
CCW = Positive
Ring
CW = Negative
Rotate Whole
Assembly CCW
Sun
Planet
Ring
Arm
+1
+1
+1
+1
Epicyclic Gears 4
The Tabular (Superposition) Method
Next, we hold the arm fixed, and rotate
whichever gear is fixed during operation
one turn clockwise.
INPUT
1
Turn
Planet
Here, we will turn the Ring clockwise one turn
(-1), holding the arm fixed.
• The Planet will turn NRing / NPlanet turns
clockwise.
• Since the Planet drives the Sun, the Sun will
turn (NRing / NPlanet) x (-NPlanet / NSun) = - NRing /
NSun turns (counter-clockwise).
Sun
• The arm doesn’t move.
Ring
We enter these motions into the second row of
the table.
Rotate Whole
Assembly CCW
Hold Arm, Rotate
Ring CW
Sun
Planet
Ring
Arm
+1
+1
+1
+1
NRing / NSun
- NRing / NPlanet
-1
0
Epicyclic Gears 5
The Tabular (Superposition) Method
Finally, we sum the motions in the first
and second rows of the table.
Rotate Whole
Assembly CCW
Hold Arm, Rotate
Ring CW
Total Motion
Now, we can
write the
relationship:
Sun
Planet
Ring
Arm
+1
+1
+1
+1
NRing / NSun
- NRing / NPlanet
-1
0
0
+1
1 + NRing / NSun 1 - NRing / NPlanet
nSun 1 
N Ring
n Arm 
1
1
N Sun
N Ring
 n Arm , or
 nSun
N Sun
If the Sun has 53 teeth and the Ring 122 teeth, the output to
input speed ratio is +1 / 3.3 , with the arm moving the same
direction as the Sun.
Epicyclic Gears 6
Planetary Gearset
OUTPUT
The configuration shown here,
with the input at the Sun and the
output at the Ring, is not
epicyclic.
It is simply a Sun driving an
internal Ring gear through a set
of three idlers.
INPUT
Sun
Planet
The gear ratio is:
nring
nsun
 N sun N planet  N sun



N planet N ring
N ring
Ring
n = speed; N = # Teeth
Where the minus sign comes from the change in direction between the two
external gears.
If the Sun has 53 teeth and the Ring 122 teeth, the ratio is -1 / 2.3 .
Epicyclic Gears 7
Another Epicyclic Example
Here are three different representations of the same gearset:





 








Given:
1.
2.
3.
4.
5.
Ring = 100 Teeth (Input)
Gear = 40 Teeth
Gear = 20 Teeth
Ring = 78 Teeth (Fixed)
Arm = Output
Always draw this
view to get
direction of
rotation correct.
See the Course Materials folder for the solution.
Epicyclic Gears 8
Epicyclic Tips
• If you encounter a gear assembly with two inputs, use
superposition. Calculate the output due to each input with the
other input held fixed, and then sum the results.
• Typically, when an input arm is held fixed, the other output to
input relationship will not be epicyclic, but be a simple product of
tooth ratios.
OUT
IN 1
IN 2
• Use the ± sign with tooth ratios to carry the direction information.
Epicyclic Gears 9
Gear Loading
• Once you determine the rotational speeds of the gears in
a train, the torque and therefore the tooth loading can be
determined by assuming a constant power flow through the
train.
• Power = Torque x RPM, so
Torque = Power / RPM
• If there are n multiples of a component (such as the 3
idlers in the planetary gearset example) , each component
will see 1/n times the torque based on the RPM of a single
component.
• From the torque, T, compute the tangential force on the
teeth as Wt = T/r = 2T/D , where D is the pitch diameter.
Epicyclic Gears 10
The Tabular Method - Example 2
First, we lock the assembly and rotate
everything one turn CCW
Rotate Whole
Assembly CCW
Hold Arm, Rotate
Fixed Ring CW
Ring 1
Gear 2
Gear 3
Ring 4
Arm 5
+1
+1
+1
+1
+1
Total Motion


Next, we hold the arm fixed, and turn the
fixed component (Ring 4) one turn CW, to
fill in row two.



Epicyclic Gears 11
The Tabular Method - Example 2
First, we lock the assembly and
rotate everything one turn CCW
Rotate Whole
Assembly CCW
Hold Arm, Rotate
Fixed Ring CW
Ring 1
Gear 2
Gear 3
Ring 4
Arm 5
+1
+1
+1
+1
+1
Total Motion





Next, we hold the arm fixed, and
turn the fixed component (Ring 4)
one turn CW, to fill in row two.
Epicyclic Gears 12
The Tabular Method - Example 2
We turn Ring 4 one turn CW (-1).


• Ring 4 drives Gear 3, which turns + N4/N3 x (-1)
= - N4/N3 rotations.
• Gear 2 is on the same shaft as Gear 3, so it
also turns - N4/N3 rotations.



• Gear 2 drives Ring 1, which turns
+ N2/N1 x n2 = + N2/N1 x (- N4/N3) = - N2N4/N1N3
rotations.
• The arm was fixed, so it does not turn.
We enter these motions into row two of the table:
Rotate Whole Assembly
CCW
Hold Arm, Rotate Ring
CW
Ring 1
Gear 2
Gear 3
Ring 4
Arm 5
+1
+1
+1
+1
+1
- N2N4/N1N3
- N4/N3
- N4/N3
-1
0
Epicyclic Gears 13
The Tabular Method - Example 2
Then we add the two rows to get the total motion
Rotate Whole Assembly
CCW
Hold Arm, Rotate Ring
CW
Total Motion
And we can
write the
relationship:
Ring 1
Gear 2
Gear 3
Ring 4
Arm 5
+1
+1
+1
+1
+1
- N2N4/N1N3
- N4/N3
- N4/N3
-1
0
0
+1
1 - N2N4/N1N3 1 - N4/N3 1 - N4/N3
nRing1 1 
n Arm 
N2 N4
 n Arm , or
N1 N 3
1
1
N2N4
 nRing1





N1N 3
Epicyclic Gears 14
The Tabular Method - Example 2
For our example
Ring 1:
Gear 2:
Gear 3:
Ring 4:
We compute:
N1 = 100 Teeth
N2 = 40 Teeth
N3= 20 Teeth
N4 = 78 Teeth
n Arm 




1
N2N4
 nRing1
N1N 3
n Arm 

1
1
1
40  78
 nRing1
100  20
1
1
n Arm 
 nRing1 
 nRing1
1 1.56
0.56
n Arm   1.786 nRing1
The output arm rotates almost twice as fast as
the input ring, and in the opposite direction.
• Output direction is dependent on the
numbers of teeth on the gears!
Epicyclic Gears 15