Reduction Gears
need:
propellers
waterjet pump
low speed diesels
medium speed diesels
steam turbine
gas turbine
60-300 rpm
300-1500 rpm
70-250 rpm
350-1200 rpm
6000-9000 rpm
3600-15000 rpm
(larger @ lower rpm)
reduction gears make conversion.
some history
ref: Marine Engineering Chapter IX Reduction Gears, by Gary P. Mowers, (SNAME) page 325 ff and
others
19th - 20th century (1890-1910) ships propelled by reciprocating steam engines - direct drive
1904 - study by consulting engineers George Melville Adm (Ret.) and John Alpine
George Melville was Chief Bureau of Steam Engineering and in 1899 President of ASME see
study: - Problem - steam engine succeeding reciprocating engine:
"If one could devise a means of reconciling, in a practical manner, the necessary high speed of revolution of the
turbine with the comparatively low rate of revolution required by an efficient propeller, the problem would be solved
and the turbine would practically wipe out the reciprocating engine for the propulsion of ships. The solution of this
problem would be a stroke of great genius." Ref: Mar. Eng.
First gear generally attributed to Pierre DeLaval in 1892. Parsons (cavitation) and George Westinghouse
developed prototypes and installed gears:
1910 - 15,000 shp with geared turbine drive
1940 - 100,000 shp with geared turbine drive
1917 - double reduction introduced
Development has been evolutionary - few step advances
- single to double
- welding in construction of gear wheel as and casing
- higher hardness pinion and gear materials => higher tooth load
Many types of gears are used (defined) and there is an extensive nomenclature associated with gear definitions.
One source: (formerly available free via registration via:
http://www.agma.org/Content/NavigationMenu/EducationTraining/OnlineEducation/default.htm )
AGMA Gear Nomenclature, Definitions of Terms with Symbols
ANSUAGMA 1012-F90
(Revision of AGMA 112.05)
[Tables or other self-supporting sections may be quoted or extracted in their entirety. Credit lines
should read: Extracted from AGMA 1012-F90, Gear Nomenclature Terms, Definitions, Symbols and
Abbreviations, with the permission of the publisher, American Gear Manufacturers Association, 1500
King Street, Suite 201, Alexandria, Virginia 22314 ] Availability changed to require registration in course.
I have copy from previous registration when it was free.
See: handout from Marine Engineering, (on web site)
11/29/2006
1
Single Reduction
gear_ratio = R =
R=
Single reduction
input
pinion
diameter_of_gear
+
diameter_of_pinion
number_teeth_gear
number_teeth_pinion
=
rpm_pinion
rpm_gear
=
NP
output
NG
bull
gear
+
Double Reduction (locked train 50% each drive)
R1 =
dg1
dp1
R2 =
dg2
Double reduction
R = R1 ⋅R2
dp2
dp 2
dg 2
dp 1
dg 1
dg 1
dg 1
dp+ 2
dp+2
dp 1
+
input
output
dg 2
Epicyclic gear summary
various combinations can be used with this
system
Epicyclic gear
ring
+
planets
sun
+
cage
cage
Type
Fixed
Input
Output
Ratio
+
+
Normal range
Planetary
ring
sun
cage
RR/RS + 1
3:1 - 12:1
Star
cage
sun
ring
(-) RR/RS
(-) 2:1 - 11:1
Solar
sun
ring
cage
RS/RR + 1
RS = radius_sun
RR = radius_ring
Application: Planetary gears used in high speed as tooth loading
reduced by multiplicity of planet gears. Also can provide
counter-rotation.
RP = radius_planet
11/29/2006
1.2:1 - 1.7:1
2
Wr*Rr
ring
Wp*Rp
to show above ratios consider: R = pitch_diameter
subscripts ...
W = angular_velocity
s = sun
WT = tangential_load
p = planet
Wc
Wr
planet
Ws
Wc*Rr
Wp
Wp*Rp+Wc*Rs
Ws*Rs
sun
r = ring
n = rpm
c = carrier( cage)
ref: Gear Drive Systems; Design and
Application, P. Lynwander TJ184.L94 1983
N.B. in this development Ws and Wc are
1 = primary
for compound
systems
2 = secondary
clockwise and Wr and Wp are ccw
Ws⋅Rs
1. point on pitch diameter of sun gear has tangential velocity
2. point on sun gear pitch diameter meshing with planet gear pitch diameter has tangential velocity
Wp ⋅Rp + Wc⋅Rs
first term from rotation of planet
second from rotation of carrier about center of carrier and sun
Ws⋅Rs = Wp ⋅Rp + Wc⋅Rs
mesh =>
Wr⋅Rr = Wp ⋅Rp − Wc⋅Rr
3. similarly at ring ...
(
(
)
Ws⋅ Rs = Wr⋅ Rr + Wc⋅ Rr + Rs
Planetary arrangement ... input sun,
or .. for solving below ...
output carrier (cage),
(
fixed ring
Ws
)
Ws⋅Rs = Wc⋅ Rr + Rs
Wr = 0
Star arrangement ... input sun, output ring,
Wc
Solar arrangement ... input ring
Ws
(
Wr
output carrier (cage)
)
0 = Wr⋅ Rr + Wc⋅ Rr + Rs
Wr
Wc
=
Rr + Rs
Rs
=
Rr
Rs
+ 1
fixed carrier (cage)
Ws⋅Rs = Wr⋅Rr
Wc = 0
11/29/2006
)
Ws⋅Rs = Wp ⋅Rp + Wc⋅Rs = Wr⋅Rr + Wc⋅Rr + Wc⋅Rs = Wr⋅Rr + Wc⋅ Rr + Rs
combining ...
Ws = 0
Wr⋅Rr + Wc⋅Rr = Wp ⋅Rp
or ...
=−
=
Rr
but these are in opposite directions
can use for reversing
Rs
fixed sun
Rr + Rs
⎛ Rs
= −⎜
⎝ Rr
Rr
3
⎞
+ 1⎟
⎠
but in figure Wr and Wc are
opposite rotation => this is the
same actual rotation in result
defined mcd here for
symbolic calculation,
actually below
k 1 :=
Hertz stress 1 − ν1
2
2
k 2 :=
π⋅E1
1 − ν 2
2
π⋅E2
a brief intro to some details. see Timoshenko, Theory of
Elasticity ... page 418 ff for more specifics
It can be shown that the width of contact between two parallel cylinders given: elliptical loading, q o , etc.
b :=
(
)
4 ⋅ P_p ⋅ k 1 + k 2 ⋅R1 ⋅R2
this is the solution to the
width of contact given:
elliptical loading, q o , etc.
R1 + R2
where ...
P_pr =
1
2
⋅π⋅b⋅q o
R1
(236)
q o = max_pressure_elliptical_distribution
b = half_width_of_rectangular_contact_area
q o :=
2 ⋅ P_pr
P_pr = P' =
π⋅bb
kn =
L
load
1 − νn
b
(235)
length
2
all n
π⋅En
R2
(236)
elliptical loading details
b := 2 p_over_Pmax( x ) :=
1−
2
⎛x⎞
⎜ ⎟
⎝b⎠
x := −b , −b + 0.01 .. b
elliptical loading numerical plot
2
p ( xx) := P_max⋅ 1 −
want
⌠
⎮
⌡
⎛ xx ⎞
⎜ ⎟
⎝ bb ⎠
2
pressure
symbolically
bb
p ( xx) dxx = load
− bb
let ...
1
xx( θ , bb) := bb⋅cos( θ)
xx( θ , bb) → (−bb) ⋅ sin( θ )
d
0
2
dθ
1
0
1
2
distance
then ...
p ( θ ) := P_max⋅ sin( θ ) ⌠
⎮
⌡
bb
− bb
0
⌠
p ( xx) dxx = ⎮ p ( xx) ⋅ ( −bb⋅ sin( θ ))
dθ
⌡
π
therefore ..
0
⌠
1
⎮ p ( θ ) ⋅ ( −bb⋅sin( θ )) dθ → ⋅π⋅P_max⋅bb
⌡
2
π
elliptical loading details
11/29/2006
4
and with interpretation on
per unit length basis
P_pr =
1
2
⋅π⋅b⋅ q o
P_pr
qo →
substitution for b and solve for q o in terms of
1
load per unit length P_pr...
⎡
⎛ 1 − ν 2 1 − ν 2⎞
R2 ⎥⎤
1
2 ⎟
⎢
⎜
π⋅ P_pr⋅
⎢
⎜ π⋅E + π⋅E ⎟ ⋅R1⋅ R + R ⎥
1
2⎦
1
2 ⎠
⎣
⎝
or ...
2 ⋅ P_pr
qo =
π⋅
qo =
(
2
π
R1 + R2
⋅P_pr⋅
(
R1 + R2
2
)
4 ⋅ P_pr⋅ k 1 + k 2 ⋅R1 ⋅R2
P_pr ⋅4
=
π
2
P_pr
⋅
⋅
(
)
4 ⋅ P_pr⋅ k 1 + k 2 ⋅R1 ⋅R2
R1 + R2
P_pr
π
)
4 ⋅ P_pr⋅ k 1 + k 2 ⋅ R1 ⋅ R2
=
2
2
R1 + R2
⋅
=
(k1 + k2)⋅R1⋅R2
P_pr
π
R1 + R2
⋅
⎛ 1 − ν 2 1 − ν 2⎞
2 ⎟
1
⎜
⎜ π⋅E + π⋅E ⎟ ⋅R1 ⋅R2
1
2 ⎠
⎝
2
=
π
R1 + R2
⎛ 1 − ν 2 1 − ν 2⎞
(240)
⎜
2 ⎟
1
+
⋅R ⋅R
⎜ E
E2 ⎟ 1 2
⎝ 1
⎠
same material E1 = E2, and ... ν = 0.3 ...
1
qo =
P_pr
π
⋅
R1 + R2
⎛ 1 − ν2 ⎞
⎟ ⋅R ⋅R
⎝ E ⎠ 1 2
=
P_pr ⋅ E
π
2⎜
⋅
(
R1 + R2
2
)
=
2 1 − ν ⋅R1 ⋅R2
2
1
P_pr ⋅ E R1 + R2
⎡
⎤
⋅
⎢
2 ⎥
R1 ⋅R2
π
π⋅2
1
ν
−
⎣
⎦
(
)
1
ν := 0.3
2
1
⎡
⎤ = 0.418
⎢
2 ⎥
⎣ π⋅2 1 − ν ⎦
(
)
q o = 0.418⋅
Marine Engineering ...
S=
P_pr ⋅ E R1 + R2
⋅
R1 ⋅R2
π
(241)
r1 + r2
P
0.175⋅ ⋅E⋅
L
r1 ⋅r2
page 330 between (10) and (11)
S = maximum_compressive_stress⋅ psi
P
L
Wt
Fe
unit version
=
P
L
hp
Wt
= 126051
=
Wt
Fe
=
π
⋅
(
1
2 1−ν
2
)
= 0.175
tangential_tooth_load
⋅
lbf
effective_face_width_at_pitch_diameter in
hp = horse_power_transmitted_per_mesh
π⋅rpmpinion⋅d pinion⋅Fe
Fe
11/29/2006
1
= loading_per_inch_length = P_pr
P d1 + d2
⋅
L d 1 ⋅d 2
maximum stress ~
E, r straight forward
hp in horsepower
hp
rpmpinion⋅d pinion⋅Fe
5
rpm in min-1
dpinion in inches
Fe in length; carries to result
result ...
lbf
Fe_unit
replace ...
d1 = dg
d2 = dp
dg
+ 1
Wt d p
⋅
=
dg
Fe
Wt d g + d p
⋅
=
Fe d g ⋅ d p
maximum stress ~
motivates parameter K factor
maximum stress ~
Wt R + 1
⋅
=
Fe d g
Wt 1 R + 1
⋅ ⋅
Fe d p
R
K=
as ...
Wt R + 1
⋅
Fe d p ⋅R
R=
dg
dp
units of pressure: kPa. psi, N/m^2 etc.
K
Wt
R
hp
= K⋅ d p ⋅
=
Fe
R+1
π⋅n p ⋅d p ⋅Fe
some observations ...
⎛ Fe ⎞
=C
⎜d ⎟
p
⎝ ⎠ max
for double helical pinion gear
substitute F e = C*d p into
3
dp =
above ...
dp =
π⋅K⋅n p ⋅Fe
2 ≤ C ≤ 2.5
⋅⎜
⎟
π⋅K⋅C⋅n p ⎝ R ⎠
3
⎡ hp ⋅ ⎛ R + 1 ⎞⎤
⎢ π⋅K⋅C⋅n ⎜⎝ R ⎟⎠⎥
p
⎣
⎦
classification societies limit to 300 psi
⎟
⎝ R ⎠
550
π
⋅60⋅12 = 126051
hp
100psi = 689.5
K used in design 100 - 1000 psi
highest in hardened and ground aircraft engine gears
R + 1⎞
lbf ⋅ ft 60sec 12in
⋅
⋅
sec
ft
min
1
dp =
⋅ ⎛⎜
1.5 for epicyclic
unit conversion
⎛R + 1⎞
hp
hp
2
=>
kN
1000psi = 6895
2
m
300psi = 2068
kN
2
m
kN
2
m
Marine Engineering suggests the first approximation for the gear ratio for the second reduction be taken as ...
Roverall − 1
another Navy study (ref: Prof
Carmichael) suggests
volume ... for solid gear
volume_bull_gear =
total_volume =
11/29/2006
hp
3
π
π⋅K⋅n p
1
Roverall
vol =
d p ⋅Fe = d p ⋅C =
locked train
for double reduction
Roverall
2
Roverall + 3
articulated
⋅ ⎛⎜
π
4
3
for triple reduction
2
⋅d ⋅Fe
R + 1⎞
⎟
⎝ R ⎠
π
π 2 2
π 2 hp ⎛ R + 1 ⎞
2
⋅d g ⋅Fe = ⋅R ⋅d p ⋅Fe = ⋅R ⋅
⋅⎜
⎟
4
4
4
π⋅K⋅n p ⎝ R ⎠
(
)
(
π
hp ⎛ R + 1 ⎞ 2
2
2
⋅d p ⋅Fe⋅ R + 1 = ⋅
⋅⎜
⎟⋅ R + 1
4
4 π⋅K⋅n p ⎝ R ⎠
6
)
in terms of diameter of pinion
n p = R⋅ n g
or substituting
total_volume =
π
(
R + 1⎞ 2
⋅ R +1
⎜
2 ⎟
4 π⋅K⋅n g
R
⎝
⎠
⋅
hp
⋅⎛
)
in terms of diameter of bull gear
overall volume ... increases with
1. power increases (per torque path
2. rpmgear decreases
3. K decreases
4. R increases
data for plot
$/hp
nominal cost plot showing trends
single reduction
double reduction
triple reduction
0
10
20
30
40
overall reduction ratio R
11/29/2006
7
50
60