we have a trim board = 1*4 or 1*6
all boards are white.
we need to design a machine that can do things to the board.
first thing screw some screws. on the door frame, you may have a piece of trim
and the door frame about 36" up if you have a door jamb.
we could put brackets on the trim.
the board we got to stack one against another and they're being shoved into this
machine so we need to build a machine over here between the boards to stop
and start the board and puts in a bracket. it might have bracket here to screw
here screw there. ( same as the figure) between we need a motion control stuff.
part of that we going need a coupler nip rollers (coupler, motor,..). will squeezing
the board and one of them probably has a motor and one is an ipoare and we've
got some friction right here so we can drive the boards. then registration sensor,
and we need to sensor that can detect the board. then we will have basically a
plate we going to amount have two screwdrivers into this plate. we need motion
go up and down. and each one has to exist motor, Dc motor ( to perform better
torque control in it). ( 1/2 hp brushed motor)
two screws ( 2" and .5") during travel 2.22" and 0.75"
-constant force
- fix time 1.2 to driver screw
- when sensor red should move 36" and stop.
- accy +1.5mm
-cycle time 5 second.
- diameter the roller > 2.5"
- contact whole board.
- pushed with 3lb force from the beginning.
- need to take care of rolling and the tension of rolling.
- length of board 72"-90"
-force 1lb from the middle to the end.
- between the rolling can be bushed, 2 board.
- screw collaboration should be 1.2second to drive screw.
-power supply 110, 230... no limitation.
- the screwdrivers = 20lb.
Introduction to Motor Sizing
INTRODUCTION TO
MOTOR SIZING
Nick Repanich
Adjunct Research Professor
Department of Mechanical and Mechatronic Engineering
and Sustainable Manufacturing
California State University, Chico
Chico, CA 95929-0789
(530) 520-2548
Introduction to Motor Sizing
1
SEVEN STEPS OF SIZING AND SELECTION
Step 1:
Develop the torque and inertia equations that model the system mechanics.
1a: Draw/diagram the system to establish the relative location of the load mechanics.
1b: Develop the acceleration (Ta=), friction (Tf=), gravity (Tg=), and thrust (Tth=) torque
equations. Since Ta= (J)(α), this will also involve developing the inertial model (Jload=).
Step 2:
•
Determine the load motion profile(s) and calculate peak values.
2 X 2 xa
=
tm
ta
- a simple triangular profile minimizes torque (lowers cost), but uses higher speeds
First find the max velocity reached (V) in a triangular move. If OK, use it.
V=
•
If V is too high for any reason, then find the optimized acceleration time (ta) in a trapezoidal
X
move using overall move time, distance & the new constrained velocity.
ta = tm −
V
- trapezoidal profiles are useful where rated motor torque drops with speed
- for steppers, set maximum speed where the motor changes from the constant torque range
into the constant power range, otherwise reduction may be necessary.
V
•
If there are multiple move profiles, find the worst-case acceleration (a), where
a=
ta
Step 3:
Calculate the mass moment of inertias of the load mechanics.
• Anything that moves with the motor is part of the inertia (J).
Step 4:
Determine the peak torque (Tp) at maximum speed excluding motor inertia.
•
determine worst case combination of Ta , Tf , Tg , Tth
( T peak = Ta ± T f ± Tg ± Tth )
•
calculate power required to move the load (to find a system in the right power range)
Step 5:
Choose an approximate motor/drive system.
•
find a motor/drive with more than the required speed and about double the power
•
look at speed-torque curves and price list simultaneously
- with servos, use most of the speed available or you waste the power available
Step 6:
Determine the peak torque at speed including motor inertia.
• calculate RMS torque to evaluate motor heating issues (necessary for servo systems)
• give 50-100% torque margin (only use lower margins if measurable mechanics exist)
• check for 10-20% velocity margin
Step 7:
Optimize the system.
•
Check the load-to-motor inertia ratio, and compare to the machine’s stiffness to the
performance desired. A ratio higher than 10:1 is an indicator of potential instability
problems with non-stiff systems. (Note: It is not the cause of the problem.)
•
Check the need for a power dump circuit (high inertia ratios or vertical load)
•
It may be necessary to adjust mechanics, add reduction, and start over. Remember, iteration
is crucial to successful design!
Introduction to Motor Sizing
Direct Drive (Rotary) Formulas and Examples
MOTION PROFILE FORMULAS
for triangular profiles:
V=
for trapezoidal profiles:
2 X 2 xa
=
tm
ta
ta = tm −
Where:
X
V
V
=
maximum velocity
X
=
total move distance
tm =
total move time
xa =
acceleration distance
ta
=
acceleration time
Note: V is used here rather that ω, as these formulas work with either linear or rotary units.
GENERAL TORQUE & POWER FORMULAS
T peak = Ta ± T f ± Tg ± Tth
Torque Equations:
T12 t1 + T22 t 2 + T32 t 3 + T42 t 4
t1 + t 2 + t 3 + t 4
Motor torque required to accelerate the inertial load
Motor torque required to overcome the frictional forces
Motor torque required to overcome the gravitational forces load
Motor torque required to overcome any additional thrust forces
Torque to accelerate the load from zero speed to max speed (Tf + Ta)
Torque to keep the motor moving once it reaches max speed (Ta = 0)
Torque required to decelerate from max speed to a stop (Ta - Tf)
Torque required while motor is sitting still at zero speed
time spent accelerating the load
time spent while motor is turning at constant speed
time spent decelerating the load
time spent while motor is at rest
Trms =
Where: Ta
Tf
Tg
Tth
T1
T2
T3
T4
t1
t2
t3
t4
=
=
=
=
=
=
=
=
=
=
=
=
Torque Unit Conversions:
kg ⋅ m
N= 2
s
2
∴
kg ⋅ cm 2 ⎛ 1 m ⎞
⎜
⎟ = N ⋅ m
s 2 ⎝ 100 cm ⎠
Power Unit Conversions:
2π (N ⋅ m )(rpm ) (oz ⋅ in)(rps ) (in ⋅ lb)(rpm ) ( ft ⋅ lb)(rpm ) ( ft ⋅ lb)(rps )
Watts = 2π (N ⋅ m )(rps ) =
=
=
=
=
60
22.52
84.45
7.038
0.1173
1 hp = 746 W
2
Introduction to Motor Sizing
Direct Drive (Rotary) Formulas and Examples
DIRECT DRIVE (ROTARY) FORMULAS
Inertia (Step 3):
Solid Cylinder
Hollow Cylinder
Where ρ, the density is known, use
πLρR 4
J load =
2
Where mass and radius are known use
mR 2
J load =
2
m = πLρR 2
Where ρ, the density, is known use
πLρ 4
J load =
R2 − R14
2
Where m, the mass, is known use
(
J load =
)
m 2
R1 + R22
2
(
(
)
m = πLρ R22 − R12
)
Torque (Step 2, 4-7)
⎛ ω ⎞
Ta = (J load + J cp + J m )⎜⎜ m ⎟⎟
⎝ t a ⎠
T f = (F )(R )
Where:
Jcp
Jm
ta
ωm
F
R
=
=
=
=
=
=
inertia of the coupler (g·cm2)
inertia of the motor's rotor (g·cm2)
acceleration time (s)
maximum motor velocity reached (rad/s)
frictional force (N)
radius at which the frictional force acts (cm)
Note 1: When using Imperial units, specifically inertia in convenient units of oz·in2, the
acceleration torque formula must be modified to correct the units. The units of oz·in2 for inertia
are not proper inertial units, yet they are quite convenient to calculate inertia based on real world
data. To use the acceleration torque formula with inertia in oz·in2, divide by the gravitational
constant g (1/386 in/s2), which converts oz·in2 to proper units of oz·in·s2 as follows:
⎛ ω ⎞
1
(
Ta =
J load + J cp + J m )⎜⎜ m ⎟⎟
in
⎝ t a ⎠
386 2
s
Note 2: Though the correct symbol for mass moment of inertia is “I”, typically in mechatronics
we (incorrectly) use the symbol “J” for inertia, as “I” is also used for electrical current.
3
Introduction to Motor Sizing
Direct Drive (Rotary) Formulas and Examples
DIRECT DRIVE WITH REDUCTION
Gear Drive Inertia Formulas:
m R2
J load = load 2load
2N
or
4
πLload ρ load Rload
J load =
2N 2
J gear1 =
J gear 2 =
Where:
Jx
m
R
N
L
ρ
2
mgear1 Rgear
1
2N 2
2
m gear 2 Rgear
2
2
=
=
=
=
=
=
inertia "as seen by the motor"
mass
radius
reduction ratio (R1/R2)
length
density
Note: Most gearhead manufacturers provide the reflected inertia of the reducer in their user
guide. This will help eliminate the need to calculate the individual gear inertias.
⎛ J
⎞⎛ ω N ⎞
Ta = ⎜ load + J r + J m ⎟⎜⎜ m ⎟⎟
⎝ e
⎠⎝ t a ⎠
Tf =
(F )(R )
eN
Where:
e
N
Jr
Jm
ωm
ta
F
R
= efficiency of transmission (reduction)
= reduction ratio
= effective inertia of reducer, gear box, belts and pulleys, or gears
= inertia of the motor's rotor
= maximum motor velocity reached
= acceleration time
= frictional force
= radius at which the frictional force acts
4
Introduction to Motor Sizing
Direct Drive (Rotary) Formulas and Examples
EXAMPLE #1 - Fluted-Bit Cutting Machine
Parameters for Axis 4:
•
•
•
•
•
•
•
•
Move distance
Move time
Maximum velocity allowed
Length of chuck
Diameter of steel chuck
Length of bit
Diameter of steel bit
Torque required during cutting
V=
Step 1:
0.2
0.15
20
4.0
3.0
14
0.8
0.050
rev
s
rps
cm
cm
cm
cm
N·m
2 X (2)(0.2)
=
= 2.67 rps
tm
0.15
(OK, since not above "knee" on most stepper curves)
Step 2:
Step 3:
J bit
⎛ ω ⎞
Ta = (J bit + J chuck + J motor )⎜⎜ m ⎟⎟
⎝ t a ⎠
4
π
g ⎛ 0.8 cm ⎞
2
= (14 cm )(7.75 3 )⎜
⎟ = 4.36 g ⋅ cm
2
cm ⎝ 2 ⎠
J chuck =
π
2
4
(4 cm )(7.75
g ⎛ 3.0 cm ⎞
2
)
⎟ = 246.5 g ⋅ cm
3 ⎜
cm ⎝ 2 ⎠
5
Introduction to Motor Sizing
Direct Drive (Rotary) Formulas and Examples
Step 4:
2
2
cm
m
⎛ 2.67 rps ⎞
Ta = 4.36 g ⋅ cm + 246.5 g ⋅ cm (2π )⎜
⎟ = 56,112 g ⋅ 2 = 0.0056 kg ⋅ 2 = 0.0056 N ⋅ m
s
s
⎝ 0.15 s / 2 ⎠
(
2
2
)
Tf = 0.050 N·m
Tpeak = Ta + Tf = 0.0056 + 0.050 N·m = 0.0556 N·m @ 2.67 rps
P = 2π (0.0556 N ⋅ m)(2.67 rps) = 0.93 W
2
Jload = 4.36 + 246.5 = 250.86 g·cm
Step 5: Try Nema 23 half-stack step motor,
Step 6:
Jm = 0.070 kg·cm2 = 70 g·cm2
cm 2
⎛ 2.67 ⎞
Ta = (4.36 + 246.5 + 70 )(2π )⎜
⎟ = 71,770 g ⋅ 2 = 0.0072 N ⋅ m
s
⎝ 0.15/ 2 ⎠
Tpeak = Ta + Tf = 0.0072 + 0.050 N·m = 0.0572 N·m @ 2.67 rps
Step 7:
Inertia Ratio =
Torque Margin (@speed) =
Jl
250.86 g ⋅ cm 2
=
= 3.6 (fine)
Jm
70 g ⋅ cm 2
Tavailable − T peak
T peak
=
0.26 N ⋅ m - 0.0572 N ⋅ m
= 355 % (fine)
0.00572 N ⋅ m
6
Introduction to Motor Sizing
Direct Drive (Rotary) Formulas and Examples
PROBLEM #1 - Grinding Of Bicycle Rim Braking Surface
Parameters of Axis 3:
•
•
•
•
•
•
•
•
•
•
Outside diameter of steel bicycle rim
33.0 cm
Move distance
1/36 rev
Move time
0.10 s
Inertia of rim-holding fixture
1686 kg·cm2
Width of rim
4.5 cm
Inside diameter of rim
31.5 cm
Friction torque during grinding
1.6 N·m
Dwell time between moves
0.05 s
Velocity is not limited in any way
This is not the final grinding move, but is the worst case move, making
repeated passes over the weld.
Answer: approximately 20 N·m @ 0.56 rps
7
Introduction to Motor Sizing
Leadscrew Formulas and Examples
LEADSCREW FORMULAS
J load =
Inertia (Step 3)
Where:
m
L
R
ρ
p
=
=
=
=
=
m
(2πp )2
J screw =
πLρR 4
2
mass (kg)
length (cm)
radius (cm)
density (g/cm3)
pitch of screw (revs/unit length) (inverse of lead)
The formula for Jload converts linear mass into the rotational inertia as reflected to the motor
shaft by the lead screw.
⎛ J
⎞⎛ ω ⎞
Ta = ⎜ load + J screw + J cp + J m ⎟⎜⎜ m ⎟⎟
⎝ e
⎠⎝ t a ⎠
Torque (Step 2, 4-7)
T f = Tbr +
Where:
Jscrew
Tbr
F
p
e
ωm
g
Jcp
Jm
ta
Tg
µd
=
=
=
=
=
=
=
=
=
=
=
=
F
2πpe
Tg =
mg
2πpe
F = µ d mg
inertia of lead screw or ball screw (kg·cm2)
breakaway torque of nut on screw (N·m)
force or thrust required
pitch of screw (revs/unit length)
efficiency of nut and screw
maximum motor speed (rev/s). (watch critical speed of screw)
acceleration due to gravity
inertia of the coupling (kg·cm2)
inertia of the motor's rotor (kg·cm2)
acceleration time (s)
torque required to overcome gravity (vertical)
coefficient of dynamic friction
8
Introduction to Motor Sizing
Leadscrew Formulas and Examples
EXAMPLE #3 - "Smart" Battery Inspection
Parameters for Axis 1:
•
•
•
•
•
•
•
•
•
•
Length of steel leadscrew
Pitch of leadscrew
Efficiency of leadscrew
Radius of leadscrew
Move distance
Move time
Maximum allowed velocity
Weight of load
Coefficient of friction
Breakaway Torque
36 in
5
0.65
0.5 in
1.5 in
0.5 s
5 in/s
50 lbs.
0.01
25 oz·in
9
Introduction to Motor Sizing
Leadscrew Formulas and Examples
Step 1:
V=
(2) (1. 5in)
= 6 in / sec, too high
0.5sec
ta = tm −
V=5
Step 2:
Ta =
in
rev
rev
×5
= 25
sec
in
sec
1 # Jload
& 2πV
+ Jleadscrew + Jmotor + Jcp (
%
386$ e
' ta
Jload =
Step 3:
X
1.5
= 0. 5 −
= 0.2 sec
V
5
W
( 2 πp )
2
=
50(16)
( 2π5)
2
= 0. 81 oz ⋅in2
πLρ R4 π
4
= (36)(4. 48)(. 5) = 15.8 oz ⋅ in2
2
2
Jleadscrew =
Jcp = 0
Ta =
1 ! . 81
2 π (25 )
+ 15 .8 $&
= 34.69 oz⋅ in @ 25 rps
#
386 " . 65
% .2
Step 4:
T f = T br +
F
0.01(50 )(16 )
= 25 +
= 25.39 oz⋅ in
2πp e
2 π 5(.65 )
Tpeak = 34.69 + 25.39 = 60.08 oz·in
hp =
Step 5:
!
(60. 08)25
watts $
= 0. 0894# 746
& = 66. 7 watts
16, 800
hp %
"
Choose motor with about 0.2 hp
Try S83-93, Jm = 6.70
10
Introduction to Motor Sizing
Step 6:
Ta =
Leadscrew Formulas and Examples
1 ! .81
2π( 25)
+ 15. 8 + 6.70$&
= 48. 3 oz ⋅ in
#
386" .65
% .2
T peak = 48. 3 + 25. 39 = 73. 7 oz ⋅ in @ 25 rps
Jl
16.61
=
= 2. 48, O.K.
Jm
6. 7
Step 7:
Pick S83-93
11
Introduction to Motor Sizing
Tangential Formulas and Examples
TANGENTIAL DRIVES FORMULAS
Inertia Formulas:
2
J load = ml Rdp
J dp = J fp =
m p R p2
2
=
πL p ρR p4
2
2
dp
J b = mb R
Where:
Jdp
Jfp
Jb
Rdp
ml
mp
mb
= inertia of the driving pulley
= inertia of the follow pulley
= inertia of the belt
= radius of the driving pulley
= mass of the load
= mass of the pulley
= mass of the belt
12
Introduction to Motor Sizing
Appendix A
Belt Drive Torque Formulas:
Direct Drive:
With a gearhead:
⎛ J + J belt + J mechanics
⎞⎛ ω ⎞
Ta = ⎜⎜ load
+ J dp + J m ⎟⎟⎜⎜ m ⎟⎟
eb
⎝
⎠⎝ t a ⎠
T f = FR dp
⎛ J
⎞⎛ ω m N gh
J dp
+ J belt + J mechanics
⎟⎜
Ta = ⎜ load
+
+
J
+
J
gh
m
2
2
⎜
⎟⎜ t a
e
e
N
e
N
b gh
gh
gh
gh
⎝
⎠⎝
FRdp
Tf =
e gh N gh
Where:
ta
eb
egh
Ngh
Jmechanics
Rdp
Jgh
ωm
Jm
F
= acceleration time
= efficiency of belt over pulleys
= efficiency of gearhead
= reduction ratio of gearhead
= inertia of other moving masses
= radius of driving pulley
= effective inertia of reducer
= maximum motor speed
= inertia of the motor's rotor
= force or thrust or friction forces at the drive pulley
⎞
⎟⎟
⎠
13
Appendix A
Introduction to Motor Sizing
Rack & Pinion Torque Formulas:
Note: In the following formulas, the rack is stationary and the motor moves with the load. If the
rack moves and the motor is stationary, then use the belt and pulley formulas, where the rack acts
essentially like the belt.
⎛ J load + J m + J p ⎞⎛ ω m ⎞
⎟⎜
⎟
Ta = ⎜
Direct Drive:
⎜
⎟⎜ t ⎟
e
g
⎝
⎠⎝ a ⎠
FR p
Tf =
eg
With gearhead on motor:
⎛ J load + J p J gh + J m
Ta = ⎜
+
⎜ e N 2 e
eg
gh
⎝ g
FR p
Tf =
e g N 2 e gh
⎞⎛ ω m N ⎞
⎟⎜
⎟
⎟⎜ t a ⎟
⎠
⎠⎝
Where:
Jload
eg
Jp
Rp
F
Jm
ωm
ta
Jgh
egh
N
=
=
=
=
=
=
=
=
=
=
=
inertia of moving load
efficiency of transmission between rack and pinion
inertia of pinion gear and shafts and couplers and outboard bearings (if any)
effective radius of pinion gear
force, thrust or friction
inertia of the motor's rotor
maximum motor speed
acceleration time
effective inertia of reducer
efficiency of reducer
reduction ratio
14
Appendix A
Introduction to Motor Sizing
APPENDIX A
STANDARD END FIXTURING METHODS
The method of end fixturing has a direct effect on critical speed, column load bearing capacity,
and system stiffness. Four common methods are shown below.
Fixed-Free
One end held in a duplex (preloaded) bearing,
the other end is free
Fixed-Simple
One end held in a duplex (preloaded)
bearing, one end held in a single bearing.
Simple-Simple
Both ends held in a single bearing.
Fixed-Fixed
Both ends held in duplex (preloaded)
bearings.
DENSITIES OF COMMON
MATERIALS
Material
oz/in3 gm/cm3
Aluminum Alloys
1.54
2.8
Brass, Bronze
4.80
8.6
Copper
5.15
8.9
Plastics
0.64
1.1
Steel (carbon,alloys,stainless) 4.48
7.8
Hard Wood
0.46
0.80
KINEMATIC EQUATIONS
(straight-line motion w/ constant acceleration)
Equation
Contains
x vx ax t
ü ü ü
vx = vx + axt
0
x = x0 + 12 v x0 + v x t
ü
x = x0 + v x0 t + 12 a x t 2
ü
(
2
2
)
v x = v x0 + 2a x (x − x0 )
ü
ü
ü
ü
ü
ü
ü
COEFFICIENTS OF FRICTION
Materials (dry contact unless noted)
µs
µd
Steel on Steel
0.74 0.58
Steel on Steel (lubricated)
0.23 0.15
Aluminum on Steel
0.61 0.45
Copper on Steel
0.22
Brass on Steel
0.19
Teflon on Steel
0.04 0.04
Round rails w/ball bearings
0.002 0.002
Linear guides - radius groove
0.003 0.002
Linear guides - gothic arch (depends 0.008 0.004
on load, pre-load & type)
to .05 to .02
15
Appendix A
Introduction to Motor Sizing
BALL SCREW FORMULAS
Critical Speed
⎛ S ⋅ d × 10 6 ⎞
⎟⎟
N c = ⎜⎜
2
f
⋅
L
⎝ s
⎠
Nc
d
L
fs
S
=
=
=
=
=
=
=
=
critical speed (rpm)
root diameter of screw (cm)
distance between bearings (cm)
Safety factor (recommend 1.25)
end support conditions factor
4.22
fixed - free
18.85
fixed - simple
27.31
fixed – fixed
Note: The ‘S’ term carries units rev ⋅ cm /min. This
equation is based on the natural frequency of the rod,
and acquires its time component from that derivation.
=
10 6 inches
(Fm / dynamic load )3
For Equivalent Load
(
⎛ 14.03 × 10 6 ⋅ S ⋅ d 4 ⎞
⎟⎟
Fcl = ⎜⎜
L2
⎝
⎠
Fcl = maximum load (lb)
S = end fixity factor
= .250 one end fixed one free
= 1.0 both ends supported
(simple)
= 2.0 one fixed one simple
= 4.0 both fixed
d = root diameter of screw
L = distance between ball nut and
load-carrying bearing
Note: The ‘S’ term carries hidden units that make
the formula simpler and easier to use.
Backdriving Torque
(mainly used to determine holding brake torque)
Tb =
=
=
=
=
For Other then Rated (Dynamic) Load
10 6 inches
Life =
(operating load/dynamic load )3
3
3
Fm = 3 Y1 (F1 ) + …Yn (Fn )
Column Load
Tb
F
l
e
Life Expectancy
F ×l ×e
2π
torque required to backdrive
axial load
lead of screw
efficiency of screw
)
Fm = equivalent load
Fn = a particular increment of load
Yn = the portion of a cycle (sub
cycles) of a particular
increment of load expressed as
a decimal, i.e. the sum of the
sub cycles must equal one.
Example if L1 is applied for
20% of the cycle, L2 is
applied for 30% of the cycle
and L3 is applied for 50% of
the cycle, then the associated
Y values are Y1 =.2, Y2 =
.3, Y3 = .5.
16
X-Axis
Z-Axis
Completo
Complelo
Attempted (F3 of PIN) Altompled (S of P/N)
Name:
Grade:
Sketch of Layout
Linear Bearings Selected
Drive Mechanism Selected
Motor Sizing incl. amplifier
Motor Sizing RMS Calcs
Electrical Load Calcs
Coupler / Shaft Load Calcs
Sensor Selected
Accuracy Predicted
Name:
X-Axis Z-Axis
Complete
completo
Altempted (F8 or PIN) Altempted (FS of P/N)
Grade:
work
on
sensor
New Aplication.
company name fance Regestration
two surowes driver
moulon contray
1st
stuff
i put
3
$736"
1
lxu
vies
1x 6
319
meto
(nip rollers)
mechaum
can
do things for the board
36" up
The
door
board
The
Stor & stop
puł Pracket
2 exis Motor
A
{beteer corge control
I bp brushed
2
3
dc
Motor
hed}
motor
Z
محقر القاسمي المساهم
F
Two scrows
2.22
Travel
-0.75 travel
move it up & down
Et constant force
during travel
fix Time
o
when sensor red should move
słop.
1.2 to drive scrow
//
36
and
2
sohnall Goles en
e pois
Two scrows
2.22
Travel
-0.75 Travel
move ił up & down
-constant force
during cravel
TE
fix Time
w
1.2 o drive scrow
a
O
when sensor
red should move 36 and
1207
stop.
a accy + 1,5 mm
cycle time
5 second
every ss
diameter the roller > 2.5 in
Contact whole board
pushed withe 316
Need
Take care
of rooling
and
the tenilen of rolling
lingon of the board
72" - 40"
force 116
- bushed force jill knell is
bełt ween
rolling can bushed
2 board
من المرايا
To
(
the rolling

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