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Formulas + Calculations for Optimum Selection of a Stepmotor 1.1. Determination of Load Torque MLoad the required drive torque of a Spindle Drive is determined by the sum of the load torques and the required acceleration torque.
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0% found this document useful (0 votes)
500 views32 pagesSelection PaP1 1
Formulas + Calculations for Optimum Selection of a Stepmotor 1.1. Determination of Load Torque MLoad the required drive torque of a Spindle Drive is determined by the sum of the load torques and the required acceleration torque.
Uploaded by
RofochoCopyright
© Attribution Non-Commercial (BY-NC)
We take content rights seriously. If you suspect this is your content, claim it here.
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/ 32
berger berger
motoren elecGronii
Formulas + Calculations
for Optimum Selection of a Stepmotor
yy es
peacen tare
BERGERLELE 1
1
4aa
o
(2)
12
General Formulas
Vismire.
sme
Spindie Drive
Determination of Load Torque M, og
FE ra
Frapie + Fworingtoas * Ferestroce + FetovingForce
‘The required drive torque of a spindle drive is determined by the sum of the load torques and
the required acceleration torque.
Mucor = Miows + Mecsas (Nem
Moa =F (ata + ils [Nem]
Total force on nut IN]
Spindle pitch fom]
Spindle bearing mean radius [om]
Spindle bearing friction coefficient
Gear ratio = Motor! Meena
ee
tana |i i
1 = alee = Efficiency costficient of converting M into an axial force
Experience Values
1 = 0.9for bail bearing spindles (see figure}
= 0.3forsteel spindles with bronze nut
fp-Hp = 0.016 cmtor roller bearing
fq lg = 0.15 .om for steel/bronze friction bearings
Frrastroes
At10% Prestress andh= Smm: approx. 11to 15N
At 20% Prestressandh= 5mm: approx.22to 30N
At 10% Prestress andh= 10mm: approx. 40to 60N
At 20% Prestrass andh = 10 mm: approx. 80 to 120 N @
Determination of Total Load F on Spindte Nut
a) Vertically Acting Forces:
Fay G+Fy+Fre
IN)
G = Weight of carriage and structure (kal
F, = Moving Force IN}
Fre = Prestress when using springloaded counter nut
" Friction coefficient
Gos 10m N
m = Mass [kg]
Values for u Dry Lubricated
Steel on steel 0,18 0,12
Steel oncastiron oi 040
Steal on bronze ont 0.10
Axial Guide
Rolling friction, Rollers - 0,005 4)
(4ay
(4b)
(4c)
b) ,,Non-vertical” Spindle
Fa Feet FutFe 2 ™
F, = Gina +4 cose) ifmoving up
F, = G(sing —p-cosa) ifmoving down
For upward mation
+Fy+G (sin@ +4 -cose) N)
NI 14
(5)
a)
©
m™
(Tay
8
(9)
Determination of Moments of Inertia
The total moment of inertia Jye. ig the sum of the moments of inertia of all masses in
rotatory and translatory motion.
at = Stor Stans tkgem?}
Jen = Totalexternal J referenced to motor shaft
Jit = Rotatory moment of inertia (full cylinder)
rans. = Translatory moment of inertia.
Jor 7 Soa + Sat Ikgom"}
2) Rotatory Moment of Inertia J,..~Full Cylinder
eon? Het TL Ikgom4]
r= Radiusofspindle cm
L = Length cm
¥ = Specificweight — kg/em?
Stee! 7 = 7,85: 10 kg/em?
Aluminum y = 2,7» 105kg/em?
Brass 7 = 8,4 - 10-3kg/om?
For Steel
[kgern?]
For Aluminum
Yor = 2.7 10 aL | kgoma
b) Translatory Moment of Inertia Jans
Janse (32)? tkgom}
m = Moved mass in kg
h = Spindte pitch inom
ta reduction gear is used, the external moment of inertia J,x is reduced by the square
of the gear ratio.
Yon = rar tSians? Ikgom?]
Ez 1.2
1.24
1254
09)
6)
(20)
4.242
(18)
(2)
(22)
Rack and Pinion Drive
Horizontally moved mass driven by rack and pinion.
=~+_ >, Rack
Pinion
Stepmot
Total Torques
‘The motor must provide the following total torques:
a) Acceleration of weight G, including rack
b) Acceleration of pinion
} Acceleration of rotor
d) Overcoming the friction
Moments of inertia
The following formula is used to calculate the rotatory moment of inertia equivalent to the weigt
Jeg = Gt? [kgom2]
Weight in kg
r= Radiysincm
Jy = Ferree Ley tkgem’] Soy
door log + Yap tq, kgem?]
Moy = Me M, [Nom]
Moos Guor [Nom] 10
1.3
134
1344
19)
©
134.2
(18)
(23)
(2y
Drum Drive
Lifting a Waight by Cabl
Total Torques
a} Acceleration of the weight
) Acceleration of the drum
¢} Lifting of the Weight G
Cable Drum Stepmotor
Moments of inertia
eq = Equivatent moment of inertia
‘The same formulas as used for the rack and pinion drive apply.
Jeg = Ger? ikgem?}
Youn = okey tego
Acceleration and Load Torques,
Mos (Nom)
M, = Torque required for accelerating the system.
Mo = Gr [Nom]
M, = Torque required for lifting the weight
Mor = Ma+M, [Nem]
Mg: = Total torque required tor lifting the weight 14 Additional Formulas
1.4.1 Start-Stop Operation
{10)
{102) foun ~ 199 « Ve E]
vi. = Total Moment of Inertia {kgom?]
consisting of Je, + Unter
f=
step frequency {Hz]
step angle (Degrees)
a
T,_ = Division angle of rotor teeth
(for S-phase motors: 7.2")
Ma = My Ger (Nem Drum drive
;
g
a
2
5 1.4.4 Advance, Speed and Power
Distance Increment
cr) as=zy | tom
h Spindle pitch in cm
Z_ = Number of steps per revolution
i = Gearratio
‘Advance Speed
(13) veds-tot¢ lem/s] (for spindle dives)
zi = te ene
{= Step frequency [s*]
wiZei = in
(ay ft = 22) sq for rotatory drives)
‘Advance ina defined time t
(14) sads-ft fem)
t = timeins
co) [min
1 ol
Powerfor rotation
(16) P =0,00105 M-n mM
Min Nem
in min
Pos 1g time for short distances
ea | aye S% | oy
Positioning frequency for short distances
Tay, 44, = Positioning time (s]
5) | at, - ae
{26) eh fea 43, = Positioning distance [or
41, = Max.positoning trea. [s
Suuon™ Acosleration distance [
= Senmme 48 | for
(28) || Sener 7 Sante (om) 8, = Acceleration distance ir 145
ay
Determination of the Moment of Inertia J for Arbitary Bodies by
Means of Measurement
Procedure,
‘The body is freely suspended by two strings attached to two fixes
paints. Then itis brought into rotational oscillation about the
‘skotched center ine, Tha moment of inertia is found using the
previously determined mass m of the body and the distances a,b
and hin the following formuta.
J=25-T2m loom}
J = Moment of inertia in gem?
T = Duration of period in sec.
Fixeaponts
Swings
b= Distances in om 14
1.4.6 Howto Find the Optimum Stepmotor
Aids: 1. “Formulasand Calculations for Optimum Selection of a Stepmotor” (this brochure)
2. BERGER Catalog “5-Phase Stepmotor System”
Determi
ion of motor
External moment of Inertia known?
ox fkgem?}
No
oo
Use formulas
5,6, 7, 8, 9,19, 20
Trial selection of motor from Catalog No. 250/...
"5-Phase Stepmotor System”
Max. allowable external moments of inertia,
/,. Terque [Nem] of the machine tobe driven known?
Myotr= Mood + Macon: Nom]
Use formulas
1,2,3, 4,10
M [em]
required values
Use formulas 10-16
Trial selection of motor
compatible with final
selection? 24
244
24.2
(2)
@
Calculation Examples
Calculation: Horizontal Spindle Drive
SS
LE LLLL LTT tae
ES
ez
Known and Required Values
Known Values Required Values
G=1000N Voue= 12 m/min =20 m/s.
F, =250N Positioning time for 10mm=0.58
B04 Resolution: 0.01 mm
9 =09
Spindle diameter 35 mm
Spindlelength =~ 800mm
Spindiepitchh = = Ss mm
Traveldistance 700 mm
Sought: the correct stepmotor
Required Torque
1
mer(ctoeen) } pen
Fo=WGth + Fre WN]
F 0,1: 1000N+250N=350N
95cm
2-314 08
M, = 36 Nem .
ma.=a50n( +0018 om) =96tlem
‘The calculation example doos not include gear ratio ( ) and Fora 213
©)
(6)
(8)
6)
244
1a)
245
6
Existing Moments of Inertia
Jen rot Jtrans [kgm
dot 0,5 t-18- Le 7 Ekgom?]
7,88 ka
Jrot = 05° 3,14 - (1,75 cm)" - 80cm
ag = 9,25 kgem?
10°
ne
smn (fs) em
2-314
oars = 0,83 kgem?
trans = 100 kg ( ) Ikgom?]
Yea = deot + dtrans
Jou — 9,25 kgom? + 0,63 kgom? = 9,88 kgcm? = 10 kgom?
ya = 10 kgom?
Required Operating Frequency
vie fomis}
te = _20cm/s-500Steps/Rev. fg
05 om
1 = 200008"
Values Determining the Motor Size
1M. = 36.Nem
2 Jen = 10 kgem?
3.1 = 20000Hz 24.6 Determination of Motor Size
5-Phase Stepmotor Data Overview
‘The motor data are contained
in our Catalog No. 2507...
»5-Phase Stepmotor System«
in the following arrangement.
“Size 110"
MOTOR MODEL > ROM 5117/50 ROM 5122/50
‘Step angle [Degrees] O72 O72
Maximum torque [Nem] 700 1000
Holding Torque, excited [Nem] 750 1100
Max. Power (W] ate HZ Fig ge
Moment of inertia of rotor [kgom?) 115
Motor Characteristics (Constant-Current Operation / Standard Winding)
Motor Model RDM 51117/50
Ry=030 ly=5A
ne ar’ T
Torque (Noa)
3
Constant Curent
5A/Phase
100
$iat-siop|
00 5001000)
a8
ff 1 (Stepar
1200 6000 12000
Irn“
2 oo 20 Maximum Permissible External Moments of Inertia
Motors of Series 511 ../50
,
‘ent ROM SHI7/50
T= 8AiPhase
eo
»
Fulstep0.72
©
cI
:
|
|
»
®
rn in
i
om ®
‘Motor size determined form curves: RDM 51117/50
Motor data see Catalog 250 "5-Phase Stepmotor Syster
7,5 kgom?
Myr (2t 20 kHz) —M,_INem]
150 Nem=36 Nem=114 Nem
zee
i
nud 247
(6a)
mn)
21.8
en
21.9
Determination of Acceleration Time.
Forlinear acceleration and deceleration, the acceleration and deceleration times are equal,
Jnct* Yana + Sot = 17.5 kgom?
Qr-asf
i. To 560° MM, Tor |S
4 -0,72° - 20000 Hz
‘360° - 4 Nem « 107
= 17.5 kgom?
t, 7 0986s ~ 0295
ty = Time for acceleration
tg ~ Time for constant speed
fp = Time for deceleration
ty > Total travel time
1 pal
20000 Fv = =~
Steps for acceleration
‘Steps for constant speed
Steps for deceleration
fear tel rat f
ree Ete
Sn
‘ (wed '
Distance Traveled in Total Travel Time
ft,
> Distance in steps)
8,= 20000 0,39 § _ 3900 Steps
a2
During acceleration phase = 9800 Steps =
During deceleration phase = 9900 Steps = sp
‘Sum for acceleration and deceleration = 7800 Steps
Total distance traveled s,..= 700 mm 70 000 Steps
Total Travel Time
fe = tet theth fs)
= Sia 7 (Sa + 8g)
t ' {s]
1. — 70000 = (3900 + 3800
. 20000
= atts -
htt +t, Is)
ty, = 0,99 + 3.11 + 0.99
tye = 3,808 24.10. Verification of Required Values
2.10.4 Positioning Time tps]
St, required0.5 8
Sot" 49 te
Sa toe
(24) At, {s)
4s, = 10mm = 10m
28) Sato = 5, [Steps] 48 = 54-4
Sate = 9800 Steps gpe'SOT-~ = 990m
bt, = O898-1oM _ org
pS "39 om
2110.2 Smallest Distance Increment per Step
48 required0.01 mm
12) As = > tem
4s = 08
BOGET = 0.001 cm = 0.01 mm {|
\
2.2
224
2.2.2
(2)
(de)
Calculation: Vertical Spindle Drive (Lifting Force)
Known and Required Values.
32 Known Values
A G = 150kg
EXZZZZ722TRA DPIC 7 209
Ea .
EA w= O4
EA Spindle diameter 63mm
EA length 10m
A pitch 10mm
Ea
A
Gearratio i=20:1
Required Values.
Positioning time 10mmin 1 ¢
Resolution < 0.01 mm
Stepmotor
Required Torque
= 1
MOTE (atatnt) + Nem
F = G+(sing+u-cos@) IN]
90°
1
cose = 0
ata
sin « 22
22.3
@)
(6)
®
Fo = 1500N (1 + 01-0)
Fo = 1500N
tom 4
My s800N (72% + 9) 1 (Nom)
M. = 1500N (0,177 + 0,015) 3 = 14,4 Nom
Existing Moments of Inertia
sn + (4a *San)
Sot = O,5