linear applications

7/28/2019 Linear Applications

1/14

Trapezoidal

Move Distance x = 100 mm

Max Velocity v = 400 mm/sec

Acceleration rate a r= 0.200 G

Deceleration rate d r= 0.500 G

Acc. Due to Gravity G G = 9.8E+03 mm/s2

Acceleration a = 2.0E+03

Deceleration d = 4.9E+03

Accel Time T1 a t = 0.204

Decel Time T3 d t = 0.082

Accel Distance a x = 40.789

Decel Distance d x = 16.315

Const Vel Distance c x = 42.896

Const Vel Time T2 c t = 0.107

Move Time TTOTAL t = 0.393

Time Distance Velocity

0 0 0

0.204 40.789 400

0.311 83.685 400

0.393 100 0

mm/s2 = a r G [Eqn. 1]

mm/s2 = d r G [Eqn. 2]

sec = v

a

[Eqn. 3]

sec = v

d

[Eqn. 4]

mm = v2

2 a

[Eqn. 5]

mm = v2

2 d

[Eqn. 6]

mm = x - (a x + d x) [Eqn. 7]

sec = c x

v

[Eqn. 8]

sec = a t + d t + c t [Eqn. 9]

Most nutop spee

There ar(a), and ocombinati

As an extop speedeceleratimm. Simisubtractinvelocity.mm, andthe veloci

0

50

100

150

200

250

300

350

400

450

0 20 40 60 80 100 120

mm/s

mm

Distance

450

Time
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2/14

0

50

100

150

200

250

300

350

400

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45

mm/s

Seconds


3/14

eric questions about positioning applications boil down to one of two types: How long does it take to g; at time t; point x; etc.)? The next question is what is the required torque to perform the profile.

four physical quantities associated with motion calculations: position (x); its first derivative, velocity (v);f course, time (t). These quantities are related by several familiar equations, which can in turn be expaion. The full set of possible combinations follows.

mple, consider the simple move profile of Figure 20. A positioning table makes a move of 100 mm (0.1of 400 mm/sec (0.4 m/sec), and then decelerating at 0.5 G. Since one G is 9.8 m/s2, the acceleratio

ion occurs at 4.9 m/s2. Using Eqn. 8 (t=v/a), the accel time is found to be 0.204 second; using Eqn. 10,ilarly, the decel time and distance are found to be 0.082 second and 16.3 mm, respectively. The constag the sum of the accel and decel distances from the overall move size of 100 mm, leading to a 42.9 msing Eqn. 15, this phase has a duration of 0.107 second, for an overall move time of 0.393 second. If t

the acceleration and deceleration values were unchanged, then the stage would not be capable of reacity profile would be triangular, rather than the trapezoidal shape shown.


4/14


5/14

et there? and How fast am I going (at

; its second derivative, accelerationded by substitution to cover any

meter), accelerating at 0.2 G to aoccurs at 1.96 m/s2, and the

the accel distance is found to be 40.8nt velocity distance is found by

length of the move at constanthe move distance was less than 57.1hing the 400 mm/sec top speed, and


6/14

Triangular

x = 100 mm

t = 0.393 sec

v average(sec) = 254.4529

v average(min) = 15267.18

v peak(sec) = 508.9059

v peak(min) = 30534.35

mm/sec = x

t

[Eqn. 1]

mm/min = 60 vaverage(sec) [Eqn. 2]

mm/sec = 2 vaverage(sec) [Eqn. 3]

mm/min = 2 vaverage(min) [Eqn. 4]
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7/14

Time Distance Velocity verage Vel

0 0 0 254.4529

0.1965 50 508.9059 254.4529

0.393 100 0 254.4529

0

100

200

300

400

500

600

0 20 40 60 80 100 120

mm/s

mm

Distance

0

100

200

300

400

500

600

0 0.1 0.2 0.3 0.4 0.5

mm/s

Seconds

Time


8/14

Mass m = 0 kg

Linear velocity v = 20 mm/sec

Lead of screw L = 4.234 mm/rev

Lead screw efficiency h s = 78%

Lead screw friction coefficient mo = 0

Sliding surface friction coefficient m = 0

External force FA = 800 N

Lead screw tilt angle a DEG = 0

a = 0 Radians

Force F = 800

Load Torque TL = 691.1405

Gear ratio i = 9

Gear efficiency hG= 81%

Motor torque TM = 94.80666

Motor speed SM = 2550.779

Speed at output shaft NG = 283.4199

N = FA + m (sin[a] + m cos[a]) [Eqn. 1]

mNm = F L

2 p h s

+ mo L

2 p

[Eqn. 2]

mNm = TL

i hG

[Eqn. 3]

rpm = NG i [Eqn. 4]

rpm = v 60

L

[Eqn. 5]
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9/14

Mass m = 0.5 kg

Acceleration a = 1.961 m/s

Coefficient of friction m = 0

Lead of screw L = 4.234 mm/rev

Efficiency of screw h s = 78%

Acc. Force = 0.9805

Acc. Torque = 0.847079

]

N = m a [Eqn. 6]

mNm = [Acc. Force + m m] L

2 p h s

[Eqn. 7]
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10/14

Minimum Fixity

Dmin = 21.336 mm

SL = 6.35 mm

L = 1051.214 mm

Fs = 80%

K = 1.21E+08Fixed-Free

Simple-Simple

Fixed-Simple

Fixed-Fixed

Cssec = 254 mm/sec 0.36 1 1.47 2.23

Csmin = 15240 mm/min

Fe = 1.29E+00

Angular velocity 2400

Dmin = Root diameter

SL = LeadL = Length of screw between supports

Fs = Safety Factor

K = Constant

Cs = Critical Speed (linear)

Fe = End Fixity

= Csmin L2

K Fs Dmin SL

[Eqn. 1]

rpm = 1

SL

Csmin [Eqn. 3]
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11/14

Dmin = 21.336 mm

SL = 6.35 mm

L = 1050.214 mm

Fe = 1.47 Fixed-Simple

Fs = 80%

K = 1.21E+08

Csmin = 17465.42 mm/min

Cssec = 291.0904 mm/sec

Angular velocity 2750.46

= Fe K Fs Dmin SL

L2

[Eqn. 2]

= Csmin

60

[Eqn. 4]

rpm = 1

SL

Csmin [Eqn. 5]


12/14

Minimum Fixity

Dmin = 21.336 mm Dmin =

L = 1050.2138 mm L =

Fs = 80% Fe =

K = 9.864E+03 Fs =

Pc = 2965.327 kgf K =Pc = 29079.924 N Pc =

Fe = 2.00 Pc =

Dmin =

L =

Fe =

Fixed-

Free

Simple-

Simple

Fixed-

Simple

Fixed-

Fixed Fs =

0.25 1 2 4 K =

Pc =

Dmin = Root diamter

L = Length of screw between supports

Fe = End Fixity

Fs = Safety Factor

K = Constant

Pc = Cloumn strength of screw

= Pc / K Fs Dmin4

L2

[Eqn. 1]
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13/14

21.336 mm

1050.2138 mm

2.00 Fixed-Simple

80%

9.864E+032965.327 kgf

29079.924 N

0.84 in

41.347 in

2.00 Fixed-Simple

80%

1.403E+07

6537.42606 lbs = Fe K Fs Dmin4

L2

[Eqn. 2]
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14/14

LO = 6.5 lbs

NDLC = 2500 lbs

K = 1250 psi

P = 3.25 psi

DO = 1 in

SO = 381.9719 rpm

V = 100 ft/min

PV= 325

LO =Actual operating load

NDLC = Nut dynamic load capacity

K = 1250 psi

P = Pressure

DO = Outside diameter of the screw

SO = Operating speed

V = Velocity

PV= PV value

= LO

NDLC

K [Eqn. 1]

= DO p SO

12

[Eqn. 2]

= P V [Eqn. 3]

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