nc and cnc
TRANSCRIPT
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Continuous-path
The continuous-path control system is also
known as the contouring system. It involves
simultaneous motion control of two or more
axes. The contouring system is more
complex because each axis of motionrequires separate position and velocity
loops. The contouring along a predefined
tool path is implemented by means of
interpolation, in which the systemgenerates a set of intermediate data points
between given coordinate positions
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CNC Interpolation
An interpolator provides two functions:
• It computes individual axis velocities to drive the tool along theprogrammed path at the given feed rate.
• It generates intermediate coordinate positions along the
programmed path. There are five types of interpolation: linear,
circular, helical, parabolic, and cubic.
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Circular Interpolation
The interpolator computes the axial velocity components and
produces a sequence of reference pulses for each control axisof motion. The advantage of circular interpolation is its ability
to generate an arc in a single program block. In some NC
controls, circular interpolation is limited to a 90°arc in a single
block.
The information required for programming a circular
interpolation includes:
(1) coordinates of the start point and end point,(2) radius of the arc or coordinates of the arc centre, and
(3) direction in which the tool is to proceed (CW or CCW).
The circular interpolation is limited to the two-axis plane.
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Helical Interpolation
It combines the two-axis circular interpolation
and a linear interpolation along the third axis.
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Parabolic Interpolation
Parabolic interpolation uses three noncollinear points to
approximate free-form curves.
It can be used to cut either planar or spatial curves. It isprimarily used in mold and die making, where free-form
designs are preferred over precisely defined shapes.
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Cubic Interpolation
It is of the third order and can be used to
generate complex tool path for machining
complicated shapes such as automobile sheetmetal dies with a relatively small number of
programmed points. However, it is very
complex and requires considerable computing
power and a large memory.
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• , , . " ( )
α .
.
(). .
(/).
sn360=α
sn
pm n A =
m A pn
Thisimagecann otcurrently bedisplayed.
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The motor shaft is generally connected to the lead screw through a
gearbox, so:
: angle of lead screw rotation (degrees).
: Gear ratio.
: Rotational speed of motor (rpm).
N : Rotational speed lead screw (rpm).
N
N
A
A
r
mm
g ==
g
p
r
n A =
A
gr m N
Where
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Linear movement of work table is given by:
: X-axis position relative to the starting position (mm, in).: Pitch of lead screw (mm/rev, in/rev).
: Number of lead screw revolutions.
The number of pulses required to achieve a specified x-position
increment is given by (using the preceding relationships):
360
pA x =
x p
360
A
p
xr Ar An
ggm p
360
α α α
===
p
n xr n
sg
p =
p
r xn
p
r xn
gsg
p ==
α
360
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Control pulses are generated at a certain frequency, which drives
the worktable, where:
: Rotational speed of lead screw (rpm).
: Pulses train frequency ( Hz, pulses/s).
: Steps per revolution or pulses per revolution.
: Gear ratio .
gs
p
r n f N 60=
N
p f
sn
gr
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The work table travel speed in the direction of lead screw axis is:
: (mm/min, in /min)
: Table feed rate (mm/min, in /min).
: lead screw pitch (mm/rev, in/rev).
Np f vr t
==
pr f t v
gs
p
r n
f N 60=
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•
pr nv f gst
p
60=
pr n f gsr
60=
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•
6 .
5 1
48 .
250 500 / .
Motor lead screw
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•
)
)
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)
.
°=== 150006
250*360360
p
x A
puslesr A
p
r n
n
gg
p
s
100005.7
)5)(15000(**360
5.748
360360
=
°
===
°==°
=
α α
α
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)
Hz pr nv
f
RPM N N
N r
RPM p
v N
gst p
mm
g
t
333.333)6(60
5)48(500
60
7.416333.83*5
333.836
500
===
==
===
→=
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• .
•
. ,
α : (deg/slot) ns : number of slots in disk
n p
: # of pulses sensed by the encoder &emitted
A : angle of rotation of the encoder shaftα
α
An
n
p
s
=
=
360
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The pulse train generated by the encoder is compared with position and
feed rate specified in the pant program , and the difference is used by the
MCU to drive a servomotor , which drives the worktable .
Closed-loop NC is good for milling and turning because of the reactionary
force that resists the movement of the table .
s
pr t
s
p
n
f p f v
n
n p x
**60
*
==
= x : worktable pos.
p : lead screw pitch(mm/rev)
vt : worktable velocity (mm/min)
fr : feederate (mm/min)
fp : frequency of the pulse train
(Hz,pulse / sec )
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An NC work table operates by closed-loop positioning system consisting of
a servo-motor, lead screw and optical encoder. The lead screw has a pitch
= 6 mm and is coupled to the motor shaft with a gear ratio 5:1 (5 turns ofthe drive motor for each turn of lead screw). The optical encoder generates
48 pulses/rev. of its output shaft. The encoder output shaft is coupled to the
lead screw with a 4:1 reduction (4 turns of the encoder shaft for each turn
of the lead screw). The table has been programmed to move a distance
300 mm at a feed rate = 600 mm/min.
Determine i) how many pulses should be received by the control system
to verify that the table has moved 300 mmii) the pulse rate of the encoder and
iii) the drive motor speed that corresponds to the specified feed rate.
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122
*
1
−=
=
B
gs
LCR
r n
pCR CR1 : Electromechanical (mm)
CR2 : Computer control system (mm)
L : Axis range (mm)
B : Number of bits in the devoted bit storage register
Both equations can be used for open or closed loop.
},max{ 21 CRCRCR =
Standard-deviation
* Typical value of CR is 0.0025
mm
→+= σ 3
2
CR Accuracy
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