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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

      mm

    g  ==

    g

     p

    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

    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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