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    Cnc_polygon.svg    
    
    
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    CNC lets cut material directly from computer design file (dxf, stl, g-code ...)This ensures precision, reproducibility, shape-complexity and automation.    The 3-axis CNC can process:- 2.5D : xy-path at z constant- 3D: xyz-path (if ad-hoc reamer and particular path)    Cutting technology:- laser cutter (Only 2.5D: cutting and engraving)- water jet (Only 2.5D)- reamer / mechanical (2.5D and 3D depending on shape and reamer shape)    Minimal curve radius constraint:- laser and water-jet implies no specific constraint- For mechanical cutter, inner curve must have a curve radius bigger than the reamer radius.    
    
    
    
    curve (C)    point (A) of (C)    tangent (T) of (C) in (A)    osculating circle of (C) in (A)    curve radius (r) of (C) in (A)     
    
    
    
    
    So inner corner can not be cut with reamer.They must be replaced by inner curve.Tight inner curve must be smoothed to respect the minimal curve radius constraint.    
    outer corner    outer curve    inner curve    inner corner    
    
    
    
    2D path    Possible 2D shape with a reamer of radius (r)    
    
    reamer of radius (r)    
    
    
    
    
    
    
    If you want a perfect fitting between two coplanar shapes,then outer corners and outer curves must be rounded to get a minimum curve radius bigger than the reamer radius.For a perfect fitting, the coplanar shapes must be complementary.    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    If two 2D shapes are not coplanar, then outer corners can not be rounded by 2.5D cnc.    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    For fitting not coplanar shapes, we need to enlarge inner corners    
    
    
    
    
    
    
    
    
    
    
    
    
    ideal shape    
    
    
    
    required shape    
    details of smoothed corners (for fitting two coplanar shapes):    
    
    ideal path    
    reamer of radius (r)    
    
    
    
    
    
    
    
    final path    
    
    A    E    F    O    (D1), (D2) : two straight linesA : intersection of (D1) and (D2)(C) : circle or radius (r) tangent to (D1) and (D2)E : intersection of (C) and (D1)F : intersection of (C) and (D2)O : the center of (C)(EAF)=a is the angle between (D1) and (D2)    (D1)    (D2)    (C)    
        (C) is tangent to (D1), so (D1) is perpendicular to (EO)(C) is tangent to (D2), so (D2) is perpendicular to (FO)FO=EO=r, So O belongs to the bisector of (EAF).We have: AF=AE and (FA0)=(EAO)=a/2AEO is right triangle in E.tan(EAO) = OE/AEAE = r/tan(a/2)sin(EAO) = OE/AOAO = r/sin(a/2)    G    H    Knowing Gx,Gy,Ax,Ay,Hx,Hy, we want to calculate: a.(xAG) = atan((Gy-Ay)/(Gx-Ax))(xAH) = atan((Hy-Ay)/(Hx-Ax))a=(EAF)=(GAH)=(xAH)-(aAG)a=atan((Hy-Ay)/(Hx-Ax))-atan((Gy-Ay)/(Gx-Ax))Other method with the law of cosines c²=a²+b²-2*a*b*cos(C)In the triangle GHA:h=AG=sqrt((Gx-Ax)²+(Gy-Ay)²)g=AH=sqrt((Hx-Ax)²+(Hy-Ay)²)a=GH=sqrt((Hx-Gx)²+(Hy-Gy)²)a=(GAH)=acos((h²+g²-a²)/(2*g*h))    
    x    Knowing Gx,Gy,Ax,Ay,Hx,Hy,a we want to calculate: Ex,Ey,Fx,Fy.Ex=Ax+(Gx-Ax)*AE/AG=Ax+(Gx-Ax)*r/(tan(a/2)*sqrt((Gx-Ax)²+(Gy-Ay)²))    
    
    
    
    
    
    
    
    A    E    F    O    G    H    
    x    I    (C)    I is the intersection of (C) and (AO)(D3) is the straight line perpendicular to (AO) and including I.K is the intersection of (D3) and (D1)L is the intersection of (D3) and (D1)    
    
    L    K    
    A    I    
    
    L    K    
    (D3)    
    
    (D1)    (D2)    J    The triangles KAI and IAL are similar so AL=AK.(LAI)=(IAK)=a/2AI=AO-IO=r/sin(a/2)-r=r*(1-sin(a/2))/sin(a/2)AK=AI/cos(a/2)=r*(1-sin(a/2))/(sin(a/2)*cos(a/2))=r*(1-sin(a/2))*2/sin(a)AJ=AK+AL=(AI+IL)+(AI+IK)=2*AIAI=(AK+AL)/2Kx=Ax+(Gx-Ax)*AK/AG    Knowing Gx,Gy,Ax,Ay,Hx,Hy,a we want to calculate: Ix, Iy.    With E,I and F, we define the arc than can be build with a reamer of radius r.    details of enlarged corners (for fitting two not-coplanar shapes):    
    
    
    
    
    
    
    
    
    
    a<pi/2    A    E    F    A    E    F    A    E    F    
    
    a>pi/2    a=pi/2    (C1)    O    O    O    P    K    L    Let's consider three points A, G and H.(D1) is the bisector of (GAH).O is a point of (D1) such as AO=r(C1) is the circle of center O and radius rE is the intersection of (C1) and (AG)F is the intersection of (C1) and (AH)(D2) is the straight line perpendicular to (D1) and including OK and L are the intersection of (D2) with (C1)     (D1)        G    H    Let's calculate AE:OA=OE=r.We define I, the orthogonal projection of O on (AE).AI=EI because AEO is isosceles in O.AI=AO/cos(a/2)=r*cos(a/2)AE=2*r*cos(a/2)    
    
    
    
    
    
    
    
    
    
    M    N    (C1)    (C2)    K    L    (D2)            (D2)    (D3)    (D3) is the straight line perpendicular to (D1) and such that the length MN is equal to 2*r with M the intersection of (D3) and (AG) and N the intersection of (D3) and (AH).P is the intersection of (D3) and (D1).    (D1)        G    G    H    H    AM=r/sin(a/2)    
    
    
    
    
    A    O    P    K    L    
    
    
    
    
    
    M    N    (D2)    (D3)    (D1)    G    H    
    
    
    
    
    R    S    T    U    
    V    
    W    R is the middle of [AM]S is the middle of [AN]V is the intersection of (D2) and (AH)W is the intersection of (D) and (AG)    AK=AR-AS+(AV+AW)/2AR=AS=r/(2*sin(a/2))AV=AW=r/cos(a/2)    
    
    
    
    
    3-axis CNC    
    
    Osculating circle    
    
    Coplanar fitting    
    Not coplanar shapes    
    Not coplanar fitting    
    Detailed coplanar fitting    
    Arc third point for coplanar fitting    
    Detailed arc third point for coplanar fitting    
    Not coplanar fitting with obtuse angle    
    Not coplanar fitting with right angle    
    Not coplanar fitting with acute angle    
    Detailed not coplanar fitting with acute angle  
