Computer Aided Engineering Design
DIFFERENTIAL GEOMETRY OF CURVES 83 Figure P3.2 a b c Q C B P A O ρ ρ ρ ρ ρ Plot the curvature and torsion of the Viviani’s curv ...
Chapter 4 Design of Curves Shapes are created by curves in that a surface, such as the rooftop of a car, the fuselage of an airc ...
DESIGN OF CURVES 85 slopes via interpolation or curve fitting. Since the aim is to use low order parametric segments, two models ...
86 COMPUTER AIDED ENGINEERING DESIGN u dd d u j j 2 12 =1 = 3 3 + = 3.61 + 4.47 10.91 = 0.74, = 1 Σ Eq. (4.2) for the four data ...
DESIGN OF CURVES 87 4.1 Ferguson’s or Hermite Cubic Segments A cubic Ferguson’s segment is designed like an arch with two end po ...
88 COMPUTER AIDED ENGINEERING DESIGN Similar treatment can be employed for y(u) and z(u) starting with the cubic form in Eq. (4. ...
DESIGN OF CURVES 89 From Eq. (4.9), T(u) = ru(u) = UM 1 G with Ti = citi and Ti+1 = ci+1ti+1. Therefore T( ) = [ 1] 0000 6–63 3 ...
90 COMPUTER AIDED ENGINEERING DESIGN respective independent parameters, that is, a point on the composite curve belongs to r(1)( ...
DESIGN OF CURVES 91 (pi,qi,ri) and (pi+1,qi+1,ri+1). Also, r(2) (u 2 ) is modeled with end points (xi+1,yi+1,zi+1), (xi+2,yi+2,z ...
92 COMPUTER AIDED ENGINEERING DESIGN wheren + 1 is the number of data points. Eq. (4.17) suggests that for a cubic composite cur ...
DESIGN OF CURVES 93 r 33 ( ) = [ 33 32 1] 33 2–21 1 –3 3 –2 –1 0010 1000 32 6–1 10 1–1 uuuu , 0 1u ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ...
94 COMPUTER AIDED ENGINEERING DESIGN To obtain C^2 continuous curve, Eq. (4.17) may be employed to get the intermediate slopes. ...
DESIGN OF CURVES 95 To maintain the same parametric (cubic) form, a linear relation between u and v is sought. An advantage is t ...
96 COMPUTER AIDED ENGINEERING DESIGN Also d d d du du d uu du du ji uu ujiu rv rv rv r r () = () () = ( – ) () = ( – ) ( ) v v v ...
DESIGN OF CURVES 97 (12, 6) and the end tangents are given by Ti+2 = (–8, 5) and Ti+3 = (7, –7). Blend a curve between B andC to ...
98 COMPUTER AIDED ENGINEERING DESIGN r(u) = (1 – 3u^2 + 2u^3 )Pi + (3u^2 – 2u^3 )Pi+1 + (u – 2u^2 + u^3 )Ti + (–u^2 + u^3 )Ti+1 ...
DESIGN OF CURVES 99 In similar triangles EBD and CBA, since D is the mid point of AB, the ratio BE/BD = BC/BA =^12 which implies ...
100 COMPUTER AIDED ENGINEERING DESIGN Note that for α =^12 , the coefficient of u^3 is zero and r(u) is of degree 2. In fact, we ...
DESIGN OF CURVES 101 andc lie on the parabola as well. To find the tangents at a and c, Eq. (4.27) may be differentiated with re ...
102 COMPUTER AIDED ENGINEERING DESIGN j= 1, i = 0; bbb bb 01 = (1 – )uu 00 + 10 = (1 – )uu 01 + j = 1, i = 1: bbb bb 11 = (1 – ) ...
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