Computer Aided Engineering Design
DESIGN OF CURVES 123 [x(t),y(t),z(t)] is termed as the rational Bézier segment. Note that Br tin() are barycentric, that is, for ...
124 COMPUTER AIDED ENGINEERING DESIGN P PPP () = (1 – ) + 2 (1 – ) + (1 – ) + 2 (1 – ) + (^20122) t 22 twttt twttt (4.67) wherew ...
DESIGN OF CURVES 125 |OD | = | OP 0 | sin θ and so | | = | | – | | = | | 1 sin DP 11 OP OD OP 0 – sin θ ⎛ θ ⎝ ⎞ ⎠ Also, |DP | = ...
126 COMPUTER AIDED ENGINEERING DESIGN Figure 4.24 Circular arcs designed with rational Bézier segments for different positions o ...
DESIGN OF CURVES 127 Exercises Consider a parametric cubic curve r(u) where rPPPP( ) = uF F F 10 + 21 + 30 ′′ + F 41 0 ≤u≤ 1 w ...
128 COMPUTER AIDED ENGINEERING DESIGN (b) If the end tangents have the magnitudes as α and β, show some results of the variation ...
DESIGN OF CURVES 129 r P P P 1 2 222 111 000 0 1 2 ( ) = [uuu 1] abc abc abc ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ r P P P 2 2 ...
Chapter 5 Splines A natural way of designing a curve (or a surface) is to first sketch a general contour of the curve (or surfac ...
SPLINES 131 segment between two consecutive ducks to be a thin simply supported beam across which the bending moment varies line ...
132 COMPUTER AIDED ENGINEERING DESIGN 5.2 Why Splines? The motivation is to develop Bernstein polynomials like basis functions Ψ ...
SPLINES 133 be known. For the second span, t 1 ≤t≤t 2 ,Φ 1 (t 1 ) = y 1 and Φ 1 (t 2 ) = y 2 are known. The remaining two condit ...
134 COMPUTER AIDED ENGINEERING DESIGN Φi′′ i i i i i i i i ii i i i i ii i ii i i ii i i i tt ttt h ttt h tt h tt h tt h tt h tt ...
SPLINES 135 (ii) Built-in(clamped) end: Where the first derivatives at t 0 and tn are specified as Φ 00 ′()t = g 0 or Φn′–1() = ...
136 COMPUTER AIDED ENGINEERING DESIGN 5.4 B-Splines (Basis-Splines) In Section 5.3, generation of a cubic polynomial spline Φ(t) ...
SPLINES 137 Φ 0 (t) = a 0 + a 1 t + a 2 t^2 + a 3 t^3 Noting that Φi(0) = ΦΦii′′(0) = (0) = 0,a 0 = a 1 = a 2 = 0 so that Φ 0 (t ...
138 COMPUTER AIDED ENGINEERING DESIGN Mm,i(t) ti–4 ti–3 ti–2 ti–1 ti t Figure 5.6 Schematic of a B-spline basis function of orde ...
SPLINES 139 y 2 =pn–1(x 2 ) = α 0 + α 1 (x 2 – x 0 ) + α 2 (x 2 – x 0 ) (x 2 – x 1 ) ⇒ α 2 20 21 21 10 10 =^1 (– ) xx – ...
140 COMPUTER AIDED ENGINEERING DESIGN some intermediate point on the curve, the first divided differences y[xs+1,xs+ 2 ] = yy xx ...
SPLINES 141 Fork= 2, y[xj, xj+ 1 , xj+ 2 ]= () () () +1 +1 +2 +2 y wx y wx y wx j j j j j ′′ ′j Here,w(x) = (x–xj)(x–xj+ 1 ) ...
142 COMPUTER AIDED ENGINEERING DESIGN Note that f(t) = f′(t) = f′′(t) =... = fm–2(t) = 0 at t = 0. However, fm–1(t = 0+) = (m–1) ...
«
3
4
5
6
7
8
9
10
11
12
»
Free download pdf