Saturday, August 13, 2005

Why Bezier Spline is so useful?

[KEYWORD: Bezier spline, Computer-based design, Bezier curve, Parametric equations for bezier spline, formula]

Bezier splines are considered to be useful for computer-assisted design work because of several characteristics. First, with a little practice, you can usually manipulate the curve into something close to a desired shape.

Second, the Bezier spline is very well controlled. In some splines, the curve does not pass through any of the points that define the curve. The Bezier spline is always anchored at the two end points. (This is one of the assumptions that is used to derive the Bezier formulas.) Also, some forms of splines have singularities where the curve veers off into infinity. In computer-based design work, this is rarely desired. The Bezier curve never does this; indeed, it is always bounded by a four-sided polygon (called a "convex hull") that is formed by connecting the end points and control points.

Third, another characteristic of the Bezier spline involves the relationship between the end points and the control points. The curve is always tangential to and in the same direction as a straight line draw from the begin point to the first control point. (This is visually illustrated by the Bezier program.) Also, the curve is always tangential to and in the same direction as a straight line drawn from the second control point to the end point. These are two other assumptions used to derive the Bezier formulas.

Fourth, the Bezier spline is often aesthetically pleasing. I know this is a subjective criterion, but I'm not the only person who thinks so.

Prior to the 32-bit versions of Windows, you'd have to create your own Bezier splines using the Polyline function. You would also need knowledge of the following parametric equations for the Bezier spline. The begin point is (x0, y0), and the end point is (x3, y3). The two control points are (x1, y1) and (x2, y2). The curve is drawn for values of t ranging from 0 to 1:

x(t) = (1 - t)3 x0 + 3t (1 - t)2 x1 + 3t2 (1 - t) x2 + t3 x3

y(t) = (1 - t)3 y0 + 3t (1 - t)2 y1 + 3t2 (1 - t) y2 + t3 y3

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