P0
P3
P1
P2
Q0
Q1
Q2
R0
R1
B (t0)
MATH 164, Spring 2001 Due Date: Name(s):
Honors Project 8a: Bezier Curves
Bezier curves (such as the one to the right) are frequently used in
computer drawing programs. A curves such as this is drawn by locating
numerous points along it and "connecting the dots" (the assumption
being that the eye cannot distinguish between a smooth curve and
a many-sided polygonal arc). In this project we describe an algorithm
-- the deCastlejau algorithm -- for constructing Bezier curves.
Narrative
To construct a Bezier curve segment of degree n we begin by selecting n + 1 points in the plane. The
first and last points are the endpoints of the curve segment; the intermediate points help "guide the curve",
but they usually do not lie on it. The points n + 1 are called control points of the Bezier curve, and the
segments joining them in order form its control polygon. In this project we consider the case n = 3, although
everything we say is true if n = 2 or n > 3.
Suppose the points we have selected are P0(x0, y0), . . . , P3(x3, y3). The deCastlejau algorthm for finding
the point on the curve that corresponds to the parameter value t = t0, 0  t0  1, is based on linear
interpolation, and, indeed, applying it repeatedly: A parametrization of the segment P0P1 is given by
x = (1 - t)x0 + tx1, y = (1 - t)y0 + ty1,
where t  [0, 1]. We can abbreviate the above notation by writing (1 - t)P0 + tP1, meaning we perform the
indicated operation on both components. For a given parameter value t0 we find a point Q0 on P0P1 that
is t0 of the distance from P0 to P1:
Q0 = (1 - t0)P0 + t0P1.
(This process is known as linear interpolation.) Similarly we find points on P1P2 and P2P3:
Q1 = (1 - t0)P1 + t0P2 and Q2 = (1 - t0)P2 + t0P3.
Then we interpolate again using the same parameter value t0 to find points R0 and R1 on Q0Q1 and Q1Q2:
R0 = (1 - t0)Q0 + t0Q1 and R1 = (1 - t0)Q1 + t0Q2.
The point B(t0) on the curve we are constructing, is
B(t0) = (1 - t0)R0 + t0R1.
Tasks
1. Choose 4 points in the plane and carry out this construction for the values t0 = 1/3 and t0 = 2/3.
2. Show that the expanded and simplified formula for B(t) is
B(t) = (1 - t)3
P0 + 3t(1 - t)2
P1 + 3t2
(1 - t)P2 + t3
P3.
Confirm that the curve starts at P0 when t = 0 and ends at P3 when t = 1.
3. a) Compute the derivative B (t).
b) Find the derivatives at t = 0 and t = 1.
c) Find the lines tangent to the curve at the beginning and ending points.
d) Show that the tangent line at S is the line through R0 and R1.
4. Carry out the deCastlejau construction for the case n = 2, with 3 points.
5. Find the formula for B(t) in the case n = 2.
6. Show that if n = 2 then B(1/2) lies midway between P1 and M, the midpoint of segment P0P2. This
point is called the shoulder point of the curve.
7. Show that for n = 2 the curve is an arc of a parabola from P0 to P2. (Hint: Consider eliminating the
parameter t.)
Contributor: Dr. Richard Patterson
Department of Mathematical Sciences, IUPUI