Curves in 3D - 3D Math Primer for Graphics and Game Development - page 674

/ / d e g r e e o f t h e c u r v e by one .

f o r ( i n t i = 0 ; i < n ; ++ i ) {

p o i n t s [ i ] = p o i n t s [ i ] ∗ ( 1 . 0 f − t ) + p o i n t s [ i +1] ∗ t ;

}

}

/ / R e s u l t i s now i n t h e f i r s t s l o t .

r e t u r n p o i n t s [ 0 ] ;

}
Listing 13.1
Evaluating a point on a B ezier curve using de Casteljau's algorithm

This gives us a method for locating a point at any given t through re-
peated interpolation, but it doesn't directly give us a closed form expression
to calculate the point in terms of the control points. We emphasize that
such a closed form expression is often not needed, but let's derive it in
monomial form anyway. We're looking for a polynomial grouped by powers
of t. We'll work our way up from the linear and quadratic cases to the cu-
bic. Section 13.4.2 presents a general pattern leading us to the expression
for arbitrary degree curves.
The linear case comes straight from the recurrence relation without any
real work:

b i (t) = b i ,
b i (t) = (1 − t)[ b i (t)] + t[ b i+1 (t)]
= (1 − t) b i + t b i+1
= b i + t( b i+1 − b i ).

Applying one more level gives us a quadratic polynomial:

b i (t) = (1 − t)[ b i (t)] + t[ b i+1 (t)]
= (1 − t)[ b i + t( b i+1 − b i )] + t[ b i+1 + t( b i+2 − b i+1 )]
= [ b i + t( b i+1 − b i )] − t[ b i + t( b i+1 − b i )]
+ t[ b i+1 + t( b i+2 − b i+1 )]
= b i + t( b i+1 − b i ) − t b i − t 2 ( b i+1 − b i )
+ t b i+1 + t 2 ( b i+2 − b i+1 )

− 2 b i ) + t 2 ( b i

= b i + t(2 b i+1 − 2 b i+1 + b i+2 ).

In other words, quadratic Bezier curves, which have three control points,
can be expressed in monomial form as

p (t) = b 0 (t) = b 0 + t(2 b 1 − 2 b 0 ) + t 2 ( b 0

− 2 b 1 + b 2 ). (13.17)

Quadratic Bezier curve
in monomial form

Before we do the last round of interpolation to get the cubic curve, let's
take a closer look at the quadratic expression in Equation (13.17). This