
Notes for April 5 class  Introduction to splines
Splines
There are many reasons we might want to create
smooth controllable curves in computer graphics.
Perhaps we might want to create an organic shape,
or animate something along a continuous path.
As you can see on the right, we can do this
by breaking down our smooth curve into simpler
pieces.
If we think of the spline curve as a path of motion,
each of these pieces must match its neighbors in both
position and rate, which means that for each
coordinate x and y, we need four values:
a position at both the start and the end,
and a rate (or derivative) at both the start and the end.
The smallest order polynomial that can
satisfy four constraints is a cubic.
So we describe both the x and y coordinates of
each piece using
parametric cubic polynomials in parameter t:
x = a_{x}t^{3} +
b_{x}t^{2} +
c_{x}t + d_{x}
y = a_{y}t^{3} +
b_{y}t^{2} +
c_{y}t + d_{y}
where (a_{x}, b_{x}, c_{x}, d_{x})
and (a_{y}, b_{y}, c_{y}, d_{y})
are constant valued polynomial coefficients,
and t varies from t = 0 to t = 1 along
the curve.



Cubic splines
Although it is technically true that
cubic splines can be designed by tweaking
their polynial coefficients, in practice
that doesn't usually work out very well.
As you can see in the example on the right,
there is no intuitive connection between
the shape of a cubic polynomial and the values of its
four coefficients 
in this case 7.7, 11.7, 5.0 and 0.6, respectively,
for the
t^{3},
t^{2}, t and constant term.
For that reason, we need a better way to specify cubic spline curves.
Rather than use
t^{3},
t^{2}, t and 1 as our four basis functions,
we can use four different basis functions that have
a more intuitive geometric meaning.
In the following sections we will look at two
different examples of such alternate basis functions.



Hermite splines, part 1
We can choose four basis functions
that give us independent control over the position
at t = 0 and t = 1,
as well as the rate of change
at t = 0 and t = 1.
This is called the Hermite basis,
after the french mathematician who devised it.
If we want a curve with position A at t = 0,
position B at t = 1,
rate C at t = 0,
and rate D at t = 1,
we can use the four functions to the right
to compute the cubic polynomial we are looking for.


Position
2t^{3}  3t^{2} + 1
2t^{3} + 3t^{2}

Rate
t^{3}  2t^{2} + t
t^{3}  t^{2}

Hermite splines, part 2
Because these four hermite basis polynomials never change,
and the cubic polynomial we want is just a weighted sum of those four polynomials,
we can express this weighted sum as a multiplication of the weights
by a matrix, which we call the Hermite matrix.
In other words, the expression:
A (2t^{3}  3t^{2} + 1) +
B (2t^{3} + 3t^{2}) +
C (t^{3}  3t^{2} + t) +
D (t^{3}  t^{2})
can be expressed as a matrix vector product:
a
b
c
d


=




A
B
C
D

to convert positions and rates at the two ends into the desired cubic polynomial:
at^{3} + bt^{2} + ct + d.

Bezier splines, part 1
Artists and designers often find it more convenient
to create splines by moving points around, rather
than needing to deal with derivatives.
The Bezier spline enables designers
of spline curves
to work this way.
Bezier splines work by repeated linear interpolation.
For example, the image to the right shows a simplified
version of a Bezier spline, using three key points
to specify a parametric quadratic spline.
Note that points along the curve are
found as a linear interpolation of linear interpolations.
We first find points along the edges AB and BC
by linear interpolations (to get the points represented as blue dots):
(1t) A + t B
(1t) B + t C
and then we interpolate again (to get the point represented as a red dot):
P =
(1t) (
(1t) A + t B
)
+
t
(
(1t) B + t C
)
If we multiply out all the terms, we get:
A (1t)^{2} + 2 B (1t) t + C t^{2}
Note that the weights of the coefficients (1 2 1) follow Pascal's triangle.



Bezier splines, part 2
Now it becomes easier to see what is going on with the full
parametric cubic Bezier spline,
which uses four key points:
The basic setup is a
linear interpololation of
linear interpolations of
linear interpolations.
So we start with the points in blue:
P = (1t) A + t B
Q = (1t) B + t C
R = (1t) C + t D
When the first and second terms are linearly interpolated,
we get the two dots in violet:
S = (1t) P + t Q
T = (1t) Q + t R
Finally we linearly interpolate these two points:
(1t) S + t T
When we multiply everything out, writing the equation in terms of
our original four key points A,B,C and D,
the weights form the next level (1 3 3 1) of Pascal's triangle:
A (1t)^{3} + 3 B (1t)^{2} t + 3 C (1t) t^{2} + D t^{3}



Bezier splines, part 3
We can multiply out the terms of the above polynomial,
and regroup by powers of t, to get:
(A + 3B  3C + D) t^{3} +
(3A  6B + 3C) t^{2} +
(3A + 3B) t +
D
This makes it easy to see that,
as was the case for Hermite splines, the Bezier spline has a characteristic
matrix, which
can be used to translate the above polynomial until the
standard cubic polynomial, with coefficients (a,b,c,d):
a
b
c
d


=




A
B
C
D

One powerful property of Bezier splines is that the
direction between A and B determines the direction of the
spline curve at t=0, and the
direction between C and D determines the direction of the
spline curve at t=1.
This makes it easy to match up splines endtoend,
so that the resulting composite curve has a continuous derivative.

No new homework yet. I will give homework on Thursday.

 