Download Parametric Curves

Curves
Interpolation
• Bresenham’s line drawing algorithm is
nothing more than a form of
interpolation
– Given two points, find the location of points
in between them
– The function (in this case it’s a line) must
pass through the given points, by definition
Curves
• We’ll look at two techniques for
generating curves
– Interpolation
– Approximation
• We’ll also see that interpolation is very
restrictive when considering parametric
curves
• Approximation is a much better approach
Parametric Curves
• Similar to what we saw with lines
x  x  x  x t
y  y  y  y t
0
0
1
1
0
0
0  t 1
• The parameter t is used to map out a set of
(x, y) pairs that represent the line
Parametric Curves
• In the case of a curve we concern the
parametric function is of the form
– Q(u) = (x(u), y(u), z(u)) in 3 dimension
• The derivative of Q(u) is of the form
– Q’(u) = (x’(u), y’(u), z’(u))
• Significance of the derivative?
– It is the tangent vector to a given location (u) on
the curve
Derivative
Parametric Curves
• In the case of object drawing, u is a
spatial parameter
• Q(0) is the start of the curve and Q(1) is
the end, as was the case for the
parametric line
Parametric Curves
• Curvature
ρ
Osculating circle of radius ρ
touches the curve at exactly
1 point
• Curvature k = 1/ ρ
• The higher the curvature, the more the curve
bends at the given point
Parametric Curves
• From calculus, a function f is continuous at a
value x0 if
lim f ( x )  f ( x0)
x  x0
• In layman’s terms, we can draw the curve
without ever lifting our pen from the drawing
surface
• f(x) is continuous over an interval (a,b) if it is
continuous for every point in the interval
• We call this C0 continuity
Parametric Curves
Continuous over (a,b) – C0
a
b
Continuous over (a,b) – C0
b
a
Parametric Curves
• From calculus, a function’s derivative f’ is
continuous at a value x0 if
lim f ' ( x )  f ' ( x0)
x  x0
• In layman’s terms, there are no “sharp” changes
in direction
• f’(x) is continuous over an interval (a,b) if it is
continuous for every point in the interval
• We call this C1 (tangential) continuity
Parametric Curves
Continuous derivative over (a,b) – C1
a
b
Discontinuous derivative over (a,b) – not C1
b
a
Parametric Curves
• When we need to join two curves at a
single point we can guarantee C1
continuity across the joint
– Just make sure that the tangents of the two
curves at the join are of equal length and
direction
• If the tangents at the joint are of
identical direction but differing lengths
(change in curvature) then we have G1
continuity
Lagrange Polynomials
• To generate a function that passes
through every specified point, the type
of function depends on the number of
specified points
– Two points → linear function
– Three points → quadratic function
– Four points → cubic function
• Generating such functions makes use of
Lagrange polynomials
Lagrange Polynomials
• The general
form
is
n
P(t )   Pk Ln ,k t 
k 0
(t  t 0)(t  t1)...(t  t k 1)(t  t k 1)...(t  t n)
L t   (  )(  )...( 
t t t t t t
(t  t )
L t    (  )
t t
n,k
k
0
k
n
n,k
i  0 ,i  k
1
i
k
i
• Let’s look at an example
k
)(t k  t k 1)...(t k  t n )
k 1
Lagrange Polynomials
• For two points:

t 
t
P(t ) 
1

t 
t

0
t  t  P t  t  P
0
1
0
1
1
0
• For the starting point (t0=0) and ending point
(t1=1)
1
0
t
t
P(t ) 

P
0
0  1
1  0 P1
 

P(t )  (1  t ) P0  (t ) P1

Lagrange Polynomials
• For three points:

 t t  t 
t
P(t ) 
1
2

 t t  t 
t

0
2

 t t  t 
t

0
1
t  t t  t  P t  t t  t  P t  t t  t  P
0
1
0
2
0
1
0
1
2
1
2
0
2
2
• And it only gets worse for larger numbers of
points
• Suffice it to say, this isn’t the most optimum
way to draw curves
– Too many operations per point
– Too complex if the artist decides to change the
curve
2
A Better Way
• The problem with Lagrange polynomials lies
in the fact that we try to make the curve pass
through all of the specified points
• A better way is to specify points that control
how the curve passes from one point to the
next
• We do so by specifying a cubic function
controlled by four points
• The four points are called boundary
conditions
Hermite Boundary Conditions
P0
P’0
• Two points
• Two tangent vectors
P1
P’1
Cubic Functions
This is why it’s called a “cubic” function
• Generalized form
Q(u )  a
u b u c u  D
3
2
• Derivative
Q' (u )  3a
u  2b u  c
2
• Our goal is to “solve” these equations in “closed form”
so that we can generate a series of points on the
curve
Cubic Functions
• There are four unknown values in the
equation
– a, b, c, and D (remember, a, b, c, and D are
vectors in x, y, z)
• We need to use these equations to generate
values of x, y, and z along the curve
• We can generate a closed form solution
(solve the equations for x, y, and z) since we
have four boundary conditions
u = 0 → Q(0) = P0 and Q’(0) = P’0
u = 1 → Q(1) = P1 and Q’(1) = P’1
Solution of Equations
• Go to the white board…
Implementation
•
So, all you have to do to generate a curve is to implement this vector
equation:
'
'
a  2 P0  2 P1  P0  P1
b  3P0  3P1  2 P0'  P1'
c  P0'
D  P0
Q (u )  au 3  bu 2  cu  D
•
by stepping 0 ≤ u ≤ 1
P0, P1, P0’ and P1’ are vectors in x, y, z so there are really 12
coefficients to be computed and you’ll be implementing 3 equations for
Q(u)
Qx (u ), Q y (u ), Qz (u )
Implementation
• Note that you’ll have to estimate the
step size for u or…(any ideas?)
• …use your Bresenham code to draw
short straight lines between the points
you generate on the curve (to fill gaps)
• There is no trick (that I’m aware of)
comparable to the Bresenham approach
Bezier Curves
• Similar derivation to Hermite
• Different boundary conditions
– Bezier uses 2 endpoints and 2 control
points (rather than 2 endpoints and 2
slopes)
Bezier Curves
Control points
Endpoints
Implementation
•
So, all you have to do to generate a curve is to implement this vector
equation:
a   P0  3P1  3P2  P3
b  3P0  6 P1  3P2
c  3P0  3P1
D  P0
Q (u )  au 3  bu 2  cu  D
•
by stepping 0 ≤ u ≤ 1
P0, P1, P2 and P3 are vectors in x, y, z so there are really 12 coefficients
to be computed and you’ll be implementing 3 equations for Q(u)
Qx (u ), Q y (u ), Qz (u )
Hermite vs. Bezier
• Hermite is easy to control continuity at the endpoints
when joining multiple curves to create a path
– But difficult to control the “internal” shape of the curve
• Bezier is easy to control the “internal” shape of the
curve
– But a little more (not much) difficult to control continuity at
the endpoints when joining multiple curves to create a path
• Bottom line is, when creating a path you have to be
very selective about endpoints and adjacent control
points (Bezier) or tangent slopes (Hermite)
Result
• Go to the demo program…
• My code generates a Hermite curve that
is of C1 continuity
– As I generate new segments along the
curve I join them by keeping the adjoining
tangent vectors equal