Curve Modeling. Dr. S.M. Malaek Assistant: M. Younesi. Parametric Polynomials. Parametric Polynomials. For interpolating n point we need a polynomial of degree n-1 . Parametric Polynomials. Polynomial interpolation has several disadvantages :
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Dr. S.M. Malaek
Assistant: M. Younesi
Polynomial interpolation has several disadvantages:
the Spline Curve
f(u) = ( u, -u2, 0 )
g(v) = ( v, v2, 0 )
f'(u) = ( 1, -2u, 0 ) ,f''(u) = ( 0, -2, 0 )
g'(u) = ( 1, 2v, 0 ) ,g''(v) = ( 0, 2, 0 )
Two curve segments may be C1 continuous and even curvature continuous, but not C2 continuous
Parametric Continuity Conditions
f(u) = A + u(B - A)
g(v) = B + v(C - B)
Two line segments:
f'(u) = B - A
g'(v) = C - B
Is it strange?
Re-parameterizing the curve segments may overcome the problem
F(u) = A + u(B - A)/ | B - A |
G(v) = B + v(C - B)/ | C - B |
uis in the range of 0 and | B - A | and v is in the range of 0 and | C – B|
f(u) = ( -cos(u2 PI/2), sin(u2 PI/2), 0 )
g(v) = ( sin(v2 PI/2), cos(v2 PI/2), 0 )
At (0,1,0) = f(1) = g(0). ) C0)
f'(u) = ( PI u sin(u2 PI/2), PI u cos(u2 PI/2), 0 )
f''(u) = ( PI2 u2 cos(u2 PI/2), -PI2 u2 sin(u2 PI/2), 0 )
f'(u) × f''(u) = ( 0, 0, -PI3u3 )
| f'(u) | = PI u
| f'(u) × f''(u) | = PI3u3
k(u) = 1
g'(v) = ( PI v cos(v2 PI/2), -PI v sin(v2 PI/2), 0 )
g''(v) = ( -PI2 v2 cos(v2 PI/2), -PI2 v2 cos(v2 PI/2), 0 )
g'(v) × g''(v) = ( 0, 0, -PI3u3 )
| g'(v) | = PI v
| g'(v) × g''(v) | = PI3v3
k(v) = 1
Let u2 = p in f(u) and let v2 = q in g(v). The new equations are:
f(p) = ( -cos(p PI/2), sin(p PI/2), 0 )
g(q) = ( sin(q PI/2), cos(q PI/2), 0)
Their derivatives are
f'(p) = ( (PI/2) sin(p PI/2), (PI/2) cos(p PI/2), 0 )
f''(p) = ( (PI/2)2 cos(p PI/2), -(PI/2)2 sin(p PI/2), 0 )
g'(q) = ( (PI/2) cos(q PI/2), -(PI/2) sin(q PI/2), 0 )
g''(q) = ( -(PI/2)2 sin(q PI/2), -(PI/2)2 cos(q PI/2), 0 )
Therefore, after changing variables, both f'(1) and g'(0) are equal to ( PI/2, 0, 0 ) and hence they are C1continuous. Moreover, both f''(1) and g''(0) are equal to ( 0, -(PI/2)2, 0 ) and hence they are C2 continuous! They are also curvature continuous because they have curvature 1 everywhere (remember they are circles).
The tangent vectors point to the opposite directions, and, as a result, they are not G1 at the joining point
C1(1) = C2(0)
ULinear Spline Interpolation
Cubic splines are used to:
Cubic splines are more flexible for modeling arbitrary curve shapes.
Methods for setting the boundary conditions for cubic interpolation splines:
If we have n+1 control points to fit, then:
3 control points (n=2)
2 curve sections
4×2 polynomial coefficients to be determined.
At eachn-1interior control points, we have 4boundary condition:
At P1 interior control:
We still needtwo more conditions to be able to determine values for all coefficients.
Set the second derivatives at p0and pn to 0
P3Natural Cubic Splines
Add a control pointP-1and a control point Pn+1.
Given a set of control points:
Dpk and Dpk+1 are the values for the parametric derivations (slope of the curve) at control points Pk and Pk+1
The derivative of the point function:
We can write the constraints
in a matrix form:
Solving this equation for the polynomial coefficients:
MH, the Hermit matrix, is the inverse of the boundary constraint matrix.
Hermit Blending Functions
For most problems in modeling, it is more useful togenerate splinecurve withoutrequiring input values for curve slopes.
For a cardinal spline, the value for the slope at a control point is calculatedfrom the coordinates of the two adjacent control points.