[英]How to find knot vector if control points are known for a NURBS curve?
我有一組控制點
pts = [[849, 1181],
[916, 1257],
[993, 1305],
[1082,1270],
[1137,1181],
[1118,1055],
[993,1034],
[873,1061],
[849, 1181]]
我有生成一個開結矢量的邏輯:
/*
Subroutine to generate a B-spline open knot vector with multiplicity
equal to the order at the ends.
c = order of the basis function
n = the number of defining polygon vertices
nplus2 = index of x() for the first occurence of the maximum knot vector value
nplusc = maximum value of the knot vector -- $n + c$
x() = array containing the knot vector
*/
knot(n,c,x)
int n,c;
int x[];
{
int nplusc,nplus2,i;
nplusc = n + c;
nplus2 = n + 2;
x[1] = 0;
for (i = 2; i <= nplusc; i++){
if ( (i > c) && (i < nplus2) )
x[i] = x[i-1] + 1;
else
x[i] = x[i-1];
}
}
另一個用於生成周期性結矢量的方法:
/* Subroutine to generate a B-spline uniform (periodic) knot vector.
c = order of the basis function
n = the number of defining polygon vertices
nplus2 = index of x() for the first occurence of the maximum knot vector value
nplusc = maximum value of the knot vector -- $n + c$
x[] = array containing the knot vector
*/
#include <stdio.h>
knotu(n,c,x)
int n,c;
int x[];
{
int nplusc,nplus2,i;
nplusc = n + c;
nplus2 = n + 2;
x[1] = 0;
for (i = 2; i <= nplusc; i++){
x[i] = i-1;
}
}
但是,我需要生成一個[0,1]范圍內的非均勻結向量
上述算法導致統一的結向量。
請建議是否有任何方法可以做到這一點。 如果代碼在python中會更好
結向量(均勻或不均勻)是NURBS曲線定義的一部分。 因此,實際上您可以定義自己的非均勻結矢量,只要該結矢量遵循基本規則即可:
1)結點數=控制點數+順序
2)所有結值必須不減小。 即,k [i]≤k[i + 1]。
對於具有9個控制點的示例,您可以具有非均勻結矢量,例如[0,0,0,0,a,b,c,d,e,1,1,1,1],其中0.0 <a <對於3度B樣條曲線= b <= c <= d <= e <1.0 當然,為a,b,c,d和e選擇不同的值將導致曲線具有不同的形狀。
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