Question

http://cairographics.org/samples/curve_to/“rel=“nofollow noreferer”

double x=25.6,  y=128.0;
double x1=102.4, y1=230.4,
       x2=153.6, y2=25.6,
       x3=230.4, y3=128.0;

cairo_move_to (cr, x, y);
cairo_curve_to (cr, x1, y1, x2, y2, x3, y3);

cairo_set_line_width (cr, 10.0);
cairo_stroke (cr);

cairo_set_source_rgba (cr, 1, 0.2, 0.2, 0.6);
cairo_set_line_width (cr, 6.0);
cairo_move_to (cr,x,y);   cairo_line_to (cr,x1,y1);
cairo_move_to (cr,x2,y2); cairo_line_to (cr,x3,y3);
cairo_stroke (cr);

can generate the curve and two pink lines.

enter image description here

但这需要4点(x,y),(x1,y1),(x2,y2),(x3,y3)

If I only have x,y and x3, y3 (start and end points of the curve), is there any math formula to generate those pink lines without knowing x1,y1 and x2,y2?

http://www.ohchr.org。

就我以下列方式产生曲线的情况而言。

cairo_move_to (cr, x, y);
cairo_curve_to (cr, x, y3, x3, y, x3, y3);

Answer 1

仅提出以下几点:

start with your two known points, (x₁,y₁) and (x₃,y₃):

“entergraph

join the two lines:

“entergraph

create P₂ as halfway between P₁ and P₃:

enter image description here

now rotate P₃ 90° clockwise:

“entergraph

do with same with P₄, create it halfway between P₁ and P₃:

enter image description here

rotate P₄ 90° clockwise:

“entergraph

Now you have your four points, and can draw your bézier curve:

enter image description here

中点可计算为:

<>P<>><>><>>>><>>>>>> = (x<1>/sub>+x<>3)/2 , (y/sub><3)/2

double x1=25.6,  y1=128.0;
double x3=153.6, y3=25.6;

double xm = (x1+x3)/2;
double ym = (y1+y3)/2;

//rotate Pm by 90degrees around p1 to get p2
double x2 = -(ym-y1) + y1;
double y2 =  (xm-x1) + x1;

//rotate Pm by 90degrees around p3 to get p4
double x4 = -(ym-y3) + y3;
double y4 =  (xm-x3) + x3;

Answer 2

并非除非你提供某种制约,可用来确定汇线的位置。两个终点本身只能界定一个直线部分。

Answer 3

粉碎线是两端的病媒。如果没有这些病媒,这两点之间的“接触”只是一条直线(除非你们有其他一些界定信息)。

如果你没有(x1,y1)和(x2,y2),你就可以将(x3,y3)作为汇线的终点从(x,y)到反之。他们ll在你黑线上,应该直线。

如果曲线由功能加以界定,则计算衍生物,因为你接近尾声,沿这一角度划出细微线。

友情链接