Question

I m以超文本5的形式从事游戏。

I want is draw an S-shaped cubic bezier curve between two points, but I m looking for a way to calculate the coordinates of the control points so that the curve itself is always the same length no matter how close those points are, until it reaches the point where the curve becomes a straight line.

Answer 1

This is solvable numerically. I assume you have a cubic bezier with 4 control points. at each step you have the first (P0) and last (P3) points, and you want to calculate P1 and P2 such that the total length is constant.

加上这一限制,就会取消一定程度的自由,这样我们就有1人离开(从4起起就确定了终点(-2),而持续时间是另一个――1)。因此,你需要就此作出决定。

啤酒曲线是在0到1年之间界定的聚合物,你需要把元素(2d?)的根基结合起来。对于一立方米苯,这意味着六分之六的rt,而该wo子不知道如何解决。但是,如果你知道所有其他控制点(或知道依赖某些其他限制),那么你就可以为这一限制制定预估的数值。

Answer 2

是否确实有必要使曲线成为灵丹妙药? 减去两个总长度不变的圆环。你们将永远获得S-shape。

安装两个循环弧:

<>D是终点之间的滑坡距离。让我们永远保持下去。 www.un.org/Depts/DGACM/index_french.htm

b = sqrt(D*sin(C/4)/4 - (D^2)/16)

如果是正确的,那么如果某人有某种不同的话,我就没有听说过的话。

http://www.un.org。在处理正确公式和检查时,你也应考虑我获得的消极解决办法。

b = -sqrt(D*sin(C/4)/4 - (D^2)/16)

Answer 3

Here s a working example in SVG that s close to correct:
http://phrogz.net/svg/constant-length-bezier.xhtml

“entergraph

I experimentally determined that when the endpoints are on top of one another the handles should be
desiredLength × cos(30°)
away from the handles; and (of course) when the end points are at their greatest distance the handles should be on top of one another. Plotting all ideal points looks sort of like an ellipse:

“Graph

The blue line is the actual ideal equation, while the red line above is an ellipse approximating the ideal. Using the equation for the ellipse (as my example above does) allows the line to get about 9% too long in the middle.

Here相关文本:

// M is the MoveTo command in SVG (the first point on the path)
// C is the CurveTo command in SVG:
//   C.x is the end point of the path
//   C.x1 is the first control point
//   C.x2 is the second control point
function makeFixedLengthSCurve(path,length){
  var dx   = C.x - M.x, dy = C.y - M.y;
  var len  = Math.sqrt(dx*dx+dy*dy);
  var angle = Math.atan2(dy,dx);
  if (len >= length){
    C.x  = M.x + 100 * Math.cos(angle);
    C.y  = M.y + 100 * Math.sin(angle);
    C.x1 = M.x; C.y1 = M.y;
    C.x2 = C.x; C.y2 = C.y;
  }else{
    // Ellipse of major axis length and minor axis length*cos(30°)
    var a = length, b = length*Math.cos(30*Math.PI/180);
    var handleDistance = Math.sqrt( b*b * ( 1 - len*len / (a*a) ) ); 
    C.x1 = M.x + handleDistance * Math.sin(angle);
    C.y1 = M.y - handleDistance * Math.cos(angle);
    C.x2 = C.x - handleDistance * Math.sin(angle);
    C.y2 = C.y + handleDistance * Math.cos(angle);
  }
}

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