如何创建一条线的头距其末端一定距离的线-Java/AWT [英] How to create a line whose head is some distance from its end - java / awt

问题描述

我有分数

A(x1，y1)

A(x1, y1)

B(x2，y2)

我需要画这条线的箭头，但不要画在最后.它必须与末端有一定距离.这东西会做什么?

I need to draw the arrow head of this line but not on the end. It must be some distance from the end. What way would this thing do?

我有:

    ---------------------->

我需要:

    ----------------->-----

查看图片:

这是有角度的:

谢谢你们的帮助.这是另一个.

Thank you guys for helping. This is for another.

让我们创建主要功能并绘制一条线和箭头:

Let's create main function and draw a line and arrowhead:

   private void drawLineWithArrowHead(Point from, Point to, Graphics2D graphics){
       Polygon arrowHead = new Polygon();
       arrowHead.addPoint( 0,6);
       arrowHead.addPoint( -6, -6);
       arrowHead.addPoint( 6,-6);int y1,y2,x1,x2;
       x1=from.getPosX();
       y1=from.getPosY();
       x2=to.getPosX();
       y2=to.getPosY();
       Line2D.Double line = new Line2D.Double(x1,y1,x2,y2);
       graphics.draw(line);

让我们加载旧的仿射变换并从行中获取新的仿射:

Lets load old affine transform and get new from line:

   AffineTransform tx, old_tx = graphics.getTransform();
   tx = calcAffineTransformation(line);

一些数学:

   double dx = (x2-x1), dy = (y2-y1);
   double len = Math.sqrt(dx*dx + dy*dy);
   double udx = dx/len, udy = dy/len;
   double cordx = x2  - (size-5) * udx, cordy = y2  - (size-5) * udy;
   double r_cordx = x2  - (size+3) * udx, r_cordy = y2  - (size+3) * udy;

现在放置箭头:

   tx.setToIdentity(); // null transform to origin
   double angle = Math.atan2(line.y2 - line.y1, line.x2 - line.x1);
   !! important !! must firstly translate secondly rotate
   tx.translate( cordx,  cordy ); // setup of cord of arrowhead
   tx.rotate((angle - Math.PI / 2d)); // head rotate
   graphics.setTransform(tx); // set transform for graphics
   graphics.fill(arrowHead);
   graphics.setTransform(old_tx); // get original transform back

CalcAffineTransformation函数获取行的位置并旋转:

CalcAffineTransformation function to get line position and rotate:

   private AffineTransform calcAffineTransformation(Line2D.Double line) {
       AffineTransform transformation = new AffineTransform();
       transformation.setToIdentity();
       double angle = Math.atan2(line.y2 - line.y1, line.x2 - line.x1);
       transformation.translate(line.x2, line.y2);
       transformation.rotate((angle - Math.PI / 2d));
       return transformation;
   }

仅此而已.这就是代码的作用:

Thats all. And this is what code do:

推荐答案

您的行具有方向矢量

 dx, dy = (x2 - x1), (y2 - y1)

它的长度是

 len = sqrt(dx*dx + dy*dy)

单位方向向量为

udx, udy = dx/len, dy/len

从末端到D点的距离(据我所知，这是箭头的头部):

Point at distance D from the end (this is head point of arrow, as I understand):

x3, y3 = x2 - D * udx, y2 - D * udy

您还需要其他东西来建造箭头吗?

do you need something else to build arrow?

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