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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