用比例因子绘制点的图形
我使用以下内容集中缩放图表上的一个框：
var x1 = (this.width - (this.image.width * this.scale)) / 2 + this.origin.x;
var y1 = (this.height - (this.image.height * this.scale)) / 2 + this.origin.y;
var x2 = (this.image.width * this.scale);
var y2 = (this.image.height * this.scale);
context.drawImage(this.image, x1, y1, x2, y2);
现在我需要能够缩放该图表上的单个点。该点具有原点 x / y、点 x / y 以及当前级别的比例因子 (this.scale)。如何将此比例因子转换为具有给定比例的盒子上的点？
I am centrally scaling a box on a graph with the following:
var x1 = (this.width - (this.image.width * this.scale)) / 2 + this.origin.x;
var y1 = (this.height - (this.image.height * this.scale)) / 2 + this.origin.y;
var x2 = (this.image.width * this.scale);
var y2 = (this.image.height * this.scale);
context.drawImage(this.image, x1, y1, x2, y2);
Now I need to be able to scale a single point on that graph. The point has an origin x / y, point x / y, and a scale factor (this.scale) at the current level. How can I translate this scale factor into a point on the box with the given scale?
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评论（1）
我认为你的问题是 - 一个点已经给出了坐标
x,y并且你需要想出新的 x 和 y 坐标a, b以便它在缩放框中的位置与整个图中的位置相同。该点与原点水平距离为
x个单位，您需要它距离新原点为x/scale个单位。因此，
a = x/scale + x1（因为新原点的坐标为x1, y1），类似地，该点垂直于原点，并且您需要它距新原点为
y/scale单位。和
b = y/scale + y1我想我可能误解了新原点的位置（框的左角，但如果我简单地替换
x1和y1与新原点的坐标）I think your question is - a point has given coordinates
x,yand you need to come up with the new x and y co-ordinatesa, bso that its in the same place within the scaled box as it was in the whole graph.The point is
xunits away horizontally from the origin, and you need it to bex/scaleunits from the new origin.So,
a = x/scale + x1( because the new origin has coordinatesx1, y1)similarly, the point is
yunits vertically from the origin, and you need it to bey/scaleunits from the new origin.and
b = y/scale + y1I think i may have misunderstood where the new origin is (the left corner of the box, but if I have then simply replace
x1andy1with the coordinates of the new origin)