我想弄清楚正确的触发。 eq./function确定以下内容： 两个方向向量 （已确定） 之间的角度变化（单位为DEGREES） ，表示两条线段。 这用于SHAPE RECOGTNITION（由用户在屏幕上手绘）的上下文中。
基本上，
a）如果用户绘制（粗糙）形状，例如圆形，椭圆形或矩形等，组成该形状的线条被细分为20点（xy对）。
b）我有这些线段的每一个的DirectionVector。
c）所以开始一个LINE SEGMENT（x0，y0），就是前一行的END点（以形成一个像矩形的闭合形状，比方说）。
那么，我的问题是，给定上下文（即确定多边形的类型）， 如何找到两个DIRECTION VECTORS之间的角度变化（可用作x和y的两个浮点值）？
我见过很多不同的触发器。 方程式，我正在寻求澄清。
非常感谢大家！
I am trying to figure out the correct trig. eq./function to determine the following: The Angle-change (in DEGREES) between two DIRECTION VECTORS(already determined), that represent two line-segment. This is used in the context of SHAPE RECOGTNITION (hand drawn by user on screen).
SO basically,
a) if the user draws a (rough) shape, such as a circle, or oval, or rectangle etc - the lines that makes up that shape are broken down in to say.. 20 points(x-y pairs).
b) I have the DirectionVector for each of these LINE SEGMENTS.
c) So the BEGINNING of a LINE SEGMENT(x0,y0), will the END points of the previous line(so as to form a closed shape like a rectangle, let's say).
SO, my question is , given the context(i.e. determinign the type of a polygon), how does one find the angle-change between two DIRECTION VECTORS(available as two floating point values for x and y) ???
I have seen so many different trig. equations and I'm seeking clarity on this.
Thanks so much in advance folks!
最满意答案
如果（x1，y1）是第一个方向向量而（x2，y2）是第二个方向向量，则它成立：
cos（alpha）=（x1 * x2 + y1 * y2）/（sqrt（x1 * x1 + y1 * y1）* sqrt（x2 * x2 + y2 * y2））
sqrt表示平方根。
查看http://en.wikipedia.org/wiki/Dot_product
特别是“几何表示”一节。
If (x1,y1) is the first direction vector and (x2,y2) is the second one, it holds:
cos( alpha ) = (x1 * x2 + y1 * y2) / ( sqrt(x1*x1 + y1*y1) * sqrt(x2*x2 + y2*y2) )
sqrt means the square root.
Look up http://en.wikipedia.org/wiki/Dot_product
Especially the section "Geometric Representation".
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