英文:
Computing the angle of three points
问题
我有三个点，每个点都有一个 Bar_Index 和 Value。
我试图计算三个点之间的角度，以使点 C 是中心的一个（在图中的顶部），而 1 和 2 是两侧的点（点C的左侧和右侧）。
我不确定为什么我的计算出的角度如此小，根据图示，它应该要大得多（大于20度）。有人知道我漏掉了什么吗？
这是代码，
Cx = bar_index_0
Cy = Value_0
P0x = bar_index_1
P0y = Value_1
P1x = bar_index_2
P1y = Value_2
P0C = math.sqrt(math.pow((P0x-Cx),2)+math.pow((P0y-Cy),2))
P1C = math.sqrt(math.pow((P1x-Cx),2)+math.pow((P1y-Cy),2))
P0P1 = math.sqrt(math.pow((P1x-P0x),2)+math.pow((P1y-P0y),2))
top = (P1C*P1C + P0C*P0C - P0P1*P0P1)
down = 2*P1C*P0C
Angle = (math.todegrees(math.acos(top/down)))
此外，如果有人对计算三角形内的三个角有更好的建议，我将很高兴听到它。
英文:
I have three points, each point has a Bar_Index and Value.
I am trying to compute the angle between the tree points in a way that point C is the one in the center (on the top in the fig) and 1 and 2 are the ones on the sides (left and right of point C).
I am not sure why my computed angle in degree is so small, based on the figure it should be much larger (more than 20 degree).Does anyone know what I am missing?
Here is the code,
Cx = bar_index_0
Cy = Value_0
P0x = bar_index_1
P0y = Value_1
P1x = bar_index_2
P1y = Value_2
P0C = math.sqrt(math.pow((P0x-Cx),2)+math.pow((P0y-Cy),2))
P1C = math.sqrt(math.pow((P1x-Cx),2)+math.pow((P1y-Cy),2))
P0P1 = math.sqrt(math.pow((P1x-P0x),2)+math.pow((P1y-P0y),2))
top = (P1C*P1C + P0C*P0C - P0P1*P0P1)
down = 2*P1C*P0C
Angle = (math.todegrees(math.acos(top/down)))
Also, if anyone has a better suggestion to compute the three angles inside the triangle I will be happy to hear it.
答案1
得分: 1
在向量数学中，角度是((p0-C)点乘(p1-C)) / (len(p0-C)*len(p1-C))。
因此，尝试：
P0Cx = P0x - Cx
P0Cy = P0y - Cy
P1Cx = P1x - Cx
P1Cy = P1y - Cy
dot = P0Cx * P1Cx + P0Cy * P1Cy
div = P0C * P1C
angle = math.acos(dot / div)
英文:
In vector math, the angle is ((p0-C) dot (p1-C)) / (len(p0-C)*len(p1-C)).
So, try:
P0Cx = P0x-Cx
P0Cy = P0y-Cy
P1Cx = P1x-Cx
P1Cy = P1y-Cy
dot = P0Cx*p1Cx + P0Cy*P1Cy
div = P0C*P1C
angle = math.acos(dot/div)
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