从3个点计算定向旋转角 [英] Calculation of directional rotation angles from 3 points
问题描述
我想做的是找到由三个连续点形成的线之间的旋转角度.这些是连续的点，因此旋转方向很重要.我的输入是一对坐标序列.
What I'm trying to do is find the angle of rotation between lines formed by three consecutive points. These are sequential points, so the direction of rotation matter. My input is a sequence of pair of coordinates.
我想要的输出是每个点的旋转角度，其中该点将充当该角度的顶点.该角度应在1到360之间，其中负数表示向左旋转，正数表示向右旋转.
My desired output is the rotation angle of each point, where the point would be acting as the vertex of the angle. This angle would be between 1 and 360, where a negative number indicates a rotation to the left, and a positive number a rotation to the right.
我已经为此苦苦挣扎了好几个星期，但是我终于比以往任何时候都更接近解决方案了.我编写了以下脚本，并将其与程序地理空间建模工具(GME)的路径"功能的输出进行了比较.
I've been struggling with this for weeks, but I'm finally closer to the solution than ever before. I wrote the following script and compared it to the output of the "pathmetrics" function of the program Geospatial Modelling Tool(GME).
coords=coords=[[283907,971700],[284185,971634],[284287,971507],[284275,971608],[283919,971761],[284311,971648],[284277,971637],[284280,971651],[284174,971649],[283909,971701],[283941,971700],[284294,971518],[284288,971517],[284315,971539],[284250,971505]
print "A"+"\t"+"B"+"\t"+"C"+"\t"+"orientation"+"\t"+"angle"+"\t"+"bearing AB"+"\t"+"bearing BC"
for a in coords:
A=a
indexA=coords.index(a)
B=coords[indexA+1]
C=coords[indexA+2]
##Find the bearings of AB and BC
AB=[B[0]-A[0],B[1]-A[1]] #find the extreme of vector AB
BearAB=math.atan2(AB[0],AB[1]) #use arctan2 to find the angle
ABBearDeg=math.degrees(BearAB) #in degrees
if ABBearDeg<0: #if negative, add 360 in order to obtain the angle in a clockwise direction
ABBearDeg=360+ABBearDeg #Bearing AB
BC=[C[0]-B[0],C[1]-B[1]] #Do the same for points BC
BearBC=math.atan2(BC[0],BC[1])
BCBearDeg=math.degrees(BearBC)
if BCBearDeg<0:
BCBearDeg=360+BCBearDeg #Bearing BC
##Find the angle between the lines
alfa=BCBearDeg-ABBearDeg #Obtain the difference between the bearing angles
if abs(alfa)>180: #If greater than 180
if alfa<0: #and negative
angle=(360+alfa) #Direction of rotation is right and angle is obtained by adding 360
print format(A)+"\t"+format(B)+"\t"+format(C)+"\t"+"right"+"\t"+format(angle)+"\t"+format(round(ABBearDeg,2))+"\t"+format(round(BCBearDeg,2))
else: #If positive
angle=alfa-360 #Direction of rotation is left and angle is obtained by substracting 360
print format(A)+"\t"+format(B)+"\t"+format(C)+"\t"+"left"+"\t"+format(angle)+"\t"+format(ABBearDeg)+"\t"+format(round(BCBearDeg,2))
else: #If the difference was less than 180, then the rotation angle is equal to it
angle=alfa
if angle<0: #If negative, left rotation
print format(A)+"\t"+format(B)+"\t"+format(C)+"\t"+"left"+"\t"+format(angle)+"\t"+format(ABBearDeg)+"\t"+format(round(BCBearDeg,2))
else: #If positive, right rotation
print format(A)+"\t"+format(B)+"\t"+format(C)+"\t"+"right"+"\t"+format(angle)+"\t"+format(ABBearDeg)+"\t"+format(round(BCBearDeg,2))
虽然许多结果一致，但其他结果却不一致.
While many of the results coincide, others don't.
My results GME results
A B C orientation angle bearing AB bearing BC GMEangle GMEbearing AB GMEbearing BC
[283907, 971700] [284185, 971634] [284287, 971507] right 37.8750 103.3553 141.2300 37.8750 103.3553 141.2303
[284185, 971634] [284287, 971507] [284275, 971608] left -148.0060 141.2303 353.2200 -148.0060 141.2303 353.2243
[284287, 971507] [284275, 971608] [283919, 971761] left -59.9675 353.2243 293.2600 -68.9673 353.2243 284.2570
[284275, 971608] [283919, 971761] [284311, 971648] right 172.8236 293.2600 106.0800 96.6181 284.2570 106.0804
[283919, 971761] [284311, 971648] [284277, 971637] right 145.9916 106.0804 252.0700 145.9916 106.0804 252.0721
[284311, 971648] [284277, 971637] [284280, 971651] right 120.0227 252.0700 12.0900 120.0227 252.0721 12.0948
[284277, 971637] [284280, 971651] [284174, 971649] left -103.1757 12.0948 268.9200 -103.1757 12.0948 268.9191
[284280, 971651] [284174, 971649] [283909, 971701] right 12.1828 268.9191 281.1000 -131.4097 268.9191 137.5094
[284174, 971649] [283909, 971701] [283941, 971700] right 170.6880 281.1000 91.7900 85.2053 137.5094 9.4623
[283909, 971701] [283941, 971700] [284294, 971518] right 25.4848 91.7899 117.2700 146.4722 9.4623 85.0429
[283941, 971700] [284294, 971518] [284288, 971517] right 143.2629 117.2748 260.5400 123.0283 85.0429 260.5377
[284294, 971518] [284288, 971517] [284315, 971539] right 150.2887 260.5400 50.8300 150.2887 260.5377 50.8263
[284288, 971517] [284315, 971539] [284250, 971505] left -168.4394 50.8263 242.3900 -147.5925 50.8263 263.2338
(抱歉，表格凌乱；我无法根据需要上传图片)
(sorry for the messy table; I couldn't upload an image as I wanted)
我已经能够准确指出发生错误的位置，但是我无法弄清楚为什么会发生错误，因为它严格取决于我无法控制的预设公式.因此，不同之处在于(有时)我对向量方位的计算与GME计算的结果有所不同.奇怪的是，它仅在某些时候发生，我不知道是什么触发了它.
I've been able to pin-point at with point the error occurs, but I can't figure out why it happens because it depends strictly on pre-set formulas I have no control of. So, the difference is that (SOMETIMES) my calculations of the bearing of the vectors differs from the one calculated by GME. The weird part is that it happens only sometimes, and I have no idea what triggers it.
有什么想法吗?
如果您知道任何其他方法来计算结合运动方向的线之间的角度，请告诉我.一切正常，我都可以.
If you know any other way of calculating angles between lines that incorporate the direction of the movement, do let me know. Anything that works will be okay with me.
谢谢!!!!
推荐答案
您拥有点A，B和C.根据您的描述，我假设B是旋转点.也就是说，向量BA转换为BC.
You have the points A, B and C. I'm assuming from your description that B is the rotation point. That is, the vector BA transforms into BC.
- 创建向量BA(A-B)和BC(C-B).
- 计算向量的角度(使它们为正)及其差
- 矢量之间的角度差就是旋转角度.
使用numpy可简化此操作.像这样:
Using numpy makes this easy. Like this:
In [1]: import numpy as np
In [2]: A = np.array([283907, 971700])
In [3]: B = np.array([284185, 971634])
In [4]: C = np.array([284287, 971507])
In [5]: BA = A - B
In [6]: BC = C - B
In [7]: BA
Out[7]: array([-278, 66])
In [8]: BC
Out[8]: array([ 102, -127])
In [9]: s = np.arctan2(*BA)
In [10]: if s < 0:
....: s += 2*np.pi
....:
In [11]: s
Out[11]: 4.9454836529138948
In [12]: e = np.arctan2(*BC)
In [13]: if e < 0:
e += 2*np.pi
....:
In [14]: e
Out[14]: 2.4649341681747883
In [15]: delta = e - s
In [16]: np.degrees(delta)
Out[16]: -142.12501634890182
In [17]: delta
Out[17]: -2.4805494847391065
正角是逆时针方向.
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