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How can I find the angle between two vectors that can range from 0 rad to 2 rad, instead of the inside angle with the arccos function

I'm making a python script right now that is trying to find the length of an arc, where it given this information:
center of arc: x1, y1
start point of arc: x2, y2
end point of arc: x3, y3
direction, cw, ccw
so far I have been able to successfully calculate the radius, and I tried calculating the angle using the equation:
But for any arcs that have an angle greater than 1*pi or 180 degrees, it returns the incorrect (but correct) inside angle.
What is the correct equation knowing the radius and these three points that I can use to find the value of the angle of the arc from 0 rad/degrees to 360 degrees/2pi radians, going in either the clockwise or counterclockwise direction (it can be either or and I need to be able to calculate for both scenarios)
Code:
# code to find theta
aVector = np.array([x1 - x2, y1 - y2])
bVector = np.array([x1 - x3, y1 - y3])
aMag = np.linalg.norm(aVector)
bMag = np.linalg.norm(aVector)
theta = np.arcos(np.dot(aVector, bVector) / (aMag * bMag))
as you can see here, I'm using arccos which to my dismay only outputs 0-180 degrees
Solution/Working code:
# equation for angle using atan2
start = math.atan2(y2 - y1, x2 - x1)
end = math.atan2(y3 - y1, x3 - x1)
if gcodeAnalysis[tempLineNum][4] == "G3": # going CW
start, end = end, start
tau = 2.0 * math.pi
theta = math.fmod(math.fmod(end - start, tau) + tau, tau)
Working Values:
X1 = 0.00048399999999998444
Y1 = 0.0002720000000007161
X2 = 0.378484
Y2 = -14.694728
X3 = 3.376
Y3 = -14.307
Proper result/value
Theta = 6.077209477545957
Assume this arc was done CCW

As you noticed, the range of math.acos is [0, pi], making it rather useless for telling you the relative directions of the vectors. To get full circular information about a pair of angles, you can use math.atan2. While regular math.atan has a range of [-pi/2, pi/2], atan2 splits the inputs into two parts and returns an angle in the range (-pi, pi]. You can compute the angles relative to any reference, not necessarily relative to each other:
start = math.atan2(y2 - y1, x2 - x1)
end = math.atan2(y3 - y1, x3 - x1)
Now you can use some common formulae to find the difference between the angles in whatever direction you want. I've implemented some of these in a small utility library I made called haggis. The specific function you want is haggis.math.ang_diff_pos.
First, the "manual" computation:
if direction == 'cw':
start, end = end, start
tau = 2.0 * math.pi
angle = math.fmod(math.fmod(end - start, tau) + tau, tau)
If you want to use my function, you can do
if direction == 'cw':
start, end = end, start
angle = ang_diff_pos(start, end)
All of these operations can be easily vectorized using numpy if you find yourself dealing with many points all at once.

You can use the cross product of the two vector to determine if the two vector need to rotate clock or counter-clock wise.
See code below:
import numpy as np
from numpy import linalg as LA
x1 = 0
y1 = 0
x2 = 2
y2 = 0
x3 = 2
y3 = -2
dir = 'ccw' # or ccw
v1 = np.array([x2-x1,y2-y1])
v2 = np.array( [x3-x1,y3-y1])
# if the cross product is positive, then the two vector need to rotate counter clockwise
rot = np.cross(v1,v2)
vdir = 'ccw' if rot >0 else 'cw'
r = (v1[0]*v2[0]+v1[1]*v2[1])/(LA.norm(v1)*LA.norm(v2))
deg = np.arccos(r)/np.pi*180
if vdir != dir:
deg = 360 -deg
print(deg)

How do I construct a point from x,y, line and angle?

I am making a geometry interface in python (currently using tkinter) but I have stumbled upon a major problem: I need a function that is able to return a point, that is at a certain angle with a certain line segment, is a certain length apart from the vertex of the angle. We know the coordinates of the points of the line segment, and also the angle at which we want the point to be. I have attached an image below for a more graphical view of my question.
The problem: I can calculate it using trigonometry, where
x, y = vertex.getCoords()
endx = x + length * cos(radians(angle))
endy = y + length * sin(radians(angle))
p = Point(endx, endy)
The angle I input is in degrees. That calculation is true only when the line segment is parallel to the abscissa. But the sizes of the angles I get back are very strange, to say the least. I want the function to work wherever the first two points are on the tkinter canvas, whatever the angle is. I am very lost as to what I should do to fix it. What I found out: I get as output a point that when connected to the vertex, makes a line that is at the desired angle to the abscissa. So it works when the first arm(leg, shoulder) of the angle is parallel to the abscissa, then the function runs flawlessly (because of cross angles) - the Z formation. As soon as I make it not parallel, it becomes weird. This is because we are taking the y of the vertex, not where the foot of the perpendicular lands(C1 on the attached image). I am pretty good at math, so feel free to post some more technical solutions, I will understand them
EDIT: I just wanted to make a quick recap of my question: how should I construct a point that is at a certain angle from a line segment. I have already made functions that create the angle in respect to the X and Y axes, but I have no idea how i can make it in respect to the line inputted. Some code for the two functions:
def inRespectToXAxis(vertex, angle, length):
x, y = vertex.getCoords()
newx = x + length * cos(radians(angle))
newy = y + length * sin(radians(angle))
p = Point(abs(newx), abs(newy))
return p
def inRespectToYAxis(vertex, length, angle):
x, y = vertex.getCoords()
theta_rad = pi / 2 - radians(angle)
newx = x + length * cos(radians(angle))
newy = y + length * sin(radians(angle))
p = Point(newx, newy)
return p

Seems you want to add line segment angle to get proper result. You can calculate it using segment ends coordinates (x1,y1) and (x2,y2)
lineAngle = math.atan2(y2 - y1, x2 - x1)
Result is in radians, so apply it as
endx = x1 + length * cos(radians(angle) + lineAngle) etc

Calculating angle of line crossing two points - how to scale correctly?

I have two points and would like to calculate the angle of the line crossing these points in degrees.
I calculated the angle like so:
import numpy as np
p1 = [0, 0.004583285714285714]
p2 = [1, 0.004588714285714285]
x1 = p1[0]
y1 = p1[1]
x2 = p2[0]
y2 = p2[1]
angle = np.rad2deg(np.arctan2(y1 - y2, x2 - x1))
print(angle)
As expected, the angle is a very small negative number (a small downward slope in relation to the X plane):
-0.00031103423163937605
If I plot this, you will see what I mean:
plt.ylim([0,1]) # making y axis range the same as X (a full unit)
plt.plot([x1, x2], [y1, y2])
Clearly the angle of that line is a very small number because the Y values are so small.
I know the lowest y number in this plot is 0.00458 and the highest is 0.00459.
I'm having trouble coming up with the way to scale this properly so that I can obtain this angle instead:
Which is closer to -35 degrees or so (visually).
How can I get the angle a person would see if the chart was plotted with the Y axis ranging only between those min and max values above?
Of course all plots are just for illustration - I'm trying to calculate just the raw angle number given two points and the min and max values for the Y axis.

Solved it, turns out was exceedingly simple and I'm not sure why I was having trouble with it (or why folks seem not to understand the question ¯\ (ツ)/¯ ).
The angle I was looking for can be obtained by
yRange = yMaxValue - yMininumValue
scaledY1 = y1 / yRange
scaledY2 = y2 / yRange
angle = np.rad2deg(np.arctan2(scaledY1 - scaledY2, x2 - x1))
Which for the values posted in the question, result in -28.495638618242538

Understanding opencv's decomposeHomographyMat outputs

I'm trying to find the angle required to move my camera so it's directly in front of an object. If my camera is looking at the object at a 30 degree angle from the left, then my script should return 30 degrees. I'm using cv2.decomposeHomographyMat to find a rotation matrix which works fine. There are 4 solutions returned from this function, so in my script I am outputting 4 angles. Of these angles, there are only two unique angles. My problem is I don't know which of these two angles is correct.
I know the decomposeHomographyMat returns four possible solutions, but shouldn't the angles be the same? I also found the coordinates of my points projected on a 2D plane, but I wasn't sure what to do with this information in regards to finding which angle is correct (here pts3D are the 2D points of the object taken from the camera image with a 0 added for the z column making it 3D pts):
for i in range(len(Ts)):
projectedPts = cv2.projectPoints(pts3D, Rs[i], Ts[i], CAM_MATRIX, DIST_COEFFS)[0][:,0,:]
Here is a snippet from my code. Maybe I am incorrectly determining the angles from the rotation matrix? In my example below, y1 and y2 will be the same angle, and y3 and y4 will be the same angle. Can someone help explain how I determine which angle is the correct angle, and why there are two different angles returned?
def rotationMatrixToEulerAngles(R):
sy = math.sqrt(Rs[0][0] * R[0][0] + R[1][0] * R[1][0])
singular = sy < 1e-6
if not singular :
x = math.atan2(R[2][1] , R[2][2])
y = math.atan2(-R[2][0], sy)
z = math.atan2(R[1][0], R[0][0])
else :
x = math.atan2(-R[1][2], R[1][1])
y = math.atan2(-R[2][0], sy)
z = 0
return np.rad2deg(y)
H, status = cv2.findHomography(corners, REFPOINTS)
output = cv2.warpPerspective(frame, H, (800,800))
# Rs returns 4 matricies, we use the first one
_, Rs, Ts, Ns = cv2.decomposeHomographyMat(H, CAM_MATRIX)
y1 = rotationMatrixToEulerAngles(Rs[0])
y2 = rotationMatrixToEulerAngles(Rs[1])
y3 = rotationMatrixToEulerAngles(Rs[2])
y4 = rotationMatrixToEulerAngles(Rs[3])
Thanks!

Calculate the volume of a 3D polyhedron with Python?

I am trying to figure out the best way of calculating the volume of a 3D polyhedron with Python, and I'm hoping there is a simple solution out there, which I can't seem to find.
Example polyhedron
I did find this post that describes calculating the area of a planar polygon in 3D space, but that doesn't seem to help.

If you only have convex polyhedrons you can use the QHull binding of scipy.spatial.ConvexHull.
import numpy as np
from scipy.spatial import ConvexHull
points = np.array([[....]]) # your points
volume = ConvexHull(points).volume
Additionally, with the module Delaunay you can triangulate your passed points into tetrahedra for other stuff..

Is your polygon such that you can find a point inside so that you can connect every vertex to the point without crossing a face? If so, you can subdivide each face into triangles. You can do this easily by letting one vertex of a face be a pivot point and drawing lines from the other vertices to the pivot vertex. For instance, a pentagon gets divided into three triangles that fan from a common vertex. Each triangle will form a tetrahedron (a 3-sided pyramid) with the point inside. You can then add up the volumes of all of the tetrahedra for each face. The following is for a convex polyhedron that surrounds the origin (x=0,y=0,z=0). It assumes that there is a list of faces f, and each face has a list of vertices v.
def volume(self):
''' calculate volume of polyhedron '''
vol = 0.
for f in self.f: # the faces
n = len(f.v)
v2 = f.v[0] # the pivot of the fan
x2 = v2.x
y2 = v2.y
z2 = v2.z
for i in range(1,n-1): # divide into triangular fan segments
v0 = f.v[i]
x0 = v0.x
y0 = v0.y
z0 = v0.z
v1 = f.v[i+1]
x1 = v1.x
y1 = v1.y
z1 = v1.z
# Add volume of tetrahedron formed by triangle and origin
vol += math.fabs(x0 * y1 * z2 + x1 * y2 * z0 \
+ x2 * y0 * z1 - x0 * y2 * z1 \
- x1 * y0 * z2 - x2 * y1 * z0)
return vol/6.