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
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
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!
Let's say you have two arrays of data values from a calculation, that you can model with a continuos, differentiable function each. Both "lines" of data points intersect at (at least) one point and now the question is whether the functions behind these datasets are actually crossing or anticrossing.
The image below shows the situation, where I know (from the physics behind it) that at the upper two "contact points" the yellow and green lines actually should "switch color", whereas at the lower one both functions go out of each others way:
To give an easier "toy set" of data, take this code for example:
import matplotlib.pyplot as plt
import numpy as np
x=np.arange(-10,10,.5)
y1=[np.absolute(i**3)+100*np.absolute(i) for i in x]
y2=[-np.absolute(i**3)-100*np.absolute(i) for i in x][::-1]
plt.scatter(x,y1)
plt.scatter(x,y2,color='r')
plt.show()
Which should produce the following image:
Now how could I extrapolate whether the trend behind the data is crossing (so the data from the lower left continues to the upper right) or anti-crossing (as indicated with the colors above, the data from the lower left continues to the lower right)?
So far I was able to find the "contact point" between these to datasets by looking at the derivative of the Difference between them, roughly like this:
closePoints=np.where(np.diff(np.diff(array_A - array_B) > 0))[0] + 1
(which probably would be faster to evaluate with something like scipy's cKDTree).
Should I go on and (probably very inefficiently) check the derivative on both sides of the intersection? Or can I somehow check if the extrapolation of the data on the left side fits better to crossing or anticrossing?
I understood your problem as:
You have two sequences of points in a 2D plane.
The true curves can be approximated by straight lines between consecutive points of the sequences.
You want to know how often and where the two curves intersect (not only come into contact but really cross each other) (polygon intersection).
A potential solution is:
You look at each combination of a line segment of one curve with a line segment of another curve.
Combinations where the bounding boxes of the line segments have an overlap can potentially contain intersection points.
You solve a linear equation system to compute if and where an intersection between two lines occurs
In case of no solution to the equation system the lines are parallel but not overlapping, dismiss this case
In case of one solution check that it is truly within the segments, if so record this crossing point
In case of infinitely many intersections the lines are identical. This is also no real crossing and can be dismissed.
Do this for all combinations of line segments and eliminate twin cases, i.e. where the two curves intersect at a segment start or end
Let me give some details:
How to check if two bounding-boxes (rectangles) of the segments overlap so that the segments potentially can intersect?
The minimal x/y value of one rectangle must be smaller than the maximal x/y value of the other. This must hold for both.
If you have two segments how do you solve for intersection points?
Let's say segment A has two points (x1, y1) and (x2, y2) and segment B has two points (x2, y3) and (x4, y4).
Then you simply have two parametrized line equations which have to be set equal:
(x1, y1) + t * (x2 - x1, y2 - y1) = (x3, y3) + q * (x4 - x3, y4 - y3)
And you need to find all solutions where t or q in [0, 1). The corresponding linear equation system may be rank deficient or not solvable at all, best is to use a general solver (I chose numpy.linalg.lstsq) that does everything in one go.
Curves sharing a common point
Surprisingly difficult are cases where one point is common in the segmentation of both curves. The difficulty lies then in the correct decision of real intersection vs. contact points. The solution is to compute the angle of both adjacent segments of both curves (gives 4 angles) around the common point and look at the order of the angles. If both curves come alternating when going around the equal point then it's an intersection, otherwise it isn't.
And a code example based on your data:
import math
import matplotlib.pyplot as plt
import numpy as np
def intersect_curves(x1, y1, x2, y2):
"""
x1, y1 data vector for curve 1
x2, y2 data vector for curve 2
"""
# number of points in each curve, number of segments is one less, need at least one segment in each curve
N1 = x1.shape[0]
N2 = x2.shape[0]
# get segment presentation (xi, xi+1; xi+1, xi+2; ..)
xs1 = np.vstack((x1[:-1], x1[1:]))
ys1 = np.vstack((y1[:-1], y1[1:]))
xs2 = np.vstack((x2[:-1], x2[1:]))
ys2 = np.vstack((y2[:-1], y2[1:]))
# test if bounding-boxes of segments overlap
mix1 = np.tile(np.amin(xs1, axis=0), (N2-1,1))
max1 = np.tile(np.amax(xs1, axis=0), (N2-1,1))
miy1 = np.tile(np.amin(ys1, axis=0), (N2-1,1))
may1 = np.tile(np.amax(ys1, axis=0), (N2-1,1))
mix2 = np.transpose(np.tile(np.amin(xs2, axis=0), (N1-1,1)))
max2 = np.transpose(np.tile(np.amax(xs2, axis=0), (N1-1,1)))
miy2 = np.transpose(np.tile(np.amin(ys2, axis=0), (N1-1,1)))
may2 = np.transpose(np.tile(np.amax(ys2, axis=0), (N1-1,1)))
idx = np.where((mix2 <= max1) & (max2 >= mix1) & (miy2 <= may1) & (may2 >= miy1)) # overlapping segment combinations
# going through all the possible segments
x0 = []
y0 = []
for (i, j) in zip(idx[0], idx[1]):
# get segment coordinates
xa = xs1[:, j]
ya = ys1[:, j]
xb = xs2[:, i]
yb = ys2[:, i]
# ax=b, prepare matrices a and b
a = np.array([[xa[1] - xa[0], xb[0] - xb[1]], [ya[1] - ya[0], yb[0]- yb[1]]])
b = np.array([xb[0] - xa[0], yb[0] - ya[0]])
r, residuals, rank, s = np.linalg.lstsq(a, b)
# if this is not a
if rank == 2 and not residuals and r[0] >= 0 and r[0] < 1 and r[1] >= 0 and r[1] < 1:
if r[0] == 0 and r[1] == 0 and i > 0 and j > 0:
# super special case of one segment point (not the first) in common, need to differentiate between crossing or contact
angle_a1 = math.atan2(ya[1] - ya[0], xa[1] - xa[0])
angle_b1 = math.atan2(yb[1] - yb[0], xb[1] - xb[0])
# get previous segment
xa2 = xs1[:, j-1]
ya2 = ys1[:, j-1]
xb2 = xs2[:, i-1]
yb2 = ys2[:, i-1]
angle_a2 = math.atan2(ya2[0] - ya2[1], xa2[0] - xa2[1])
angle_b2 = math.atan2(yb2[0] - yb2[1], xb2[0] - xb2[1])
# determine in which order the 4 angle are
if angle_a2 < angle_a1:
h = angle_a1
angle_a1 = angle_a2
angle_a2 = h
if (angle_b1 > angle_a1 and angle_b1 < angle_a2 and (angle_b2 < angle_a1 or angle_b2 > angle_a2)) or\
((angle_b1 < angle_a1 or angle_b1 > angle_a2) and angle_b2 > angle_a1 and angle_b2 < angle_a2):
# both in or both out, just a contact point
x0.append(xa[0])
y0.append(ya[0])
else:
x0.append(xa[0] + r[0] * (xa[1] - xa[0]))
y0.append(ya[0] + r[0] * (ya[1] - ya[0]))
return (x0, y0)
# create data
def data_A():
# data from question (does not intersect)
x1 = np.arange(-10, 10, .5)
x2 = x1
y1 = [np.absolute(x**3)+100*np.absolute(x) for x in x1]
y2 = [-np.absolute(x**3)-100*np.absolute(x) for x in x2][::-1]
return (x1, y1, x2, y2)
def data_B():
# sine, cosine, should have some intersection points
x1 = np.arange(-10, 10, .5)
x2 = x1
y1 = np.sin(x1)
y2 = np.cos(x2)
return (x1, y1, x2, y2)
def data_C():
# a spiral and a diagonal line, showing the more general case
t = np.arange(0, 10, .2)
x1 = np.sin(t * 2) * t
y1 = np.cos(t * 2) * t
x2 = np.arange(-10, 10, .5)
y2 = x2
return (x1, y1, x2, y2)
def data_D():
# parallel and overlapping, should give no intersection point
x1 = np.array([0, 1])
y1 = np.array([0, 0])
x2 = np.array([-1, 3])
y2 = np.array([0, 0])
return (x1, y1, x2, y2)
def data_E():
# crossing at a segment point, should give exactly one intersection point
x1 = np.array([-1,0,1])
y1 = np.array([0,0,0])
x2 = np.array([0,0,0])
y2 = np.array([-1,0,1])
return (x1, y1, x2, y2)
def data_F():
# contacting at one segment point, should give no intersection point
x1 = np.array([-1,0,-1])
y1 = np.array([-1,0,1])
x2 = np.array([1,0,1])
y2 = np.array([-1,0,1])
return (x1, y1, x2, y2)
x1, y1, x2, y2 = data_F() # select the data you like here
# show example data
plt.plot(x1, y1, 'b-o')
plt.plot(x2, y2, 'r-o')
# call to intersection computation
x0, y0 = intersect_curves(x1, y1, x2, y2)
print('{} intersection points'.format(len(x0)))
# display intersection points in green
plt.plot(x0, y0, 'go')
plt.show() # zoom in to see that the algorithm is correct
I tested it extensively and should get most (all) border cases right (see data_A-F in code). Some examples:
Some Comments:
The assumption about the line approximation is crucial. Most true curves might only be to some extent be approximable to lines locally. Because of this places where the two curves come close but to not intersect with a distance in the order of the distance of consecutive sampling points of your curve - you may obtain false positives or false negatives. The solution is then to either use more points or to use additonal knowledge about the true curves. Splines might give a lower error rate but also require more computations, better sampling of the curves would be preferable then.
Self-intersection is trivially included when taking two times the same curve and let them intersect
This solution has the additional advantage that it isn't restricted to curves of the form y=f(x) but it's applicable to arbitrary curves in 2D.
You could use a spline interpolation for the difference function g(x) = y1(x) - y(2). Finding the minimum of the square g(x)**2 would be a contact or crossing point. Looking at the first and second derivative you could decide if it is a contact point( g(x) has minimum, g'(x)==0, g''(x) != 0) or a crossing point (g(x) is a stationary point, g'(x)==0, g''(x)==0).
The following code searches for a minimum of g(x)**2 in constrained interval and then plot the derivatives. The use of a constrained interval is to find multiple points successively by excluding intervals in which previous points were.
import matplotlib.pyplot as plt
import numpy as np
import scipy.optimize as sopt
import scipy.interpolate as sip
# test functions:
nocrossingTest = True
if nocrossingTest:
f1 = lambda x: +np.absolute(x**3)+100*np.absolute(x)
f2 = lambda x: -np.absolute(x**3)-100*np.absolute(x)
else:
f1 = lambda x: +np.absolute(x**3)+100*x
f2 = lambda x: -np.absolute(x**3)-100*x
xp = np.arange(-10,10,.5)
y1p, y2p = f1(xp), f2(xp) # test array
# Do Interpolation of y1-y2 to find crossing point:
g12 = sip.InterpolatedUnivariateSpline(xp, y1p - y2p) # Spline Interpolator of Difference
dg12 = g12.derivative() # spline derivative
ddg12 = dg12.derivative() # spline derivative
# Bounded least square fit to find minimal distance
gg = lambda x: g12(x)*g12(x)
rr = sopt.minimize_scalar(gg, bounds=[-1,1]) # search minium in Interval [-1,1]
x_c = rr['x'] # x value with minimum distance
print("Crossing point is at x = {} (Distance: {})".format(x_c, g12(x_c)))
fg = plt.figure(1)
fg.clf()
fg,ax = plt.subplots(1, 1,num=1)
ax.set_title("Function Values $y$")
ax.plot(xp, np.vstack([y1p,y2p]).T, 'x',)
xx = np.linspace(xp[0], xp[-1], 1000)
ax.plot(xx, np.vstack([f1(xx), f2(xx)]).T, '-', alpha=0.5)
ax.grid(True)
ax.legend(loc="best")
fg.canvas.draw()
fg = plt.figure(2)
fg.clf()
fg,axx = plt.subplots(3, 1,num=2)
axx[0].set_title("$g(x) = y_1(x) - y_2(x)$")
axx[1].set_title("$dg(x)/dx$")
axx[2].set_title("$d^2g(x)/dx^2$")
for ax,g in zip(axx, [g12, dg12, ddg12]):
ax.plot(xx, g(xx))
ax.plot(x_c, g(x_c), 'ro', alpha=.5)
ax.grid(True)
fg.tight_layout()
plt.show()
The difference function show that the difference is not smooth: