CODEWITHSUNDEEP
pythonnumpysympy
Shailender Joseph
Shailender Joseph

Reputation: 1

Solving for intersection points between parametric curves

The parametric equations of the two curves are as follows:

Curve1: r(t) = (2(t-sin(t)),2(1 -cos(t)))

Curve2: s(t) = (2t - sin(t),2 - cos(t))

I need to find the points of intersection in the region [0,4π].

I was able to plot the graph for the mentioned region and observed 4 points of intersection. But I am not able to determine the exact points of intersection.

For non-parametric equations, fsolve from sympy can be used, but the curves which are given in their parametric forms, I am not able to find a workaround. 

t = np.arange(-0.25*np.pi,4.25*np.pi,0.01)
rx = np.zeros(len(t))
ry = np.zeros(len(t))
for i in t:
   rx = 2*(t - np.sin(t))
   ry = 2*(1 - np.cos(t))
sx = np.zeros(len(t))
sy = np.zeros(len(t))
for i in t:
   sx = 2*t - np.sin(t)
   sy = 2 - np.cos(t)
plt.plot(rx,ry)
plt.plot(sx,sy)

Upvotes: 0

Views: 1361

Answers (2)

asmeurer
asmeurer

Reputation: 91500

The two curves don't intersect at the same time (that would be points where sin(t) = cos(t) = 0, which has no solutions). So you really want to know when 

R = (2*t1 - 2*sin(t1), 2 - 2*cos(t1))
S = (2*t2 - sin(t2), 2 - cos(t2))

intersect.

This is two equations with two unknowns, so it is straightforward to solve with sympy.nsolve. You have to fiddle with the starting values a bit to find the ones that converge to different solutions. If you know what they are roughly from the graph that is the best place to start. 

>>> t1, t2 = symbols('t1 t2')
>>> R = (2*t1 - 2*sin(t1), 2 - 2*cos(t1))
>>> S = (2*t2 - sin(t2), 2 - cos(t2))
>>> nsolve([R[0] - S[0], R[1] - S[1]], [t1, t2], [1, 1])
Matrix([
[ 1.09182358380672],
[0.398264297579454]])
>>> nsolve([R[0] - S[0], R[1] - S[1]], [t1, t2], [5, 5])
Matrix([
[5.19136172337286],
[5.88492100960013]])
>>> nsolve([R[0] - S[0], R[1] - S[1]], [t1, t2], [7, 7])
Matrix([
[7.37500889098631],
[6.68144960475904]])
>>> nsolve([R[0] - S[0], R[1] - S[1]], [t1, t2], [10, 10])
Matrix([
[11.4745470305524],
[12.1681063167797]])

Upvotes: 0

smichr
smichr

Reputation: 19047

For a given x you can find t for each curve and see if the corresponding y are the same. You can step over the x range with some grid looking for such locations where the thee curves hit and use bisection to zero in on a more precise x. Since you can't solve the parametrix x(t) - x for t, nsolve will have to be used to find an approximate t. Something like this finds values for your 4 roots (confirmed graphically) after correcting your OP equation for Curve1 to be the same as in the code you wrote afterwards.

f = lambda xx: a[1].subs(t, tt)-b[1].subs(t,nsolve(b[0]-xx,tlast))
tlast = 0 # guess for t for a given xx to be updated as we go
tol = 1e-9  # how tight the bounds on x must be for a solution
dx = 0.1
for ix in range(300):
 xx = ix*dx
 tt=nsolve(a[0]-xx,tlast)
 y2 = f(xx)
 if ix != 0 and yold*y2 < 0 and tt<4*pi:
   tlast = tt  # updating guess for t
   # bisect for better xx now that bounding xx are found
   x1 = xx-dx
   x2 = xx
   y1 = yold
   while x2 - x1 > tol:
     xm = (x1 + x2)/2
     ym = f(xm)
     if ym*y1 < 0:
       y2 = ym
       x2 = xm
     elif ym != 0:
       y1 = ym
       x1 = xm
     else:
       break
   print(xm)  # a solution
 yold = y2

I don't know of a more automated way to do this in SymPy.

Upvotes: 1

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