Best-Fit without point interpolation

Question

I have two sets of data. One is nominal form. The other is actual form. The problem is that when I wish to calculate the form error alone. It's a big problem when the two sets of data isn't "on top of each other". That gives errors that also include positional error.

Both curves are read from a series of data. The nominal shape (black) is made up from many different size radius that are tangent to each other. Its the leading edge of an airfoil profile.

I have tried various methods of "Best-Fit" I've found both here and on where ever google took me. But the problem is that they all smooth my "actual" data. So it get modified and is not keeping it's actual form.

Is there any function in scipy or any other python lib that "simply" can fit my two curves together without altering the actual shape? I wish for the green curve with red dots to lie as much as possible on top of the black.

Might it be possible to calculate the center of gravity of both curves and then move the actual curve in x and y depending on the value difference from the center point? It might not be the ultimate solution, but it would get closer?

Answer 1

Here is a solution assuming that the nominal form can be described as a conic, i.a as solution of the equation ax^2 + by^2 + cxy + dx + ey = 1. Then, a least square fit can be applied to find the coefficients (a, b, c, d, e).

import numpy as np
import matplotlib.pylab as plt

# Generate example data
t = np.linspace(-2, 2.5, 25)
e, theta = 0.5, 0.3  # ratio minor axis/major & orientation angle major axis
c, s = np.cos(theta), np.sin(theta)
x = c*np.cos(t) - s*e*np.sin(t)
y = s*np.cos(t) + c*e*np.sin(t)

# add noise:
xy = 4*np.vstack((x, y))
xy += .08 *np.random.randn(*xy.shape) + np.random.randn(2, 1)


# Least square fit by a generic conic equation
# a*x^2 + b*y^2 + c*x*y + d*x + e*y = 1
x, y = xy
x = x - x.mean()
y = y - y.mean()

M = np.vstack([x**2, y**2, x*y, x, y]).T
b = np.ones_like(x)
# solve M*w = b
w, res, rank, s = np.linalg.lstsq(M, b, rcond=None)
a, b, c, d, e = w

# Get x, y coordinates for the fitted ellipse:
# using polar coordinates
# x = r*cos(theta), y = r*sin(theta)
# for a given theta, the radius is obtained with the 2nd order eq.:
# (a*ct^2 + b*st^2 + c*cs*st)*r^2 + (d*ct + e*st)*r - 1 = 0
# with ct = cos(theta) and st = sin(theta)

theta = np.linspace(-np.pi, np.pi, 97)
ct, st = np.cos(theta), np.sin(theta)

A = a*ct**2 + b*st**2 + c*ct*st
B = d*ct + e*st
D = B**2 + 4*A
radius = (-B + np.sqrt(D))/2/A

# Graph
plt.plot(radius*ct, radius*st, '-k', label='fitted ellipse');
plt.plot(x, y, 'or', label='measured points');
plt.axis('equal'); plt.legend();
plt.xlabel('x'); plt.ylabel('y');