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Update/add some rudimentary comments

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This commit is contained in:
James Godfrey-Kittle 2016-07-28 12:32:29 -07:00
parent f7b9ebba06
commit c0b6ed92e3
1 changed files with 7 additions and 2 deletions
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9
Lib/cu2qu/__init__.py
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@ -79,6 +79,8 @@ def split_cubic_into_two(p0, p1, p2, p3):
def split_cubic_into_three(p0, p1, p2, p3, _27=1/27):
# we define 1/27 as a keyword argument so that it will only be evaluated
# only once but still in the scope of this function
mid1 = (8*p0 + 12*p1 + 6*p2 + p3) * _27
deriv1 = (p3 + 3*p2 - 4*p0) * _27
mid2 = (p0 + 6*p1 + 12*p2 + 8*p3) * _27
@ -148,10 +150,11 @@ def cubic_farthest_fit(p0, p1, p2, p3, tolerance):
def cubic_approx_spline(cubic, n, tolerance):
"""Approximate a cubic bezier curve with a spline of n quadratics.
Returns None if no quadratic approximation is found at or below the
tolerance.
Returns None if no quadratic approximation is found which lies entirely
within a distance `tolerance` from the original curve.
"""
# special-case single-segment spline using intersection of ab and cd
if n == 1:
q1 = calc_intersect(*cubic)
if q1 is None:
@ -167,6 +170,7 @@ def cubic_approx_spline(cubic, n, tolerance):
return None
return c0, q1, c3
# calculate the spline of quadratics
spline = [cubic[0]]
if n == 2:
segments = split_cubic_into_two(*cubic)
@ -178,6 +182,7 @@ def cubic_approx_spline(cubic, n, tolerance):
spline.append(cubic_approx_control(segments[i], i / (n - 1)))
spline.append(cubic[3])
# determine whether the spline is within the tolerance error
for i in range(1, n + 1):
if i == 1:
q0, q1, q2 = spline[0], spline[1], (spline[1] + spline[2]) * .5