Result of running tools/sanitize_source_files.py (which was added in https://codereview.appspot.com/6465078/)
This CL is part II of IV (I broke down the 1280 files into 4 CLs).
Review URL: https://codereview.appspot.com/6474054
git-svn-id: http://skia.googlecode.com/svn/trunk@5263 2bbb7eff-a529-9590-31e7-b0007b416f81
diff --git a/experimental/Intersection/LineQuadraticIntersection.cpp b/experimental/Intersection/LineQuadraticIntersection.cpp
index c3e6d23..f269b71 100644
--- a/experimental/Intersection/LineQuadraticIntersection.cpp
+++ b/experimental/Intersection/LineQuadraticIntersection.cpp
@@ -3,13 +3,13 @@
#include "LineUtilities.h"
#include "QuadraticUtilities.h"
-/*
+/*
Find the interection of a line and quadratic by solving for valid t values.
From http://stackoverflow.com/questions/1853637/how-to-find-the-mathematical-function-defining-a-bezier-curve
-"A Bezier curve is a parametric function. A quadratic Bezier curve (i.e. three
-control points) can be expressed as: F(t) = A(1 - t)^2 + B(1 - t)t + Ct^2 where
+"A Bezier curve is a parametric function. A quadratic Bezier curve (i.e. three
+control points) can be expressed as: F(t) = A(1 - t)^2 + B(1 - t)t + Ct^2 where
A, B and C are points and t goes from zero to one.
This will give you two equations:
@@ -17,7 +17,7 @@
x = a(1 - t)^2 + b(1 - t)t + ct^2
y = d(1 - t)^2 + e(1 - t)t + ft^2
-If you add for instance the line equation (y = kx + m) to that, you'll end up
+If you add for instance the line equation (y = kx + m) to that, you'll end up
with three equations and three unknowns (x, y and t)."
Similar to above, the quadratic is represented as
@@ -29,24 +29,24 @@
Using Mathematica, solve for the values of t where the quadratic intersects the
line:
- (in) t1 = Resultant[a*(1 - t)^2 + 2*b*(1 - t)*t + c*t^2 - x,
+ (in) t1 = Resultant[a*(1 - t)^2 + 2*b*(1 - t)*t + c*t^2 - x,
d*(1 - t)^2 + 2*e*(1 - t)*t + f*t^2 - g*x - h, x]
- (out) -d + h + 2 d t - 2 e t - d t^2 + 2 e t^2 - f t^2 +
+ (out) -d + h + 2 d t - 2 e t - d t^2 + 2 e t^2 - f t^2 +
g (a - 2 a t + 2 b t + a t^2 - 2 b t^2 + c t^2)
(in) Solve[t1 == 0, t]
(out) {
{t -> (-2 d + 2 e + 2 a g - 2 b g -
- Sqrt[(2 d - 2 e - 2 a g + 2 b g)^2 -
+ Sqrt[(2 d - 2 e - 2 a g + 2 b g)^2 -
4 (-d + 2 e - f + a g - 2 b g + c g) (-d + a g + h)]) /
(2 (-d + 2 e - f + a g - 2 b g + c g))
},
{t -> (-2 d + 2 e + 2 a g - 2 b g +
- Sqrt[(2 d - 2 e - 2 a g + 2 b g)^2 -
+ Sqrt[(2 d - 2 e - 2 a g + 2 b g)^2 -
4 (-d + 2 e - f + a g - 2 b g + c g) (-d + a g + h)]) /
(2 (-d + 2 e - f + a g - 2 b g + c g))
}
}
-
+
Using the results above (when the line tends towards horizontal)
A = (-(d - 2*e + f) + g*(a - 2*b + c) )
B = 2*( (d - e ) - g*(a - b ) )
@@ -57,19 +57,19 @@
And solve accordingly in Mathematica:
- (in) t2 = Resultant[a*(1 - t)^2 + 2*b*(1 - t)*t + c*t^2 - g'*y - h',
+ (in) t2 = Resultant[a*(1 - t)^2 + 2*b*(1 - t)*t + c*t^2 - g'*y - h',
d*(1 - t)^2 + 2*e*(1 - t)*t + f*t^2 - y, y]
- (out) a - h' - 2 a t + 2 b t + a t^2 - 2 b t^2 + c t^2 -
+ (out) a - h' - 2 a t + 2 b t + a t^2 - 2 b t^2 + c t^2 -
g' (d - 2 d t + 2 e t + d t^2 - 2 e t^2 + f t^2)
(in) Solve[t2 == 0, t]
(out) {
{t -> (2 a - 2 b - 2 d g' + 2 e g' -
- Sqrt[(-2 a + 2 b + 2 d g' - 2 e g')^2 -
+ Sqrt[(-2 a + 2 b + 2 d g' - 2 e g')^2 -
4 (a - 2 b + c - d g' + 2 e g' - f g') (a - d g' - h')]) /
(2 (a - 2 b + c - d g' + 2 e g' - f g'))
},
{t -> (2 a - 2 b - 2 d g' + 2 e g' +
- Sqrt[(-2 a + 2 b + 2 d g' - 2 e g')^2 -
+ Sqrt[(-2 a + 2 b + 2 d g' - 2 e g')^2 -
4 (a - 2 b + c - d g' + 2 e g' - f g') (a - d g' - h')])/
(2 (a - 2 b + c - d g' + 2 e g' - f g'))
}
@@ -80,7 +80,7 @@
B = 2*(-(a - b ) + g'*(d - e ) )
C = ( (a ) - g'*(d ) - h' )
*/
-
+
class LineQuadraticIntersections : public Intersections {
public:
@@ -156,7 +156,7 @@
}
protected:
-
+
double findLineT(double t) {
const double* qPtr;
const double* lPtr;