#include "geometry.h"
#include <math.h>
-#include "circle.h"
-#include "dimension.h"
-#include "line.h"
+#include <stdio.h>
+#include "global.h"
#include "mathconstants.h"
+// This is unused
+#if 0
Point Geometry::IntersectionOfLineAndLine(Point p1, Point p2, Point p3, Point p4)
{
// Find the intersection of the lines by formula:
return Point(px / d, py / d, 0);
}
+#endif
// Returns the parameter of a point in space to this vector. If the parameter
//
Point Geometry::RotatePointAroundPoint(Point point, Point rotationPoint, double angle)
{
-// Vector v = Vector(point, rotationPoint);
Vector v = Vector(rotationPoint, point);
double px = (v.x * cos(angle)) - (v.y * sin(angle));
double py = (v.x * sin(angle)) + (v.y * cos(angle));
}
+void Geometry::Intersects(Object * obj1, Object * obj2)//, double * tp/*= 0*/, double * up/*= 0*/, double * vp/*= 0*/, double * wp/*= 0*/)
+{
+ Global::numIntersectPoints = Global::numIntersectParams = 0;
+
+ if ((obj1->type == OTLine) && (obj2->type == OTLine))
+ CheckLineToLineIntersection(obj1, obj2);
+ else if ((obj1->type == OTCircle) && (obj2->type == OTCircle))
+ CheckCircleToCircleIntersection(obj1, obj2);
+}
+
+
/*
Intersecting line segments:
An easier way:
*/
-#if 0
-// Finds the intesection between two objects (if any)
-bool Geometry::Intersects(Object * obj1, Object * obj2, double * t, double * s)
+// Finds the intersection between two lines (if any)
+void Geometry::CheckLineToLineIntersection(Object * l1, Object * l2)//, double * tp, double * up)
{
-}
-#endif
+ Global::numIntersectPoints = Global::numIntersectParams = 0;
+ Vector r(l1->p[0], l1->p[1]);
+ Vector s(l2->p[0], l2->p[1]);
+ Vector v1 = l2->p[0] - l1->p[0]; // q - p
#if 0
-// Finds the intersection between two lines (if any)
-int Geometry::Intersects(Line * l1, Line * l2, double * tp/*= 0*/, double * up/*= 0*/)
-{
- Vector r(l1->position, l1->endpoint);
- Vector s(l2->position, l2->endpoint);
- Vector v1 = l2->position - l1->position; // q - p
-// Vector v2 = l1->position - l2->position; // p - q
-//printf("l1: (%lf, %lf) (%lf, %lf), l2: (%lf, %lf) (%lf, %lf)\n", l1->position.x, l1->position.y, l1->endpoint.x, l1->endpoint.y, l2->position.x, l2->position.y, l2->endpoint.x, l2->endpoint.y);
+ Vector v2 = l1->p[0] - l2->p[0]; // p - q
+printf("l1: (%lf, %lf) (%lf, %lf), l2: (%lf, %lf) (%lf, %lf)\n", l1->p[0].x, l1->p[0].y, l1->p[1].x, l1->p[1].y, l2->p[0].x, l2->p[0].y, l2->p[1].x, l2->p[1].y);
+#endif
double rxs = (r.x * s.y) - (s.x * r.y);
double t, u;
{
double qpxr = (v1.x * r.y) - (r.x * v1.y);
-//printf(" --> R x S = 0! (q - p) x r = %lf\n", qpxr);
-//printf(" -->(q - p) . r = %lf, r . r = %lf\n", v1.Dot(r), r.Dot(r));
-//printf(" -->(p - q) . s = %lf, s . s = %lf\n", v2.Dot(s), s.Dot(s));
-//printf(" -->(q - p) . s = %lf, (p - q) . r = %lf\n", v1.Dot(s), v2.Dot(r));
+#if 0
+printf(" --> R x S = 0! (q - p) x r = %lf\n", qpxr);
+printf(" -->(q - p) . r = %lf, r . r = %lf\n", v1.Dot(r), r.Dot(r));
+printf(" -->(p - q) . s = %lf, s . s = %lf\n", v2.Dot(s), s.Dot(s));
+printf(" -->(q - p) . s = %lf, (p - q) . r = %lf\n", v1.Dot(s), v2.Dot(r));
+#endif
// Lines are parallel, so no intersection...
if (qpxr != 0)
- return 0;
+ return;
#if 0
//this works IFF the vectors are pointing in the same direction. everything else
return 0;
#else
// Check to see which endpoints are connected... Four possibilities:
- if (l1->position == l2->position)
+ if (l1->p[0] == l2->p[0])
t = 0, u = 0;
- else if (l1->position == l2->endpoint)
+ else if (l1->p[0] == l2->p[1])
t = 0, u = 1.0;
- else if (l1->endpoint == l2->position)
+ else if (l1->p[1] == l2->p[0])
t = 1.0, u = 0;
- else if (l1->endpoint == l2->endpoint)
+ else if (l1->p[1] == l2->p[1])
t = 1.0, u = 1.0;
else
- return 0;
+ return;
#endif
}
else
5. Otherwise, the two line segments are not parallel but do not intersect.
*/
// Return parameter values, if user passed in valid pointers
+#if 0
if (tp)
*tp = t;
return 1;
return 0;
+#else
+ Global::intersectParam[0] = t;
+ Global::intersectParam[1] = u;
+
+ // If the parameters are in range, we have overlap!
+ if ((t >= 0) && (t <= 1.0) && (u >= 0) && (u <= 1.0))
+ Global::numIntersectParams = 1;
+#endif
}
+void Geometry::CheckCircleToCircleIntersection(Object * c1, Object * c2)
+{
+ // Set up global vars
+ Global::numIntersectPoints = Global::numIntersectParams = 0;
+
+ // Get the distance between the centers of the circles
+ Vector centerLine(c1->p[0], c2->p[0]);
+ double d = centerLine.Magnitude();
+ double clAngle = centerLine.Angle();
+
+ // If the distance between centers is greater than the sum of the radii or
+ // less than the difference between the radii, there is NO intersection
+ if ((d > (c1->radius[0] + c2->radius[0]))
+ || (d < fabs(c1->radius[0] - c2->radius[0])))
+ return;
+
+ // If the distance between centers is equal to the sum of the radii or
+ // equal to the difference between the radii, the intersection is tangent
+ // to both circles.
+ if ((d == (c1->radius[0] + c2->radius[0]))
+ || (d == fabs(c1->radius[0] - c2->radius[0])))
+ {
+ Global::intersectPoint[0].x = c1->p[0].x + (cos(clAngle) * c1->radius[0]);
+ Global::intersectPoint[0].y = c1->p[0].y + (sin(clAngle) * c1->radius[0]);
+ Global::numIntersectPoints = 1;
+ return;
+ }
+
+ // Use the Law of Cosines to find the angle between the centerline and the
+ // radial line on Circle #1
+ double a = acos(((c1->radius[0] * c1->radius[0]) + (d * d) - (c2->radius[0] * c2->radius[0])) / (2.0 * c1->radius[0] * d));
+
+ // Finally, find the points of intersection by using +/- the angle found
+ // from the centerline's angle
+ Global::intersectPoint[0].x = c1->p[0].x + (cos(clAngle + a) * c1->radius[0]);
+ Global::intersectPoint[0].y = c1->p[0].y + (sin(clAngle + a) * c1->radius[0]);
+ Global::intersectPoint[1].x = c1->p[0].x + (cos(clAngle - a) * c1->radius[0]);
+ Global::intersectPoint[1].y = c1->p[0].y + (sin(clAngle - a) * c1->radius[0]);
+ Global::numIntersectPoints = 2;
+}
+
+
+#if 0
// Finds the intersection between two lines (if any)
int Geometry::Intersects(Line * l1, Dimension * d1, double * tp/*= 0*/, double * up/*= 0*/)
{
// Find the distance from the center of c1 to the perpendicular chord
// (which contains the points of intersection)
+ // [N.B.: This is derived from Pythagorus by using the unknown distance
+ // from the center line to the point where the two radii coincide as
+ // a common unknown to two instances of the formula.]
double x = ((d * d) - (c2->radius * c2->radius) + (c1->radius * c1->radius))
/ (2.0 * d);
// Find the the length of the perpendicular chord
// should we just do common trig solves, like AAS, ASA, SAS, SSA?
// Law of Cosines:
-// c^2 = a^2 + b^2 -2ab*cos(C)
+// c² = a² + b² - 2ab * cos(C)
// Solving for C:
-// cos(C) = (c^2 - a^2 - b^2) / -2ab = (a^2 + b^2 - c^2) / 2ab
+// cos(C) = (c² - a² - b²) / -2ab = (a² + b² - c²) / 2ab
// Law of Sines:
// a / sin A = b / sin B = c / sin C
// Use law of sines to find 2nd & 3rd angles
// sin A / a = sin B / b
// sin B = (sin A / a) * b
+// B = arcsin( sin A * (b / a))
+// ??? ==> B = A * arcsin(b / a)
+/*
+Well, look here:
+sin B = sin A * (b / a)
+sin B / sin A = b / a
+arcsin( sin B / sin A ) = arcsin( b / a )
+
+hmm... dunno...
+*/
+
double angle2 = asin(s2 * (sin(angle1) / s1));
double angle3 = asin(s3 * (sin(angle1) / s1));
*a3 = angle3;
}
+
+Point Geometry::GetPointForParameter(Object * obj, double t)
+{
+ if (obj->type == OTLine)
+ {
+ // Translate line vector to the origin, then add the scaled vector to
+ // initial point of the line.
+ Vector v = obj->p[1] - obj->p[0];
+ return obj->p[0] + (v * t);
+ }
+
+ return Point(0, 0);
+}
+
projects / architektonas / blobdiff