X-Git-Url: http://shamusworld.gotdns.org/cgi-bin/gitweb.cgi?a=blobdiff_plain;f=src%2Fgeometry.cpp;h=1ec93db7e1430b2752cc4e77a2b2d9d30ce61aeb;hb=7f3a6b11585376eecd80979ec3da2346c5314d88;hp=058dbfa14c06895ce659db9310ad778e57f7c925;hpb=c85958d34fac175452fe420ff24ea82f26d6f1f3;p=architektonas

diff --git a/src/geometry.cpp b/src/geometry.cpp
index 058dbfa..1ec93db 100644
--- a/src/geometry.cpp
+++ b/src/geometry.cpp
@@ -18,6 +18,7 @@
 #include "circle.h"
 #include "dimension.h"
 #include "line.h"
+#include "mathconstants.h"
 
 
 Point Geometry::IntersectionOfLineAndLine(Point p1, Point p2, Point p3, Point p4)
@@ -215,7 +216,7 @@ int Geometry::Intersects(Line * l1, Dimension * d1, double * tp/*= 0*/, double *
 }
 
 
-// Finds the intesection(s) between a line and a circle (if any)
+// Finds the intersection(s) between a line and a circle (if any)
 int Geometry::Intersects(Line * l, Circle * c, double * tp/*= 0*/, double * up/*= 0*/, double * vp/*= 0*/, double * wp/*= 0*/)
 {
 #if 0
@@ -263,7 +264,16 @@ I'm thinking a better approach to this might be as follows:
 	// Now we have to check for intersection points.
 	// Tangent case: (needs to return something)
 	if ((distance == c->radius) && (t >= 0.0) && (t <= 1.0))
+	{
+		// Need to set tp & up to something... !!! FIX !!!
+		if (tp)
+			*tp = t;
+
+		if (up)
+			*up = Vector(c->position, p).Angle();
+
 		return 1;
+	}
 
 	// The line intersects the circle in two points (possibly). Use Pythagorus
 	// to find them for testing.
@@ -300,3 +310,116 @@ I'm thinking a better approach to this might be as follows:
 }
 
 
+// Finds the intersection(s) between a circle and a circle (if any)
+// There can be 0, 1, or 2 intersections.
+// Returns the angles of the points of intersection in tp thru wp, with the
+// angles returned as c1, c2, c1, c2 (if applicable--in the 1 intersection case,
+// only the first two angles are returned: c1, c2).
+int Geometry::Intersects(Circle * c1, Circle * c2, double * tp/*= 0*/, double * up/*= 0*/, double * vp/*= 0*/, double * wp/*= 0*/, Point * p1/*= 0*/, Point * p2/*= 0*/)
+{
+	// Get the distance between centers. If the distance plus the radius of the
+	// smaller circle is less than the radius of the larger circle, there is no
+	// intersection. If the distance is greater than the sum of the radii,
+	// there is no intersection. If the distance is equal to the sum of the
+	// radii, they are tangent and intersect at one point. Otherwise, they
+	// intersect at two points.
+	Vector centerLine(c1->position, c2->position);
+	double d = centerLine.Magnitude();
+//printf("Circle #1: pos=<%lf, %lf>, r=%lf\n", c1->position.x, c1->position.y, c1->radius);
+//printf("Circle #2: pos=<%lf, %lf>, r=%lf\n", c2->position.x, c2->position.y, c2->radius);
+//printf("Distance between #1 & #2: %lf\n", d);
+
+	// Check to see if we actually have an intersection, and return failure if not
+	if ((fabs(c1->radius - c2->radius) > d) || ((c1->radius + c2->radius) < d))
+		return 0;
+
+	// There are *two* tangent cases!
+	if (((c1->radius + c2->radius) == d) || (fabs(c1->radius - c2->radius) == d))
+	{
+		// Need to return something in tp & up!! !!! FIX !!! [DONE]
+		if (tp)
+			*tp = centerLine.Angle();
+
+		if (up)
+			*up = centerLine.Angle() + PI;
+
+		return 1;
+	}
+
+	// Find the distance from the center of c1 to the perpendicular chord
+	// (which contains the points of intersection)
+	double x = ((d * d) - (c2->radius * c2->radius) + (c1->radius * c1->radius))
+		/ (2.0 * d);
+	// Find the the length of the perpendicular chord
+// Not needed...!
+	double a = sqrt((-d + c2->radius - c1->radius) * (-d - c2->radius + c1->radius) * (-d + c2->radius + c1->radius) * (d + c2->radius + c1->radius)) / d;
+
+	// Now, you can use pythagorus to find the length of the hypotenuse, but we
+	// already know that length, it's the radius! :-P
+	// What's needed is the angle of the center line and the radial line. Since
+	// there's two intersection points, there's also four angles (two for each
+	// circle)!
+	// We can use the arccos to find the angle using just the radius and the
+	// distance to the perpendicular chord...!
+	double angleC1 = acos(x / c1->radius);
+	double angleC2 = acos((d - x) / c2->radius);
+
+	if (tp)
+		*tp = centerLine.Angle() - angleC1;
+
+	if (up)
+		*up = (centerLine.Angle() + PI) - angleC2;
+
+	if (vp)
+		*vp =  centerLine.Angle() + angleC1;
+
+	if (wp)
+		*wp = (centerLine.Angle() + PI) + angleC2;
+
+	if (p1)
+		*p1 = c1->position + (centerLine.Unit() * x) + (Vector::Normal(Vector(), centerLine) * (a / 2.0));
+
+	if (p2)
+		*p2 = c1->position + (centerLine.Unit() * x) - (Vector::Normal(Vector(), centerLine) * (a / 2.0));
+
+	return 2;
+}
+
+
+// should we just do common trig solves, like AAS, ASA, SAS, SSA?
+// Law of Cosines:
+// c^2 = a^2 + b^2 -2ab*cos(C)
+// Solving for C:
+// cos(C) = (c^2 - a^2 - b^2) / -2ab = (a^2 + b^2 - c^2) / 2ab
+// Law of Sines:
+// a / sin A = b / sin B = c / sin C
+
+// Solve the angles of the triangle given the sides. Angles returned are
+// opposite of the given sides (so a1 consists of sides s2 & s3, and so on).
+void Geometry::FindAnglesForSides(double s1, double s2, double s3, double * a1, double * a2, double * a3)
+{
+	// Use law of cosines to find 1st angle
+	double cosine1 = ((s2 * s2) + (s3 * s3) - (s1 * s1)) / (2.0 * s2 * s3);
+
+	// Check for a valid triangle
+	if ((cosine1 < -1.0) || (cosine1 > 1.0))
+		return;
+
+	double angle1 = acos(cosine1);
+
+	// Use law of sines to find 2nd & 3rd angles
+// sin A / a = sin B / b
+// sin B = (sin A / a) * b
+	double angle2 = asin(s2 * (sin(angle1) / s1));
+	double angle3 = asin(s3 * (sin(angle1) / s1));
+
+	if (a1)
+		*a1 = angle1;
+
+	if (a2)
+		*a2 = angle2;
+
+	if (a3)
+		*a3 = angle3;
+}
+