#include "circle.h"
#include "dimension.h"
#include "line.h"
+#include "mathconstants.h"
Point Geometry::IntersectionOfLineAndLine(Point p1, Point p2, Point p3, Point p4)
}
-// 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
// 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.
}
+// 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;
+}
+
projects / architektonas / blobdiff