//
#include "geometry.h"
+#include <math.h>
+#include <stdio.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
// is between 0 and 1, the normal of the vector to the point is on the vector.
-double Geometry::ParameterOfLineAndPoint(Point lp1, Point lp2, Point point)
+// Note: lp1 is the tail, lp2 is the head of the line (vector).
+double Geometry::ParameterOfLineAndPoint(Point tail, Point head, Point point)
{
// Geometric interpretation:
// The parameterized point on the vector lineSegment is where the normal of
// the perpendicular lies beyond the 1st endpoint. If pp > 1, then the
// perpendicular lies beyond the 2nd endpoint.
- Vector lineSegment = lp1 - lp2;
+ Vector lineSegment = head - tail;
double magnitude = lineSegment.Magnitude();
- Vector pointSegment = point - lp2;
+ Vector pointSegment = point - tail;
double t = lineSegment.Dot(pointSegment) / (magnitude * magnitude);
return t;
}
-Point Geometry::MirrorPointAroundLine(Point point, Point p1, Point p2)
+Point Geometry::MirrorPointAroundLine(Point point, Point tail, Point head)
{
// Get the vector of the intersection of the line and the normal on the
// line to the point in question.
- double t = ParameterOfLineAndPoint(p1, p2, point);
- Vector v = Vector(p1, p2) * t;
+ double t = ParameterOfLineAndPoint(tail, head, point);
+ Vector v = Vector(tail, head) * t;
- // Get the point normal to point to the line passed in (p2 is the tail)
- Point normalOnLine = p2 + v;
+ // Get the point normal to point to the line passed in
+ Point normalOnLine = tail + v;
// Make our mirrored vector (head - tail)
Vector mirror = -(point - normalOnLine);
return mirroredPoint;
}
+
+//
+// point: The point we're rotating
+// rotationPoint: The point we're rotating around
+//
+Point Geometry::RotatePointAroundPoint(Point point, Point rotationPoint, double angle)
+{
+ 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));
+
+ return Vector(rotationPoint.x + px, rotationPoint.y + py, 0);
+}
+
+
+double Geometry::Determinant(Point p1, Point p2)
+{
+ return (p1.x * p2.y) - (p2.x * p1.y);
+}
+
+
+int Geometry::Intersects(Object * obj1, Object * obj2, double * tp/*= 0*/, double * up/*= 0*/, double * vp/*= 0*/, double * wp/*= 0*/)
+{
+ if ((obj1->type == OTLine) && (obj2->type == OTLine))
+ return CheckLineToLineIntersection(obj1, obj2, tp, up);
+
+ return 0;
+}
+
+
+/*
+Intersecting line segments:
+An easier way:
+Segment L1 has edges A=(a1,a2), A'=(a1',a2').
+Segment L2 has edges B=(b1,b2), B'=(b1',b2').
+Segment L1 is the set of points tA'+(1-t)A, where 0<=t<=1.
+Segment L2 is the set of points sB'+(1-s)B, where 0<=s<=1.
+Segment L1 meet segment L2 if and only if for some t and s we have
+tA'+(1-t)A=sB'+(1-s)B
+The solution of this with respect to t and s is
+
+t=((-b?'a?+b?'b?+b?a?+a?b?'-a?b?-b?b?')/(b?'a?'-b?'a?-b?a?'+b?a?-a?'b?'+a?'b?+a?b?'-a?b?))
+
+s=((-a?b?+a?'b?-a?a?'+b?a?+a?'a?-b?a?')/(b?'a?'-b?'a?-b?a?'+b?a?-a?'b??+a?'b?+a?b?'-a?b?))
+
+So check if the above two numbers are both >=0 and <=1.
+*/
+
+
+// Finds the intersection between two lines (if any)
+int Geometry::CheckLineToLineIntersection(Object * l1, Object * l2, double * tp, double * up)
+{
+ 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
+ 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;
+
+ if (rxs == 0)
+ {
+ double qpxr = (v1.x * r.y) - (r.x * v1.y);
+
+#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;
+
+#if 0
+//this works IFF the vectors are pointing in the same direction. everything else
+//is fucked!
+ // If (q - p) . r == r . r, t = 1, u = 0
+ if (v1.Dot(r) == r.Dot(r))
+ t = 1.0, u = 0;
+ // If (p - q) . s == s . s, t = 0, u = 1
+ else if (v2.Dot(s) == s.Dot(s))
+ t = 0, u = 1.0;
+ else
+ return 0;
+#else
+ // Check to see which endpoints are connected... Four possibilities:
+ if (l1->p[0] == l2->p[0])
+ t = 0, u = 0;
+ else if (l1->p[0] == l2->p[1])
+ t = 0, u = 1.0;
+ else if (l1->p[1] == l2->p[0])
+ t = 1.0, u = 0;
+ else if (l1->p[1] == l2->p[1])
+ t = 1.0, u = 1.0;
+ else
+ return 0;
+#endif
+ }
+ else
+ {
+ t = ((v1.x * s.y) - (s.x * v1.y)) / rxs;
+ u = ((v1.x * r.y) - (r.x * v1.y)) / rxs;
+ }
+/*
+Now there are five cases (NOTE: only valid if vectors face the same way!):
+
+1. If r × s = 0 and (q − p) × r = 0, then the two lines are collinear. If in addition, either 0 ≤ (q − p) · r ≤ r · r or 0 ≤ (p − q) · s ≤ s · s, then the two lines are overlapping.
+
+2. If r × s = 0 and (q − p) × r = 0, but neither 0 ≤ (q − p) · r ≤ r · r nor 0 ≤ (p − q) · s ≤ s · s, then the two lines are collinear but disjoint.
+
+3. If r × s = 0 and (q − p) × r ≠ 0, then the two lines are parallel and non-intersecting.
+
+4. If r × s ≠ 0 and 0 ≤ t ≤ 1 and 0 ≤ u ≤ 1, the two line segments meet at the point p + t r = q + u s.
+
+5. Otherwise, the two line segments are not parallel but do not intersect.
+*/
+ // Return parameter values, if user passed in valid pointers
+ if (tp)
+ *tp = t;
+
+ if (up)
+ *up = u;
+
+ // If the parameters are in range, we have overlap!
+ if ((t >= 0) && (t <= 1.0) && (u >= 0) && (u <= 1.0))
+ return 1;
+
+ return 0;
+}
+
+
+#if 0
+// Finds the intersection between two lines (if any)
+int Geometry::Intersects(Line * l1, Dimension * d1, double * tp/*= 0*/, double * up/*= 0*/)
+{
+ Line l2(d1->position, d1->endpoint);
+ return Intersects(l1, &l2, tp, up);
+}
+
+
+// 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
+ Vector center = c->position;
+ Vector v1 = l->position - center;
+ Vector v2 = l->endpoint - center;
+ Vector d = v2 - v1;
+ double dr = d.Magnitude();
+ double determinant = (v1.x * v2.y) - (v1.y * v2.x);
+
+ double discriminant = ((c->radius * c->radius) * (dr * dr)) - (determinant * determinant);
+
+ if (discriminant < 0)
+ return false;
+
+
+
+ return true;
+#else
+/*
+I'm thinking a better approach to this might be as follows:
+
+-- Get the distance of the circle's center from the line segment. If it's
+ > the radius, it doesn't intersect.
+-- If the parameter is off the line segment, check distance to endpoints. (Not sure
+ how to proceed from here, it's different than the following.)
+ [Actually, you can use the following for all of it. You only know if you have
+ an intersection at the last step, which is OK.]
+-- If the radius == distance, we have a tangent line.
+-- If radius > distance, use Pythagorus to find the length on either side of the
+ normal to the spots where the hypotenuse (== radius' length) contact the line.
+-- Use those points to find the parameter on the line segment; if they're not on
+ the line segment, no intersection.
+*/
+ double t = ParameterOfLineAndPoint(l->position, l->endpoint, c->position);
+//small problem here: it clamps the result to the line segment. NOT what we want
+//here! !!! FIX !!! [DONE]
+ Vector p = l->GetPointAtParameter(t);
+ double distance = Vector::Magnitude(c->position, p);
+
+ // If the center of the circle is farther from the line than the radius, fail.
+ if (distance > c->radius)
+ return 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.
+ double offset = sqrt((c->radius * c->radius) - (distance * distance));
+//need to convert distance to paramter value... :-/
+//t = position on line / length of line segment, so if we divide the offset by length,
+//that should give us what we want.
+ double length = Vector::Magnitude(l->position, l->endpoint);
+ double t1 = t + (offset / length);
+ double t2 = t - (offset / length);
+
+//need to find angles for the circle...
+ Vector cp1 = l->position + (Vector(l->position, l->endpoint) * (length * t1));
+ Vector cp2 = l->position + (Vector(l->position, l->endpoint) * (length * t2));
+ double a1 = Vector(c->position, cp1).Angle();
+ double a2 = Vector(c->position, cp2).Angle();
+
+//instead of this, return a # which is the # of intersections. [DONE]
+ int intersections = 0;
+
+ // Now check for if the parameters are in range
+ if ((t1 >= 0) && (t1 <= 1.0))
+ {
+ intersections++;
+ }
+
+ if ((t2 >= 0) && (t2 <= 1.0))
+ {
+ intersections++;
+ }
+
+ return intersections;
+#endif
+}
+
+
+// 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;
+}
+#endif
+
+// 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
+// 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));
+
+ if (a1)
+ *a1 = angle1;
+
+ if (a2)
+ *a2 = angle2;
+
+ if (a3)
+ *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