#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:
- // px = (x1y2 - y1x2)(x3 - x4) - (x1 - x2)(x3y4 - y3x4)
- // py = (x1y2 - y1x2)(y3 - y4) - (y1 - y2)(x3y4 - y3x4)
- // d = (x1 - x2)(y3 - y4) - (y1 - y2)(x3 - x4) = 0 if lines are parallel
- // Intersection is (px / d, py / d)
-
- double d = ((p1.x - p2.x) * (p3.y - p4.y)) - ((p1.y - p2.y) * (p3.x - p4.x));
-
- // Check for parallel lines, and return sentinel if so
- if (d == 0)
- return Point(0, 0, -1);
-
- double px = (((p1.x * p2.y) - (p1.y * p2.x)) * (p3.x - p4.x))
- - ((p1.x - p2.x) * ((p3.x * p4.y) - (p3.y * p4.x)));
- double py = (((p1.x * p2.y) - (p1.y * p2.x)) * (p3.y - p4.y))
- - ((p1.y - p2.y) * ((p3.x * p4.y) - (p3.y * p4.x)));
-
- 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.
// Note: lp1 is the tail, lp2 is the head of the line (vector).
}
-void Geometry::Intersects(Object * obj1, Object * obj2)//, double * tp/*= 0*/, double * up/*= 0*/, double * vp/*= 0*/, double * wp/*= 0*/)
+void Geometry::Intersects(Object * obj1, Object * obj2)
{
Global::numIntersectPoints = Global::numIntersectParams = 0;
CheckLineToLineIntersection(obj1, obj2);
else if ((obj1->type == OTCircle) && (obj2->type == OTCircle))
CheckCircleToCircleIntersection(obj1, obj2);
+ else if ((obj1->type == OTLine) && (obj2->type == OTCircle))
+ CheckLineToCircleIntersection(obj1, obj2);
+ else if ((obj1->type == OTCircle) && (obj2->type == OTLine))
+ CheckLineToCircleIntersection(obj2, obj1);
}
// Finds the intersection between two lines (if any)
-void Geometry::CheckLineToLineIntersection(Object * l1, Object * l2)//, double * tp, double * up)
+void Geometry::CheckLineToLineIntersection(Object * l1, Object * l2)
{
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
- 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);
-#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;
-#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;
t = 1.0, u = 1.0;
else
return;
-#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 0
- 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;
-#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
}
// 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])))
+ if (d == (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;
}
+ else if (d == fabs(c1->radius[0] - c2->radius[0]))
+ {
+ double sign = (c1->radius[0] > c2->radius[0] ? +1 : -1);
+ Global::intersectPoint[0].x = c1->p[0].x + (cos(clAngle) * c1->radius[0] * sign);
+ Global::intersectPoint[0].y = c1->p[0].y + (sin(clAngle) * c1->radius[0] * sign);
+ Global::numIntersectPoints = 1;
+ return;
+ }
+/*
+ c² = a² + b² - 2ab·cos µ
+2ab·cos µ = a² + b² - c²
+ cos µ = (a² + b² - c²) / 2ab
+*/
// 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));
}
-#if 0
-// Finds the intersection between two lines (if any)
-int Geometry::Intersects(Line * l1, Dimension * d1, double * tp/*= 0*/, double * up/*= 0*/)
+//
+// N.B.: l is the line, c is the circle
+//
+void Geometry::CheckLineToCircleIntersection(Object * l, Object * c)
{
- Line l2(d1->position, d1->endpoint);
- return Intersects(l1, &l2, tp, up);
-}
+ // Set up global vars
+ Global::numIntersectPoints = Global::numIntersectParams = 0;
+ // Step 1: Find shortest distance from center of circle to the infinite line
+ double t = ParameterOfLineAndPoint(l->p[0], l->p[1], c->p[0]);
+ Point p = l->p[0] + (Vector(l->p[0], l->p[1]) * t);
+ Vector radial = Vector(c->p[0], p);
+ double distance = radial.Magnitude();
-// 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);
+ // Step 2: See if we have 0, 1, or 2 intersection points
- double discriminant = ((c->radius * c->radius) * (dr * dr)) - (determinant * determinant);
+ // Case #1: No intersection points
+ if (distance > c->radius[0])
+ return;
+ // Case #2: One intersection point (possibly--tangent)
+ else if (distance == c->radius[0])
+ {
+ // Only intersects if the parameter is on the line segment!
+ if ((t >= 0.0) && (t <= 1.0))
+ {
+ Global::intersectPoint[0] = c->p[0] + radial;
+ Global::numIntersectPoints = 1;
+ }
- if (discriminant < 0)
- return false;
+ return;
+ }
-
+ // Case #3: Two intersection points (possibly--secant)
- 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;
+ // So, we have the line, and the perpendicular from the center of the
+ // circle to the line. Now figure out where the intersection points are.
+ // This is a right triangle, though do we really know all the sides?
+ // Don't need to, 2 is enough for Pythagoras :-)
+ // Radius is the hypotenuse, so we have to use c² = a² + b² => a² = c² - b²
+ double perpendicularLength = sqrt((c->radius[0] * c->radius[0]) - (distance * distance));
- if (up)
- *up = Vector(c->position, p).Angle();
+ // Now, find the points using the length, then check to see if they are on
+ // the line segment
+ Vector lineUnit = Vector(l->p[0], l->p[1]).Unit();
+ Point i1 = p + (lineUnit * perpendicularLength);
+ Point i2 = p - (lineUnit * perpendicularLength);
- return 1;
- }
+ // Now we have our intersection points, next we need to see if they are on
+ // the line segment...
+ double u = ParameterOfLineAndPoint(l->p[0], l->p[1], i1);
+ double v = ParameterOfLineAndPoint(l->p[0], l->p[1], i2);
- // 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))
+ if ((u >= 0.0) && (u <= 1.0))
{
- intersections++;
+ Global::intersectPoint[Global::numIntersectPoints] = i1;
+ Global::numIntersectPoints++;
}
- if ((t2 >= 0) && (t2 <= 1.0))
+ if ((v >= 0.0) && (v <= 1.0))
{
- intersections++;
+ Global::intersectPoint[Global::numIntersectPoints] = i2;
+ Global::numIntersectPoints++;
}
-
- 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)
- // [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
-// 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:
// c² = a² + b² - 2ab * cos(C)
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