+//
// geometry.cpp: Algebraic geometry helper functions
//
// Part of the Architektonas Project
#include "geometry.h"
#include <math.h>
-#include "circle.h"
-#include "dimension.h"
-#include "line.h"
+#include <stdio.h>
+#include "global.h"
#include "mathconstants.h"
-
-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);
-}
-
-
// 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).
double magnitude = lineSegment.Magnitude();
Vector pointSegment = point - tail;
double t = lineSegment.Dot(pointSegment) / (magnitude * magnitude);
+
return t;
}
+double Geometry::DistanceToLineFromPoint(Point tail, Point head, Point point)
+{
+ // Interpretation: given a line in the form x = a + tu, where u is the
+ // unit vector of the line, a is the tail and t is a parameter which
+ // describes the line, the distance of a point p to the line is given by:
+ // || (a - p) - ((a - p) . u) u ||
+ // We go an extra step: we set the sign to reflect which side of the line
+ // it's on (+ == to the left if head points away from you, - == to the
+ // right)
+ Vector line(tail, head);
+ Vector u = line.Unit();
+ Vector a_p = tail - point;
+ Vector dist = a_p - (u * (a_p).Dot(u));
+
+ double angle = Vector::Angle(tail, point) - line.Angle();
+
+ if (angle < 0)
+ angle += TAU;
+
+ return dist.Magnitude() * (angle < HALF_TAU ? +1.0 : -1.0);
+}
Point Geometry::MirrorPointAroundLine(Point point, Point tail, Point head)
{
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(point, rotationPoint);
-// Vector v = Vector(rotationPoint, point);
+ 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);
}
+void Geometry::Intersects(Object * obj1, Object * obj2)
+{
+ 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);
+ else if ((obj1->type == OTLine) && (obj2->type == OTCircle))
+ CheckLineToCircleIntersection(obj1, obj2);
+ else if ((obj1->type == OTCircle) && (obj2->type == OTLine))
+ CheckLineToCircleIntersection(obj2, obj1);
+}
/*
Intersecting line segments:
*/
-#if 0
-// Finds the intesection between two objects (if any)
-bool Geometry::Intersects(Object * obj1, Object * obj2, double * t, double * s)
-{
-}
-#endif
-
-#if 0
// Finds the intersection between two lines (if any)
-int Geometry::Intersects(Line * l1, Line * l2, double * tp/*= 0*/, double * up/*= 0*/)
+void Geometry::CheckLineToLineIntersection(Object * l1, Object * l2)
{
- 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);
+ 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
+
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));
-
// 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
-//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->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;
-#endif
+ return;
}
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;
+ 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))
- return 1;
-
- return 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);
+ Global::numIntersectParams = 1;
}
-
-// 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*/)
+void Geometry::CheckCircleToCircleIntersection(Object * c1, Object * c2)
{
-#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);
+ // Set up global vars
+ Global::numIntersectPoints = Global::numIntersectParams = 0;
- 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();
+ // 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();
- return 1;
- }
+ // 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;
- // 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 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]))
{
- intersections++;
+ 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;
}
-
- if ((t2 >= 0) && (t2 <= 1.0))
+ else if (d == fabs(c1->radius[0] - c2->radius[0]))
{
- intersections++;
+ 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;
}
- return intersections;
-#endif
+/*
+ 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));
+
+ // 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;
}
-
-// 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*/)
+//
+// N.B.: l is the line, c is the circle
+//
+void Geometry::CheckLineToCircleIntersection(Object * l, Object * c)
{
- // 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);
+ // Set up global vars
+ Global::numIntersectPoints = Global::numIntersectParams = 0;
- // 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;
+ // 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();
- // 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();
+ // Step 2: See if we have 0, 1, or 2 intersection points
- if (up)
- *up = centerLine.Angle() + PI;
+ // 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;
+ }
- return 1;
+ return;
}
- // 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);
+ // Case #3: Two intersection points (possibly--secant)
- if (tp)
- *tp = centerLine.Angle() - angleC1;
+ // 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 = (centerLine.Angle() + PI) - angleC2;
+ // Now, find the intersection points using the length...
+ Vector lineUnit = Vector(l->p[0], l->p[1]).Unit();
+ Point i1 = p + (lineUnit * perpendicularLength);
+ Point i2 = p - (lineUnit * perpendicularLength);
- if (vp)
- *vp = centerLine.Angle() + angleC1;
+ // 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);
- 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));
+ if ((u >= 0.0) && (u <= 1.0))
+ {
+ Global::intersectPoint[Global::numIntersectPoints] = i1;
+ Global::numIntersectPoints++;
+ }
- return 2;
+ if ((v >= 0.0) && (v <= 1.0))
+ {
+ Global::intersectPoint[Global::numIntersectPoints] = i2;
+ Global::numIntersectPoints++;
+ }
}
-#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)
+// 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);
+}
+
+Point Geometry::Midpoint(Line * line)
+{
+ return Point((line->p[0].x + line->p[1].x) / 2.0,
+ (line->p[0].y + line->p[1].y) / 2.0);
+}
+
+/*
+How to find the tangent of a point off a circle:
+
+ • Calculate the midpoint of the point and the center of the circle
+ • Get the length of the line segment from point and the center divided by two
+ • Use that length to construct a circle with the point at the center and the
+ radius equal to that length
+ • The intersection of the two circles are the tangent points
+
+Another way:
+
+ • Find angle between radius and line between point and center
+ • Angle +/- from line (point to center) are the tangent points
+
+*/
+void Geometry::FindTangents(Object * c, Point p)
+{
+ // Set up global vars
+ Global::numIntersectPoints = Global::numIntersectParams = 0;
+
+ // Get the distance between the point and the center of the circle, the
+ // length, and the angle.
+ Vector centerLine(c->p[0], p);
+ double d = centerLine.Magnitude();
+ double clAngle = centerLine.Angle();
+
+ // If the point is on or inside the circle, there are no tangents.
+ if (d <= c->radius[0])
+ return;
+
+ // Use 'cos(a) = adjacent / hypotenuse' to get the tangent angle
+ double a = acos(c->radius[0] / d);
+
+ // Finally, find the points of intersection by using +/- the angle found
+ // from the centerline's angle
+ Global::intersectPoint[0].x = c->p[0].x + (cos(clAngle + a) * c->radius[0]);
+ Global::intersectPoint[0].y = c->p[0].y + (sin(clAngle + a) * c->radius[0]);
+ Global::intersectPoint[1].x = c->p[0].x + (cos(clAngle - a) * c->radius[0]);
+ Global::intersectPoint[1].y = c->p[0].y + (sin(clAngle - a) * c->radius[0]);
+ Global::numIntersectPoints = 2;
+}
+
+/*
+The extangents can be found by collapsing the circle of smaller radius to a point, and diminishing the larger circle by the radius of the smaller, then treating it like the point-circle case (you have to translate the tangent back out by the length of the radius of the smaller circle once found).
+
+Not sure what the analogous method for finding the intangents are... :-/
+Looks like the intangent can be found by augmenting the larger circle by the smaller radius, and doing the center of the smaller as the point-circle case again, only translating the line back by the radius of the smaller.
+*/
+void Geometry::FindTangents(Object * c1, Object * c2)
+{
+ // Set up global vars
+ Global::numIntersectPoints = Global::numIntersectParams = 0;
+
+ // Find the larger and smaller of the two:
+ Object * cLg = c1, * cSm = c2;
+
+ if (c2->radius[0] > c1->radius[0])
+ cLg = c2, cSm = c1;
+
+ // Get the distance between the point and the center of the circle, the
+ // length, and the angle.
+ Vector centerLine(cLg->p[0], cSm->p[0]);
+ double d = centerLine.Magnitude();
+ double clAngle = centerLine.Angle();
+
+ // If one circle is completely inside the other, there are no tangents.
+ if ((d + cSm->radius[0]) <= cLg->radius[0])
+ return;
+
+ // Subtract the radius of the smaller from the larger, and use the point-circle method to find the extangent:
+ double a = acos((cLg->radius[0] - cSm->radius[0]) / d);
+
+ // Finally, find the points of intersection by using +/- the angle found
+ // from the centerline's angle
+ Global::intersectPoint[0].x = cLg->p[0].x + (cos(clAngle + a) * cLg->radius[0]);
+ Global::intersectPoint[0].y = cLg->p[0].y + (sin(clAngle + a) * cLg->radius[0]);
+ Global::intersectPoint[1].x = cSm->p[0].x + (cos(clAngle + a) * cSm->radius[0]);
+ Global::intersectPoint[1].y = cSm->p[0].y + (sin(clAngle + a) * cSm->radius[0]);
+
+ Global::intersectPoint[2].x = cLg->p[0].x + (cos(clAngle - a) * cLg->radius[0]);
+ Global::intersectPoint[2].y = cLg->p[0].y + (sin(clAngle - a) * cLg->radius[0]);
+ Global::intersectPoint[3].x = cSm->p[0].x + (cos(clAngle - a) * cSm->radius[0]);
+ Global::intersectPoint[3].y = cSm->p[0].y + (sin(clAngle - a) * cSm->radius[0]);
+ Global::numIntersectPoints = 4;
+
+ // If the circles overlap, there are no intangents.
+ if (d <= (cLg->radius[0] + cSm->radius[0]))
+ return;
+
+ // Add the radius of the smaller from the larger, and use the point-circle method to find the intangent:
+ a = acos((cLg->radius[0] + cSm->radius[0]) / d);
+
+ // Finally, find the points of intersection by using +/- the angle found
+ // from the centerline's angle
+ Global::intersectPoint[4].x = cLg->p[0].x + (cos(clAngle + a) * cLg->radius[0]);
+ Global::intersectPoint[4].y = cLg->p[0].y + (sin(clAngle + a) * cLg->radius[0]);
+ Global::intersectPoint[5].x = cSm->p[0].x + (cos(clAngle - a) * cSm->radius[0]);
+ Global::intersectPoint[5].y = cSm->p[0].y + (sin(clAngle - a) * cSm->radius[0]);
+
+ Global::intersectPoint[6].x = cLg->p[0].x + (cos(clAngle - a) * cLg->radius[0]);
+ Global::intersectPoint[6].y = cLg->p[0].y + (sin(clAngle - a) * cLg->radius[0]);
+ Global::intersectPoint[7].x = cSm->p[0].x + (cos(clAngle + a) * cSm->radius[0]);
+ Global::intersectPoint[7].y = cSm->p[0].y + (sin(clAngle + a) * cSm->radius[0]);
+ Global::numIntersectPoints = 8;
+}
+
+//
+// Parameter 1: point in question
+// Parameter 2, 3: points we are comparing to
+//
+Point Geometry::NearestTo(Point point, Point p1, Point p2)
+{
+ double l1 = Vector::Magnitude(point, p1);
+ double l2 = Vector::Magnitude(point, p2);
+
+ return (l1 < l2 ? p1 : p2);
+}
projects / architektonas / blobdiff