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import {Adder} from "d3-array";
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import {cartesian, cartesianCross, cartesianNormalizeInPlace} from "./cartesian.js";
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import {abs, asin, atan2, cos, epsilon, epsilon2, halfPi, pi, quarterPi, sign, sin, tau} from "./math.js";
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function longitude(point) {
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return abs(point[0]) <= pi ? point[0] : sign(point[0]) * ((abs(point[0]) + pi) % tau - pi);
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}
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export default function(polygon, point) {
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var lambda = longitude(point),
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phi = point[1],
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sinPhi = sin(phi),
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normal = [sin(lambda), -cos(lambda), 0],
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angle = 0,
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winding = 0;
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var sum = new Adder();
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if (sinPhi === 1) phi = halfPi + epsilon;
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else if (sinPhi === -1) phi = -halfPi - epsilon;
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for (var i = 0, n = polygon.length; i < n; ++i) {
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if (!(m = (ring = polygon[i]).length)) continue;
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var ring,
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m,
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point0 = ring[m - 1],
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lambda0 = longitude(point0),
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phi0 = point0[1] / 2 + quarterPi,
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sinPhi0 = sin(phi0),
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cosPhi0 = cos(phi0);
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for (var j = 0; j < m; ++j, lambda0 = lambda1, sinPhi0 = sinPhi1, cosPhi0 = cosPhi1, point0 = point1) {
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var point1 = ring[j],
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lambda1 = longitude(point1),
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phi1 = point1[1] / 2 + quarterPi,
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sinPhi1 = sin(phi1),
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cosPhi1 = cos(phi1),
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delta = lambda1 - lambda0,
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sign = delta >= 0 ? 1 : -1,
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absDelta = sign * delta,
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antimeridian = absDelta > pi,
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k = sinPhi0 * sinPhi1;
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sum.add(atan2(k * sign * sin(absDelta), cosPhi0 * cosPhi1 + k * cos(absDelta)));
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angle += antimeridian ? delta + sign * tau : delta;
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// Are the longitudes either side of the point’s meridian (lambda),
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// and are the latitudes smaller than the parallel (phi)?
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if (antimeridian ^ lambda0 >= lambda ^ lambda1 >= lambda) {
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var arc = cartesianCross(cartesian(point0), cartesian(point1));
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cartesianNormalizeInPlace(arc);
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var intersection = cartesianCross(normal, arc);
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cartesianNormalizeInPlace(intersection);
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var phiArc = (antimeridian ^ delta >= 0 ? -1 : 1) * asin(intersection[2]);
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if (phi > phiArc || phi === phiArc && (arc[0] || arc[1])) {
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winding += antimeridian ^ delta >= 0 ? 1 : -1;
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}
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}
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}
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}
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// First, determine whether the South pole is inside or outside:
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//
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// It is inside if:
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// * the polygon winds around it in a clockwise direction.
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// * the polygon does not (cumulatively) wind around it, but has a negative
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// (counter-clockwise) area.
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//
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// Second, count the (signed) number of times a segment crosses a lambda
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// from the point to the South pole. If it is zero, then the point is the
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// same side as the South pole.
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return (angle < -epsilon || angle < epsilon && sum < -epsilon2) ^ (winding & 1);
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}
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Block a user