All files S2LatLngRect.ts

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import {S1Interval} from "./S1Interval";
import {R1Interval} from "./R1Interval";
import {S2LatLng} from "./S2LatLng";
import {S2Region} from "./S2Region";
import {S2} from "./S2";
import {S2Point} from "./S2Point";
import {S1Angle} from "./S1Angle";
import {S2Cell} from "./S2Cell";
import {S2EdgeUtil} from "./S2EdgeUtil";
import {S2Cap} from "./S2Cap";
import {Decimal} from './decimal';
export class S2LatLngRect implements S2Region {
  constructor(public lat:R1Interval, public lng:S1Interval) {
 
  }
 
  static fromLatLng(lo:S2LatLng, hi:S2LatLng):S2LatLngRect {
    return new S2LatLngRect(
        new R1Interval(
            lo.latRadians,
            hi.latRadians
        ),
        new S1Interval(
            lo.lngRadians,
            hi.lngRadians
        )
    );
  }
 
 
  /** The canonical empty rectangle */
  public static  empty():S2LatLngRect {
    return new S2LatLngRect(R1Interval.empty(), S1Interval.empty());
  }
 
  /** The canonical full rectangle. */
  public static  full():S2LatLngRect {
    return new S2LatLngRect(S2LatLngRect.fullLat(), S1Interval.full());
  }
 
  /** The full allowable range of latitudes. */
  public static fullLat() {
    return new R1Interval(-S2.M_PI_2, S2.M_PI_2);
  }
 
 
  /**
   * Construct a rectangle from a center point (in lat-lng space) and size in
   * each dimension. If size.lng is greater than 360 degrees it is clamped,
   * and latitudes greater than +/- 90 degrees are also clamped. So for example,
   * FromCenterSize((80,170),(20,20)) -> (lo=(60,150),hi=(90,-170)).
   */
  public static  fromCenterSize(center:S2LatLng, size:S2LatLng) {
    return S2LatLngRect.fromPoint(center).expanded(size.mul(0.5));
  }
 
  /** Convenience method to construct a rectangle containing a single point. */
  public static  fromPoint(p:S2LatLng):S2LatLngRect {
    // assert (p.isValid());
    return S2LatLngRect.fromLatLng(p, p);
  }
 
  /**
   * Convenience method to construct the minimal bounding rectangle containing
   * the two given points. This is equivalent to starting with an empty
   * rectangle and calling AddPoint() twice. Note that it is different than the
   * S2LatLngRect(lo, hi) constructor, where the first point is always used as
   * the lower-left corner of the resulting rectangle.
   */
  public static fromPointPair(p1:S2LatLng, p2:S2LatLng):S2LatLngRect {
    // assert (p1.isValid() && p2.isValid());
    return new S2LatLngRect(R1Interval.fromPointPair(p1.latRadians, p2
        .latRadians), S1Interval.fromPointPair(p1.lngRadians, p2.lngRadians));
  }
 
  /**
   * Return a latitude-longitude rectangle that contains the edge from "a" to
   * "b". Both points must be unit-length. Note that the bounding rectangle of
   * an edge can be larger than the bounding rectangle of its endpoints.
   */
  public static  fromEdge(a:S2Point, b:S2Point):S2LatLngRect {
    // assert (S2.isUnitLength(a) && S2.isUnitLength(b));
    let r = S2LatLngRect.fromPointPair(S2LatLng.fromPoint(a), S2LatLng.fromPoint(b));
 
    // Check whether the min/max latitude occurs in the edge interior.
    // We find the normal to the plane containing AB, and then a vector "dir" in
    // this plane that also passes through the equator. We use RobustCrossProd
    // to ensure that the edge normal is accurate even when the two points are
    // very close together.
    const ab = S2.robustCrossProd(a, b);
    const dir = S2Point.crossProd(ab, new S2Point(0, 0, 1));
    const da = dir.dotProd(a);
    const db = dir.dotProd(b);
    if (da.times(db).gte(0)) {
      // Minimum and maximum latitude are attained at the vertices.
      return r;
    }
    // Minimum/maximum latitude occurs in the edge interior. This affects the
    // latitude bounds but not the longitude bounds.
    const absLat = Decimal.acos(ab.z.dividedBy(ab.norm()).abs());
    if (da.lt(0)) {
      return new S2LatLngRect(new R1Interval(r.lat.lo, absLat), r.lng);
    } else {
      return new S2LatLngRect(new R1Interval(-absLat, r.lat.hi), r.lng);
    }
  }
 
  /**
   * Return true if the rectangle is valid, which essentially just means that
   * the latitude bounds do not exceed Pi/2 in absolute value and the longitude
   * bounds do not exceed Pi in absolute value.
   *
   */
  public isValid():boolean {
    // The lat/lng ranges must either be both empty or both non-empty.
    return (this.lat.lo.abs().lte(S2.M_PI_2) && this.lat.hi.abs().lte(S2.M_PI_2)
    && this.lng.isValid() && this.lat.isEmpty() == this.lng.isEmpty());
  }
 
  public lo():S2LatLng {
    return new S2LatLng(this.lat.lo, this.lng.lo);
  }
 
  public hi():S2LatLng {
    return new S2LatLng(this.lat.hi, this.lng.hi);
  }
 
  /**
   * Return true if the rectangle is empty, i.e. it contains no points at all.
   */
  public isEmpty():boolean {
    return this.lat.isEmpty();
  }
 
// Return true if the rectangle is full, i.e. it contains all points.
  public isFull():boolean {
    // console.log(this.lat.toString());
    // console.log(S2LatLngRect.fullLat().toString());
    return this.lat.equals(S2LatLngRect.fullLat()) && this.lng.isFull();
  }
 
  /**
   * Return true if lng_.lo() > lng_.hi(), i.e. the rectangle crosses the 180
   * degree latitude line.
   */
  public  isInverted():boolean {
    return this.lng.isInverted();
  }
 
  /** Return the k-th vertex of the rectangle (k = 0,1,2,3) in CCW order. */
  public  getVertex(k:number):S2LatLng {
    // Return the points in CCW order (SW, SE, NE, NW).
    switch (k) {
      case 0:
        return this.lo();
      case 1:
        return new S2LatLng(this.lat.lo, this.lng.hi);
      case 2:
        return this.hi();
      case 3:
        return new S2LatLng(this.lat.hi, this.lng.lo);
      default:
        throw new Error("Invalid vertex index.");
    }
  }
 
  /**
   * Return the center of the rectangle in latitude-longitude space (in general
   * this is not the center of the region on the sphere).
   */
  public  getCenter():S2LatLng {
    return new S2LatLng(this.lat.getCenter(), this.lng.getCenter());
  }
 
  /**
   * Return the minimum distance (measured along the surface of the sphere)
   * from a given point to the rectangle (both its boundary and its interior).
   * The latLng must be valid.
   */
  public getDistanceLL(p:S2LatLng):S1Angle {
    // The algorithm here is the same as in getDistance(S2LagLngRect), only
    // with simplified calculations.
    const a = this;
    Iif (a.isEmpty()) {
      throw new Error();
    }
    Iif (!p.isValid()) {
      throw new Error('point is not valid');
    }
 
 
    if (a.lng.contains(p.lngRadians)) {
      return new S1Angle(
          Decimal.max(
              0.0,
              Decimal.max(
                  p.latRadians.minus(a.lat.hi),
                  a.lat.lo.minus(p.latRadians)
              )
          )
      );
    }
 
    let interval = new S1Interval(a.lng.hi, a.lng.complement().getCenter());
    let aLng = a.lng.lo;
    Eif (interval.contains(p.lngRadians)) {
      aLng = a.lng.hi;
    }
 
    const lo = new S2LatLng(a.lat.lo, aLng).toPoint();
    const hi = new S2LatLng(a.lat.hi, aLng).toPoint();
    let loCrossHi = new S2LatLng(0, aLng.minus(S2.M_PI_2)).normalized().toPoint();
    return S2EdgeUtil.getDistance(p.toPoint(), lo, hi, loCrossHi);
  }
 
  /**
   * Return the minimum distance (measured along the surface of the sphere) to
   * the given S2LatLngRect. Both S2LatLngRects must be non-empty.
   */
  public  getDistanceLLR(other:S2LatLngRect):S1Angle {
    const a = this;
    const b = other;
    if (a.isEmpty()) {
      throw new Error();
    }
    if (b.isEmpty()) {
      throw new Error();
    }
 
 
    // First, handle the trivial cases where the longitude intervals overlap.
    if (a.lng.intersects(b.lng)) {
      if (a.lat.intersects(b.lat)) {
        return new S1Angle(0);  // Intersection between a and b.
      }
 
      // We found an overlap in the longitude interval, but not in the latitude
      // interval. This means the shortest path travels along some line of
      // longitude connecting the high-latitude of the lower rect with the
      // low-latitude of the higher rect.
      let lo, hi;
      if (a.lat.lo.gt(b.lat.hi)) {
        lo = b.lat.hi;
        hi = a.lat.lo;
      } else {
        lo = a.lat.hi;
        hi = b.lat.lo;
      }
      return new S1Angle(hi.radians().minus(lo.radians()));
    }
 
    // The longitude intervals don't overlap. In this case, the closest points
    // occur somewhere on the pair of longitudinal edges which are nearest in
    // longitude-space.
    let aLng, bLng;
    const loHi = S1Interval.fromPointPair(a.lng.lo, b.lng.hi);
    const hiLo = S1Interval.fromPointPair(a.lng.hi, b.lng.lo);
    if (loHi.getLength().lt(hiLo.getLength())) {
      aLng = a.lng.lo;
      bLng = b.lng.hi;
    } else {
      aLng = a.lng.hi;
      bLng = b.lng.lo;
    }
 
    // The shortest distance between the two longitudinal segments will include
    // at least one segment endpoint. We could probably narrow this down further
    // to a single point-edge distance by comparing the relative latitudes of the
    // endpoints, but for the sake of clarity, we'll do all four point-edge
    // distance tests.
    let aLo = new S2LatLng(a.lat.lo, aLng).toPoint();
    let aHi = new S2LatLng(a.lat.hi, aLng).toPoint();
    let aLoCrossHi = new S2LatLng(0, aLng.radians().minus(S2.M_PI_2)).normalized().toPoint();
    let bLo = new S2LatLng(b.lat.lo, bLng).toPoint();
    let bHi = new S2LatLng(b.lat.hi, bLng).toPoint();
    let bLoCrossHi = new S2LatLng(0, bLng.radians().minus(S2.M_PI_2)).normalized().toPoint();
 
    return S1Angle.min(S2EdgeUtil.getDistance(aLo, bLo, bHi, bLoCrossHi),
        S1Angle.min(S2EdgeUtil.getDistance(aHi, bLo, bHi, bLoCrossHi),
            S1Angle.min(S2EdgeUtil.getDistance(bLo, aLo, aHi, aLoCrossHi),
                S2EdgeUtil.getDistance(bHi, aLo, aHi, aLoCrossHi))));
  }
 
  /**
   * Return the width and height of this rectangle in latitude-longitude space.
   * Empty rectangles have a negative width and height.
   */
  public  getSize():S2LatLng {
    return new S2LatLng(this.lat.getLength(), this.lng.getLength());
  }
 
  /**
   * More efficient version of Contains() that accepts a S2LatLng rather than an
   * S2Point.
   */
  public containsLL(ll:S2LatLng):boolean {
    // assert (ll.isValid());
    return (this.lat.contains(ll.latRadians) && this.lng.contains(ll.lngRadians));
 
  }
 
  /**
   * Return true if and only if the given point is contained in the interior of
   * the region (i.e. the region excluding its boundary). The point 'p' does not
   * need to be normalized.
   */
  public interiorContainsP(p:S2Point):boolean {
    return this.interiorContainsLL(S2LatLng.fromPoint(p));
  }
 
  /**
   * More efficient version of InteriorContains() that accepts a S2LatLng rather
   * than an S2Point.
   */
  public interiorContainsLL(ll:S2LatLng):boolean {
    // assert (ll.isValid());
    return (this.lat.interiorContains(ll.latRadians) && this.lng
        .interiorContains(ll.lngRadians));
  }
 
  /**
   * Return true if and only if the rectangle contains the given other
   * rectangle.
   */
  public containsLLR(other:S2LatLngRect):boolean {
    return this.lat.containsI(other.lat) && this.lng.containsI(other.lng);
  }
 
  /**
   * Return true if and only if the interior of this rectangle contains all
   * points of the given other rectangle (including its boundary).
   */
  public interiorContainsLLR(other:S2LatLngRect):boolean {
    return (this.lat.interiorContainsI(other.lat) && this.lng
        .interiorContainsI(other.lng));
  }
 
  /** Return true if this rectangle and the given other rectangle have any
   points in common. */
  public intersectsLLR(other:S2LatLngRect):boolean {
    return this.lat.intersects(other.lat) && this.lng.intersects(other.lng);
  }
 
  /**
   * Returns true if this rectangle intersects the given cell. (This is an exact
   * test and may be fairly expensive, see also MayIntersect below.)
   */
  public intersects(cell:S2Cell):boolean {
    // First we eliminate the cases where one region completely contains the
    // other. Once these are disposed of, then the regions will intersect
    // if and only if their boundaries intersect.
 
    if (this.isEmpty()) {
      return false;
    }
    if (this.containsP(cell.getCenter())) {
      return true;
    }
    if (cell.contains(this.getCenter().toPoint())) {
      return true;
    }
 
    // Quick rejection test (not required for correctness).
    if (!this.intersectsLLR(cell.getRectBound())) {
      return false;
    }
 
    // Now check whether the boundaries intersect. Unfortunately, a
    // latitude-longitude rectangle does not have straight edges -- two edges
    // are curved, and at least one of them is concave.
 
    // Precompute the cell vertices as points and latitude-longitudes.
    const cellV:S2Point[] = [];
    const cellLl:S2LatLng[] = new S2LatLng[4];
    for (let i = 0; i < 4; ++i) {
      cellV[i] = cell.getVertex(i); // Must be normalized.
      cellLl[i] = S2LatLng.fromPoint(cellV[i]);
      if (this.containsLL(cellLl[i])) {
        return true; // Quick acceptance test.
      }
    }
 
    for (let i = 0; i < 4; ++i) {
      const edgeLng = S1Interval.fromPointPair(
          cellLl[i].lngRadians, cellLl[(i + 1) & 3].lngRadians);
      if (!this.lng.intersects(edgeLng)) {
        continue;
      }
 
      const a = cellV[i];
      const b = cellV[(i + 1) & 3];
      if (edgeLng.contains(this.lng.lo)) {
        if (S2LatLngRect.intersectsLngEdge(a, b, this.lat, this.lng.lo)) {
          return true;
        }
      }
      if (edgeLng.contains(this.lng.hi)) {
        if (S2LatLngRect.intersectsLngEdge(a, b, this.lat, this.lng.hi)) {
          return true;
        }
      }
      if (S2LatLngRect.intersectsLatEdge(a, b, this.lat.lo, this.lng)) {
        return true;
      }
      if (S2LatLngRect.intersectsLatEdge(a, b, this.lat.hi, this.lng)) {
        return true;
      }
    }
    return false;
  }
 
  /**
   * Return true if and only if the interior of this rectangle intersects any
   * point (including the boundary) of the given other rectangle.
   */
  public  interiorIntersects(other:S2LatLngRect):boolean {
    return (this.lat.interiorIntersects(other.lat) && this.lng
        .interiorIntersects(other.lng));
  }
 
  public addPoint(p:S2Point):S2LatLngRect {
    return this.addPointLL(S2LatLng.fromPoint(p));
  }
 
// Increase the size of the bounding rectangle to include the given point.
// The rectangle is expanded by the minimum amount possible.
  public  addPointLL(ll:S2LatLng):S2LatLngRect {
    const newLat = this.lat.addPoint(ll.latRadians);
    const newLng = this.lng.addPoint(ll.lngRadians);
    return new S2LatLngRect(newLat, newLng);
  }
 
  /**
   * Return a rectangle that contains all points whose latitude distance from
   * this rectangle is at most margin.lat, and whose longitude distance from
   * this rectangle is at most margin.lng. In particular, latitudes are
   * clamped while longitudes are wrapped. Note that any expansion of an empty
   * interval remains empty, and both components of the given margin must be
   * non-negative.
   *
   * NOTE: If you are trying to grow a rectangle by a certain *distance* on the
   * sphere (e.g. 5km), use the ConvolveWithCap() method instead.
   */
  public  expanded(margin:S2LatLng):S2LatLngRect {
    // assert (margin.latRadians >= 0 && margin.lngRadians >= 0);
    Iif (this.isEmpty()) {
      return this;
    }
    return new S2LatLngRect(
        this.lat
            .expanded(margin.latRadians)
            .intersection(
                S2LatLngRect.fullLat()
            ),
        this.lng.expanded(margin.lngRadians)
    );
  }
 
  /**
   * Return the smallest rectangle containing the union of this rectangle and
   * the given rectangle.
   */
  public union(other:S2LatLngRect):S2LatLngRect {
    return new S2LatLngRect(this.lat.union(other.lat), this.lng.union(other.lng));
  }
 
  /**
   * Return the smallest rectangle containing the intersection of this rectangle
   * and the given rectangle. Note that the region of intersection may consist
   * of two disjoint rectangles, in which case a single rectangle spanning both
   * of them is returned.
   */
  public  intersection(other:S2LatLngRect):S2LatLngRect {
    const intersectLat = this.lat.intersection(other.lat);
    const intersectLng = this.lng.intersection(other.lng);
    if (intersectLat.isEmpty() || intersectLng.isEmpty()) {
      // The lat/lng ranges must either be both empty or both non-empty.
      return S2LatLngRect.empty();
    }
    return new S2LatLngRect(intersectLat, intersectLng);
  }
 
//
// /**
//  * Return a rectangle that contains the convolution of this rectangle with a
//  * cap of the given angle. This expands the rectangle by a fixed distance (as
//  * opposed to growing the rectangle in latitude-longitude space). The returned
//  * rectangle includes all points whose minimum distance to the original
//  * rectangle is at most the given angle.
//  */
// public S2LatLngRect convolveWithCap(/*S1Angle*/ angle) {
//   // The most straightforward approach is to build a cap centered on each
//   // vertex and take the union of all the bounding rectangles (including the
//   // original rectangle; this is necessary for very large rectangles).
//
//   // Optimization: convert the angle to a height exactly once.
//   S2Cap cap = S2Cap.fromAxisAngle(new S2Point(1, 0, 0), angle);
//
//   S2LatLngRect r = this;
//   for (int k = 0; k < 4; ++k) {
//     S2Cap vertexCap = S2Cap.fromAxisHeight(getVertex(k).toPoint(), cap
//         .height());
//     r = r.union(vertexCap.getRectBound());
//   }
//   return r;
// }
 
  /** Return the surface area of this rectangle on the unit sphere. */
  public area():decimal.Decimal {
    if (this.isEmpty()) {
      return S2.toDecimal(0);
    }
 
    // This is the size difference of the two spherical caps, multiplied by
    // the longitude ratio.
    //TODO: check if this.lat.hi & this.lat.lo is radians. 
    return this.lng.getLength().times(Decimal.sin(this.lat.hi).minus(Decimal.sin(this.lat.lo)).abs());
  }
 
  /** Return true if two rectangles contains the same set of points. */
 
  public equals(that:any):boolean {
    Iif (!(that instanceof S2LatLngRect)) {
      return false;
    }
    return this.lat.equals(that.lat) && this.lng.equals(that.lng);
  }
 
  /**
   * Return true if the latitude and longitude intervals of the two rectangles
   * are the same up to the given tolerance (see r1interval.h and s1interval.h
   * for details).
   */
  public approxEquals(other:S2LatLngRect, EmaxError:number = 1e-15):boolean {
    return (this.lat.approxEquals(other.lat, maxError) && this.lng.approxEquals(
        other.lng, maxError));
  }
 
// //////////////////////////////////////////////////////////////////////
// S2Region interface (see {@code S2Region} for details):
 
 
  public clone():S2Region {
    return new S2LatLngRect(this.lat, this.lng);
  }
 
 
  public getCapBound():S2Cap {
    // We consider two possible bounding caps, one whose axis passes
    // through the center of the lat-long rectangle and one whose axis
    // is the north or south pole. We return the smaller of the two caps.
 
    Iif (this.isEmpty()) {
      return S2Cap.empty();
    }
 
    let poleZ, poleAngle;
    if (this.lat.lo.plus(this.lat.hi).lt(0)) {
      // South pole axis yields smaller cap.
      poleZ = -1;
      poleAngle = this.lat.hi.plus(S2.M_PI_2);
    } else {
      poleZ = 1;
      poleAngle = this.lat.lo.neg().plus(S2.M_PI_2);
    }
 
    const poleCap = S2Cap.fromAxisAngle(new S2Point(0, 0, poleZ), new S1Angle(poleAngle));
 
    // For bounding rectangles that span 180 degrees or less in longitude, the
    // maximum cap size is achieved at one of the rectangle vertices. For
    // rectangles that are larger than 180 degrees, we punt and always return a
    // bounding cap centered at one of the two poles.
    const lngSpan = this.lng.hi.minus(this.lng.lo);
    Eif (S2.IEEEremainder(lngSpan, 2 * S2.M_PI).gte(0)) {
      Eif (lngSpan.lt(2 * S2.M_PI)) {
        let midCap = S2Cap.fromAxisAngle(this.getCenter().toPoint(), new S1Angle(0));
        for (let k = 0; k < 4; ++k) {
          midCap = midCap.addPoint(this.getVertex(k).toPoint());
        }
        Eif (midCap.height.lt(poleCap.height)) {
          return midCap;
        }
      }
    }
    return poleCap;
  }
 
 
  public  getRectBound():S2LatLngRect {
    return this;
  }
 
 
  public /*boolean*/ containsC(cell:S2Cell):boolean {
    // A latitude-longitude rectangle contains a cell if and only if it contains
    // the cell's bounding rectangle. (This is an exact test.)
    return this.containsLLR(cell.getRectBound());
  }
 
  /**
   * This test is cheap but is NOT exact. Use Intersects() if you want a more
   * accurate and more expensive test. Note that when this method is used by an
   * S2RegionCoverer, the accuracy isn't all that important since if a cell may
   * intersect the region then it is subdivided, and the accuracy of this method
   * goes up as the cells get smaller.
   */
 
  public /*boolean*/ mayIntersectC(cell:S2Cell):boolean {
    // This test is cheap but is NOT exact (see s2latlngrect.h).
    return this.intersectsLLR(cell.getRectBound());
  }
 
  /** The point 'p' does not need to be normalized. */
  public /*boolean*/ containsP(p:S2Point):boolean {
    return this.containsLL(S2LatLng.fromPoint(p));
  }
 
  /**
   * Return true if the edge AB intersects the given edge of constant longitude.
   */
  private static /*boolean*/ intersectsLngEdge(a:S2Point, b:S2Point,
                                               lat:R1Interval, lng:decimal.Decimal|number) {
    // Return true if the segment AB intersects the given edge of constant
    // longitude. The nice thing about edges of constant longitude is that
    // they are straight lines on the sphere (geodesics).
 
 
    return S2.simpleCrossing(a, b, new S2LatLng(lat.lo, lng)
        .toPoint(), new S2LatLng(lat.hi, lng).toPoint());
  }
 
  /**
   * Return true if the edge AB intersects the given edge of constant latitude.
   */
  private static /*boolean*/ intersectsLatEdge(a:S2Point, b:S2Point, lat:number|decimal.Decimal,
                                               lng:S1Interval) {
    // Return true if the segment AB intersects the given edge of constant
    // latitude. Unfortunately, lines of constant latitude are curves on
    // the sphere. They can intersect a straight edge in 0, 1, or 2 points.
    // assert (S2.isUnitLength(a) && S2.isUnitLength(b));
 
    // First, compute the normal to the plane AB that points vaguely north.
    let z = S2Point.normalize(S2.robustCrossProd(a, b));
    if (z.z.lt(0)) {
      z = S2Point.neg(z);
    }
 
    // Extend this to an orthonormal frame (x,y,z) where x is the direction
    // where the great circle through AB achieves its maximium latitude.
    const y = S2Point.normalize(S2.robustCrossProd(z, new S2Point(0, 0, 1)));
    const x = S2Point.crossProd(y, z);
    // assert (S2.isUnitLength(x) && x.z >= 0);
 
    // Compute the angle "theta" from the x-axis (in the x-y plane defined
    // above) where the great circle intersects the given line of latitude.
    let sinLat = Decimal.sin(lat);
    if (sinLat.abs().gte(x.z)) {
      return false; // The great circle does not reach the given latitude.
    }
    // assert (x.z > 0);
    const cosTheta = sinLat.dividedBy(x.z);
    const sinTheta = cosTheta.pow(2).neg().plus(1).sqrt(); // Math.sqrt(1 - cosTheta * cosTheta);
    const theta = Decimal.atan2(sinTheta, cosTheta);
    // Math.atan2(sinTheta, cosTheta);
 
    // The candidate intersection points are located +/- theta in the x-y
    // plane. For an intersection to be valid, we need to check that the
    // intersection point is contained in the interior of the edge AB and
    // also that it is contained within the given longitude interval "lng".
 
    // Compute the range of theta values spanned by the edge AB.
    const abTheta = S1Interval.fromPointPair(Decimal.atan2(
        a.dotProd(y), a.dotProd(x)), Decimal.atan2(b.dotProd(y), b.dotProd(x)));
 
    if (abTheta.contains(theta)) {
      // Check if the intersection point is also in the given "lng" interval.
      const isect = S2Point.add(S2Point.mul(x, cosTheta), S2Point.mul(y,
          sinTheta));
      if (lng.contains(Decimal.atan2(isect.y, isect.x))) {
        return true;
      }
    }
    if (abTheta.contains(theta.neg())) {
      // Check if the intersection point is also in the given "lng" interval.
      const intersection = S2Point.sub(S2Point.mul(x, cosTheta), S2Point.mul(y, sinTheta));
      if (lng.contains(Decimal.atan2(intersection.y, intersection.x))) {
        return true;
      }
    }
    return false;
 
  }
 
  public allVertex() {
    return [
      this.getVertex(0),
      this.getVertex(1),
      this.getVertex(2),
      this.getVertex(3)
    ]
  }
 
  public toGEOJSON():any {
    return {
      type: 'Feature',
      geometry: {
        type: 'Polygon',
        coordinates: [this.allVertex().concat(this.getVertex(0)).map(v => [parseFloat(v.lngDegrees.toFixed(5)), parseFloat(v.latDegrees.toFixed(5))])],
      },
      properties: {}
 
    }
  }
 
  public toString():string {
    return "[Lo=" + this.lo().toString() + ", Hi=" + this.hi().toString() + "]";
  }
 
 
}