使用谷歌地图绘制不规则的同心圆

基本上,将圆计算为x,y=(cos(a),sin(a)),然后将其(两项)乘以一个半径,该半径是角度的适当函数。我不太了解Javascript,也不太了解谷歌地图,所以我会用Python来做这件事,希望这能让我明白。

from pylab import *

def Rscale(a):
    if a>3*pi/2:  # lower right, and then work CW around the circle
        return 1.
    elif a>pi:  # lower left
        return .9
    elif a>pi/2:   # upper left
        return .8
    else:       # upper right
        return 1.

def step_circle(R):
    return array([(R*Rscale(a))*array([cos(a), sin(a)]) for a in arange(0, 2*pi, .001)])

for R in (.5, .7, .9):  # make three concentric circles
    c = step_circle(R)
    plot(c[:,0], c[:,1])

show()

这给了

我真的无法理解你的草图,所以我只是猜测了数字。此外,我将最右侧的两个象限设置为相同,因为这就是你的情节,但这当然是可选的。

我想通了。这是最后的代码。也许它可以重构一点?

// Returns points for a wind field for a cyclone. Requires
// a LatLon centre point, and an array of wind radii, starting
// from the northeast quadrant (NEQ), i.e., [200, 200, 150, 175]
//
// Returns points to be used in a GPolyline object.
function pointsForWindQuadrant(centrePoint, radii){
  if(radii.length != 4){ return false; }

  var points = [];
  var angles = [0, 90, 180, 270];

  // For each angle 0, 90, 180, 270...
  for(a = 0; a < angles.length; a++){
    // For each individual angle within the range, create a point...
    for(i = angles[a]; i <= angles[a] + 90; i++){
      var point = centrePoint.destPoint(i, radii[a] * 1.85); // Radius should be in nautical miles from NHC
      points.push(new google.maps.LatLng(point.lat, point.lon));
    }
  }

  // Add the first point again, to be able to close the GPolyline
  var point = centrePoint.destPoint(0, radii[0] * 1.85);
  points.push(new google.maps.LatLng(point.lat, point.lon));

  return points;
}

结果如下: