d3.js. 旋转地球仪的酒吧

Question

我正在尝试用这个例子中的条形创建旋转地球仪.你可以在这里看到我的例子.一切顺利,直到酒吧超越地平线.我不知道如何在行星的另一边从底部切割条形.有人可以建议我怎么做吗？

 /*
 * Original code source
 * http://codepen.io/teetteet/pen/Dgvfw
 */

var width = 400;
var height = 400;
var scrollSpeed = 50;
var current = 180;

var longitudeScale = d3.scale.linear()
  .domain([0, width])
  .range([-180, 180]);

var planetProjection = d3.geo.orthographic()
  .scale(200)
  .rotate([longitudeScale(current), 0])
  .translate([width / 2, height / 2])
  .clipAngle(90);
var barProjection = d3.geo.orthographic()
  .scale(200)
  .rotate([longitudeScale(current), 0])
  .translate([width / 2, height / 2])
  .clipAngle(90);

var path = d3.geo.path()
  .projection(planetProjection);

var svg = d3.select("body").append("svg")
  .attr("width", width)
  .attr("height", height);

d3.json("https://dl.dropboxusercontent.com/s/4hp49mvf7pa2cg2/world-110m.json?dl=1", function(error, world) {
  if (error) throw error;

  var planet = svg.append("path")
    .datum(topojson.feature(world, world.objects.land))
    .attr("class", "land")
    .attr("d", path);

  d3.csv("https://dl.dropboxusercontent.com/s/v4kn2hrnjlgx1np/data.csv?dl=1", function(error, data) {
    if (error) throw error;

    var max = d3.max(data, function(d) {
      return parseInt(d.Value);
    })

    var lengthScale = d3.scale.linear()
      .domain([0, max])
      .range([200, 250])

      var bars = svg.selectAll(".bar")
        .data(data)
        .enter()
        .append("line")
        .attr("class", "bar")
        .attr("stroke", "red")
        .attr("stroke-width", "2");

    function bgscroll() {

      current += 1;

      planetProjection.rotate([longitudeScale(current), 0]);
      barProjection.rotate([longitudeScale(current), 0]);

      planet.attr("d", path);

      bars.attr("x1", function(d) {
         return planetProjection([d.Longitude, d.Latitude])[0];
       }).attr("y1", function(d) {
         return planetProjection([d.Longitude, d.Latitude])[1];
       }).attr("x2", function(d) {
         barProjection.scale(lengthScale(d.Value));
         return barProjection([d.Longitude, d.Latitude])[0];
       }).attr("y2", function(d) {
         barProjection.scale(lengthScale(d.Value));
         return barProjection([d.Longitude, d.Latitude])[1];
       });
    }

//    bgscroll();
     setInterval(bgscroll, scrollSpeed);  
  })
})

Answer 1

为了剪掉地平线上的条形，我们添加了一个以地球 2D 中心为中心的遮罩，并带有它的半径。然后我们只在底部边缘穿过地平线时应用这个掩码（通过跟踪经度）。

创建掩码

// get the center of the circle
var center = planetProjection.translate();
// edge point
var edge = planetProjection([-90, 90])
// radius
var r = Math.pow(Math.pow(center[0] - edge[0], 2) + Math.pow(center[1] - edge[1], 2), 0.5);

svg.append("defs")
    .append("clipPath")
    .append("circle")
    .attr("id", "edgeCircle")
    .attr("cx", center[0])
    .attr("cy", center[1])
    .attr("r", r)

var mask = svg.append("mask").attr("id", "edge")
mask.append("rect")
    .attr("x", 0)
    .attr("y", 0)
    .attr("width", "100%")
    .attr("height", "100%")
    .attr("fill", "white");
mask.append("use")
    .attr("xlink:href", "#edgeCircle")
    .attr("fill", "black");

敷面膜

.... bars ....
.attr("mask", function (d) {
    // make the range from 0 to 360, so that it's easier to compare
    var longitude = Number(d.Longitude) + 180;
    // +270 => -90 => the position of the left edge when the center is at 0
    // -value because a rotation to the right => left edge longitude is reducing
    // 360 because we want the range from 0 to 360
    var startLongitude = 360 - ((longitudeScale(current) + 270) % 360);
    // the right edge is start edge + 180
    var endLongitude = (startLongitude + 180) % 360;
    if ((startLongitude < endLongitude && longitude > startLongitude && longitude < endLongitude) ||
        // wrap around
        (startLongitude > endLongitude && (longitude > startLongitude || longitude < endLongitude)))
        return null;
    else
        return "url(#edge)";
});

我们也可以通过测量距离来做到这一点。

小提琴 - http://jsfiddle.net/gp3wvm8o/