如何在经纬度和半径转换为的LatLngBounds在Android的谷歌地图API第2版？ [英] How to convert a LatLng and a radius to a LatLngBounds in Android Google Maps API v2?

问题描述

我相信这是近期谷歌地图API V2的限制。他们最近添加到画一个圈在地面上的能力 - 但如果你想要放置的摄像头，使得它显示了整个圈子，存在没有办法做到这一点。

I believe this is a limitation of the recent Google Maps API v2. They have recently added the ability to draw a Circle on the ground - but if you want to position the camera such that it shows the entire Circle, there exists no way to do so.

可调用CameraUpdateFactory＃newLatLngBounds（边界，填充）其中范围是的LatLngBounds和填充是以像素为单位的距离。的问题是，没有办法来创建一个经纬度和一个半径成的LatLngBounds

One can call CameraUpdateFactory#newLatLngBounds(bounds, padding) where "bounds" is a LatLngBounds and "padding" is a distance in pixels. The issue is that there is no way to create a LatLng and a radius into a LatLngBounds.

该构造的LatLngBounds只需要2经纬度实例，并生成一个矩形，其中这些是西北和东南角落。

The constructor for LatLngBounds only takes 2 LatLng instances and generates a rectangle where these are the NW and SE corners.

推荐答案

就像 Risadinha 提到的，你可以很容易地实现与 Android的地图-utils的。只需添加：

Just like Risadinha mentioned, you can easily achieve that with android-maps-utils. Just add:

compile 'com.google.maps.android:android-maps-utils:0.3.4'

你的摇篮依赖关系，并使用以下code：

to your gradle dependencies, and use the following code:

public LatLngBounds convertCenterAndRadiusToBounds(LatLng center, double radius) {
    LatLng southwest = SphericalUtil.computeOffset(center, radius * Math.sqrt(2.0), 225);
    LatLng northeast = SphericalUtil.computeOffset(center, radius * Math.sqrt(2.0), 45);
    return new LatLngBounds(southwest, northeast);
}

编辑：

我们的目标是计算两个点（<$ C C $> LatLngs ）：西南和东北。从的<一的Javadoc href="http://googlemaps.github.io/android-maps-utils/javadoc/com/google/maps/android/SphericalUtil.html#computeOffset-LatLng-double-double-"><$c$c>SphericalUtil你可以阅读 225 和 45 是标题值，和半径*的Math.sqrt（2.0）是距离。

Our goal is to calculate two points (LatLngs): southwest and northeast. From the javadoc of the SphericalUtil you can read that 225 and 45 are heading values, and the radius * Math.sqrt(2.0) is the distance. Further explanation of the values in the picture below:

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