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问题描述

我知道的输入和输出,但我只是不知道如何和为什么它的工作原理。

这code被用于给定一个最小和最大经度/纬度(正方形),其中包含的点的集合,决定对谷歌地图的最大缩放级别,将仍显示所有的点。原作者已经一去不复返了,所以我不知道有些数字甚至对(即6371和8)。认为这是一个难题= D

  INT mapdisplay = 322; //高度和宽度的元件的分,其中包含在地图
双DIST =(6371 * Math.acos(Math.sin(min_lat / 57.2958)* Math.sin(max_lat / 57.2958)+
            (Math.cos(min_lat / 57.2958)* Math.cos(max_lat / 57.2958)* Math.cos((max_lon / 57.2958) - (min_lon / 57.2958)))));

双变焦= Math.floor(8  - 将Math.log(1.6446 * DIST /的Math.sqrt(2 *(mapdisplay * mapdisplay)))/将Math.log(2));

如果(为NumPoints == 1 ||((min_lat == max_lat)及及(min_lat == max_lat))){
    变焦= 11;
}
 

解决方案

一些数字可以很容易地解释

和再缩放级别的两倍长,每个步骤,即增加缩放级别被一个半区的大小的屏幕上。

 变焦= 8  - 日志(系数* DIST)/日志(2)= 8  -  log_2(系数* DIST)
=> DIST = 2 ^(8变焦)/因子
 

从数字,我们发现,变焦八级相当于276.89公里的距离。

I know what the input and outputs are, but I'm just not sure how or why it works.

This code is being used to, given a min and max longitude/latitude (a square) that contains a set of points, determine the maximum zoom level on Google Maps that will still display all of those points. The original author is gone, so I'm not sure what some of these numbers are even for (i.e. 6371 and 8). Consider it a puzzle =D

int mapdisplay = 322; //min of height and width of element which contains the map
double dist = (6371 * Math.acos(Math.sin(min_lat / 57.2958) * Math.sin(max_lat / 57.2958) +
            (Math.cos(min_lat / 57.2958) * Math.cos(max_lat / 57.2958) * Math.cos((max_lon / 57.2958) - (min_lon / 57.2958)))));

double zoom = Math.floor(8 - Math.log(1.6446 * dist / Math.sqrt(2 * (mapdisplay * mapdisplay))) / Math.log (2));

if(numPoints == 1 || ((min_lat == max_lat)&&(min_lat == max_lat))){
    zoom = 11;
}
解决方案

Some numbers can be explained easily

And again the zoom level doubles the size with each step, i.e. increase the zoomlevel by one halfs the size on the screen.

zoom = 8 - log(factor * dist) / log(2) = 8 - log_2(factor * dist)
=> dist = 2^(8-zoom) / factor

From the numbers we find that zoom level eight corresponds to a distance of 276.89km.

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09-06 06:43