将lat / long转换为像素X& Y坐标 [英] Convert lat/long to pixel X&Y co-ordinates

问题描述

这类问题似乎已被多次询问，但没有一个解决方案让我接近我需要的答案。

This type of question seems to have been asked numerous times but none of the solutions posted get me anywhere near the answer I need.

我有这张Northland地图，新西兰和我想将Lat / Long值映射到此图像上的x / y（像素）值（http://i.stack.imgur.com/3ok4O.gif - 这是400px square）

I have this map of Northland, New Zealand and I'd like to map Lat/Long values to x/y (pixel) values on this image (http://i.stack.imgur.com/3ok4O.gif - which is 400px square)

现在我不知道这个图像的精确纬度/长度边界，但我确实有一些示例数据对于知道他们正在做什么的人有用。

Right now I do not know the precise lat/long boundaries of this image, but I do have some example data which should be useful to someone who knows what they are doing.

  LAT              LONG            X                Y
  -35.3989854471   173.504676819   192.92777494     196.760481649
  -35.2647882735   174.121499062   271.426291418    176.82865668
  -35.3131641432   173.89125824    242.099305271    183.945780963

我现在收到的数据只有Lat / long，所以我需要能够在应用程序中生成x和y值。

The data I'm receiving now is only Lat/long, so I need to be able to produce x and y values within the application.

任何人都可以帮助m e写下创建这些x / y值所需的方法/逻辑。

Can anybody help me write the required method/logic for creating these x/y values.

谢谢！

推荐答案

在Javascript ....

in Javascript....

var lat=-35.3989854471;
var long=173.504676819;

var imageNorthLat=???;
var imageSouthLat=???;

var imageWestLong=???;
var imageEastLong=???;

var imageLongPixels=400;
var imageLatPixels=400;

var pixelsPerLat=imageLatPixels/(imageNorthLat-imageSouthLat);
var pixelsPerLong=imageLongPixels/(imageEastLong-imageWestLong);

var xPixelPosition=(long-imageWestLong)*pixelsPerLong;
var yPixelPosition=Math.abs(lat-imageNorthLat)*pixelsPerLat;

我没有尝试运行它，所以可能需要一些调试，而x和y像素位置必须添加左上角或地图的x和y位置。但你可以看到这个想法。当然，我可能还不明白你真正要问的是什么。

I didn't try to run this, so there's probably a bit of debugging necessary, and the x and y pixel positions would have to have added to them the x and y positions of the top left or your map. But you can see the idea. And, of course, I may not have understood what you're really asking.

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