简单的地图投影
我有一张 400x400 的地图，大约代表 250x250km 的区域，因为我想以纬度/经度的形式投影 GPS 坐标。
考虑到精度不是很重要（几公里的误差是可以容忍的），有什么简单的公式或算法可以进行投影并转换为像素坐标吗？如果有的话，我会遇到什么错误？ 或者我真的错了，没有简单的方法来达到我需要的精度？
注意：
- 我读过有关 PROJ.4 的内容，但我宁愿不使用任何外部库，因为该程序必须在小型设备中运行
- 我在地图上没有任何校准数据，但我可以使用在线地图自行校准。
- 从这里我记录了我知道如何将纬度/经度转换为 x/y/z 坐标。但我不知道如何用 Z 拨号
I've a map of 400x400 that approximatively represents an area of 250x250km in that I want to project a GPS coordinate in form of Lat/Lon.
Taking in account that precision is not very important(errors of some km are tolerable) there is any easy formula or algorithm to make the projection and translate to a pixel coordinate? If there is one, what error can I expect?
Or I'm really wrong and there not easy way for the precision that I need?
Notes:
- I readed about PROJ.4 but I prefer to don't use any external library because the program has to run in small devices
- I haven't any calibration data on the map but I can calibrate it myself using an online map.
- From here I documented a little and I know how to convert the lat/lon to x/y/z coordinates. But I don't know how to dial with the Z
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通常，这是使用 转换矩阵 和使用 墨卡托投影。
这是开始。
虽然它不是 java，但有一个名为 OpenHeatMap 的开源项目，它在其源代码中执行此操作。这可能是一个值得一看的好地方（特别是 maprender/src/maprender.mxml 中的 setLatLonViewingArea、setLatLonToXYMatrix、mercatorLatitudeToLatitude）。
希望这有帮助！
Usually, this is done using a transformation matrix and using a Mercarator projection.
Here is a good place to start.
Although it isn't java, there is an open source project called OpenHeatMap which does this within its source code. This might be a good place to look (specifically the setLatLonViewingArea, setLatLonToXYMatrix, mercatorLatitudeToLatitude in maprender/src/maprender.mxml).
Hope this helps!
GIS人员可能会为此向我扔石头，但假设你不是高纬度地区，你可以计算出地图对角的纬度/经度来获取边界框，选择一个角作为你的原点，以GPS 坐标与原点之间的差异，然后进行简单的乘法将其缩放为像素，然后绘制该点。
我过去曾在我正在玩的地图程序中使用过它，并且我大约位于北纬 39 度线处。如果它不必非常准确，也不必太靠近杆子（不过，对于 250 公里的正方形，你必须靠近杆子才会发生严重错误），这将是最快且最有效的。最简单的。
The GIS people will probably stone me for this, but assuming you're not a a high latitude, you could just figure out the lat/lon of diagonal corners of your map to get the bounding box, pick a corner as your origin, take the difference between your GPS coordinate and the origin, then a simple multiplication to scale that to pixels, then draw the point.
I've used this in the past for a map program I was playing with, and I'm at about the 39th parallel. If it doesn't have to be dead accurate, and not too close to a pole (Though, for a 250km square, you'd have to be close to a pole for gross errors to happen), this would be the quickest and the easiest.