C++ 如何将 Lat ,long 转换为 XY 坐标系（例如 UTM），然后将其映射到我的图像的像素空间
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how to Convert Lat ,long to an XY coordinate system (e.g. UTM) and then map this to pixel space of my Image
提问by Verve Innovation
I have a very small area map , which I downloaded from Openstreet map(PNG) and also its OSM(.osm) file which contains its Lat ,long.
我有一个非常小的区域地图，我从 Openstreet 地图（PNG）下载了它，还有它的 OSM（.osm）文件，其中包含它的经纬度。
Now I want to convert Lat ,long to an XY coordinate system (e.g. UTM) and then map this to pixel space of my Image which is of size (600 x 800 ). I know its a two way process ,like to know how to do this . Thank you
现在我想将 Lat ,long 转换为 XY 坐标系（例如 UTM），然后将其映射到我的 Image 大小（600 x 800）的像素空间。我知道它的双向过程，想知道如何做到这一点。谢谢
采纳答案by Brian Webster
GPS Coordinates to Pixels
GPS 坐标到像素
- Assuming this map does not cross prime meridian
- Assuming pixel 0,0 is upper left, and pixel 600,800 is lower right.
Assuming map is Northern Hemisphere Only (no part of map is southern hemisphere)
- Determine the left-most longitude in your 800x600 image (X)
- Determine the east-most longitude in your 800x600 image (Y)
- Determine Longitude-Diff (Z = Y - X)
- Determine north-most latitude in your 800x600 image (A)
- Determine south-most latitude in your 800x600 image (B)
- Determine Longitude-Diff (C = A - B)
- 假设这张地图不跨越本初子午线
- 假设像素 0,0 在左上角，像素 600,800 在右下角。
假设地图仅是北半球（地图的任何部分都不是南半球）
- 确定 800x600 图像中最左侧的经度 (X)
- 确定 800x600 图像中的最东经度 (Y)
- 确定经度差 (Z = Y - X)
- 确定 800x600 图像中最北端的纬度 (A)
- 确定 800x600 图像中最南端的纬度 (B)
- 确定经度差 (C = A - B)
Given a Latitude and Longitude, to determine which pixel they clicked on:
给定纬度和经度，以确定他们点击了哪个像素：
- J = Input Longitude
K = Input Latitude
Calculate X-pixel
XPixel = CInt(((Y - J) / CDbl(Z)) * 800)
Calculate Y-pixel
YPixel = CInt(((A - K) / CDbl(C)) * 600)
- J = 输入经度
K = 输入纬度
计算 X 像素
XPixel = CInt(((Y - J) / CDbl(Z)) * 800)
计算 Y 像素
YPixel = CInt(((A - K) / CDbl(C)) * 600)
UTM
UTM
Here is a cartographic librarythat should help with GPS to UTM conversions
这是一个制图库，应该有助于 GPS 到 UTM 的转换
