Android - 在自定义地图上绘制GPS坐标 [英] Android - plotting gps coordinate on custom map

问题描述

我在我的应用程序中有一个活动，其中有用作地图的图像。如果图片与谷歌地图网格对齐，那么当我使用地图的左上角和右下角时，我可以从谷歌地图在线获得，然后我可以将用户gps变成屏幕上的x和y。但是，如果地图不是网格对齐，并且其角度超过了数值，我的数学回报会将用户的位置画到屏幕外。显然，我错过了如何处理地图角度的部分，所以如果任何人都可以阐明如何做到这一点，那将是超级有益的。

我知道我将不得不以弧度计算角度并做一些转换，但我不知道从哪里开始。这是我用来得到一个X和Y到目前为止绘制在画布上。

  public double [] getPositionGPS（Location upperLeft，location lowerRight，Location current）{
 
尝试{
 
 double hypotenuse = upperLeft.distanceTo（current）; 
 double bearing = upperLeft.bearingTo（current）; 
 double currentDistanceY = Math.cos（方位* Math.PI / OneEightyDeg）*斜边; 
 //标记位置的百分比
 double totalHypotenuse = upperLeft.distanceTo（lowerRight）; 
 double totalDistanceY = totalHypotenuse * Math.cos（upperLeft.bearingTo（lowerRight）* Math.PI / OneEightyDeg）; 
 double currentPixelY = currentDistanceY / totalDistanceY * ImageSizeH; 
 
 double hypotenuse2 = upperLeft.distanceTo（current）; 
 double bearing2 = upperLeft.bearingTo（current）; 
 double currentDistanceX = Math.sin（bearing2 * Math.PI / OneEightyDeg）* hypotenuse2; 
 //标记位置的百分比
 double totalHypotenuse2 = upperLeft.distanceTo（lowerRight）; 
 double totalDistanceX = totalHypotenuse2 * Math.sin（upperLeft.bearingTo（lowerRight）* Math.PI / OneEightyDeg）; 
 double currentPixelX = currentDistanceX / totalDistanceX * ImageSizeW; 
 
返回新的double [] {currentPixelX，currentPixelY}; 
 
} catch（Exception e）{
 
 e.printStackTrace（）; 
返回新的double [] {0,0}; 
 
 
  

解决方案

是完全错误的，但我认为这是一种获得旋转角度的方法（或者可能是4个角）：

1. 获取角度 alpha 在lowerRight和upperLeft之间。我猜这是等于 180 - lowerRight.bearingTo（upperLeft） 编辑（这个方程应该取决于屏幕的象限）
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2. 使用X，Y（屏幕）坐标获得斜边与下边缘之间的角度 theta 。我猜（再次）这个角度的正弦等于：（height）/（（height ^ 2）+（width ^ 2））^ 1/2


3. 现在旋转角度等于 theta - alpha

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I have an activity in my app where there are images I use as maps. If the image is "grid aligned" to google maps then when I use the top left and bottom right corners of the map I get from googlemaps online then I can turn the users gps into an x and y on screen. However if the map is not "grid aligned" and is at an angle than the values my math returns draws the users position off screen. Obviously I am missing the part on how to handle the angle of the map so if anyone could shed some light on how to do that it would be super helpfull.

I know I would have to figure out the angle in radians and do some conversion but I have no idea where to start. This is what I use to get an x and y to plot on a canvas so far.

public double[] getPositionGPS(Location upperLeft, Location lowerRight, Location current){

    try{

        double hypotenuse = upperLeft.distanceTo(current);
        double bearing = upperLeft.bearingTo(current);
        double currentDistanceY = Math.cos(bearing * Math.PI / OneEightyDeg) * hypotenuse;
        //                           "percentage to mark the position"
        double totalHypotenuse = upperLeft.distanceTo(lowerRight);
        double totalDistanceY = totalHypotenuse * Math.cos(upperLeft.bearingTo(lowerRight) * Math.PI / OneEightyDeg);
        double currentPixelY = currentDistanceY / totalDistanceY * ImageSizeH;

        double hypotenuse2 = upperLeft.distanceTo(current);
        double bearing2 = upperLeft.bearingTo(current);
        double currentDistanceX = Math.sin(bearing2 * Math.PI / OneEightyDeg) * hypotenuse2;
        //                           "percentage to mark the position"
        double totalHypotenuse2 = upperLeft.distanceTo(lowerRight);
        double totalDistanceX = totalHypotenuse2 * Math.sin(upperLeft.bearingTo(lowerRight) * Math.PI / OneEightyDeg);
        double currentPixelX = currentDistanceX / totalDistanceX * ImageSizeW;

        return new double[]{currentPixelX, currentPixelY};

    }catch(Exception e){

        e.printStackTrace();
        return new double[]{0,0};

    }

解决方案

I might be completely wrong but I think this is a way to get the angle of rotation (or maybe the 4 corners):

1. Get the angle alpha between lowerRight and upperLeft. I guess this is equal 180 - lowerRight.bearingTo(upperLeft) EDIT (This equation should be dependent on the quadrant of the screen)
2. Use your X,Y (screen) coordinates to get the angle theta between the hypotenuse and the lower edge. I guess (again) the sine of this angle is equal to: (height) / ((height^2) + (width^2))^1/2
3. Now the angle of rotation is equal to theta - alpha

If you want to visualize this:

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