Phil
Phil

Reputation: 36289

Rotate Matrix for bitmap - Android

I have a custom map (canvas) that I am drawing a map pointer on. I am updating it in my onLocationChanged method of my main class, however I am struggling to get the bitmap to rotate. getBearing() doesn't seem to work (at least not for me) and so I am working to find the slope between points on the map. Any help would be greatly appreciated.

public void setBearing(Point prev, Point curr){

  float slope = 1;
  if (prev.x - curr.x !=0){
     slope = (float) ((y1-y2)/(x1-x2));
     bearing = (float) Math.atan(slope);
  }

}

...

Paint p = new Paint();
Matrix matrix = new Matrix();
matrix.postRotate(bearing, coords.x, coords.y);
Bitmap rotatedImage = Bitmap.createBitmap(image, 0, 0, image.getWidth(), 
                                          image.getHeight(), matrix, true);
canvas.drawBitmap(rotatedImage, x-image.getWidth()/2, y-image.getHeight()/2, p);

Edit:

Using latitude and longitude coordinates to find a bearing is more difficult than simply between two points. This code however (modified from code found here) works well:

public void setBearing(Location one, Location two){
  double lat1 = one.getLatitude();
  double lon1 = one.getLongitude();
  double lat2 = two.getLatitude();
  double lon2 = two.getLongitude();
  double deltaLon = lon2-lon1;
  double y = Math.sin(deltaLon) * Math.cos(lat2);
  double x = Math.cos(lat1)*Math.sin(lat2) - Math.sin(lat1)*Math.cos(lat2)*Math.cos(deltaLon);
  bearing = (float) Math.toDegrees(Math.atan2(y, x));
}

Upvotes: 2

Views: 4973

Answers (3)

Nayanesh Gupte
Nayanesh Gupte

Reputation: 2775

in case you want to rotate ImageView

private void rotateImage(ImageView imageView, double angle) {

    Matrix matrix = new Matrix();
    imageView.setScaleType(ScaleType.MATRIX); // required
    matrix.postRotate((float) angle, imageView.getDrawable().getBounds()
            .width() / 2, imageView.getDrawable().getBounds().height() / 2);
    imageView.setImageMatrix(matrix);
}

Upvotes: 0

antonakos
antonakos

Reputation: 8361

To correctly compute the angle with a minimum of fuss and division by zero risks, atan2() should be preferred to atan(). The following function returns the angle relative to the x-axis of a non-zero vector from a to b:

public float getBearing(Point a, Point b) { // Valid for a != b.
    float dx = b.x - a.x;
    float dy = b.y - a.y;
    return (float)Math.atan2(dy, dx);
}

I can't give advice on how to rotate the bitmap by a given angle, since I am not familiar with your API.

Upvotes: 2

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