Below is a small code which takes input of File containing image and then tilts it by an angle. Now the problem is that: the output file has a lower resolution when compared to the input one. In my case input file was of size 5.5 MB and the output file was of 1.1 MB. Why is it?
/**
*
* @param angle Angle to be rotate clockwise. Ex: Math.PI/2, -Math.PI/4
*/
private static void TurnImageByAngle(File image, double angle)
{
BufferedImage original = null;
try {
original = ImageIO.read(image);
GraphicsConfiguration gc = getDefaultConfiguration();
BufferedImage rotated1 = tilt(original, angle, gc);
//write iamge
ImageIO.write(rotated1, getFileExtension(image.getName()), new File("temp"+" "+angle+"."+getFileExtension(image.getName())));
} catch (IOException ex) {
Logger.getLogger(RotateImage2.class.getName()).log(Level.SEVERE, null, ex);
}
}
public static GraphicsConfiguration getDefaultConfiguration() {
GraphicsEnvironment ge = GraphicsEnvironment.getLocalGraphicsEnvironment();
GraphicsDevice gd = ge.getDefaultScreenDevice();
return gd.getDefaultConfiguration();
}
public static BufferedImage tilt(BufferedImage image, double angle, GraphicsConfiguration gc) {
double sin = Math.abs(Math.sin(angle)), cos = Math.abs(Math.cos(angle));
int w = image.getWidth(), h = image.getHeight();
int neww = (int)Math.floor(w*cos+h*sin), newh = (int)Math.floor(h*cos+w*sin);
int transparency = image.getColorModel().getTransparency();
BufferedImage result = gc.createCompatibleImage(neww, newh, transparency);
Graphics2D g = result.createGraphics();
g.translate((neww-w)/2, (newh-h)/2);
g.rotate(angle, w/2, h/2);
g.drawRenderedImage(image, null);
return result;
}
Thats no surprise if you look at the code (Copy&Paste without understanding what the Code does has its drawbacks). The tilt()-Method makes extra effort (in its 3rd line) to make the image properly sized.
If you think about it, you cant expect the image to stay the same size.