Skewing a bitmap only in the vertical direction

左心房为你撑大大i 提交于 2019-12-29 04:54:47

问题


I want to skew (correct me if this is not the correct word) a bitmap so that it appears to have depth. A good way to visualize what I am asking for is how the credits of Star Wars are angled to show depth.

I have tried the following:

canvas.getMatrix().postSkew(kx,ky,px,py);

and

canvas.skew(sx,sy);

But I have not had much success. The above methods seem to always transform the bitmap into a parallelogram. Is there a way to transform the bitmap into a trapezoid instead?

Here is a snippet of code that I took from the examples that Romain pointed me to.

canvas.rotate(-mOrientation[0] + mHeading, mCenterX, mCenterY);

camera.save();

if (mReverse) {
    camera.translate(0.0f, 0.0f, mDepthZ * interpolatedTime);
} else {
    camera.translate(0.0f, 0.0f, mDepthZ * (1.0f - interpolatedTime));
}

camera.rotateX(mOrientation[1]);
camera.applyToCanvas(canvas);
canvas.drawPath(mPath, mPaint);
canvas.drawCircle(mCenterX, mCenterY, mRadius - 37, mPaint);

camera.restore();

回答1:


You cannot achieve the effect you want with skew(). However, you can use a Camera object and 3D rotations to achieve this effect. The Camera will generate a Matrix for you that you can then apply on the Canvas. Note that the result will not be perspective correct, but good enough for your purpose. This how 3D rotations are done in Honeycomb's Launcher for instance (and many other apps.)




回答2:


I spent a lot of time working on this today (ran into the same problem) and came up with the code below.

Key thing to note, you need to set preTranslate() and postTranslate() to the center (or somewhere else) of your Canvas area. It seems to mean that it uses the center of the image to apply the transformation from, instead of the upper left corner (x=0,y=0) by default. This is why you would get a parallelogram instead of what you would expect, a trapezoid (Thanks for teaching me the names of those).

The other important thing that I picked up is the Save/Restore functions on the Canvas/Camera. Basically, if you call the Rotate functions consecutively three times without restoring the state back each time, you would keep rotating your object around and around each time you draw. That might be what you want, but I certainly didn't in my case. Same applies to the canvas as you are basically applying the Matrix from the Camera object to the Canvas and it needs to be reset otherwise the same thing occurs.

Hope this helps someone, this is not well documented for beginners. Tip to anyone reading this, check out the APIDemos folder in the SDK Samples. There is a Rotate3dAnimation.java file which demonstrates this as well.

//Snippet from a function used to handle a draw
mCanvas.save(); //save a 'clean' matrix that doesn't have any camera rotation in it's matrix
ApplyMatrix(); //apply rotated matrix to canvas
Draw(); //Does drawing
mCanvas.restore(); //restore clean matrix
//    

public void ApplyMatrix() {
  mCamera.save();
  mCamera.rotateX(-66);
  mCamera.rotateY(0);
  mCamera.rotateZ(0);
  mCamera.getMatrix(mMatrix);

  int CenterX = mWidth / 2;
  int CenterY = mHeight / 2; 
  mMatrix.preTranslate(-CenterX, -CenterY); //This is the key to getting the correct viewing perspective
  mMatrix.postTranslate(CenterX, CenterY); 

  mCanvas.concat(mMatrix);
  mCamera.restore();    
}



回答3:


I don't think the "Star Wars effect" is an affine transformation, which I think are the only operations supported by Matrix.



来源:https://stackoverflow.com/questions/5354183/skewing-a-bitmap-only-in-the-vertical-direction

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