如何使用不透明度混合多个图层?

时间:2016-11-17 07:56:55

标签: javascript three.js

使用three.js,我想对单个场景应用单独的后处理效果,然后将所有这些场景组合成最终渲染。要做到这一点,我使用的是three.js Effects Composer。



const radialBlurShader = {
	uniforms: {
		"tDiffuse": { value: null },
	},
	vertexShader: `
		varying vec2 vUv;
		void main() {
			vUv = uv;
			gl_Position = projectionMatrix * modelViewMatrix * vec4(position, 1.0);
		}
	`,

	fragmentShader: `
		uniform sampler2D tDiffuse;
		varying vec2 vUv;

    const float strength = 0.7;
    const int samples = 50;

		void main() {
      vec4 col = vec4(0);

      vec2 dir = vec2(0.5) - vUv;
      for (int i = 0; i < samples; i++) {
        float amount = strength * float(i) / float(samples);

        vec4 sample = 
          texture2D(tDiffuse, vUv + dir * amount) + 
          texture2D(tDiffuse, vUv - dir * amount);

        col += sample;
      }

			gl_FragColor = 0.5 * col / float(samples);
		}
	`
};

const renderWidth = window.innerWidth;
const renderHeight = window.innerHeight;

const renderer = new THREE.WebGLRenderer({
  antialias: true,
});

renderer.setSize(renderWidth, renderHeight);
document.body.appendChild(renderer.domElement);

const camera = new THREE.PerspectiveCamera(45, renderWidth / renderHeight, 0.1, 1000);
camera.position.z = 10;

var geometry = new THREE.PlaneGeometry(1, 1);
var material = new THREE.MeshBasicMaterial({ 
  color: 0x0000ff,
  transparent: true
});

function makeEC(scene) {
  const ec = new THREE.EffectComposer(renderer);
  const rp = new THREE.RenderPass(scene, camera);
  const sp = new THREE.ShaderPass(radialBlurShader);

  rp.clearColor = 0xffffff;
  rp.clearAlpha = 0;

  sp.renderToScreen = true;
  sp.material.transparent = true;
  sp.material.blending = THREE.CustomBlending;

  ec.addPass(rp);
  ec.addPass(sp);
  return ec;
}

const scene1 = new THREE.Scene();
const mesh1 = new THREE.Mesh(geometry, material);
mesh1.position.x = 2;
scene1.add(mesh1);
ec1 = makeEC(scene1);

const scene2 = new THREE.Scene();
const mesh2 = new THREE.Mesh(geometry, material);
mesh2.position.x = -2;
scene2.add(mesh2);
ec2 = makeEC(scene2);

renderer.setClearColor(0xffffaa, 1);
renderer.autoClear = false;
renderer.clear();
ec1.render();
ec2.render();
&#13;
<script src="https://cdnjs.cloudflare.com/ajax/libs/three.js/r82/three.js"></script>
<script src="https://rawgit.com/mrdoob/three.js/r82/examples/js/shaders/CopyShader.js"></script>
<script src="https://rawgit.com/mrdoob/three.js/r82/examples/js/postprocessing/EffectComposer.js"></script>
<script src="https://rawgit.com/mrdoob/three.js/r82/examples/js/postprocessing/RenderPass.js"></script>
<script src="https://rawgit.com/mrdoob/three.js/r82/examples/js/postprocessing/ShaderPass.js"></script>
&#13;
&#13;
&#13;

我为要应用的每个后处理着色器创建一个单独的Effects Composer实例。 (在这个例子中,为了简单起见,我使用相同的径向模糊着色器两次。)

function makeEC(scene) {
    const ec = new THREE.EffectComposer(renderer);
    const rp = new THREE.RenderPass(scene, camera);
    const sp = new THREE.ShaderPass(radialBlurShader);

    rp.clearColor = 0xffffff;
    rp.clearAlpha = 0;

    sp.renderToScreen = true;
    sp.material.transparent = true;

    // defaults to SRC_ALPHA, ONE_MINUS_SRC_ALPHA
    sp.material.blending = THREE.CustomBlending;

    ec.addPass(rp);
    ec.addPass(sp);
    return ec;
}

const ec1 = makeEC(scene1);
const ec2 = makeEC(scene2);

如您所见,我将渲染过程清除为透明背景。然后着色器通道将使用典型的SRC_ALPHA,ONE_MINUS_SRC_ALPHA混合绘制到渲染缓冲区。

我的渲染代码看起来像这样

renderer.setClearColor(0xffffaa, 1);
renderer.autoClear = false;
renderer.clear();
ec1.render();
ec2.render();

但是,此过程不会正确地将图层混合在一起

这就是我得到的

enter image description here

混合前的第一遍(正确)

enter image description here

混合前的第二遍(正确)

enter image description here

两个传球按照描述混合(不正确) 太暗了

enter image description here

premultipliedAlpha禁用(不正确) 太透明了

当两个图层混合在一起时,为什么正方形太暗?

当禁用预乘的Alpha时,为什么方块太透明了?

如何将两个图层混合在一起,使它们看起来与混合前相同?

1 个答案:

答案 0 :(得分:1)

将Shader Pass blendSrc更改为ONE可解决问题。 e.g。

sp.material.blending = THREE.CustomBlending;
sp.material.blendSrc = THREE.OneFactor;

我相信这是有效的,因为径向模糊着色器的功能尤其如此。首先,渲染通道颜色缓冲区被清除为黑色透明颜色,并在其内部绘制不透明的正方形。然后,径向模糊着色器使用其周围的透明像素模糊这些不透明像素。这具有预乘任何非透明像素的alpha的效果。

此时将纹理绘制到缓冲区并不需要将源像素再次乘以alpha通道。这是正确的解释吗?