使用THREE.geometry的简单四面体

Question

从THREEjs.org/docs提供的示例中：

var geometry = new THREE.Geometry();

geometry.vertices.push(
    new THREE.Vector3( -10,  10, 0 ),
    new THREE.Vector3( -10, -10, 0 ),
    new THREE.Vector3(  10, -10, 0 )
);

geometry.faces.push( new THREE.Face3( 0, 1, 2 ) );

geometry.computeBoundingSphere();

通过指定顶点来创建三角形非常简单。 如果我想继续使用此方法制作四面体，我发现自己要重复三遍此代码，一次针对四面体的每个面，每次更改一个矢量来定义组成四面体的三角形之一...

如果定义四面体的第四个矢量，是否有一个简单的方法可以创建四面体而不必分别定义每个面？

就像是：

var geometry = new THREE.Geometry();

geometry.vertices.push(
    new THREE.Vector3( -10,  10, 0 ),
    new THREE.Vector3( -10, -10, 0 ),
    new THREE.Vector3(  10, -10, 0 ),
    new THREE.Vector3(  10, 10, 10 ),
);

geometry.faces.push( new THREE.Face3( 0, 1, 2 ) );

geometry.computeBoundingSphere();

var mesh = new THREE.Mesh( geometry, material );
scene.add( mesh )

当然，此代码不起作用，仅忽略最后一个顶点。

Answer 1

在three.js代码中查看src / extras / geometries / TetrahedronGeometry.js

将代码复制到此处以供参考：

THREE.TetrahedronGeometry = function ( radius, detail ) {

    var vertices = [ 1,  1,  1,   - 1, - 1,  1,   - 1,  1, - 1,    1, - 1, - 1];
    var indices = [ 2,  1,  0,    0,  3,  2,    1,  3,  0,    2,  3,  1];

    THREE.PolyhedronGeometry.call( this, vertices, indices, radius, detail );
};

THREE.TetrahedronGeometry.prototype = Object.create( THREE.Geometry.prototype );