项目简介
文章里有相当多的用到中学数学中的知识,推导3d的几何模型是如何绘制到2d平面中去的,最终利用推导出的结论编写代码,实现一个波纹的demo
项目地址:https://github.com/zz632893783/canvas-3d
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安装项目依赖模块
npm install
运行项目
npm run dev
从z轴观察yz平面上的点
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想象一下有这么一个三维空间(如图),有一个点B,我们从A点观察B点。那么B点在xy平面上的投影即AB的延长线与平面xy的交点C。而xy平面不就是可以看一个二维的canvas画布吗。
我们暂且将A点放在z轴,B点放在yz平面,则A点的三维坐标可以表示为
A(0,0,zA),B点的三维坐标可以表示为B(0,yB,zB)。从B点做一条垂线垂z轴于D点。
ADB与AOC是相似三角形,所以有
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变换得
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其中DB即B点的y坐标,AO即A点的z坐标,DO即B点的z坐标,所以
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这里的OC也就是C点的y坐标。
从z轴观察xz平面上的点
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同理我们从A点观察平面xz上的某一点E(xE,0,zE),ADE与AOF是相似三角形
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变换得
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从z轴观察空间内任意坐标
之前所观测的B点是位于yz平面内,E点是位于xz平面内,但是如果是空间内任意位置的点呢
其实道理都是一样的,如下如
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如果将直线BD平移到E点,直线DE平移到B点,那么将形成一个矩形DBGE,矩形DBGE在xy平面上的投影为矩形OCHF。
由于AGE与AHF相似,所以有
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并且由于ADE与AOF也是相似三角形,所以
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所以
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推导得
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其中GE也就是G点的y坐标,因为矩形DBGE是平行于xy平面的,所以它们z坐标相同,DO等价于G点的z坐标,所以对于空间内任意位置G(xG,yG,zG)
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同样的方法我们可以推导出
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变换得
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结合上两步,CH是H点的x坐标,HF是H点的y坐标,所以从轴上的点A(0,0,zA)观察空间内任意位置G(xG,yG,zG)在平面xy上的投影可表示为
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从任意位置观察空间内任意坐标
沿着x轴平移坐标系
之前的推论到从z轴观察空间内任意位置的投影了,但是实际上A点是有特殊性的,因为它是位于z轴上的某一个点,其xy坐标都为0,如果A是空间内的任意一个点,情况又如何,请看下图
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假设这个时候真正的坐标系是xy'z',而坐标系xyz是我们临时所建立的一个虚拟的坐标系,那么这个时候A点相对于坐标系xy'z'来说,坐标点可表示为A(xA,0,zA),G点依旧表示为(xG,yG,zG)
我们之前推导的相似三角形的关系,即使换了坐标系,它们的关系依然成立,所以
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变换得
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只不过这个时候 BG=xG-xA,AO与DO与之前相同
求得
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之前推导出的相似三角形关系依旧成立,所以
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变换得
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由于GE,AO,DO与之前相比都没有变化,
所以得
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与之前的推导一致,最后我们得出结论,我们沿着x轴方向移动坐标系的时候(即图中的坐标系有xy'z'移动到了xyz位置),G点在平面xy的投影H点的y坐标不会有变化,但是x坐标为
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沿着y轴平移坐标系
如下图
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假设x'yz'是真正的坐标系,沿着y轴平移得到临时坐标系xyz,推导步骤和之前的相同,这里不再赘述,直接贴结果
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也就是说当沿着y轴方向移动坐标系的时候,投影H的x坐标不会有变化,y坐标变为
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对于空间内任意位置
对于空间内任意位置,我们都可以看成是在z轴上的某一点A(0,0,zA),先经历一次x轴方向的平移(此时投影H的y坐标不变),再经历一次y轴方向的平移(此时投影H的x坐标不变),平移之前点A观察到点G的投影H坐标可表示为
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对其进行x轴方向的平移,(此时投影H的y坐标不变),H的坐标可表示为
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再对其进行y轴方向的平移,(此时投影H的x坐标不变),H的坐标可表示为
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最终结论
从空间内的任意点A(xA,yA,zA)观察空间内的任一点G(xG,yG,zG),它在xy平面内的投影H的坐标为
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首先我们尝试写一个简单的几何图形
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立方体边长为100,则A(-50,50,50),B(-50,50,-50),C(50,50,-50),D(50,50,50),E(-50,-50,50),F(-50,-50,-50),G(50,-50,-50),H(50,-50,50),假定从z轴上某一点(0,0,300)观察
<template>
<div class="cube">
<canvas ref="cube" v-bind:width="canvasWidth" v-bind:height="canvasHeight"></canvas>
</div>
</template>
<script>
export default {
data: function () {
return {
canvasWidth: 600,
canvasHeight: 400,
ctx: null,
visual: {
x: 0,
y: 0,
z: 300
},
pointMap: {
A: (-50, 50, 50),
B: (-50, 50, -50),
C: (50, 50, -50),
D: (50, 50, 50),
E: (-50, -50, 50),
F: (-50, -50, -50),
G: (50, -50, -50),
H: (50, -50, 50)
}
}
},
methods: {
init: function () {
this.ctx = this.$refs.cube.getContext('2d')
},
draw: function () {}
},
mounted: function () {
this.init()
this.draw()
}
}
</script>
绘制方法也很简单,分别绘制矩形ABCD,矩形EFGH,然后再将AE,BF,CG,DH连线即可,只不过这里的ABCDEFGH点需要换算成投影在三维坐标系xy平面上的点,运用我们之前得出的结论,我们定义一个转换坐标点的函数
transformCoordinatePoint: function (x, y, z, offsetX = this.canvasWidth / 2, offsetY = this.canvasHeight / 2) {
return {
x: (x - this.visual.x) * this.visual.z / (this.visual.z - z) + offsetX,
y: (y - this.visual.y) * this.visual.z / (this.visual.z - z) + offsetY
}
}
然后编写draw函数
draw: function () {
let point
this.ctx.clearRect(0, 0, this.canvasWidth, this.canvasHeight)
// 绘制矩形ABCD
this.ctx.beginPath()
point = this.transformCoordinatePoint(...this.pointMap.A)
this.ctx.moveTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.B)
this.ctx.lineTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.C)
this.ctx.lineTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.D)
this.ctx.lineTo(point.x, point.y)
this.ctx.closePath()
this.ctx.stroke()
// 绘制矩形EFGH
this.ctx.beginPath()
point = this.transformCoordinatePoint(...this.pointMap.E)
this.ctx.moveTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.F)
this.ctx.lineTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.G)
this.ctx.lineTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.H)
this.ctx.lineTo(point.x, point.y)
this.ctx.closePath()
this.ctx.stroke()
// 绘制直线AE
this.ctx.beginPath()
point = this.transformCoordinatePoint(...this.pointMap.A)
this.ctx.moveTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.E)
this.ctx.lineTo(point.x, point.y)
this.ctx.stroke()
this.ctx.closePath()
// 绘制直线BF
this.ctx.beginPath()
point = this.transformCoordinatePoint(...this.pointMap.B)
this.ctx.moveTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.F)
this.ctx.lineTo(point.x, point.y)
this.ctx.stroke()
this.ctx.closePath()
// 绘制直线CD
this.ctx.beginPath()
point = this.transformCoordinatePoint(...this.pointMap.C)
this.ctx.moveTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.G)
this.ctx.lineTo(point.x, point.y)
this.ctx.stroke()
this.ctx.closePath()
// 绘制直线DH
this.ctx.beginPath()
point = this.transformCoordinatePoint(...this.pointMap.D)
this.ctx.moveTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.H)
this.ctx.lineTo(point.x, point.y)
this.ctx.stroke()
this.ctx.closePath()
}
查看代码运行结果
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似乎是对的,但是有感觉怪怪的,我们尝试将立方体绕着y轴旋转
这里需要另一个数学关系的推导
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想象一下从y轴俯视yz平面,这个时候点D的位置关系如下图
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这个时候假定D点与x轴的夹角是α,圆的半径为R,将D点绕着y轴旋转β旋转至D'点,这个时候D'与x轴夹角为α+β,此时D'的x坐标为cos(α+β) R,D'的z坐标为sin(α+β)R
回一下中学时候我们学过的三角形倍角公式
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D'的x坐标cos(α+β) R=Rcosα cosβ-Rsinα sinβ
D'的z坐标sin(α+β)R=R sinαcosβ+R cosαsinβ
而R sinα就是旋转之前D点的z坐标,Rcosα就是旋转之前D点的x坐标,
D'的x坐标为x cosβ-zsinβ
D'的z坐标为z cosβ+xsinβ
将结论代入到我们的立方体的8个顶点ABCDEFGH中
对于任一点D(xD,yD,zD),其绕y轴旋转β角的时候,它的三维坐标变为(xD cosβ-zDsinβ,yD,zD cosβ+xDsinβ)
转换为代码
methods: {
init: function () {
this.ctx = this.$refs.cube.getContext('2d')
},
transformCoordinatePoint: function (x, y, z, offsetX = this.canvasWidth / 2, offsetY = this.canvasHeight / 2) {
return {
x: (x - this.visual.x) * this.visual.z / (this.visual.z - z) + offsetX,
y: (y - this.visual.y) * this.visual.z / (this.visual.z - z) + offsetY
}
},
draw: function () {
let point
this.ctx.clearRect(0, 0, this.canvasWidth, this.canvasHeight)
// 绘制矩形ABCD
this.ctx.beginPath()
point = this.transformCoordinatePoint(...this.pointMap.A)
this.ctx.moveTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.B)
this.ctx.lineTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.C)
this.ctx.lineTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.D)
this.ctx.lineTo(point.x, point.y)
this.ctx.closePath()
this.ctx.stroke()
// 绘制矩形EFGH
this.ctx.beginPath()
point = this.transformCoordinatePoint(...this.pointMap.E)
this.ctx.moveTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.F)
this.ctx.lineTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.G)
this.ctx.lineTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.H)
this.ctx.lineTo(point.x, point.y)
this.ctx.closePath()
this.ctx.stroke()
// 绘制直线AE
this.ctx.beginPath()
point = this.transformCoordinatePoint(...this.pointMap.A)
this.ctx.moveTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.E)
this.ctx.lineTo(point.x, point.y)
this.ctx.stroke()
this.ctx.closePath()
// 绘制直线BF
this.ctx.beginPath()
point = this.transformCoordinatePoint(...this.pointMap.B)
this.ctx.moveTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.F)
this.ctx.lineTo(point.x, point.y)
this.ctx.stroke()
this.ctx.closePath()
// 绘制直线CD
this.ctx.beginPath()
point = this.transformCoordinatePoint(...this.pointMap.C)
this.ctx.moveTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.G)
this.ctx.lineTo(point.x, point.y)
this.ctx.stroke()
this.ctx.closePath()
// 绘制直线DH
this.ctx.beginPath()
point = this.transformCoordinatePoint(...this.pointMap.D)
this.ctx.moveTo(point.x, point.y)
point = this.transformCoordinatePoint(...this.pointMap.H)
this.ctx.lineTo(point.x, point.y)
this.ctx.stroke()
this.ctx.closePath()
},
animationFrame: function () {
let rotationAngle = 1
window.requestAnimationFrame(() => {
for (let key in this.pointMap) {
let point = this.pointMap[key]
// 保存x,y,z坐标
let x = point[0]
let y = point[1]
let z = point[2]
// 变换后的x坐标
point[0] = x * Math.cos(rotationAngle / 180 * Math.PI) - z * Math.sin(rotationAngle / 180 * Math.PI)
// 绕y轴旋转,y左边不会发生变化
point[1] = y
// 变换后的z坐标
point[2] = z * Math.cos(rotationAngle / 180 * Math.PI) + x * Math.sin(rotationAngle / 180 * Math.PI)
}
this.draw()
this.animationFrame()
})
}
},
mounted: function () {
this.init()
this.animationFrame()
}
代码运行效果
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绘制波浪
波浪是由若干条正弦函数组成的,我们先绘制一条正弦函数
中学数学中,描述一条正弦函数的方程式 y=a*sin(b * x + c) + d,所以我们构造一个类,需要的参数也是a,b,c,d,为了确定函数的起始位置和结束位置,另外需要两个参数start,end
class Line {
constructor (a, b, c, d, start, end) {
this.a = a
this.b = b
this.c = c
this.d = d
this.start = start
this.end = end
}
}
export default Line
实际上每条正弦函数曲线并不是真正的连线,而是由于一个个点组成,我们在增加一个参数,确定每个点之间的间距,并在实例化的时候生成这些点,我们这里保存在pointList中
class Line {
constructor (a, b, c, d, start, end, gap) {
this.a = a
this.b = b
this.c = c
this.d = d
this.start = start
this.end = end
this.gap = gap
this.pointList = []
this.computePointList()
}
computePointList () {
this.pointList = []
for (let i = this.start; i <= this.end; i = i + this.gap) {
let x = i
let y = this.a * Math.sin((this.b * x + this.c) / 180 * Math.PI) + this.d
this.pointList.push({
x,
y
})
}
}
}
export default Line
在页面中,Line实例保存在lineList中,遍历lineList绘制点
<template>
<canvas class="wave" ref="wave" v-bind:width="canvasWidth" v-bind:height="canvasHeight"></canvas>
</template>
<script>
import Line from './line'
export default {
props: {},
data: function () {
return {
canvasWidth: 600,
canvasHeight: 400,
ctx: null,
visual: {
x: 0,
y: -100,
z: 1000
},
lineList: [
new Line(20, 2, 0, 0, -200, 200, 10)
]
}
},
methods: {
init: function () {
this.ctx = this.$refs.wave.getContext('2d')
},
draw: function () {
this.ctx.clearRect(0, 0, this.canvasWidth, this.canvasHeight)
this.lineList.forEach(line => {
line.pointList.forEach(item => {
this.ctx.beginPath()
this.ctx.arc(item.x + this.canvasWidth / 2, item.y + this.canvasHeight / 2, 2, 0, 2 * Math.PI)
this.ctx.closePath()
this.ctx.fill()
})
})
}
},
mounted: function () {
this.init()
this.draw()
}
}
</script>
<style lang="stylus" scoped>
.wave {
border: 1px solid;
}
</style>
看一下代码效果
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我们再试着让它动起来,波浪的运动改变的实际上是每个点的纵坐标,只要我们知道每个点距离原点的偏移量,我们就能计算出当前的纵坐标,所以我们在生成点的时候,记录偏移量,我们我们声明一个updatePointList方法用以跟新点的位置
computePointList () {
this.pointList = []
for (let i = this.start; i <= this.end; i = i + this.gap) {
let x = i
let y = this.a * Math.sin((this.b * x + this.c) / 180 * Math.PI) + this.d
let offset = i
this.pointList.push({
x,
y,
offset
})
}
}
updatePointList () {
this.pointList.forEach(item => {
item.y = this.a * Math.sin((this.b * item.x + this.c + item.offset) / 180 * Math.PI) + this.d
})
}
在页面中,我们定义一个变量lineOffset,通过调整它控制line实例的c值(也就是对直线进行平移),并不断地调用之前写好的updatePointList方法,更新点的位置
animationFrame: function () {
window.requestAnimationFrame(() => {
this.lineList.forEach(line => {
line.c = this.lineOffset
line.updatePointList()
})
this.lineOffset = this.lineOffset + 1
this.draw()
this.animationFrame()
})
}
代码运行效果
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但是这个只是二维平面的,想象一下空间中有很多条这样的直线,然后有的直线离屏幕比较近,有的离屏幕比较远,所以我们如果在三维空间中描述直线的话,我们还需要知道三维坐标系中的z坐标,除此之代直线的x,z与之前的相比并无变化
constructor (a, b, c, d, z, start, end, gap) {
this.a = a
this.b = b
this.c = c
this.d = d
this.z = z
this.start = start
this.end = end
this.gap = gap
this.pointList = []
this.computePointList()
}
我们之前已经推导过,对于任一点D(xD,yD,zD),其绕y轴旋转β角的时候,它的三维坐标变为(xDcosβ-zDsinβ,yD,zDcosβ+xDsinβ),想象一下我们直线上的每一个点,其实都是绕着y轴旋转的,旋转之后y轴的坐标不会发生变化,然后看我们原型中声明的updatePointList方法
updatePointList () {
this.pointList.forEach(item => {
item.y = this.a * Math.sin((this.b * item.x + this.c + item.offset) / 180 * Math.PI) + this.d
})
}
y轴的坐标我们之前已经写好了,我们运用(xDcosβ-zDsinβ,yD,zDcosβ+xDsinβ)推导每个点旋转β角后的坐标位置
updatePointList (rotationAngleSpeed) {
this.pointList.forEach(item => {
let x = item.x
let z = item.z
item.x = x * Math.cos(rotationAngleSpeed / 180 * Math.PI) - z * Math.sin(rotationAngleSpeed / 180 * Math.PI)
item.z = z * Math.cos(rotationAngleSpeed / 180 * Math.PI) + x * Math.sin(rotationAngleSpeed / 180 * Math.PI)
})
}
代码运行效果
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但是此时的粒子并没有沿着y轴方向移动,我们将两步结合
updatePointList (rotationAngleSpeed, visual) {
this.pointList.forEach(item => {
let x = item.x
let y = item.y
let z = item.z
item.x = x * Math.cos(rotationAngleSpeed / 180 * Math.PI) - z * Math.sin(rotationAngleSpeed / 180 * Math.PI)
item.z = z * Math.cos(rotationAngleSpeed / 180 * Math.PI) + x * Math.sin(rotationAngleSpeed / 180 * Math.PI)
item.y = this.a * Math.sin((this.b * x + this.c + item.offset) / 180 * Math.PI) + this.d
})
}
然后我们看一下运行结果
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非常的怪异,我们似乎哪里写错了
回过头来看我们的代码,波纹的左右移动实际上是靠从新计算每个点的y坐标实现,而计算y坐标我们用的函数是
item.y = this.a * Math.sin((this.b * x + this.c + item.offset) / 180 * Math.PI) + this.d
但是我们实际上每计算一次item.y的值,我们通过控制this.c来实现平移,所以除了this.c之外,
this.a * Math.sin((this.b * x + this.c + item.offset) / 180 * Math.PI) + this.d
中的 this.a,x(这里的x也就是item.x),this.b,item.offset,this.d都不应该有变化,但是我们代码中的
item.x = x * Math.cos(rotationAngleSpeed / 180 * Math.PI) - z * Math.sin(rotationAngleSpeed / 180 * Math.PI)
却在不停地变化item.x的值,所以我们需要保存一份最开始时时候的x值
computePointList () {
this.pointList = []
for (let i = this.start; i <= this.end; i = i + this.gap) {
let x = i
let y = this.a * Math.sin((this.b * x + this.c) / 180 * Math.PI) + this.d
let offset = i
this.pointList.push({
x,
y,
z: this.z,
originX: x,
offset
})
}
}
updatePointList (rotationAngleSpeed, visual) {
this.pointList.forEach(item => {
let x = item.x
let y = item.y
let z = item.z
item.x = x * Math.cos(rotationAngleSpeed / 180 * Math.PI) - z * Math.sin(rotationAngleSpeed / 180 * Math.PI)
item.z = z * Math.cos(rotationAngleSpeed / 180 * Math.PI) + x * Math.sin(rotationAngleSpeed / 180 * Math.PI)
item.y = this.a * Math.sin((this.b * item.originX + this.c + item.offset) / 180 * Math.PI) + this.d
})
}
继续看运行效果
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虽然代码是对的,但是这个时候的这些点还只是平面上的点,并没有3d效果,我们回到最开始推导出的结论
从空间内的任意点A(xA,yA,zA)观察空间内的任一点G(xG,yG,zG),它在xy平面内的投影H的坐标为
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我们顶一个个观察点
visual: {
x: 0,
y: -100,
z: 1000
}
并在每次updatePointList方法中调用它,计算这个点在平面xy上的投影位置
animationFrame: function () {
window.requestAnimationFrame(() => {
this.lineList.forEach(line => {
line.c = this.lineOffset
line.updatePointList(this.rotationAngleSpeed, this.visual)
})
this.lineOffset = this.lineOffset + 1
this.draw()
this.animationFrame()
})
}
在updatePointList函数中,我们拿到传入的视角点visual,并根据视角点计算空间内的点在平面xy上的投影,我们记为(canvasX,canvasY)
updatePointList (rotationAngleSpeed, visual) {
this.pointList.forEach(item => {
let x = item.x
let y = item.y
let z = item.z
item.x = x * Math.cos(rotationAngleSpeed / 180 * Math.PI) - z * Math.sin(rotationAngleSpeed / 180 * Math.PI)
item.z = z * Math.cos(rotationAngleSpeed / 180 * Math.PI) + x * Math.sin(rotationAngleSpeed / 180 * Math.PI)
item.y = this.a * Math.sin((this.b * item.originX + this.c + item.offset) / 180 * Math.PI) + this.d
item.canvasX = (item.x - visual.x) * visual.z / (visual.z - z)
item.canvasY = (item.y - visual.y) * visual.z / (visual.z - z)
})
}
由于我们现在是要绘制投影的坐标,所以我们的draw方法中的绘制圆点的方法需要换成(canvasX,canvasY)
draw: function () {
this.ctx.clearRect(0, 0, this.canvasWidth, this.canvasHeight)
this.lineList.forEach(line => {
line.pointList.forEach(item => {
this.ctx.beginPath()
this.ctx.arc(item.canvasX + this.canvasWidth / 2, item.canvasY + this.canvasHeight / 2, 2, 0, 2 * Math.PI)
this.ctx.closePath()
this.ctx.fill()
})
})
}
运行结果
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然后我们试着加入更多的线条
lineList: [
new Line(20, 2, 0, 0, -150, -200, 200, 10),
new Line(20, 2, 0, 0, -120, -200, 200, 10),
new Line(20, 2, 0, 0, -90, -200, 200, 10),
new Line(20, 2, 0, 0, -60, -200, 200, 10),
new Line(20, 2, 0, 0, -30, -200, 200, 10),
new Line(20, 2, 0, 0, 0, -200, 200, 10),
new Line(20, 2, 0, 0, 30, -200, 200, 10),
new Line(20, 2, 0, 0, 60, -200, 200, 10),
new Line(20, 2, 0, 0, 90, -200, 200, 10),
new Line(20, 2, 0, 0, 120, -200, 200, 10),
new Line(20, 2, 0, 0, 150, -200, 200, 10)
]
运行结果
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我们试着再对每条直线作不同的平移,我们平移直线是通过line构造函数中的参数c控制的,在animationFrame方法中
animationFrame: function () {
window.requestAnimationFrame(() => {
this.lineList.forEach((line, index) => {
line.c = this.lineOffset
line.updatePointList(this.rotationAngleSpeed, this.visual)
})
this.lineOffset = this.lineOffset + 1
this.draw()
this.animationFrame()
})
}
line.c是被赋值为this.lineOffset,所以我们看到每条直线的偏移量都是一致的,我们试着修改代码,使每条直线的偏移量不一致
animationFrame: function () {
window.requestAnimationFrame(() => {
this.lineList.forEach((line, index) => {
line.c = this.lineOffset + index * 30
line.updatePointList(this.rotationAngleSpeed, this.visual)
})
this.lineOffset = this.lineOffset + 1
this.draw()
this.animationFrame()
})
}
代码运行结果
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实际上我们还忽略了一个点,那就是点的远近大小关系,真实情况应该是离我们屏幕较近的点,看起来更大,离屏更远的点,看起来更小,而离屏幕的距离不就是z的坐标吗
我们回到最开始推论的那副图

在A点观察直线DB在平面xy内的投影OC,由相似三角形可知
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推导得
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所以假定小圆点的半斤是R,站在A(0,0,)点观测小圆点位于平面xy上投影的半径为

我们将draw方法中的代码做修改
draw: function () {
this.ctx.clearRect(0, 0, this.canvasWidth, this.canvasHeight)
this.lineList.forEach(line => {
line.pointList.forEach(item => {
this.ctx.beginPath()
// 暂且假定小圆点的原始半径是2,则投影半径可表示为
let pointSize = 2 * this.visual.z / (this.visual.z - item.z)
this.ctx.arc(item.canvasX + this.canvasWidth / 2, item.canvasY + this.canvasHeight / 2, pointSize, 0, 2 * Math.PI)
this.ctx.closePath()
this.ctx.fill()
})
})
}
运行效果
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我们不断调整实例化时候line的各个参数,最终实现效果
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到此,请记住这篇文章最重要的一个结论
从空间内的任意点A(xA,yA,zA)观察空间内的任一点G(xG,yG,zG),它在xy平面内的投影H的坐标为


如果以后还有与canvas绘制3d图形有关的文章,这个结论会一直用到
作者:zz632893783
链接:https://www.jianshu.com/p/e3ebe08dddad
来源:简书
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。
来源:oschina
链接:https://my.oschina.net/mdu/blog/4296084