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5fa619a73a
| Author | SHA1 | Date | |
|---|---|---|---|
 jinchao
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jinchao
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jinchao
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baee74ef27 | ||
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jinchao
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jinchao
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jinchao
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jinchao
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jinchao
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jinchao
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7096f66a6d |
@@ -28,6 +28,8 @@ export default defineUserConfig({
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lastUpdated: false,
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lastUpdated: false,
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repo: "https://github.com/thejinchao/thejinchao.github.io",
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repo: "https://github.com/thejinchao/thejinchao.github.io",
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colorMode: "light",
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colorMode: "light",
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editLink : false,
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contributors : false,
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}),
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}),
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head:[
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head:[
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['link', { rel: 'icon', href: 'favicon.png' }]
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['link', { rel: 'icon', href: 'favicon.png' }]
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@@ -40,7 +40,15 @@ module.exports = [
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"2025/02/DH.md",
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"2025/02/DH.md",
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"2025/02/SegmentCircle.md",
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"2025/02/SegmentCircle.md",
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"2025/02/Ellipse.md",
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"2025/02/Ellipse.md",
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"2025/02/Fibonacci.md"
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"2025/02/Fibonacci.md",
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{
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text: '匀速贝塞尔曲线运动的实现',
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collapsible: true,
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children: [
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{ text: "Part1", link: "2025/03/BezierLine01.md" },
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{ text: "Part2", link: "2025/03/BezierLine02.md" }
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]
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}
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]
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]
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},
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},
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{
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{
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@@ -1,7 +1,7 @@
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<html>
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<html>
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<head>
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<head>
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<script src="https://cdn.jsdelivr.net/npm/p5@1.4.2/lib/p5.js"></script>
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<script src="https://cdn.jsdelivr.net/npm/p5@1.4.2/lib/p5.js"></script>
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<script src="/js/bezier_app.js"></script>
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<script src="/js/bezier_quad.js"></script>
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<script src="/js/bezier_base.js"></script>
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<script src="/js/bezier_base.js"></script>
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</head>
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</head>
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<body style="margin:0px; padding:0px; overflow: hidden">
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<body style="margin:0px; padding:0px; overflow: hidden">
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@@ -0,0 +1,52 @@
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<html>
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<head>
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<script src="https://cdn.jsdelivr.net/npm/p5@1.4.2/lib/p5.js"></script>
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<script src="/js/bezier_cubic.js"></script>
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<script src="/js/bezier_base.js"></script>
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</head>
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<body style="margin:0px; padding:0px; overflow: hidden">
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<script>
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var lineSegments;
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var cubicBezierLine;
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var P0={x:10.0, y:10.0}, P1={x:100.0, y:160.0}, P2={x:350.0, y:200.0}, P3={x:300.0, y:80.0};
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function setup() {
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const searchParams = new URLSearchParams(window.location.search);
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let uniformSpeed = searchParams.get('uniformSpeed')!=0;
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canvas = createCanvas(windowWidth, windowHeight);
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lineSegments = new LineSegments();
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lineSegments.addPoint(P0.x, P0.y);
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lineSegments.addPoint(P1.x, P1.y);
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lineSegments.addPoint(P2.x, P2.y);
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lineSegments.addPoint(P3.x, P3.y);
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cubicBezierLine = new CubicBezierLine(P0, P1, P2, P3, 20, uniformSpeed)
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}
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function draw() {
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clear();
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background('white');
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noStroke();
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lineSegments.draw();
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cubicBezierLine.draw()
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}
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function mousePressed(){
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lineSegments.handleMousePressed();
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}
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function mouseDragged(){
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lineSegments.handleMouseDragged();
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}
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function mouseReleased(){
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lineSegments.handleMouseReleased();
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cubicBezierLine.updatePoints(lineSegments.points[0], lineSegments.points[1], lineSegments.points[2], lineSegments.points[3]);
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}
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</script>
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</body>
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</html>
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After Width: | Height: | Size: 19 KiB |
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After Width: | Height: | Size: 42 KiB |
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After Width: | Height: | Size: 211 KiB |
@@ -0,0 +1,116 @@
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class CubicBezierLine {
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constructor(_p0, _p1, _p2, _p3, _step, _uniformSpeedMode) {
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this.step=_step;
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this.points=[];
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this.uniformSpeedMode=_uniformSpeedMode;
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this.updatePoints(_p0, _p1, _p2, _p3);
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}
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position(t) {
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let it = 1-t;
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let x = it*it*it*this.p0.x + 3*it*it*t*this.p1.x +
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3*it*t*t*this.p2.x + t*t*t*this.p3.x;
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let y =it*it*it*this.p0.y + 3*it*it*t*this.p1.y +
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3*it*t*t*this.p2.y + t*t*t*this.p3.y;
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return {x:x, y:y}
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}
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speed(t) {
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let it = 1-t;
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let sx = -3 * this.p0.x*it*it + 3*this.p1.x*it*it -
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6*this.p1.x*it*t + 6*this.p2.x*it*t - 3*this.p2.x*t*t + 3*this.p3.x*t*t;
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let sy = -3 * this.p0.y*it*it + 3*this.p1.y*it*it -
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6*this.p1.y*it*t + 6*this.p2.y*it*t - 3*this.p2.y*t*t + 3*this.p3.y*t*t;
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return Math.sqrt(sx*sx+sy*sy);
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}
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length(t){
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//使用simpson算法的分割数
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const TOTAL_SIMPSON_STEP = 1000;
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//分割份数
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let stepCounts = Math.floor(TOTAL_SIMPSON_STEP*t);
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if(stepCounts & 1) stepCounts++; //偶数
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if(stepCounts==0) return 0.0;
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let halfCounts = stepCounts/2;
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let sum1=0.0, sum2=0.0;
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let dStep = t/stepCounts;
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for(var i=0; i<halfCounts; i++) {
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sum1 += this.speed((2*i+1)*dStep);
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}
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for(var i=1; i<halfCounts; i++) {
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sum2 += this.speed((2*i)*dStep);
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}
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return (this.speed(0.0) + this.speed(1.0)+2*sum2+4*sum1)*dStep/3.0;
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}
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length_from_table(t) {
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//通过查表获取t对应的曲线长度
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if(t<=0) return 0;
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if(t>=1.0) return this.length_table[this.length_table.length-1];
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let step=t*this.step;
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let low_bounder = Math.floor(step);
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let lerp = step-low_bounder;
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return this.length_table[low_bounder]*(1-lerp)+this.length_table[low_bounder+1]*lerp;
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}
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//根据t推导出匀速运动自变量t'的方程(使用牛顿切线法)
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invertLength(t, target_length) {
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let t1 = t, t2;
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do {
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t2 = t1 - (this.length_from_table(t1)-target_length)/this.speed(t1);
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if(abs(t1-t2)<0.00001) break;
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t1=t2;
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}while(true);
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return t2;
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}
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updatePoints(_p0, _p1, _p2, _p3) {
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this.p0=_p0;
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this.p1=_p1;
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this.p2=_p2;
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this.p3=_p3;
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let ax = this.p0.x - 2 * this.p1.x + this.p2.x;
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let ay = this.p0.y - 2 * this.p1.y + this.p2.y;
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let bx = 2 * this.p1.x - 2 * this.p0.x;
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let by = 2 * this.p1.y - 2 * this.p0.y;
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this.A = 4 * (ax * ax + ay *ay);
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this.B = 4 * (ax * bx + ay *by);
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this.C = bx * bx + by * by;
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this.points.length = 0;
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let totalLength = this.length(1);
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||||||
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//构建一个length table
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||||||
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this.length_table=[];
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for (let index = 0; index <= this.step; index++) {
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let t = index / this.step;
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this.length_table.push(this.length(t));
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}
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for (let index = 1; index < this.step; index++) {
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let t = index / this.step;
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let target_length=totalLength*t;
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if(this.uniformSpeedMode) {
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t = this.invertLength(t, target_length);
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}
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this.points.push(this.position(t));
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}
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}
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draw() {
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this.points.forEach(pt => {
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fill('green');
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ellipse(pt.x, pt.y, 5, 5);
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});
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}
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}
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@@ -1,8 +1,8 @@
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class QuadBezierLine {
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class QuadBezierLine {
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constructor(_p0, _p1, _p2, _step, _uniformSpeed) {
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constructor(_p0, _p1, _p2, _step, _uniformSpeedMode) {
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this.step=_step;
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this.step=_step;
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this.points=[];
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this.points=[];
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this.uniformSpeed=_uniformSpeed;
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this.uniformSpeedMode=_uniformSpeedMode;
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this.updatePoints(_p0, _p1, _p2);
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this.updatePoints(_p0, _p1, _p2);
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}
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}
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||||||
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||||||
@@ -12,29 +12,29 @@ class QuadBezierLine {
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|||||||
}
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}
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//Length(t) = Integrate[Speed[t], t]
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//Length(t) = Integrate[Speed[t], t]
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//Length(t_)=((2*Sqrt[A]*(2*A*t*Sqrt[C+t*(B+A*t)]+B*(Sqrt[C + t*(B + A*t)]-Sqrt[C])) +
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//Length(t_) = ((2*Sqrt[A]*((2*A*t + B)*Sqrt[A*t^2 + B*t + C] - B*Sqrt[C]) +
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// (B^2-4*A*C)(Log[B+2*Sqrt[A]*Sqrt[C]]-Log[B+2*A*t+2*Sqrt[A]*Sqrt[C+t*(B+A*t)]]))/
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// (B^2 - 4*A*C)*(Log[B + 2*Sqrt[A]*Sqrt[C]] -
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// (8* A^(3/2)));
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// Log[B + 2*A*t + 2*Sqrt[A]*Sqrt[A*t^2 + B*t + C]]))/(8*A^(3/2)))
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length(t) {
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length(t) {
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let temp1 = Math.sqrt(this.C + t * (this.B + this.A * t));
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let temp1 = Math.sqrt(this.C + t * (this.B + this.A * t));
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let temp2 = (2 * this.A * t * temp1 + this.B * (temp1 - Math.sqrt(this.C)));
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let temp2 = (2 * this.A * t + this.B) * temp1 - this.B * Math.sqrt(this.C);
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let temp3 = Math.log(this.B + 2 * Math.sqrt(this.A) * Math.sqrt(this.C));
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let temp3 = Math.log(this.B + 2 * Math.sqrt(this.A * this.C));
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let temp4 = Math.log(this.B + 2 * this.A * t + 2 * Math.sqrt(this.A) * temp1);
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let temp4 = Math.log(this.B + 2 * this.A * t + 2 * Math.sqrt(this.A) * temp1);
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let temp5 = 2 * Math.sqrt(this.A) * temp2;
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let temp5 = 2 * Math.sqrt(this.A) * temp2;
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let temp6 = (this.B * this.B - 4 * this.A * this.C) * (temp3 - temp4);
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let temp6 = (this.B * this.B - 4 * this.A * this.C) * (temp3 - temp4);
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return (temp5 + temp6) / (8 * Math.pow(this.A, 1.5));
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return (temp5 + temp6) / (8 * Math.pow(this.A, 1.5));
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||||||
}
|
}
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||||||
|
|
||||||
//X(n+1) = Xn - F(Xn)/F'(Xn)
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//u'=u-(Length(u)-len)/Speed(u)
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invertLength(t, len) {
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invertLength(u0, len) {
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let t1 = t, t2;
|
let u1 = u0, u2;
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do {
|
do {
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t2 = t1 - (this.length(t1) - len) / this.speed(t1);
|
u2 = u1 - (this.length(u1) - len) / this.speed(u1);
|
||||||
if (Math.abs(t1 - t2) < 0.000001)
|
if (Math.abs(u1 - u2) < 0.000001)
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break;
|
break;
|
||||||
t1 = t2;
|
u1 = u2;
|
||||||
} while (true);
|
} while (true);
|
||||||
return t2;
|
return u2;
|
||||||
}
|
}
|
||||||
|
|
||||||
updatePoints(_p0, _p1, _p2) {
|
updatePoints(_p0, _p1, _p2) {
|
||||||
@@ -57,8 +57,7 @@ class QuadBezierLine {
|
|||||||
for (let index = 1; index < this.step; index++) {
|
for (let index = 1; index < this.step; index++) {
|
||||||
let t = index / this.step;
|
let t = index / this.step;
|
||||||
|
|
||||||
if(this.uniformSpeed) {
|
if(this.uniformSpeedMode) {
|
||||||
//根据 L 函数的反函数，求得 l 对应的 t 值
|
|
||||||
t = this.invertLength(t, totalLength*t);
|
t = this.invertLength(t, totalLength*t);
|
||||||
}
|
}
|
||||||
|
|
||||||
@@ -1,69 +0,0 @@
|
|||||||
---
|
|
||||||
title: "匀速贝塞尔曲线运动的实现"
|
|
||||||
tags: 程序 算法
|
|
||||||
---
|
|
||||||
# 匀速贝塞尔曲线运动的实现
|
|
||||||
|
|
||||||
二次贝塞尔曲线通常以如下方式构建，给定二维平面上的固定点$P_0$, $P_1$, $P_2$，用$B(t)$表示该条曲线
|
|
||||||
$$
|
|
||||||
\boldsymbol{B}(t)=(1-t)^2 \boldsymbol{P_0}+2t(1-t) \boldsymbol{P_1}+t^2 \boldsymbol{P_2}
|
|
||||||
$$
|
|
||||||
用一个动画来演示，可以更加清楚的表明这条曲线的构建过程
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
如果$t$变量本身是线性变化的话，这条贝塞尔曲线的生成过程是并不是匀速的，通常都是两头快中间慢。
|
|
||||||
|
|
||||||
<iframe width="100%" height="270" frameborder=0 src="/html/bezier.html?uniformSpeed=0"></iframe>
|
|
||||||
|
|
||||||
可以看出中间的点较为密集，而两边则较为稀疏。
|
|
||||||
如何想要得到匀速的贝塞尔曲线运动呢？比如我们在某款游戏中设计了一条贝塞尔曲线的路径，如何实现玩家匀速在这条路径上运动呢？
|
|
||||||
首先需要求得$B(t)$相对于$t$的速度公式$s(t)$
|
|
||||||
$$
|
|
||||||
s(t)=\sqrt{B_{x}^{'}(t)^2+B_{y}^{'}(t)^2}
|
|
||||||
$$
|
|
||||||
为了简化公式，定义如下变量
|
|
||||||
$$
|
|
||||||
\begin{aligned}
|
|
||||||
\boldsymbol{a}&=\boldsymbol{P_0}-2\boldsymbol{P_1}+\boldsymbol{P_2}\\
|
|
||||||
\boldsymbol{b}&=2\boldsymbol{P_1}-2\boldsymbol{P_0}\\
|
|
||||||
A&=4(a_x^2+a_y^2)\\
|
|
||||||
B&=4(a_xb_x+a_yb_y)\\
|
|
||||||
C&=b_x^2+b_y^2
|
|
||||||
\end{aligned}
|
|
||||||
$$
|
|
||||||
计算出$s(t)$可以表达为
|
|
||||||
$$
|
|
||||||
s(t)=\sqrt{At^2+Bt+C}
|
|
||||||
$$
|
|
||||||
根据这个公式，求得贝塞尔曲线的长度公式$L(t)$为
|
|
||||||
$$
|
|
||||||
\begin{aligned}
|
|
||||||
L(t)&=\int_0^t\sqrt{Ax^2+Bx+C}dx\\
|
|
||||||
&=\frac{1}{8A^{3/2}}\biggl(2\sqrt{A}\left[2At\sqrt{At^2+Bt+C}+B\left(\sqrt{At^2+Bt+C}-\sqrt{C}\right)\right] \\
|
|
||||||
&\quad+(B^2-4AC)\left[ln(B+2\sqrt{AC})-ln\left(B+2At+2\sqrt{A}\sqrt{At^2+Bt+C}\right)\right]\biggr)
|
|
||||||
\end{aligned}
|
|
||||||
$$
|
|
||||||
特别当$t=1.0$时，$L(1.0)$就是这条曲线的总长度。
|
|
||||||
设$t'$是能够使$L$实现匀速运动的自变量，那么此时曲线长度应该满足线性增长，也就是
|
|
||||||
$$
|
|
||||||
L(t')=L(1.0)t\tag{1}
|
|
||||||
$$
|
|
||||||
也就是$t'=L^{-1}(L(1.0)t)$，由于$L(t)$函数非常复杂，直接求其逆函数的解析解几乎不可能，还好我们知道它的导数为$s(t)$，在实际使用中，可以使用[牛顿切线法](https://en.wikipedia.org/wiki/Newton%27s_method)获得$t'$的数值解。
|
|
||||||
设$L_t=L(1.0)t$，视$t'$为未知数，根据公式1有以下方程
|
|
||||||
$$
|
|
||||||
L(t')-L_t=0
|
|
||||||
$$
|
|
||||||
根据牛顿切线法，求解的迭代公式为:
|
|
||||||
$$
|
|
||||||
t'_{n+1}=t'_n-\frac{L(t'_n)-L_t}{s(t'_n)}
|
|
||||||
$$
|
|
||||||
由于$t$和$t'$相差不大，可以设$t_0=t$来开始求解，以下是修正后的匀速贝塞尔曲线
|
|
||||||
|
|
||||||
<iframe width="100%" height="270" frameborder=0 src="/html/bezier.html?uniformSpeed=1"></iframe>
|
|
||||||
|
|
||||||
上面是使用javascript实现的互动曲线，核心代码如下
|
|
||||||
|
|
||||||
@[code js :no-line-numbers](@public/js/bezier_app.js)
|
|
||||||
|
|
||||||
## End
|
|
||||||
@@ -0,0 +1,69 @@
|
|||||||
|
---
|
||||||
|
title: "匀速贝塞尔曲线运动的实现(一)"
|
||||||
|
tags: 程序 算法
|
||||||
|
---
|
||||||
|
# 匀速贝塞尔曲线运动的实现(一)
|
||||||
|
|
||||||
|
贝塞尔曲线([Bézier curve](https://en.wikipedia.org/wiki/B%C3%A9zier_curve))是一种常用的曲线，以最简单的二次贝塞尔曲线为例，通常以如下方式构建，给定二维平面上的固定点$P_0$, $P_1$, $P_2$，用$B(t)$表示该条曲线
|
||||||
|
$$
|
||||||
|
\boldsymbol{B}(t)=(1-t)^2 \boldsymbol{P_0}+2t(1-t) \boldsymbol{P_1}+t^2 \boldsymbol{P_2}, t\in[0, 1]
|
||||||
|
$$
|
||||||
|
用一个动画来演示，可以更加清楚的表明这条曲线的构建过程
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
如果$t$变量本身是线性变化的话，这条贝塞尔曲线的生成过程是并不是匀速的，通常都是两头快中间慢。
|
||||||
|
|
||||||
|
<iframe width="100%" height="270" frameborder=0 src="/html/bezier01.html?uniformSpeed=0"></iframe>
|
||||||
|
|
||||||
|
可以看出中间的点较为密集，而两边则较为稀疏。
|
||||||
|
如何得到匀速的贝塞尔曲线运动呢？比如我们在某款游戏中设计了一条贝塞尔曲线的路径，如何实现NPC匀速在这条路径上运动？
|
||||||
|
首先需要求得$B(t)$相对于$t$的速度公式$s(t)$
|
||||||
|
$$
|
||||||
|
s(t)=\sqrt{B_{x}^{'}(t)^2+B_{y}^{'}(t)^2}
|
||||||
|
$$
|
||||||
|
为了简化公式，定义如下变量
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
\boldsymbol{a}&=\boldsymbol{P_0}-2\boldsymbol{P_1}+\boldsymbol{P_2}\\
|
||||||
|
\boldsymbol{b}&=2\boldsymbol{P_1}-2\boldsymbol{P_0}\\
|
||||||
|
A&=4(a_x^2+a_y^2)\\
|
||||||
|
B&=4(a_xb_x+a_yb_y)\\
|
||||||
|
C&=b_x^2+b_y^2
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
计算出$s(t)$可以表达为
|
||||||
|
$$
|
||||||
|
s(t)=\sqrt{At^2+Bt+C}
|
||||||
|
$$
|
||||||
|
根据这个公式，求得贝塞尔曲线的长度公式$L(t)$为
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
L(t)&=\int_0^t\sqrt{Ax^2+Bx+C}dx\\
|
||||||
|
&=\frac{1}{8A^{3/2}}\biggl(2\sqrt{A}\left[(2At+B)\sqrt{At^2+Bt+C}-B\sqrt{C}\right] \\
|
||||||
|
&\quad+(B^2-4AC)\left[ln(B+2\sqrt{AC})-ln\left(B+2At+2\sqrt{A}\sqrt{At^2+Bt+C}\right)\right]\biggr)
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
特别当$t=1.0$时，$L(1.0)$就是这条曲线的总长度。
|
||||||
|
设$u$是能够使$L(u)$实现匀速运动的自变量，那么此时曲线长度应该满足随着时间$t$线性增长，也就是
|
||||||
|
$$
|
||||||
|
L(u)=L(1.0)t\tag{1}
|
||||||
|
$$
|
||||||
|
也就是$u=L^{-1}(L(1.0)t)$，由于$L(t)$函数非常复杂，直接求其逆函数的解析解几乎不可能，还好我们知道它的导数为$s(t)$，在实际使用中，可以使用[牛顿切线法](https://en.wikipedia.org/wiki/Newton%27s_method)获得$u$的数值解。
|
||||||
|
视$u$为未知数，根据公式1有以下方程
|
||||||
|
$$
|
||||||
|
F(u)=L(u)-L(1.0)t=0
|
||||||
|
$$
|
||||||
|
根据牛顿切线法，求解$u$的迭代公式为:
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
u_{n+1}&=u_n-\frac{F(u_n)}{F'(u_n)}\\&=u_n-\frac{L(u_n)-L(1.0)t}{s(u_n)}
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
由于$t$和$u$相差不大，可以设$u_0=t$来开始求解，以下是修正后的匀速贝塞尔曲线
|
||||||
|
|
||||||
|
<iframe width="100%" height="270" frameborder=0 src="/html/bezier01.html?uniformSpeed=1"></iframe>
|
||||||
|
|
||||||
|
上面是使用javascript实现的互动曲线，核心代码如下
|
||||||
|
|
||||||
|
@[code js :no-line-numbers](@public/js/bezier_quad.js)
|
||||||
@@ -0,0 +1,23 @@
|
|||||||
|
---
|
||||||
|
title: "匀速贝塞尔曲线运动的实现(二)"
|
||||||
|
tags: 程序 算法
|
||||||
|
---
|
||||||
|
# 匀速贝塞尔曲线运动的实现(二)
|
||||||
|
|
||||||
|
实际工程应用中最为常见的是三次贝塞尔曲线，也就是下面这种用4个控制点生成的曲线
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
三次贝塞尔曲线的一个方便之处在于可以将相邻的两个控制点之间的连线视作控制点的“切线”，进而便于使用者编辑，因此在很多软件中的贝塞尔曲线编辑器都是使用的三次贝塞尔曲线。比如下面这段曲线，其实就是由几段三次贝塞尔曲线组成的。
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
三次贝塞尔曲线的公式为：
|
||||||
|
$$
|
||||||
|
\boldsymbol{B}(t)=(1-t)^3 \boldsymbol{P_0}+3t(1-t)^2 \boldsymbol{P_1}+3(1-t)t^2 \boldsymbol{P_2}+t^3\boldsymbol{P_3}, t\in[0, 1]
|
||||||
|
$$
|
||||||
|
虽然仍然可以按照相同的思路去实现匀速运动，但是由于三次贝塞尔曲线的长度计算已经非常复杂，根本无法通过对速度进行积分得到解析解，更别说通过反函数去求解匀速需要的自变量了。 因此在实际计算中，一般也只能通过提前建立一个曲线的长度查询表辅助运算。 这个长度查询表可以利用一般的数值积分的方式建立，比如[辛普森积分法](https://en.wikipedia.org/wiki/Simpson%27s_rule)
|
||||||
|
下面是一个JavaScript实现的互动范例，代码可以直接查看这个<a href="/html/bezier02.html" target="_blank">页面</a>的<a href="/js/bezier_cubic.js" target="_blank">JS源码</a>
|
||||||
|
|
||||||
|
<iframe width="100%" height="270" frameborder=0 src="/html/bezier02.html?uniformSpeed=1"></iframe>
|
||||||
|
|
||||||
+2
-1
@@ -14,4 +14,5 @@
|
|||||||
* [如何计算线段和圆的交点](/blog/2025/02/SegmentCircle.md)
|
* [如何计算线段和圆的交点](/blog/2025/02/SegmentCircle.md)
|
||||||
* [一道数学趣题](/blog/2025/02/Ellipse.md)
|
* [一道数学趣题](/blog/2025/02/Ellipse.md)
|
||||||
* [斐波那契数列和1/89](/blog/2025/02/Fibonacci.md)
|
* [斐波那契数列和1/89](/blog/2025/02/Fibonacci.md)
|
||||||
* [匀速贝塞尔曲线运动实现](/blog/2025/03/BezierLine.md)
|
* 匀速贝塞尔曲线运动实现
|
||||||
|
* [Part1](/blog/2025/03/BezierLine01.md), [Part2](/blog/2025/03/BezierLine02.md)
|
||||||
|
|||||||
@@ -13,6 +13,8 @@
|
|||||||
* [如何计算线段和圆的交点](/blog/2025/02/SegmentCircle.md)
|
* [如何计算线段和圆的交点](/blog/2025/02/SegmentCircle.md)
|
||||||
* [一道数学趣题](/blog/2025/02/Ellipse.md)
|
* [一道数学趣题](/blog/2025/02/Ellipse.md)
|
||||||
* [斐波那契数列和1/89](/blog/2025/02/Fibonacci.md)
|
* [斐波那契数列和1/89](/blog/2025/02/Fibonacci.md)
|
||||||
|
* 匀速贝塞尔曲线运动实现
|
||||||
|
* [Part1](/blog/2025/03/BezierLine01.md), [Part2](/blog/2025/03/BezierLine02.md)
|
||||||
|
|
||||||
## 开源项目
|
## 开源项目
|
||||||
* [Turbolink](https://github.com/thejinchao/turbolink)
|
* [Turbolink](https://github.com/thejinchao/turbolink)
|
||||||
|
|||||||
@@ -178,3 +178,188 @@ c+(1-c)x^2& (1-c)xy-sz&(1-c)xz+sy\\
|
|||||||
(1-c)xz-sy&(1-c)yz+sx&c+(1-c)z^2
|
(1-c)xz-sy&(1-c)yz+sx&c+(1-c)z^2
|
||||||
\end{bmatrix}
|
\end{bmatrix}
|
||||||
$$
|
$$
|
||||||
|
|
||||||
|
### 3.4 四元数
|
||||||
|
|
||||||
|
#### 3.4.1 复数
|
||||||
|
复数([Complex Number](https://simple.wikipedia.org/wiki/Complex_number))可以表达为
|
||||||
|
$$\begin{aligned}
|
||||||
|
z=a+bi\\
|
||||||
|
a,b\in\mathbb{R}, i^2=-1
|
||||||
|
\end{aligned}$$
|
||||||
|
其中$a$被成为实部(Real Part), $b$被称为虚部(Imaginary Part), 复数的基本运算规律如下，设$z_1=a_1+b_1i, z_2=a_2+b_2i$，那么
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
z_1+z_2&=(a_1+a_2)+(b_1+b_2)i\\
|
||||||
|
z_1-z_2&=(a_1-a_2)+(b_1-b_2)i\\
|
||||||
|
z_1z_2&=(a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
|
||||||
|
#### 3.4.2 四元数定义
|
||||||
|
四元数([Quaternion](https://en.wikipedia.org/wiki/Quaternion))可以视为对复数的扩充。 四元数有1个实部，3个虚部，形式如下
|
||||||
|
$$
|
||||||
|
q=s+xi+yj+zk
|
||||||
|
$$
|
||||||
|
其中$s,x,y,z\in\mathbb{R}$，$i,j,k$满足如下计算性质
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
i^2&=j^2=k^2=ijk=-1\\
|
||||||
|
ij&=k\quad ji=-k\\
|
||||||
|
jk&=i\quad kj=-i\\
|
||||||
|
ki&=j\quad ik=-j\\
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
汇总到如下表格中
|
||||||
|
| | $i$ | $j$ | $k$ |
|
||||||
|
| :--- | :--- | :--- | :--- |
|
||||||
|
| $i$ | $-1$ | $k$ | $-j$ |
|
||||||
|
| $j$ | $-k$ | $-1$ | $i$ |
|
||||||
|
| $k$ | $j$ | $-i$ | $-1$ |
|
||||||
|
|
||||||
|
四元数也可以表达成将实数部和虚数部分开的形式，对于四元数$q=s+xi+yj+zk$，设$\vec{v}=[x,y,z]^T$，则这个四元数可以表达为
|
||||||
|
$$
|
||||||
|
q=[s,\vec{v}]
|
||||||
|
$$
|
||||||
|
|
||||||
|
#### 3.4.3 四元数的基本运算
|
||||||
|
对于两个四元数$q_1=[s_1,\vec{v_1}], q_2=[s_2,\vec{v_2}]$，基本运算规律如下
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
q_1+q_2&=[s_1+s_2,\quad \vec{v_1}+\vec{v_2}]\\
|
||||||
|
q_1-q_2&=[s_1-s_2,\quad \vec{v_1}-\vec{v_2}]\\
|
||||||
|
q_1q_2&=[s_1s_2-\vec{v_1}\cdot\vec{v_2},\quad s_1\vec{v_2}+s_2\vec{v_1}+\vec{v_1}\times\vec{v_2}]
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
四元数的乘法满足满足结合律，但不满足交换律
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
(q_1q_2)q_3&=q_1(q_2q_3)\\
|
||||||
|
q_1q_2&\neq q_2q_1
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
但如果$q_1,q_2$中的两个向量部分平行，由于$\vec{v_1}\times\vec{v_2}=0$，易证
|
||||||
|
$$
|
||||||
|
q_1q_2=q_2q_1\quad(when\quad \vec{v_1}\parallel\vec{v_2})
|
||||||
|
\tag{3.4.3.1}
|
||||||
|
$$
|
||||||
|
定义四元数$q=s+xi+yj+zk$的模为
|
||||||
|
$$
|
||||||
|
\|q\|=\sqrt{s^2+x^2+y^2+z^2}
|
||||||
|
$$
|
||||||
|
如果一个四元数的模为1，那么称这个四元数为单位四元数
|
||||||
|
|
||||||
|
#### 3.4.4 四元数的共轭和逆
|
||||||
|
定义四元数$q=[s, \vec{v}]$的共轭四元数$\overline{q}$为
|
||||||
|
$$
|
||||||
|
\overline{q}=[s, -\vec{v}]
|
||||||
|
$$
|
||||||
|
共轭四元数满足如下运算
|
||||||
|
$$
|
||||||
|
q\overline{q}=\overline{q}q=\|q\|^2\tag{3.4.4.1}
|
||||||
|
$$
|
||||||
|
对于单位四元数，有$q\overline{q}=1$
|
||||||
|
定义四元数$q$的逆为$q^{-1}$，满足$qq^{-1}=1$，由公式3.4.4.1可知
|
||||||
|
$$
|
||||||
|
qq^{-1}=1=\frac{q\overline{q}}{\|q\|}
|
||||||
|
$$
|
||||||
|
所以
|
||||||
|
$$
|
||||||
|
q^{-1}=\frac{\overline{q}}{\|q\|^2}\tag{3.4.4.2}
|
||||||
|
$$
|
||||||
|
四元数的逆满足如下运算
|
||||||
|
$$
|
||||||
|
(q_1q_2)^{-1}=q_2^{-1}q_1^{-1}
|
||||||
|
$$
|
||||||
|
#### 3.4.5 四元数与向量
|
||||||
|
如果一个四元数的实部为0，那么称之为纯四元数(Pure Quaternion)，由于纯四元数仅有3个虚部，可以将一个3D向量转换为一个纯四元数。设$q_v=[0,\vec{v}]$，那么
|
||||||
|
$$\begin{aligned}
|
||||||
|
\lambda q_v&=[0, \lambda\vec{v}]\\
|
||||||
|
q_{u}\pm q_{v}&=[0, \vec{u}\pm\vec{v}]
|
||||||
|
\end{aligned}$$
|
||||||
|
由此可见，向量的线性运算，都可以使用与之对应的纯四元数来代替。但乘法则不同，设两个纯四元数$q_u=[0, \vec{u}], q_v=[0, \vec{v}]$，相乘后结果为
|
||||||
|
$$
|
||||||
|
q_uq_v=[-\vec{u}\cdot\vec{v}, \vec{u}\times\vec{v}]\tag{3.4.5.1}
|
||||||
|
$$
|
||||||
|
对于向量$\vec{v}$，记四元数$q(\theta, \vec{v})=[\cos\theta, \vec{v}\sin\theta]$，这种四元数有一些很有用的特性，首先如果$\vec{v}$是单位向量，那么$q(\theta, \vec{v})$是单位四元数，易证:
|
||||||
|
$$
|
||||||
|
\|q(\theta, \vec{v})\|=\sqrt{\cos^2\theta+\sin^2\theta(v_x^2+v_y^2+v_z^2)}=1
|
||||||
|
$$
|
||||||
|
另外
|
||||||
|
$$\begin{aligned}
|
||||||
|
q(\alpha, \vec{v})q(\theta, \vec{v})&=[\cos\alpha,\quad\vec{v}\sin\alpha][\cos\theta,\quad\vec{v}\sin\theta]\\
|
||||||
|
&=[\cos\alpha\cos\theta-\sin\alpha\sin\theta,\quad\vec{v}(\cos\alpha\sin\theta+\cos\theta\sin\alpha)]\\
|
||||||
|
&=[\cos(\alpha+\theta),\quad\vec{v}\sin(\alpha+\theta)]\\
|
||||||
|
&=q(\alpha+\theta, \vec{v})\\
|
||||||
|
q(\theta, \vec{v})^2&=q(2\theta, \vec{v})
|
||||||
|
\end{aligned}\tag{3.4.5.2}$$
|
||||||
|
|
||||||
|
#### 3.4.6 使用四元数表达旋转
|
||||||
|
观察上面推导罗德里格旋转公式过程中的公式3.3.2.5，其中
|
||||||
|
$$\begin{aligned}
|
||||||
|
\vec{\boldsymbol{n}}\times\vec{\boldsymbol{v}}&=\vec{\boldsymbol{n}}\times(\boldsymbol{v_1}+\boldsymbol{v_2})\\
|
||||||
|
&=\vec{\boldsymbol{n}}\times\vec{\boldsymbol{v_1}}+\vec{\boldsymbol{n}}\times\vec{\boldsymbol{v_2}}\\
|
||||||
|
&=\vec{\boldsymbol{n}}\times\vec{\boldsymbol{v_2}}
|
||||||
|
\end{aligned}$$
|
||||||
|
带入公式3.3.2.5,可以得到
|
||||||
|
$$
|
||||||
|
R_n(\vec{\boldsymbol{v}}_2)=\vec{\boldsymbol{v}}_2\cos(\theta)+(\vec{\boldsymbol{n}}\times\vec{\boldsymbol{v_2}})\sin(\theta)\tag{3.4.6.1}
|
||||||
|
$$
|
||||||
|
设两个纯四元数$q_n, q_v$
|
||||||
|
$$\begin{aligned}
|
||||||
|
q_n&=[0, \vec{n}]\\
|
||||||
|
q_{v2}&=[0, \vec{v_2}]
|
||||||
|
\end{aligned}\tag{3.4.6.2}
|
||||||
|
$$
|
||||||
|
根据公式3.4.5.1
|
||||||
|
$$\begin{aligned}
|
||||||
|
q_nq_{v2}&=[-\vec{n}\cdot\vec{v_2},\quad\vec{n}\times\vec{v_2}]\\
|
||||||
|
&=[0,\quad\vec{n}\times\vec{v_2}]
|
||||||
|
\end{aligned}\tag{3.4.6.3}$$
|
||||||
|
所以$q_nq_{v2}$是一个纯四元数
|
||||||
|
由于3.4.6.1中都是线性计算，将3.4.6.2和3.4.6.3代入其中，可以得到$R_n(\vec{\boldsymbol{v}}_2)$的四元数形式
|
||||||
|
$$\begin{aligned}
|
||||||
|
R_n(\vec{\boldsymbol{v}}_2)&=\vec{\boldsymbol{v}}_2\cos\theta+(\vec{\boldsymbol{n}}\times\vec{\boldsymbol{v_2}})\sin\theta\\
|
||||||
|
&= q_v\cos\theta+q_nq_{v2}\sin\theta\\
|
||||||
|
&=(\cos\theta+q_n\sin\theta)q_{v2}\\
|
||||||
|
&=[\cos\theta, \vec{n}\sin\theta]q_{v2}\\
|
||||||
|
&=q(\theta, \vec{n})q_{v2}
|
||||||
|
\end{aligned}\tag{3.4.6.4}$$
|
||||||
|
设一个新的四元数$p$
|
||||||
|
$$
|
||||||
|
p=q(\frac{\theta}{2}, \vec{n})=[\cos\frac{\theta}{2},\vec{n}\sin\frac{\theta}{2}]\tag{3.4.6.5}
|
||||||
|
$$
|
||||||
|
根据公式3.4.5.2，可以得到
|
||||||
|
$$
|
||||||
|
pp=q(\theta, \vec{n})
|
||||||
|
$$
|
||||||
|
且由于$p$是单位四元数，可以知道
|
||||||
|
$$
|
||||||
|
pp^{-1}=p\overline{p}=1
|
||||||
|
$$
|
||||||
|
将3.4.6.4带入3.3.2.4中，可以得到罗德里格旋转公式的四元数形式为
|
||||||
|
$$\begin{aligned}
|
||||||
|
R_n(\vec{\boldsymbol{v}})&=\vec{\boldsymbol{v}}_1+R_n(\vec{\boldsymbol{v}}_2)\\
|
||||||
|
&=q_{v1}+q(\theta, \vec{n})q_{v2}\\
|
||||||
|
&=p\overline{p}q_{v1}+ppq_{v2}
|
||||||
|
\end{aligned}\tag{3.4.6.6}
|
||||||
|
$$
|
||||||
|
由于$\vec{n}\parallel\vec{v_1}$，根据公式3.4.3.1，可知$\overline{p}q_{v1}=q_{v1}\overline{p}$
|
||||||
|
由于$\vec{n}\perp\vec{v_2}$，可以推断出$pq_{v2}=q_{v2}\overline{p}$，证明如下：
|
||||||
|
$$\begin{aligned}
|
||||||
|
pq_{v2}&=[\cos\frac{\theta}{2},\vec{n}\sin\frac{\theta}{2}][0, \vec{v_2}]\\
|
||||||
|
&=[0,\vec{v_2}\cos\frac{\theta}{2}+(\vec{n}\times\vec{v_2})\sin\frac{\theta}{2}]\\
|
||||||
|
q_{v2}\overline{p}&=[0, \vec{v_2}][\cos\frac{\theta}{2},-\vec{n}\sin\frac{\theta}{2}]\\
|
||||||
|
&=[0,\vec{v_2}\cos\frac{\theta}{2}-(\vec{v_2}\times\vec{n})\sin\frac{\theta}{2}]
|
||||||
|
\end{aligned}$$
|
||||||
|
带入3.4.6.6，可以得到
|
||||||
|
$$\begin{aligned}
|
||||||
|
R_n(\vec{\boldsymbol{v}})&=pq_{v1}\overline{p}+pq_{v2}\overline{p}\\
|
||||||
|
&=p(q_{v1}+q_{v2})\overline{p}\\
|
||||||
|
&=pq_v\overline{p}
|
||||||
|
\end{aligned}\tag{3.4.6.7}
|
||||||
|
$$
|
||||||
|
也就是说，对于向量$\vec{v}$，围绕单位向量$\vec{n}$旋转$\theta$，只需要构造四元数$[\cos\frac{\theta}{2}, \vec{n}\sin\frac{\theta}{2}]$，可以利用下面的等式计算旋转后的向量$\vec{v'}$
|
||||||
|
$$
|
||||||
|
[0, \vec{v'}]=[\cos\frac{\theta}{2}, \vec{n}\sin\frac{\theta}{2}][0,\vec{v}][\cos\frac{\theta}{2}, -\vec{n}\sin\frac{\theta}{2}]
|
||||||
|
$$
|
||||||
@@ -24,16 +24,86 @@
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Before Width: | Height: | Size: 12 KiB After Width: | Height: | Size: 23 KiB |
@@ -11,6 +11,7 @@
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"@vuepress/plugin-markdown-math": "^2.0.0-rc.78",
|
"@vuepress/plugin-markdown-math": "^2.0.0-rc.78",
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"@vuepress/theme-default": "^2.0.0-rc.78",
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"@vuepress/theme-default": "^2.0.0-rc.78",
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"mathjax-full": "^3.2.2",
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"mathjax-full": "^3.2.2",
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"p5": "^1.11.3",
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"sass-embedded": "^1.85.0",
|
"sass-embedded": "^1.85.0",
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"vuepress": "^2.0.0-rc.19"
|
"vuepress": "^2.0.0-rc.19"
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},
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},
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@@ -1852,6 +1852,11 @@ ora@^8.1.1:
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string-width "^7.2.0"
|
string-width "^7.2.0"
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strip-ansi "^7.1.0"
|
strip-ansi "^7.1.0"
|
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|
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|
p5@^1.11.3:
|
||||||
|
version "1.11.3"
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|
resolved "https://registry.yarnpkg.com/p5/-/p5-1.11.3.tgz#94478d5ba685cedf9f3f934a3db702155cbb36c6"
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integrity sha512-5eiyuZ3Pvo6ucr3nr8AzD+qwpoleiIEF2a2HmiO6PJM1tnbD0LmLlckZyYGJATs0tdmRjc37Muh6KPS4d1FHHg==
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parse-ms@^4.0.0:
|
parse-ms@^4.0.0:
|
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version "4.0.0"
|
version "4.0.0"
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resolved "https://registry.npmjs.org/parse-ms/-/parse-ms-4.0.0.tgz"
|
resolved "https://registry.npmjs.org/parse-ms/-/parse-ms-4.0.0.tgz"
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|
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Reference in New Issue
Block a user