Porting 3D Rose written by Wolfram Language into JavaScript

Question

I'd like to get help from Geometry / Wolfram Mathematica people. I want to visualize this 3D Rose in JavaScript (p5.js) environment.

This figure is originally generated using wolfram language by Paul Nylanderin 2004-2006, and below is the code:

Rose[x_, theta_] := Module[{
  phi = (Pi/2)Exp[-theta/(8 Pi)], 
  X = 1 - (1/2)((5/4)(1 - Mod[3.6 theta, 2 Pi]/Pi)^2 - 1/4)^2}, 
  y = 1.95653 x^2 (1.27689 x - 1)^2 Sin[phi]; 
  r = X(x Sin[phi] + y Cos[phi]); 
  {r Sin[theta], r Cos[theta], X(x Cos[phi] - y Sin[phi]), EdgeForm[]
}];

ParametricPlot3D[
  Rose[x, theta], {x, 0, 1}, {theta, -2 Pi, 15 Pi}, 
  PlotPoints -> {25, 576}, LightSources -> {{{0, 0, 1}, RGBColor[1, 0, 0]}}, 
  Compiled -> False
]

I tried implement that code in JavaScript like this below.

function rose(){
  for(let theta = 0; theta < 2700; theta += 3){
    beginShape(POINTS);
    for(let x = 2.3; x < 3.3; x += 0.02){
      let phi = (180/2) * Math.exp(- theta / (8*180));
      let X = 1 - (1/2) * pow(((5/4) * pow((1 - (3.6 * theta % 360)/180), 2) - 1/4), 2);
      let y = 1.95653 * pow(x, 2) * pow((1.27689*x - 1), 2) * sin(phi);
      let r = X * (x*sin(phi) + y*cos(phi));

      let pX = r * sin(theta);
      let pY = r * cos(theta);
      let pZ = (-X * (x * cos(phi) - y * sin(phi)))-200;
  
      vertex(pX, pY, pZ);
    }
    endShape();
  }
}

But I got this result below

Unlike original one, the petal at the top is too stretched.

I suspected that the

let y = 1.95653　*　pow(x, 2) * pow((1.27689*x - 1), 2) * sin(phi);

may should be like below...

let y = pow(1.95653*x, 2*pow(1.27689*x - 1, 2*sin(theta)));

But that went even further away from the original.

Maybe I'm asking a dumb question, but I've been stuck for several days.

If you see a mistake, please let me know. Thank you in advanse

---

Update:

I changed the x range to 0~1 as defined by the original one. Also simplified the JS code like below to find the error.

function rose_debug(){
  for(let theta = 0; theta < 15*PI; theta += PI/60){
    beginShape(POINTS);
    for(let x = 0.0; x < 1.0; x += 0.005){
      let phi = (PI/2) * Math.exp(- theta / (8*PI));
      let y = pow(x, 4) * sin(phi);
      let r = (x * sin(phi) + y * cos(phi));

      let pX = r * sin(theta);
      let pY = r * cos(theta);
      let pZ = x * cos(phi) - y * sin(phi);
      vertex(pX, pY, pZ);
    }
    endShape();
  }
}

But the result still keeps the wrong proportion↓↓↓

Also, when I remove the term "sin(phi)" in the line "let y =..." like below

let y = pow(x, 4);

then I got a figure somewhat resemble the original like below

At this moment I was starting to suspect the mistake on the original equation, but I found another article by Jorge García Tíscar(Spanish) that implemented the exact same 3D rose in wolfram language successfully.

So, now I really don't know how the original is formed by the equation

---

Update2: Solved

I followed a suggestion by Trentium (Answer No.2 below) that stick to 0 ~ 1 as the range of x, then multiply the r and X by an arbitrary number.

for(let x = 0; x < 1; x += 0.05){

r = r * 200;
X = X * 200;

Then I got this correct result looks exactly the same as the original

Simplified final code:

function rose_debug3(){
  for(let x = 0; x <= 1; x += 0.05){
    beginShape(POINTS);
    for(let theta = -2*PI; theta <= 15*PI; theta += 17*PI/2000){
      let phi = (PI / 2) * Math.exp(- theta / (8 * PI));
      let X = 1 - (1/2) * ((5/4) * (1 - ((3.6 * theta) % (2*PI))/PI) ** 2 - 1/4) ** 2;
      let y = 1.95653 * (x ** 2) * ((1.27689*x - 1) ** 2) * sin(phi);
      let r = X * (x * sin(phi) + y * cos(phi));

      if(0 < r){
        const factor = 200;
        let pX = r * sin(theta)*factor;
        let pY = r * cos(theta)*factor;
        let pZ = X * (x * cos(phi) - y * sin(phi))*factor;
        vertex(pX, pY, pZ);
      }
    }
    endShape();
  }
}

---

The reason I got the vertically stretched figure at first was the range of the x. I thought that changing the range of the x just affect the whole size of the figure. But actually, the range affects like this below.

(1): 0 ~ x ~ 1, (2): 0 ~ x ~ 1.2

(3): 0 ~ x ~ 1.5, (4): 0 ~ x ~ 2.0

(5): flipped the (4)

So far I saw the result like (5) above, didn't realize that the correct shape was hiding inside that figure.

Thank you Trentium so much for kindly helping me a lot!

Answer 1

Since this response is a significant departure from my earlier response, am adding a new answer...

In rendering the rose algorithm in ThreeJS (sorry, I'm not a P5 guy) it became apparent that when generating the points, that only the points with a positive radius are to be rendered. Otherwise, superfluous points are rendered far outside the rose petals.

(Note: When running the code snippet, use the mouse to zoom and rotate the rendering of the rose.)

<script type="module">

  import * as THREE from 'https://cdn.jsdelivr.net/npm/three@0.115.0/build/three.module.js';
  import { OrbitControls } from 'https://cdn.jsdelivr.net/npm/three@0.115.0/examples/jsm/controls/OrbitControls.js';

  //
  // Set up the ThreeJS environment.
  //
  var renderer = new THREE.WebGLRenderer();
  renderer.setSize( window.innerWidth, window.innerHeight );
  document.body.appendChild( renderer.domElement );

  var camera = new THREE.PerspectiveCamera( 45, window.innerWidth / window.innerHeight, 1, 500 );
  camera.position.set( 0, 0, 100 );
  camera.lookAt( 0, 0, 0 );

  var scene = new THREE.Scene();
  
  let controls = new OrbitControls(camera, renderer.domElement);

  //
  // Create the points.
  //
  function rose( xLo, xHi, xCount, thetaLo, thetaHi, thetaCount ){
    let vertex = [];
    let colors = [];
    let radius = [];
    for( let x = xLo; x <= xHi; x += ( xHi - xLo ) / xCount ) {
      for( let theta = thetaLo; theta <= thetaHi; theta += ( thetaHi - thetaLo ) / thetaCount ) {
        let phi = ( Math.PI / 2 ) * Math.exp( -theta / ( 8 * Math.PI ) );
        let X = 1 - ( 1 / 2 ) * ( ( 5 / 4 ) * ( 1 - ( ( 3.6 * theta ) % ( 2 * Math.PI ) ) / Math.PI ) ** 2 - 1 / 4 ) ** 2;
        let y = 1.95653 * ( x ** 2 ) * ( (1.27689 * x - 1) ** 2 ) * Math.sin( phi );
        let r = X * ( x * Math.sin( phi ) + y * Math.cos( phi ) ); 

        //
        // Fix: Ensure radius is positive, and scale up accordingly...
        //
        if ( 0 < r ) {
        
          const factor = 20;
          
          r = r * factor;
          radius.push( r );
          X = X * factor;

          vertex.push( r * Math.sin( theta ), r * Math.cos( theta ), X * ( x * Math.cos( phi ) - y * Math.sin( phi ) ) );
        }
      }
    }
    
    //
    // For the fun of it, lets adjust the color of the points based on the radius
    // of the point such that the larger the radius, the deeper the red.
    //
    let rLo = Math.min( ...radius );
    let rHi = Math.max( ...radius );
    for ( let i = 0; i < radius.length; i++ ) {
      let clr = new THREE.Color( Math.floor( 0x22 + ( 0xff - 0x22 ) * ( ( radius[ i ] - rLo ) / ( rHi - rLo ) ) ) * 0x10000 + 0x002222 );
      colors.push( clr.r, clr.g, clr.b );
    }
    
    return [ vertex, colors, radius ];
  }

  //
  // Create the geometry and mesh, and add to the THREE scene.
  //
  const geometry = new THREE.BufferGeometry();
  
  
  let [ positions, colors, radius ] = rose( 0, 1, 20, -2 * Math.PI, 15 * Math.PI, 2000 );
  
  geometry.setAttribute( 'position', new THREE.Float32BufferAttribute( positions, 3 ) );
  geometry.setAttribute( 'color', new THREE.Float32BufferAttribute( colors, 3 ) );

  const material = new THREE.PointsMaterial( { size: 4, vertexColors: true, depthTest: false, sizeAttenuation: false } );

  const mesh = new THREE.Points( geometry, material );
  scene.add( mesh );
        
  //
  // Render...
  // 
  var animate = function () {
    requestAnimationFrame( animate );
    renderer.render( scene, camera );
  };

  animate();
</script>

Couple of notables:

• When calling rose( xLo, xHi, xCount, thetaLo, thetaHi, thetaCount ), the upper range thetaHi can vary from Math.PI to 15 * Math.PI, which varies the number of petals.
• Both xCount and thetaCount vary the density of the points. The Wolfram example uses 25 and 576, respectively, but this is to create a geometry mesh, whereas if creating a point field the density of points needs to be increased. Hence, in the code the values are 20 and 2000.

Enjoy!

Answer 2

Presumably the algorithm above is referencing cos() and sin() functions that handle the angles in degrees rather than radians, but wherever using angles while employing non-trigonometric transformations, the result will be incorrect.

For example, the following formula using radians...

• phi = (Pi/2)Exp[-theta/(8 Pi)]

...has been incorrectly translated to...

• phi = ( 180 / 2 ) * Math.exp( -theta / ( 8 * 180 ) )

To test, let's assume theta = 2. Using the original formula in radians...

• phi = ( Math.PI / 2 ) * Math.exp( -2 / ( 8 * Math.PI ) )
• = 1.451 rad
• = 83.12 deg

...and now the incorrect version using degrees, which returns a different angle...

• phi = ( 180 / 2 ) * Math.exp( -2 / ( 8 * 180 ) )
• = 89.88 deg
• = 1.569 rad

A similar issue will occur with the incorrectly translated expression...

• pow( ( 1 - ( 3.6 * theta % 360 ) / 180 ), 2 )

Bottom line: Stick to radians.

P.S. Note that there might be other issues, but using radians rather than degrees needs to be corrected foremost...