1

我是Processing的新手,我一直在模拟电子运动。一切似乎都很好,直到我尝试为每个粒子添加渐变颜色。帧速率大幅下降。

这是我迄今为止尝试过的:

float a=0;
float s;
void setup()
{
  size(500,500);
  smooth();
  frameRate(500);
  colorMode(HSB,360,100,100);
  noStroke();
  ellipseMode(RADIUS);
}

void draw()
{
  background(200,0,100);
  pushMatrix();
  translate(width/2, height/2);
  rotate(radians(-18));
  for ( int r = width ; r >= 0; r = r - 1 )
  {
    s = 500*exp(-r);
    fill(202, s, 100);
    ellipse(100*cos(a), 50*sin(a), r, r);

  }
  a+=0.1;
  popMatrix();
}
4

1 回答 1

11

您的问题不是关于模拟电子运动,而是更多关于在处理中有效地绘制梯度。我看到您已经从示例 > 基础 > 颜色 > RadialGradient示例开始。请注意,示例本身运行速度很慢,可能是因为重点放在如何使用颜色 (HSB) 和绘图功能上,而不是性能。

您可以做的是使用PGraphicsPImage缓存渐变,这让您更舒服。

这是一个使用 PGraphics 的示例,如果您不习惯使用像素,这可能会更简单:

PImage e;
void setup(){
  size(500,500);
  e = getElectronImg(30,30,0,100,100);//create a cached drawing
}
void draw(){
  background(255);
  translate(width * .5, height * .5);
  float a = frameCount * .1;
  image(e,100*cos(a), 50*sin(a));
}

PImage getElectronImg(int w,int h,int hue,int satMax,int brightness){
  PGraphics electron = createGraphics(w+1,h+1);//create a PGraphics object
  electron.beginDraw();//init drawing using the same Processing drawing functions
    electron.colorMode(HSB,360,100,100);
    electron.background(0,0);//transparent bg
    electron.noStroke();
    int cx = electron.width/2;
    int cy = electron.height/2;
    for (int r = w; r > 0; --r) {
      electron.fill(hue,map(r,0,w,satMax,0),brightness);
      electron.ellipse(cx, cy, r, r);
    }
  electron.endDraw();
  return electron;
}

还值得注意的是,PGraphics 扩展了 PImage,因此可以像使用image()函数和其他 PImage 一样显示。

这是使用像素完成的相同缓存概念:

PImage e;
void setup(){
  size(500,500);
  e = getElectronImg(30,30,0,100,100);
}
void draw(){
  background(255);
  translate(width * .5, height * .5);
  float a = frameCount * .1;
  image(e,100*cos(a), 50*sin(a));
}

PImage getElectronImg(int w,int h,int hue,int satMax,int brightness){
  pushStyle();//isolate drawing styles such as color Mode
    colorMode(HSB,360,100,100);
    PImage electron = createImage(w,h,ARGB);//create an image with an alpha channel
    int np = w * h;//total number of pixels
    int cx = electron.width/2;//center on x
    int cy = electron.height/2;//center on y
    for(int i = 0 ; i < np; i++){//for each pixel
      int x = i%electron.width;//compute x from pixel index
      int y = (int)(i/electron.width);//compute y from pixel index
      float d = dist(x,y,cx,cy);//compute distance from centre to current pixel
      electron.pixels[i] = color(hue,map(d,0,cx,satMax,0),brightness,map(d,0,cx,255,0));//map the saturation and transparency based on the distance to centre
    } 
    electron.updatePixels();//finally update all the pixels
  popStyle();
  return electron;
}

当然,这将使使用更多电子变得容易。偏离真实电子运动的主题,这里有一些有趣的测试,通过对 draw() 进行微调:

void draw(){
  background(255);
  translate(width * .5, height * .5);
  for(int i = 0 ; i < 200 ; i++){
    float a = (frameCount * .025 + (i*.1));
    image(e,(100+i)*cos(a + i), (50+i)*sin(a + i));
  }
}

200

void draw(){
  background(255);
  translate(width * .5, height * .5);
  for(int i = 0 ; i < 1000 ; i++){
    float a = (frameCount * .025 + (i*.1));
    image(e,(100+(i * .25))*cos(a + i), (50+(i * .25))*sin(a + i));
  }
}

1000

void draw(){
  background(255);
  translate(width * .5, height * .5);
  scale(.25);
  for(int i = 0 ; i < 5000 ; i++){
    float a = (frameCount * .025 + (i*.1));
    image(e,sin(a) * (100+(i * .5))*cos(a + i),  (50+(i * .25))*sin(a + i));
  }
}

5K

玩得开心!

现在您可以在此处实际运行代码(使用键 1、2、3、4 更改演示):

var e,demo = 2;
function setup(){
  createCanvas(500,500);
  e = getGradientImg(30,30,0,100,100);
}
function draw(){
  background(255);
  translate(width * .5, height * .5);
  if(demo == 1){
    var a = frameCount * .1;
    image(e,100*cos(a), 50*sin(a));
  }
  if(demo == 2){
    for(var i = 0 ; i < 200 ; i++){
      var a = (frameCount * .025 + (i*.1));
      image(e,(100+i)*cos(a + i), (50+i)*sin(a + i));
    }
  }
  if(demo == 3){
    for(var i = 0 ; i < 1000 ; i++){
      var a = (frameCount * .025 + (i*.1));
      image(e,(100+(i * .25))*cos(a + i), (50+(i * .25))*sin(a + i));
    }
  }
  if(demo == 4){
    scale(.2);
    for(var i = 0 ; i < 5000 ; i++){
      var a = (frameCount * .025 + (i*.1));
      image(e,sin(a) * (100+(i * .5))*cos(a + i),  (50+(i * .25))*sin(a + i));
    }
  }
}
function keyReleased(){
  if(key === '1') demo = 1;
  if(key === '2') demo = 2;
  if(key === '3') demo = 3;
  if(key === '4') demo = 4;
}

function getGradientImg(w,h,hue,satMax,brightness){
  push();//isolate drawing styles such as color Mode
    colorMode(HSB,360,100,100);
    var gradient = createImage(w,h);//create an image with an alpha channel
    var np = w * h;//total number of pixels
    var np4 = np*4;
    var cx = floor(gradient.width * 0.5);//center on x
    var cy = floor(gradient.height * 0.5);//center on y
    gradient.loadPixels();
    for(var i = 0 ; i < np4; i+=4){//for each pixel
      var id4 = floor(i * .25);
      var x = id4%gradient.width;//compute x from pixel index
      var y = floor(id4/gradient.width);//compute y from pixel index
      var d = dist(x,y,cx,cy);//compute distance from centre to current pixel
      //map the saturation and transparency based on the distance to centre
      gradient.pixels[i]   = hue;
      gradient.pixels[i+1] = map(d,0,cx,satMax,0);
      gradient.pixels[i+2] = brightness;
      gradient.pixels[i+3] = map(d,0,cx,255,0);
    }
    gradient.updatePixels();//finally update all the pixels
  pop();
  return gradient;
}
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于 2012-12-09T14:01:37.983 回答