Question

我正致力于开发一个简单的处理应用程序，只要锤子碰到不动的形状，就会产生声音效果。但是，每当锤子对象与形状对象发生碰撞时，我都难以尝试检测代码，并且已开始采用非专业的解决方法，如在“测试”下评论的代码块中所示。任何有关创建此问题解决方案的帮助都将受到大力赞赏

import oscP5.*;
import netP5.*;

OscP5 oscP5;
NetAddress myRemoteLocation;

float bx;
float by;
int boxSizeX = 160;
int boxSizeY = 30;
boolean overBox = true;
boolean locked = false;
float xOffset = 0.0; 
float yOffset = 0.0; 
float angle = 4.70;

BeatBall b1 = new BeatBall(210,425,60, 1);

void setup()
{
  size(800, 600);
  smooth();
  frameRate(120);
  bx = width/2.0;
  by = height/2.0;

    oscP5 = new OscP5(this,12001);

  /* myRemoteLocation is a NetAddress. a NetAddress takes 2 parameters,
   * an ip address and a port number. myRemoteLocation is used as parameter in
   * oscP5.send() when sending osc packets to another computer, device, 
   * application. usage see below. for testing purposes the listening port
   * and the port of the remote location address are the same, hence you will
   * send messages back to this sketch.
   */
  myRemoteLocation = new NetAddress("127.0.0.1",12000);
}

void draw()
{ 
  background(0);

  pushMatrix();
  translate(400, 425);
  rotate(angle);
  fill(222,223,255);
  Hammer h = new Hammer(135, -67,  boxSizeY+25, boxSizeX-25, 1);
  h.displayHammer();
  rect(-25, -15, boxSizeX, boxSizeY);
  popMatrix();

  b1.displayBall();

   //Testing
   if(angle < -2.6561418 && angle > -3.043227)
     {
       background(120);
       b1.collide();
     }

     println(angle);
}



void mousePressed() 
{
  xOffset = mouseX-bx; 
  yOffset = mouseY-by; 
}

void mouseDragged()
{
    bx = mouseX-xOffset; 
    by = mouseY-yOffset; 
   angle = atan2(mouseY - 400, mouseX - 400);
}

//BEATBALL CLASS

class BeatBall {
  float x, y;
  float diameter;
  float vx = 0;
  float vy = 0;
  int id;

  BeatBall(float xin, float yin, float din, int idin) {
    x = xin;
    y = yin;
    diameter = din;
    id = idin;
  } 

  void collide() 
  {

    /* Collision Example
      float dx = Hammer.x - x;
      float dy = Hammer.y - y;
      float distance = sqrt(dx*dx + dy*dy);
      float minDist = others[i].diameter/2 + diameter/2;
      if (distance < minDist)
      { 
        float angle = atan2(dy, dx);
        float targetX = x + cos(angle) * minDist;
        float targetY = y + sin(angle) * minDist;
        float ax = (targetX - others[i].x) * spring;
        float ay = (targetY - others[i].y) * spring;
        vx -= ax;
        vy -= ay;
        others[i].vx += ax;
        others[i].vy += ay;
       */ 

      OscMessage myMessage = new OscMessage("/bubble");
      print(diameter + " ");
      myMessage.add( 1/(diameter*diameter) * 1000000); /* add an int to the osc message */

      /* send the message */
      oscP5.send(myMessage, myRemoteLocation); 

     //}
  }

  void displayBall() 
  {
    fill(191,89,0);
    ellipse(x, y, diameter, diameter);   
  }


}

//HAMMER CLASS

class Hammer {
  float x, y;
  float sizeX, sizeY;
  float vx = 0;
  float vy = 0;
  int id;

  Hammer(float xin, float yin, float sxin, float syin, int idin) {
    x = xin;
    y = yin;
    sizeX = sxin;
    sizeY = syin;
    id = idin;
  } 

   void displayHammer() 
  {
    fill(222,223,255);
    rect(x, y, sizeX, sizeY); 
  }

}

Answer 1

我已经构建了a suite of collision detection functions for Processing可能会有所帮助。

如果您可以简化事物并将对象视为圆圈，则可以使用毕达哥拉斯定理来检查它们的距离。（根据要求更新为函数。）

// variables for your objects - where are they and how big?
float ballX, ballY;
float ballRadius;
float hammerX, hammerY;
float hammerRadius;

void setup() {

  // check for a collision
  boolean hit = ballBallCollision(ballX, ballY, ballRadius, hammerX, hammerY, hammerRadius);
  if (hit) {
    // hit!
  }
  else {
    // not :(
  }
}

// a function to check for ball-ball collision
boolean ballBallCollision(float ballX, float ballY, float ballRadius, float hammerX, float hammerY, float hammerRadius) {
  // calculate distance between the objects using the Pythagorean Theorem
  float xDist = hammerX - ballX;
  float xDist = hammerY - ballY;
  float dist = sqrt( (xDist*xDist) + (yDist*yDist) );

  if (dist < ballRadius + hammerRadius) {
    return true;
  }
  return false;
}

移动和非移动物体之间的碰撞检测

1 个答案: