小车的经典解决方案 原创

发布于 2022-5-31 02:46
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@toc

一. 预备阶段

硬件:hi3861开发板等小车标配的套件,电脑

软件:VS code,openharmony 3.0,

小车的经典解决方案-开源基础软件社区

二. 开发阶段

1.目标效果

可以实现舵机托举超声波扩展板左右旋转扫描达到避障的效果。可融入于地下隧道探测与扫地机器人的开发。

2.解决方案

主要问题:超声波与探测频率(比如说在舵机左右旋转时进行多少次探测)

代码解析:超声波模块触发信号需要大于10us的高电平,预先设定获取超声波传来信号的周期为500ms,需要将PWM信号的周期也设为500ms

PWM的频率是指每秒钟信号从高电平到低电平再回到高电平的次数

参考计算公式:T=(arr+1)(psc+1)/(time)M (单位为秒)

2.1 超声波原理如下图:

小车的经典解决方案-开源基础软件社区

2.2 设备侧代码:

只例出pwm
void loop() {
  // put your main code here, to run repeatedly:
  avoidance();
}
void motorRun(int cmd,int value)
{
  analogWrite(leftPWM, value);    // Set PWM output, set speed
  analogWrite(rightPWM, value);
  switch(cmd){
    case FORWARD:
      Serial.println("FORWARD");   // signal output
      digitalWrite(leftMotor1, HIGH);
      digitalWrite(leftMotor2, LOW);
      digitalWrite(rightMotor1, HIGH);
      digitalWrite(rightMotor2, LOW);
      break;
     case BACKWARD:
      Serial.println("BACKWARD"); // signal output
      digitalWrite(leftMotor1, LOW);
      digitalWrite(leftMotor2, HIGH);
      digitalWrite(rightMotor1, LOW);
      digitalWrite(rightMotor2, HIGH);
      break;
     case TURNLEFT:
      Serial.println("TURN  LEFT");     // signal output
      digitalWrite(leftMotor1, HIGH);
      digitalWrite(leftMotor2, LOW);
      digitalWrite(rightMotor1, LOW);
      digitalWrite(rightMotor2, HIGH);
      break;
     case TURNRIGHT:
      Serial.println("TURN  RIGHT");   // signal output
      digitalWrite(leftMotor1, LOW);
      digitalWrite(leftMotor2, HIGH);
      digitalWrite(rightMotor1, HIGH);
      digitalWrite(rightMotor2, LOW);
      break;
     default:
      Serial.println("STOP");     // signal output
      digitalWrite(leftMotor1, LOW);
      digitalWrite(leftMotor2, LOW);
      digitalWrite(rightMotor1, LOW);
      digitalWrite(rightMotor2, LOW);
  }
}
void avoidance()
{
  int pos;
  int dis[3];      // long
  motorRun(FORWARD,200);
  myServo.write(90);
  dis[1]=getDistance();    //middle
 
  if(dis[1]<30)
  {
    motorRun(STOP,0);
    for (pos = 90; pos <= 150; pos += 1)
    {
      myServo.write(pos);              // tell servo to go to position in variable 'pos'
      delay(15);                       // waits 15ms for the servo to reach the position
    }
    dis[2]=getDistance(); // left
    for (pos = 150; pos >= 30; pos -= 1)
    {
      myServo.write(pos);     // tell servo to go to position in variable 'pos'
      delay(15);              // waits 15ms for the servo to reach the position
      if(pos==90)
        dis[1]=getDistance();   // middle
    }
    dis[0]=getDistance();    // right
    for (pos = 30; pos <= 90; pos += 1)

三.成果展示

详见号主哔哩哔哩~

四.结果

测试成功~

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已于2022-5-31 02:46:25修改
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