Question

所以我有一个问题，当我用超声波传感器运行我的代码，并且用电机桥接时，一个电机一直旋转，另一个电极每6秒旋转2秒但我不知道为什么。有什么帮助吗？

以下是代码：

int in1 = 2;
int in2 = 3;
int in3 = 4;
int in4 = 5;
int in5 = 6;
int in6 = 7;
int trig = 8;
int echol = 9;
int echor = 12;
int echof = 11;
long df, tf, dr, tr, dl, tl;
 void setup() {

  Serial.begin(9600);
    }

void loop() {

 pinMode (in1, OUTPUT);
 pinMode (in2, OUTPUT);
 pinMode (in3, OUTPUT);
 pinMode (in4, OUTPUT);
 pinMode (in5, OUTPUT);
 pinMode (in6, OUTPUT);
 pinMode (trig, OUTPUT);
 pinMode (echol, INPUT);
 pinMode (echor, INPUT);
 pinMode (echof, INPUT);


forward();

  digitalWrite (trig, HIGH);
  delay (0.01);
  tf = pulseIn (echof, HIGH);
  digitalWrite (trig, LOW);
  df = tf * 0.03156;

  if (df < 1.5){
  digitalWrite (trig, HIGH);
  delay (0.01);
  tr = pulseIn (echor, HIGH);
  tl = pulseIn (echol, HIGH);
  digitalWrite (trig, LOW);
  dr = tr * 0.03156;
  dl = tl * 0.03156;

   if (dr > dl) {

    right();
    delay (5000);
    forward();

  }
  else {

    left();
    delay (5000);
    forward();

  }

}

}



void forward(){
 digitalWrite (in1, HIGH);
 digitalWrite (in2, LOW);
 digitalWrite (in3, HIGH);
 digitalWrite (in4, LOW);
 digitalWrite (in5, HIGH);
 digitalWrite (in6, LOW);
}

void backward(){
  digitalWrite (in1, LOW);
  digitalWrite (in2, HIGH);
  digitalWrite (in3, LOW);
  digitalWrite (in4, HIGH);
  digitalWrite (in5, LOW);
  digitalWrite (in6, HIGH);
}

void left(){
  digitalWrite (in1, LOW);
  digitalWrite (in2, LOW);
  digitalWrite (in3, HIGH);
  digitalWrite (in4, LOW);
  digitalWrite (in5, HIGH);
  digitalWrite (in6, LOW);
}

void right(){
  digitalWrite (in1, HIGH);
  digitalWrite (in2, LOW);
  digitalWrite (in3, LOW);
  digitalWrite (in4, LOW);
  digitalWrite (in5, HIGH);
  digitalWrite (in6, LOW);
}

Answer 1

不确定所有错误，但将long乘以tl * 0.03156并将值存储在long中可能不会达到您的意图。您应该使用浮点值来包含这种计算的结果。

Answer 2

首先，您应该在Setup（）中移动引脚设置，无需在每个循环上重新初始化引脚i / o设置。

void Setup()
{
  Serial.begin(9600);
  pinMode (in1, OUTPUT);
  pinMode (in2, OUTPUT);
  pinMode (in3, OUTPUT);
  pinMode (in4, OUTPUT);
  pinMode (in5, OUTPUT);
  pinMode (in6, OUTPUT);
  pinMode (trig, OUTPUT);
  pinMode (echol, INPUT);
  pinMode (echor, INPUT);
  pinMode (echof, INPUT);
}

根据我对H-Bridge模块的了解，它们通常每个电机有3个输入，其中一次只能打开一个。我找不到与你的代码有关的任何关联....这是一个约束，所以你应该围绕它组织你的代码。它会使读取和调试更容易。 Arduino上没有调试器，因此组织代码确实有很大帮助。如果您需要更多帮助，其他人一定要更容易理解您的代码所做的事情。

void MotorA(int dir)
{
  // dir = 0 = STOP, +1 = Forward, -1 = Reverse
  digitalWrite(in1, dir > 0);     
  digitalWrite(in2, dir == 0);      // You gave no details on the module
  digitalWrite(in3, dir < 0);       // you have. the actual logic may differ...
}

void MotorB(int dir)
{
  // dir = 0 = STOP, +1 = Forward, -1 = Reverse
  digitalWrite(in4, dir > 0);
  digitalWrite(in5, dir == 0);
  digitalWrite(in6, dir < 0);
}

void Stop()
{
  MotorA(0);
  MotorB(0);
}

void Forward()
{
  MotorA(+1);
  MotorB(+1);
}

void Reverse()
{
  MotorA(-1);
  MotorB(-1);
}

void Left()
{
  MotorA(+1);
  MotorB(-1);
}

void Right()
{
  MotorA(-1);
  MotorB(+1);
}

我是否还建议您从一个更简单的循环开始，直到您的电机运行？一次添加一个功能。这将有助于您最终更快地完成项目并准备好。

void Loop()
{
  Forward()
  delay(5000);
  Stop()
  delay(1000);
  Reverse();
  delay(1000);
 // ---
}

根据编写的代码，运行带有H Bridge和Arduino的电机的Arduino代码问题

2 个答案: