How do i make it so that when an object is detected by 2 ultrasonic sensors a led turns on

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1

Im working on a arduino project for school and I cant get it so that a led turns on when an object is detected by 2 ultrasonic sensors and another led light would light up if the object is detected by only 1 ultrasonic sensor. the problem is that the led for 2 ultrasonic sensors is turned on even with no object in front of either ultrasonic sensor. I have been searching for answers but I could not find any information on the internet.

// defines pins numbers
// ledpin1 is red, ledpin2 is green
const int trigPin = 10;
const int echoPin = 9;
const int ledPin = A4;
const int ledPin2 = A5;
const int ledPin3 = A3;
const int trigPin2 = 12;
const int echoPin2 = 11;

//remember to define led pin2

// defines variables
long duration;
long duration2;
int distance;
int distance2;
int safetyDistance;
int safetyDistance2;


void setup() {
pinMode(trigPin, OUTPUT); // Sets the trigPin as an Output
pinMode(echoPin, INPUT); // Sets the echoPin as an Input
pinMode(ledPin, OUTPUT);
pinMode(ledPin3, OUTPUT);
pinMode(ledPin2, OUTPUT);
pinMode(trigPin2, OUTPUT);
pinMode(echoPin2, INPUT);
Serial.begin(9600); // Starts the serial communication
}


void loop() {
  //SENSOR1
// Clears the trigPin
digitalWrite(trigPin, LOW);
delayMicroseconds(2);

// Sets the trigPin on HIGH state for 10 micro seconds
digitalWrite(trigPin, HIGH);
delayMicroseconds(10);
digitalWrite(trigPin, LOW);

// Reads the echoPin, returns the sound wave travel time in microseconds
duration = pulseIn(echoPin, HIGH);

// Calculating the distance
distance= duration*0.034/2;

safetyDistance = distance;

//SENSOR2
// Clears the trigpin2
delay(200);
digitalWrite(trigPin2, LOW);
delayMicroseconds(2);

//Sets trigPin2 high for 10ms
digitalWrite(trigPin2, HIGH);
delayMicroseconds(2);
digitalWrite(trigPin2, LOW);

// Reads the echopin2
duration2 = pulseIn(echoPin2, HIGH);
//distance calculation
distance2= duration2 * 0.034/2;
if (safetyDistance <=30 || safetyDistance2 <=30)
{
  digitalWrite(ledPin3, HIGH);
  digitalWrite(ledPin2, LOW);
  digitalWrite(ledPin, LOW);
}
  else{
    digitalWrite(ledPin2,HIGH);
    digitalWrite(ledPin, LOW);
    digitalWrite(ledPin3, LOW);
  }
  
}
2

It I understand your question correctly, you have 3 conditions:
detect either sensor being <=30
detect neither sensor being <=30
detect both sensors being <=30

Your code addresses only 2 of these conditions. You're not addressing detect both sensors being <=30

3

OHH, Yea i see it now thanks 2112!

4

It should already be clear :roll_eyes:

You're pinging too frequently, and risk distant, late returns being misinterpreted as nearby returns.

safetyDistance2 is always zero, I believe.
(I don't know why you've got a safetyDistance variable anyway)

5

const int trigPin = 2;
const int echoPin = 3;

void setup() {
Serial.begin(9600);
pinMode(trigPin, OUTPUT);
pinMode(echoPin, INPUT);
}

void loop() {
long duration, distance;
digitalWrite(trigPin, LOW);
delayMicroseconds(2);
digitalWrite(trigPin, HIGH);
delayMicroseconds(10);
digitalWrite(trigPin, LOW);
duration = pulseIn(echoPin, HIGH);
distance = duration * 0.034 / 2; // Calculate distance in centimeters
Serial.print("Distance: ");
Serial.print(distance);
Serial.println(" cm");
delay(500); // Wait for a short period before taking the next measurement
}

6

It should already be low.

Please use code tags when posting code, however sloppy it is.

7

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