Sharp IR Remapping

Using Arduino Sensors
1

I've just purchased a Sharp IR Range Finder 20cm to 150cm. I'm trying to remap the ADC so that when nothing is in range it reads 0 and when something has approached the 20cm limit to read 100.

This is my code:

#define irSensor 1

void setup () {
  Serial.begin(9600);
}

void loop() {
  
  int irVal = analogRead(irSensor);
  int irMapVal = map(irVal, 0, 1024, 0, 100);
  Serial.println(irMapVal);
  delay(1000);
}

For some reason my maximum reading within the boundaries is 50 not 100 as I expected.

2

If you look at the specifications you will see that the output range for the GP2Y0A02 is about 0.5V to 2.75V. With a 5V analog input that's about 102 to 563. You might want to constrain the inputs before mapping:

  int irVal = analogRead(irSensor);
  int irMapVal = map(constrain(irVal,102,563), 102, 563, 0, 100);
3

johnwasser:
If you look at the specifications you will see that the output range for the GP2Y0A02 is about 0.5V to 2.75V. With a 5V analog input that's about 102 to 563. You might want to constrain the inputs before mapping:

  int irVal = analogRead(irSensor);

int irMapVal = map(constrain(irVal,102,563), 102, 563, 0, 100);

Thanks very much, thats perfect :slight_smile:

If you don't mind me asking, how did you work out the 102 to 563 range with the incoming and outgoing voltages? I know the Arduino Uno ADC is 10bit (0, 1024). Did you multiply the 0.5 and 2.75 by something?

4

The default analog input reference voltage is 5V. That means that 0V maps to 0 and 5V maps to 1024 (yes, even though the highest number you can get is 1023).

To calculate the analog input number from a voltage, divide the voltage by 5 and multiply by 1024.

0.5V / 5V = 0.1 * 1024 = 102.4
2.75V / 5V = 0.55 * 1024 = 563.2

Of course the actual numbers are truncated to integers: 102 and 563.

5

That makes perfect sense :slight_smile:

Thanks for your help, much appreciated!