Post date: Jul 02, 2010 8:31:39 PM
I borrowed the code from assignment 2.2 because it did the same thing and was much easier to look at.
int potPin = 2; // declares variables
int val = 0;
void setup() {
}
void loop() {
for (int ledPin = 13; ledPin >= 4; ledPin--) // when ledPin is at 13, decrease by one
{ // each time loop is run until ledPin=4
val = analogRead(potPin); // reads potPin and stores value in val
pinMode(ledPin, OUTPUT); // sets ledPin to output
digitalWrite(ledPin, HIGH); // turns ledPin on
delay(val); // delay for whatever val is
digitalWrite(ledPin, LOW); // turns ledPin off
delay(val); // delay for whatever val is
}
for (int ledPin = 4; ledPin <= 13; ledPin++) // when ledPin is at 4, increase by one
{ // each time loop is run until ledPin=13
val = analogRead(potPin);
pinMode(ledPin, OUTPUT);
digitalWrite(ledPin, HIGH);
delay(val);
digitalWrite(ledPin, LOW);
delay(val);
}
}
Note:
When the potentiometer is turned fully to the negative side, the LEDs appear to spaz out. If you wave the breadboard perpendicular to the row of LEDs you can see a zigzag pattern, proving that it's just your mind playing tricks on you again.