A Low Cost Approach to Graphical Waterfall

Thanks for all of your help and suggestions.. I am going to abandoned this project as i got poor results... But i hope someone will come here with clever ideas and make this whole thing work. Following is the last video of the show:

I can attach the the visualbasic files (Source Code) so that the software can be further improved to get realistic results. I am sorry the source code is not well commented and please don't ask me programming related questions:

I am attaching the Arduino Sketch for controlling the 40 solenoid valves. May be someone can improve the sketch and send me so that i can try on my hardware. Following is my email address:
engr.khalid.khattak@gmail.com

//Pin connected to ST_CP of 74HC595
int latchPin = 6;
//Pin connected to SH_CP of 74HC595
int clockPin = 13;
////Pin connected to DS of 74HC595
int dataPin = 11;
byte FirstByte;
byte SecondByte;
byte ThirdByte;
byte FourthByte;
byte FifthByte;
int val;
byte serialInArray[5]; // array for storing 5 bytes as they arrive from VB software
int serialCount = 0; // for counting the number of bytes received


void setup() {
  //Start Serial for debuging purposes	
  Serial.begin(9600);
  //set pins to output because they are addressed in the main loop
  pinMode(latchPin, OUTPUT);

}

void loop() {
  
   if (Serial.available() > 0){
       serialInArray[serialCount] = Serial.read(); // read a byte sent by processing
        serialCount++;  // increment number of bytes received

    if (serialCount > 4 ) {  
	FirstByte = serialInArray[0]; 
	SecondByte = serialInArray[1]; 
        ThirdByte = serialInArray[2];
        FourthByte = serialInArray[3];
        FifthByte = serialInArray[4];
Serial.print(FirstByte);
Serial.print(SecondByte);
Serial.print(ThirdByte);
Serial.print(FourthByte);
Serial.print(FifthByte);
Serial.println("");

     digitalWrite(latchPin, 0);
    shiftOut(dataPin, clockPin, FirstByte); 
             shiftOut(dataPin, clockPin, SecondByte);
                 shiftOut(dataPin, clockPin, ThirdByte);
                 shiftOut(dataPin, clockPin, FourthByte);  
       shiftOut(dataPin, clockPin, FifthByte);
     
   
    //return the latch pin high to signal chip that it 
    //no longer needs to listen for information
    digitalWrite(latchPin, 1);
 serialCount = 0;
 delay(20);
   }
}
}
void shiftOut(int myDataPin, int myClockPin, byte myDataOut) {
  // This shifts 8 bits out MSB first, 
  //on the rising edge of the clock,
  //clock idles low

//internal function setup
  int i=0;
  int pinState;
  pinMode(myClockPin, OUTPUT);
  pinMode(myDataPin, OUTPUT);

 //clear everything out just in case to
 //prepare shift register for bit shifting
  digitalWrite(myDataPin, 0);
  digitalWrite(myClockPin, 0);

  //for each bit in the byte myDataOut�
  //NOTICE THAT WE ARE COUNTING DOWN in our for loop
  //This means that %00000001 or "1" will go through such
  //that it will be pin Q0 that lights. 
  for (i=7; i>=0; i--)  {
    digitalWrite(myClockPin, 0);

    //if the value passed to myDataOut and a bitmask result 
    // true then... so if we are at i=6 and our value is
    // %11010100 it would the code compares it to %01000000 
    // and proceeds to set pinState to 1.
    if ( myDataOut & (1<<i) ) {
      pinState= 1;
    }
    else {	
      pinState= 0;
    }

    //Sets the pin to HIGH or LOW depending on pinState
    digitalWrite(myDataPin, pinState);
    //register shifts bits on upstroke of clock pin  
    digitalWrite(myClockPin, 1);
    //zero the data pin after shift to prevent bleed through
    digitalWrite(myDataPin, 0);
  }

  //stop shifting
  digitalWrite(myClockPin, 0);
}