[JohnFogarty]

My mentor sent me this.

That should work just fine. I was trying to reduce the number of arithmetic calculations, which would be necessary if the array/table was larger than it is. If you take the sum and subtract the oldest value and add the newest value it take a lot less time. Can you imagine adding up 100 or more entries each time?

In the final example:
What you want is a circular buffer (ring buffer) that you populate each cycle and use each cycle to get a new speed, like this:
Code:
public void getSpeed(){
samples[counter++] = Shooter_En.getRaw();
Shooter_En.reset();
if(counter>9) counter=0;
sum=0;
for(int i = 0; i <= 9; i++) sum += samples[i];
Rate = sum / (10* 0.48);
I have to assume that the first encoder value is way out of line and will skew the results since you have not reset the encoder until you read it the first time. As a matter of fact, until you have 11 samples the Rate is not really valid since you have some amount of samples that are zero since the array has not been filled yet. Would you consider this? Once you have 10 samples, a valid flag says to compute the Rate and could tell the user that it is now valid. See This;

public void getSpeed(){
samples[counter++] = Shooter_En.getRaw();
Shooter_En.reset();
if (counter >9){ counter = 0; ValidFlag = true;}

If (ValidFlag)
{
sum = 0;
for(int i = 0; i < counter; i++) sum += samples[i];
Rate = sum / (10* 0.48);
}





I first questioned the last line: Rate = sum / (10* 0.48); I thought is should be: Rate = (sum /10) / 0.48; Then realized it is the same thing. You should make it Rate = sum/4.80; to eliminate the multiplication.

Dennis