As I posted many times, this is the solution fullness = abs((Dclose(0)-Dopen(0))/(Dhigh(0)-Dlow(0)+0.00001)
It’s a solution, I prefer something which can’t yield division by zero. I don’t think it’s likely to create any weird results either. I bet your solution works too with the underlying primitives, altough without the exact knowledge of how they work it still looks like possibly erroneous code.
abs((Dclose(0)-Dopen(0))/(Dhigh(0)-Dlow(0)+0.00001) cant yield to a division by zero as High -Low is bounded to zero, so high +low + 0,0000001 is strictly positive
