Option Explicit Dim i Dim j Dim n For i=1 To GroupCount For j=1 To GroupChnCount(i) n = CNo("[" & i & "]/[" & j & "]") Call FormulaCalc("ch(" & n & ") := ch(" & n & ") - chd(1," & n & ")" ) Next Next
I know this thread is 10 years old, but it is still the first google result for "Diadem subtract by scalar". An alternative (an easier for me) method to do this is use the 3D Arithmetic channel functions and use Matrix-Scalar. Select your desired channel as the 1D matrix, enter the scalar (pay attention to the scalar's sign for subtraction), select where you store the result. Complete.
To subtract a constant value from a channel.
Analysis, basic functions Offset correction. This will do add or subtracts from all values in a channel.
to multiply use the function next to it Scale a channel.