04-10-2007 10:28 PM
04-11-2007 04:42 AM
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
Matthias Alleweldt Project Engineer / Projektingenieur | Twigeater? |
12-10-2018 11:32 AM
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.
12-10-2018 02:21 PM
Hi George,
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.
Paul