Re: Exponetial Interpolation VBA Code

On Aug 8, 6:18 am, David B <da...@xxxxxxxxxxxxxxxxxxxxxx> wrote:

You (as well as dougaj4) also make (IMO) goofy implicit
assumptions about the nature of the objects/arrays being passed in, as
well as some inefficient "dims".




David B, feel free to tell which implicit assumptions are goofy, and
why, and show us how to make the "dims" more efficient.


More to the point, the formula given in the OP gives the wrong
answers.

The code below will I believe give correct exponential interpolation:

Function ExpInterp2(D As Variant, P As Variant, V As Double) As
Variant
Dim i As Long, LogP As Double, LogPdiff As Double, Dprev As Double
Dim DDiff As Double, Logslope As Double, LogInterp As Double

If TypeName(D) = "range" Then D = D.Value
If TypeName(D) = "range" Then P = P.Value

i = 1
Do While V > D(i, 1)
i = i + 1
Loop
i = i - 1

LogP = Log(P(i, 1))
LogPdiff = Log(P(i + 1)) - LogP
Dprev = D(i, 1)
DDiff = D(i + 1, 1) - Dprev
Logslope = LogPdiff / DDiff

LogInterp = LogP + (V - Dprev) * Logslope

ExpInterp2 = Exp(LogInterp)

End Function


Goofy implicit assumptions are that D and P are single column ranges
containing values, and V is a value or a cell address containing a
value.

.

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