RE: How to make the gridlines of a chart square
- From: PY & Associates <PYAssociates@xxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Tue, 6 Sep 2005 19:58:03 -0700
I find Jon Peltier utility fasinating.
I have done three pdf files showing before and after effect, but am not
aware how to attach them to show you.
In words, when I do first time transformation (to make grid square), I
cannot predict major tick setting.
I may need to adjust the major tick setting and transform the grid again.
Would it be nice if we can control the maximum values of the axes.
"Ali Baba" wrote:
> I would like to have a chart in excel with the x-axis and y-axis ranges from
> -5 to 5 as illustrated in the link below.
> http://phpaw.com/myscrpit/milt-up/jpg/jpg-1126029383.jpg
>
> The problem is that I don't get square gridlines. I can only do this
> visually by resizing the chart. I saw Jon Peltier traffic example of making
> the gridlines square
> (http://peltiertech.com/Excel/Charts/SquareGrid.html) but I found that when
> I run it the scale changes to a value greater than 5. Also if I try to add
> titles the gridlines become not squared.
>
> Is there a way to fix the range of axes and have square gridlines?
>
> The other problem I have which you may be able to help me with is that I am
> trying to write a function to find the partial derivative of a function.
> The function can then be called from a worksheet cell almost exactly as one
> would write them normally.
>
> For example,
> we have this equation
> f(Z1, Z2, Z3) = Z1^2+Z2*Z3^3-Z3^0.5
>
> and Z1 = 2, Z2 = 4, Z3 = 6
>
> if we want to find the partial derivative with respect to Z1 then:
>
> = DerivativeZ("Z1^2+Z2*Z3^3-Z3^0.5", 2,"Z1")
> = 4
>
> with respect to Z2
> = DerivativeZ("Z1^2+Z2*Z3^3-Z3^0.5", 4,"Z2")
> = 216
>
> and so on .............
>
> So all I am doing is to substitute the values of Z2 and Z3 in the function
> and then do a normal differentiation with respect to Z1. Similarity, I
> substitute Z1 and Z3 to get the derivative with respect to Z2, .....
>
>
> I don't know what is wrong with the code. So can anybody help me to correct
> it and get it working.
>
> Thanks for any help
>
>
> ____________________________________________________
>
> Option Explicit
> Dim Z1 As Double
> Dim Z2 As Double
> Dim Z3 As Double
>
>
> Function DerivativeZ(func As String, a As Double, V As String) As Double
>
> Const h = 0.0001
> Dim n1 As Double, n2 As Double
>
> Z1 = Range("C13")
> Z2 = Range("C14")
> Z3 = Range("C15")
>
>
> Select Case UCase(Left(V, 2))
> Case Is = "Z1"
> func = Replace(func, "Z2", Z2)
> func = Replace(func, "Z3", Z3)
> n1 = (eval(func, a + (h / 2), Z1) - eval(func, a - (h / 2), Z1)) / h
> n2 = (eval(func, a + h, Z1) - eval(func, a - h, Z1)) / (2 * h)
>
> Case Is = "Z2"
> func = Replace(func, "Z1", Z1)
> func = Replace(func, "Z3", Z3)
> n1 = (eval(func, a + (h / 2), Z2) - eval(func, a - (h / 2), Z2)) / h
> n2 = (eval(func, a + h, Z2) - eval(func, a - h, Z2)) / (2 * h)
>
> Case Is = "Z3"
> func = Replace(func, "Z1", Z1)
> func = Replace(func, "Z2", Z2)
> n1 = (eval(func, a + (h / 2), Z3) - eval(func, a - (h / 2), Z3)) / h
> n2 = (eval(func, a + h, Z3) - eval(func, a - h, Z3)) / (2 * h)
>
> End Select
>
> DerivativeZ = (4 * n1 - n2) / 3
>
> End Function
>
> Function eval(funct As String, Z As Double, V As String) As Double
> eval = Evaluate(funct)
>
> Select Case UCase(Left(V, 2))
> Case Is = "Z1"
> Z1 = Z
> Case Is = "Z2"
> Z2 = Z
> Case Is = "Z3"
> Z3 = Z
> End Select
>
> End Function
>
.
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- From: Ali Baba
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