I am in the process of converting Mathcad code into VBA and am trying to figure out how to replicate the While loop, which asks the program to run the loop while TGuess < 0. At the end of the loop is an if statement to break the loop if sGuess>1/1.4 (I would attach a picture, but my reputation does not allow me to).
I have written this code in VBA, but am wondering if including the sGuess variable in the original While statement is correct, or if it could influence the output of the loop:
While TGuess < 0 And sGuess <= 1 / 1.4
kterm = (kj ^ (1 / 6)) / 26 'k term in the numerator of depth equation
epw = 3 / 5
FDepth = ((kterm * RainInt * L * CF) / sGuess ^ 0.5) ^ epw
tflow = UW_Wat * g * sGuess * FDepth 'Calc Flow Shear Stress
pflow = 7.853 * UW_Wat * (tflow / UW_Wat) ^ (3 / 2)
TGuess = pflow - pcrit 'Recalc TGuess as E-P
sGuess = sGuess + SlopeInc 'Calc new stable slope
Wend
Any input would be appreciated.
To mitigate your concern, it might be better to replace the while...wend loop with a Do While ... Loop block. You can then put your break condition where you'd have it in the corresponding Mathcad code by using something along the lines of
If sGuess > 1/1.4 Then
Exit Do
End If
Related
I have come across a issue while working in VBA . I'm supposed to write program that is Numerical integration of trapeze method (I'm not sure if It is how it's called in English) of function 100*x^99 lower limit = 0 upper limit = 1 . Cells (j,5) contains numbers (10,30,100,300,1000,3000,10000) - amount of point splits . Code seems to work but given wrong results , for amount of splits it should be around
10 - 5.000295129200607
30 - 1.786588019299606
100 - 1.0812206997600746
300 - 1.0091505687770146
1000 - 1.0008248693208752
3000 - 1.0000916650530287
10000 - 1.000008249986933
Function F(x)
F = 100 * (x ^ 99)
End Function
Sub calka()
Dim n As Single
Dim xp As Single
Dim dx As Single
Dim xk As Single
Dim ip As Single
Dim pole As Single
xp = 0
xk = 1
For j = 5 To 11
n = Cells(j, 5)
dx = (xk - xp) / n
pole = 0
For i = 1 To n - 1
pole = pole + F(xp + i * dx)
Next i
pole = pole + ((F(xp) + F(xk)) / 2)
pole = pole * dx
Worksheets("Arkusz1").Cells(j, 7) = pole
Next j
End Sub
I tried to implement same code in java and c++ and it worked flawlessly but VBA always gives me wrong results , I'm not sure if it's rounds at some point and I can disable in settings or my code is just not written right .
Apologies for low clarity It's hard for me to translate mathematic to English.
Use Doubles rather than Singles
http://www.techrepublic.com/article/comparing-double-vs-single-data-types-in-vb6/
I have two bits of code in VBA for Excel. One calculates the A-squared statistic for the Anderson-Darling test, this bit of code calculates the P value of the A-squared statistic. I am curious if there is a more concise way or more efficient way to calculate this value in VBA:
Function AndDarP(AndDar, Elements)
'Calculates P value for level of significance for the
'Anderson-Darling Test for Normality
'AndDar is the Anderson-Darling Test Statistic
'Elements is the count of elements used in the
'Anderson-Darling test statistic.
'based on calculations at
'http://www.kevinotto.com/RSS/Software/Anderson-Darling%20Normality%20Test%20Calculator.xls
'accessed 21 May 2010
'www.kevinotto.com
'kevin_n_otto#yahoo.com
'Version 6.0
'Permission to freely distribute and modify when properly
'referenced and contact information maintained.
'
'"Keep in mind the test assumes normality, and is looking for sufficient evidence to reject normality.
'That is, a large p-value (often p > alpha = 0.05) would indicate normality.
' * * *
'Test Hypotheses:
'Ho: Data is sampled from a population that is normally distributed
'(no difference between the data and normal data).
'Ha: Data is sampled from a population that is not normally distributed"
Dim M As Double
M = AndDar * (1 + 0.75 / Elements + 2.25 / Elements ^ 2)
Select Case M
Case Is < 0.2
AndDarP = 1 - Exp(-13.436 + 101.14 * M - 223.73 * M ^ 2)
Case Is < 0.34
AndDarP = 1 - Exp(-8.318 + 42.796 * M - 59.938 * M ^ 2)
Case Is < 0.6
AndDarP = Exp(0.9177 - 4.279 * M - 1.38 * M ^ 2)
Case Is < 13
AndDarP = Exp(1.2937 - 5.709 * M + 0.0186 * M ^ 2)
Case Else
AndDarP = 0
End Select
End Function
I have an equation that can be used to find the gun elevation for artillery, using the range, muzzle velocity and change in altitude in a game called Arma 3. The equation looks like this:
With g being the acceleration due to gravity (9.80665), V being the muzzle velocity, X being the range and Y being the change in altitude (called DAlt in my code).
I'm trying to convert it to a line of code so that I can make a program to calculate the elevation based on given coordinates. However I'm having trouble with it. I currently have this:
If rdoLow.Checked = True Then
Elevation = Math.Atan(((Velocity ^ 2) - (Math.Sqrt((Velocity ^ 4) - (G) * (G * (Range ^ 2) + (2 * DAlt * (Velocity ^ 2)))))) / G * Range)
Else
Elevation = Math.Atan(((Velocity ^ 2) + (Math.Sqrt((Velocity ^ 4) - (G) * (G * (Range ^ 2) + 2 * DAlt * (Velocity ^ 2))))) / G * Range)
End If
Which isn't particularly nice looking but as far as I can tell, it should work. However when I put in the values that the video I got the equation from used, I got a different answer. So there must be something wrong with my equation.
I've tried breaking it in to various parts as separate variables and calculating them, then using those variables in the overall equation, and that still didn't work and gave me an answer that was wrong in another way.
So I'm currently at a loss on how to fix it, starting to wonder if the way that vb handles long equations is different or something.
Any help is much appreciated.
You haven't given any sample data, so I can't verify that this gives the correct answer, but the last part of your equation is missing some parentheses.
Elevation = Math.Atan(((Velocity ^ 2) + Math.Sqrt((Velocity ^ 4) - (G * ((G * (Range ^ 2)) + (2 * DAlt * (Velocity ^ 2)))))) / (G * Range))
Note the parenthesis around the last G * Range.
Multiplication and division have equal precedence, so they are evaluated from left-to-right. See Operator Precedence in Visual Basic.
You were dividing everything by G, then multiplying the result by Range, whereas you needed to multiply G by Range, then divide everything by the result of that.
You can see the difference in this simple example:
Console.WriteLine(3 / 4 * 5) ' Prints 3.75
Console.WriteLine(3 / (4 * 5)) ' Prints 0.15
Out of curiosity I tried the problem. In order to have test data I found this web site, Range Tables For Mortars. I tested with the '82mm Mortar - Far' that has an initial velocity of 200 m/s. One problem I had, and don't know if I fixed it correctly, was that the first part of the equation was returning negative numbers... I also solved for the ±. To test I created a form with a button to perform the calculation, a textbox to enter the distance, and two labels to show the angles. This is what I came up with.
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
'A - launch angle
'Target
' r - range
' y - altitude
'g - gravity 9.80665 m/s^2
'v - launch speed e.g. 50 m/s
'
'
'Formula
'from - https://en.wikipedia.org/wiki/Trajectory_of_a_projectile#Angle_required_to_hit_coordinate_.28x.2Cy.29
'in parts
'parts - px
' p1 = sqrt( v^4 - g * (g * r^2 + 2 * y * v^2) )
' p2 = v^2 ± p1 note plus / minus
' p3 = p2 / g * r
'
' A = arctan(p3)
Dim Ap, Am, r, y As Double
Dim g As Double = 9.80665
Dim v As Double
Dim p1, p2p, p2m, p3p, p3m As Double
If Not Double.TryParse(TextBox1.Text, r) Then Exit Sub
y = 0
v = 200 '82mm Mortar - Far velocity
p1 = v ^ 4 - g * (g * r ^ 2 + 2 * y * v ^ 2)
If p1 < 0 Then
Debug.WriteLine(p1)
p1 = Math.Abs(p1) 'square root does not like negative numbers
End If
p1 = Math.Sqrt(p1)
'plus / minus
p2p = v ^ 2 + p1
p2m = v ^ 2 - p1
p3p = p2p / (g * r)
p3m = p2m / (g * r)
Const radiansToDegrees As Double = 180 / Math.PI
Ap = Math.Atan(p3p) * radiansToDegrees
Am = Math.Atan(p3m) * radiansToDegrees
Label1.Text = Ap.ToString("n3")
Label2.Text = Am.ToString("n3")
End Sub
Using the web site to verify the calculations the code seem correct.
Writing long formulas in a bunch of nested parentheses serves no purpose, unless you are going for confusion.
I keep getting this compile error based on the code below and I cannot figure it out for the life of me. Can you guys see what the problem is? It highlights the "p" when I go back to debug.
Thanks,
R
Function EuroBin(S, K, T, rF, sigma, n, PutCall As String)
dt = T / n: u = Exp(sigma * Sqr(dt))
d = 1 / u: p = (Exp(rF * dt) - d) / (u - d)
EuroBin = 0
For i = 0 To n
Select Case PutCall
Case "Call":
EuroBin = WorksheetFunction.Combin(n, i) * EuroBin + p^(i) * (1 - p) ^ (n - i) * WorksheetFunction.Max(S * u^(i) * d^(n - i) - K, 0)
Case "Put":
EuroBin = WorksheetFunction.Combin(n, i) * EuroBin + p^(i) * (1 - p) ^ (n - i) * WorksheetFunction.Max(K - S * u^(i) * d^(n - i), 0)
End Select
Next i
EuroBin = Exp(-rF * T) * EuroBin
End Function
For exponentiation, use
WorksheetFunction.Power(b, e)
instead of b^e, which seems to not work in VBA!
I'm saying that based on the test Debug.Print 2^2, which returns 2 2; Debug.Print WorksheetFunction.Power(2, 2) returns 4.
Exponentiation works like you expect when you use the caret ^ in 32-bit Excel VBA. It fails to work as expected only in 64-bit Excel VBA.
In 64-bit Excel VBA, the caret is the type declaration character for the LongLong variable type. So when you enter a number followed by a caret or a variable name followed by the caret, VBA assumes you are talking about a LongLong constant or variable. Having made that assumption, it then tries to parse the expression and will give you either a run-time error or an unexpected result.
If you surround your constant or variable in parentheses like (a)^b or (2)^3, you may then use the caret to perform exponentiation in 64-bit Excel VBA. That is my recommended practice as it works as expected in 32-bit Excel VBA, too.
You may also notice that doubling the caret appears to exponentiate correctly--but that is only true for whole numbers. The first caret converts your whole number into a LongLong, while the second one performs the exponentiation. But you can't count on this behavior in general, because 2.3^^3 will raise a run-time error (because 2.3 can't be a LongLong). And if your code is used in both 32-bit and 64-bit Excel, doubling the caret will produce a run-time error in 32-bit Excel. For these reasons, I recommend against the practice of doubling the caret.
I have made a program to solve a quadratic equation, and it works fine except for when the part in the square root is less than 0. When this happens I want it to display that there are no roots. This is the code I am using at the moment.
SquareRoot = System.Math.Sqrt((B ^ 2) - (4 * A * C))
If SquareRoot < 0 Then
lblRoot1.Text = "No Root"
lblRoot2.Text = "No Root"
However despite this if I run the program where this is the case, it crashes. How do I fix this? (Please can you keep the answer as simple as possible as I am still a beginner)
Your problem is that you are checking the wrong thing for negative value. The square root of a negative number is imaginary, and that is the error you get when your code crashes. Instead, move the check to the determinant, like so (this is pseudo-code only):
IF (B * B) - (4 * A * C) < 0
lblRoot1.Text = "No Root"
lblRoot2.Text = "No Root"
ELSE
SquareRoot = System.Math.Sqrt((B * B) - (4 * A * C))
For convenience you may want to create a variable to hold the determinant calculation like so:
Dim D As Integer = (B * B) - (4 * A * C) 'Use appropriate datatype instead of Integer
Why not simply use temporary variable:
Dim value = (B ^ 2) - (4 * A * C)
If value < 0 Then
' error handling
Else
Dim SquareRoot = System.Math.Sqrt(value)
End If
When you debug your program, you should see that the exception occurs in line
SquareRoot = System.Math.Sqrt((B ^ 2) - (4 * A * C))
The reason is that you are not allowed to use a negative input value to Sqrt (there is no "negative" square root, cannot be mathematically) so you need to do something like this:
Input = (B ^ 2) - (4 * A * C)
If Input >= 0 Then
SquareRoot = System.Math.Sqrt(Input)
Else
lblRoot1.Text = "No Root"
lblRoot2.Text = "No Root"