I'm trying to make a line with a fixed x1/y1, and when the user moves the cursor, its other end goes to the cursor. "cannon" is the line. When I try to use it, the line does indeed follow the cursor, but it's got a large offset and is inverted. I'm not really sure what's wrong with it, but it's probably very obvious.
Dim angle as Double = Math.Atan(Math.Abs(Cursor.Position.Y - cannon.Y1) / Math.Abs(Cursor.Position.X - cannon.X1))
Dim h As Double = Math.Sqrt((Math.Abs(Cursor.Position.X - cannon.X1) ^ 2) + Math.Abs(Cursor.Position.Y - cannon.Y1) ^ 2)
Dim a As Double = h * Math.Cos(angle)
Dim o As Double = h * Math.Sin(angle)
cannon.X2 = cannon.X1 + a
cannon.Y2 = cannon.Y1 - o
I suspect that Cursor.Position gives the mouse coordinate relative to the top left corner of your monitor - as does MousePosition.Try this..
PointToClient is a function that calculates the position of the mouse relative to your program window.
Dim angle as Double = Math.Atan(Math.Abs(Me.PointToClient(MousePosition).Y - cannon.Y1) / Math.Abs(Me.PointToClient(MousePosition).X- cannon.X1))
Dim h As Double = Math.Sqrt((Math.Abs(Me.PointToClient(MousePosition).X - cannon.X1) ^ 2) + Math.Abs(Me.PointToClient(MousePosition).Y - cannon.Y1) ^ 2)
Dim a As Double = h * Math.Cos(angle)
Dim o As Double = h * Math.Sin(angle)
cannon.X2 = cannon.X1 + a
cannon.Y2 = cannon.Y1 - o
Related
I'm trying to calculate distance in kilometers between two geographical coordinates using the haversine formula.
Code:
Dim dbl_dLat As Double
Dim dbl_dLon As Double
Dim dbl_a As Double
dbl_P = WorksheetFunction.Pi / 180
dbl_dLat = dbl_P * (dbl_Latitude2 - dbl_Latitude1)
dbl_dLon = dbl_P * (dbl_Longitude2 - dbl_Longitude1)
dbl_a = Sin(dbl_dLat / 2) * Sin(dbl_dLat / 2) + Cos(dbl_Latitude1 * dbl_P) * Cos(dbl_Latitude2 * dbl_P) * Sin(dbl_dLon / 2) * Sin(dbl_dLon / 2)
dbl_Distance_KM = 6371 * 2 * WorksheetFunction.Atan2(Sqr(dbl_a), Sqr(1 - dbl_a))
I'm testing with these coordinates:
dbl_Longitude1 = 55.629178
dbl_Longitude2 = 29.846686
dbl_Latitude1 = 37.659466
dbl_Latitude2 = 30.24441
And the code returns 20015.09, which is obviously wrong. It should be 642 km according to Yandex maps.
Where am I wrong? Are the longitude and latitude in wrong format?
As far as I can tell, the issue is that the order of arguments to atan2() varies by language. The following works* for me:
Option Explicit
Public Sub Distance()
Dim dbl_Longitude1 As Double, dbl_Longitude2 As Double, dbl_Latitude1 As Double, dbl_Latitude2 As Double
dbl_Longitude1 = 55.629178
dbl_Longitude2 = 29.846686
dbl_Latitude1 = 37.659466
dbl_Latitude2 = 30.24441
Dim dbl_dLat As Double
Dim dbl_dLon As Double
Dim dbl_a As Double
Dim dbl_P As Double
dbl_P = WorksheetFunction.Pi / 180
dbl_dLat = dbl_P * (dbl_Latitude2 - dbl_Latitude1) 'to radians
dbl_dLon = dbl_P * (dbl_Longitude2 - dbl_Longitude1) 'to radians
dbl_a = Sin(dbl_dLat / 2) * Sin(dbl_dLat / 2) + _
Cos(dbl_Latitude1 * dbl_P) * Cos(dbl_Latitude2 * dbl_P) * Sin(dbl_dLon / 2) * Sin(dbl_dLon / 2)
Dim c As Double
Dim dbl_Distance_KM As Double
c = 2 * WorksheetFunction.Atan2(Sqr(1 - dbl_a), Sqr(dbl_a)) ' *** swapped arguments to Atan2
dbl_Distance_KM = 6371 * c
Debug.Print dbl_Distance_KM
End Sub
*Output: 2507.26205401321, although gcmap.com says the answer is 2512 km. This might be a precision issue --- I think it's close enough to count as working. (Edit it might also be that gcmap uses local earth radii rather than the mean radius; I am not sure.)
Explanation
I found this description of the haversine formula for great-circle distance, which is what you are implementing. The JavaScript implementation on that page gives this computation for c:
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
In JavaScript, atan2() takes parameters y, x. However, in Excel VBA, WorksheetFunction.Atan2 takes parameters x, y. Your original code passed Sqr(dbl_a) as the first parameter, as it would be in JavaScript. However, Sqr(dbl_a) needs to be the second parameter in Excel VBA.
A comment on naming
Building on #JohnColeman's point, there are lots of ways to name variables. In this case, I would recommend using the prefixes for unit rather than for type: e.g., deg_Latitude1, RadPerDeg = Pi/180, and rad_dLat = RadPerDeg * (deg_Latitude2 - deg_Latitude1). I personally think that helps avoid unit-conversion mishaps.
My VBA code that returns the answer in feet; However 'd' is the answer in kilometers.
Imports System.Math
Module Haversine
Public Function GlobalAddressDistance(sLat1 As String, sLon1 As String, sLat2 As String, sLon2 As String) As String
Const R As Integer = 6371
Const cMetersToFeet As Single = 3.2808399
Const cKiloMetersToMeters As Integer = 1000
Dim a As Double = 0, c As Double = 0, d As Double = 0
'Convert strings to numberic double values
Dim dLat1 As Double = Val(sLat1)
Dim dLat2 As Double = Val(sLat2)
Dim dLatDiff As Double = DegreesToRadians(CDbl(sLat2) - CDbl(sLat1))
Dim dLonDiff As Double = DegreesToRadians(CDbl(sLon2) - CDbl(sLon1))
a = Pow(Sin(dLatDiff / 2), 2) + Cos(DegreesToRadians(dLat1)) * Cos(DegreesToRadians(dLat2)) * Pow(Sin(dLonDiff / 2), 2)
c = 2 * Atan2(Sqrt(a), Sqrt(1 - a))
d = R * c
'Convert kilometers to feet
Return Format((d * cKiloMetersToMeters * cMetersToFeet), "0.##").ToString
End Function
Private Function DegreesToRadians(ByVal dDegrees As Double) As Double
Return (dDegrees * PI) / 180
End Function
End Module
Introduction
I am somewhat new to using Open GL / Open TK. I have learned how to draw basic shapes, use matrices, lighting, shadowing, etc. I have a function that draws a sphere:
Private Sub drawSphere(r As Double, lats As Integer, longs As Integer)
Dim i As Integer, j As Integer
For i = 0 To lats
Dim lat0 As Double = PI * (-0.5 + CDbl(i - 1) / lats)
Dim z0 As Double = Sin(lat0)
Dim zr0 As Double = Cos(lat0)
Dim lat1 As Double = PI * (-0.5 + CDbl(i) / lats)
Dim z1 As Double = Sin(lat1)
Dim zr1 As Double = Cos(lat1)
GL.Begin(PrimitiveType.QuadStrip)
For j = 0 To longs
Dim lng As Double = 2 * PI * CDbl(j - 1) / longs
Dim x As Double = Cos(lng)
Dim y As Double = Sin(lng)
GL.Normal3(x * zr0 * r, y * zr0 * r, z0 * r)
GL.Vertex3(x * zr0 * r, y * zr0 * r, z0 * r)
GL.Normal3(x * zr1 * r, y * zr1 * r, z1 * r)
GL.Vertex3(x * zr1 * r, y * zr1 * r, z1 * r)
Next
GL.End()
Next
End Sub
I have other code that sets up the lights. I know the other code works because I have a separate function for drawing an STL object:
Dim texture As UInteger() = New UInteger(0) {}
Dim i As Integer = 0
If stl_table.Items.Count > 0 Then
find_center_of_part()
GL.Begin(PrimitiveType.Triangles)
GL.Color3(part_color.R, part_color.G, part_color.B)
Do Until i + 4 >= stl_table.Items.Count
GL.Normal3(Convert.ToSingle(stl_table.Items.Item(i).SubItems(0).Text), Convert.ToSingle(stl_table.Items.Item(i).SubItems(1).Text), Convert.ToSingle(stl_table.Items.Item(i).SubItems(2).Text))
GL.Vertex3(stl_table.Items.Item(i + 1).SubItems(0).Text - avgx, stl_table.Items.Item(i + 1).SubItems(1).Text - avgy, stl_table.Items.Item(i + 1).SubItems(2).Text - avgz)
GL.Vertex3(stl_table.Items.Item(i + 2).SubItems(0).Text - avgx, stl_table.Items.Item(i + 2).SubItems(1).Text - avgy, stl_table.Items.Item(i + 2).SubItems(2).Text - avgz)
GL.Vertex3(stl_table.Items.Item(i + 3).SubItems(0).Text - avgx, stl_table.Items.Item(i + 3).SubItems(1).Text - avgy, stl_table.Items.Item(i + 3).SubItems(2).Text - avgz)
i = i + 4
Loop
GL.End()
End If
This second function basically imports a CAD STL file and draws it as triangles. The normal vectors are simply an input from the CAD file (so they are already computed). This method's lighting works perfectly fine which makes me know my lighting code is correct.
Problem
The problem is that my sphere is not getting light correctly. I know through testing that this is due to my normal vectors.
With the current code, my sphere looks like this:
There is a "spot" of light which makes me think that is simply one of the quadstrips having the normal correct.
Does anybody have any suggestions on setting up the normal vectors correctly inside my function? Also before anybody suggests it, I can't use GLU or GLUT for what I am trying to accomplish, which is why I need the sphere function.
I am making an arc with a triangle fan. The triangle fan has to have set points for each of the vertices to make the arc shape. I have found a multitude of docs regarding DrawArc, but that is not what I am after, and cannot find anything on creating "x number of points" across the arc from point A to point B.
It has been several years since my last trig class, so I am hoping someone has an idea of how to increment the x/y location of the points between A-B. Here is what I have so far:
Dim points As Integer = 5 ' the number of points between top and right
Dim radius as Integer = 25
' Center point
Dim cx As Integer = loc.X + (size.Width - (radius))
Dim cy As Integer = loc.Y + thickness
' Top point
Dim x1 As Integer = loc.X + (size.Width - (radius))
Dim y1 As Integer = loc.Y
' Right point
Dim x2 As Integer = loc.X + (size.Width)
Dim y2 As Integer = loc.Y + radius
Dim trifan As New VertexArray(PrimitiveType.TrianglesFan)
trifan .Append(New Vertex(New Vector2f(cx, cy), col2)) ' Center point
trifan .Append(New Vertex(New Vector2f(x1, y1), col1)) ' Top point
For i = 1 To points
' append other points here...
Next
trifan .Append(New Vertex(New Vector2f(x2, y2), col1)) ' Right point
I'm just posting this here so it will have an answer for others who come across it. I had a lot of help with this (as recommended) here: https://math.stackexchange.com/questions/1789110/plot-points-on-an-arc
For i = 0 To points - 1
Dim x As Double
Dim y As Double
Dim t As Double = (PI / 2) * (i / points - 1)
x = cx + Cos(t) * (radius)
y = cy + Sin(t) * (radius)
trifan1.Append(New Vertex(New Vector2f(x, y), col1))
Next
I found a great script to arrange objects (shapes) into a circle here:
Aligning Shapes in a Circle using VBA, Microsoft Community
Sub Test()
Call AlignShapesInCircle(720 / 2, 540 / 2, 100, ActiveWindow.Selection.ShapeRange)
End Sub
Function AlignShapesInCircle(x As Single, y As Single, r As Single, shprng As ShapeRange)
'x,y = center point of the circle
'r = radius of the circle
'shprng = the shape selection that needs to be arranged
Dim angle As Single
Dim currentangle As Single
Dim x1 As Single
Dim y1 As Single
Dim i As Integer
currentangle = 0
angle = 360 / shprng.count
For currentangle = 0 To 359 Step angle
i = i + 1
x1 = r * Cos(D2R(currentangle))
y1 = r * Sin(D2R(currentangle))
shprng(i).Left = x + x1
shprng(i).Top = y + y1
Next
End Function
Function D2R(Degrees) As Double
D2R = Degrees / 57.2957795130823
End Function
Function R2D(Radians) As Double
R2D = 57.2957795130823 * Radians
End Function
Now I want the shapes to rotate so that if I use arrows the tip will always show towards the center.
I have to introduce a line here:
shprng(i).Left = x + x1
shprng(i).Top = y + y1
shprng(i).Rotation = ???
Any ideas where I could find the proper formula?
Silly - figured it out - it was easier than I thought. Don't need any SIN and COS which frightened me - just:
shprng(i).Rotation = (360 / (shprng.Count)) * (i - 1)
When I generate random numbers, I sometimes get (not always) the following error:
Run-time error '13': type mismatch.
on line Z = Sqr(time) * Application.NormSInv(Rnd()) (and the end of the second for loop).
Why do I get this error?
I think it has something to do with the fact that it contains Rnd().
Sub asiancall()
'defining variables
Dim spot As Double
Dim phi As Integer
Dim rd_cont As Double
Dim rf_cont As Double
Dim lambda As Double
Dim muY As Double
Dim sigmaY As Double
Dim vol As Double
Dim implied_vol As Double
Dim spotnext As Double
Dim time As Double
Dim sum As Double
Dim i As Long
Dim mean As Double
Dim payoff_mean As Double
Dim StDev As Double
Dim K As Double
Dim Egamma0 As Double
Dim mulTv As Double
Dim prod As Double
Dim U As Double
Dim Pois As Double
Dim Q As Double
Dim Z As Long
Dim gamma As Double
Dim payoff As Double
Dim payoff_sum As Double
Dim secondmoment As Double
Dim j As Long
Dim N As Long
Dim mu As Double
Dim sum1 As Double
'read input data
spot = Range("B3")
rd_cont = Range("C5")
rf_cont = Range("C4")
muY = Range("B17")
sigmaY = Range("B18")
lambda = Range("B16")
K = Range("F33")
implied_vol = Range("F35")
N = Range("F34")
vol = Range("B6")
'calculations
sum_BS = 0
payoff_BS = 0
mean_BS = 0
secondmoment_BS = 0
For j = 1 To N
spotnext = spot
spotnext_BS = spot
time = 0
sum1 = 0
time = 184 / (360 * 6)
For i = 1 To 6
' 'Merton uitvoeren
Egamma0 = Exp(muY + sigmaY * sigmaY * 0.5) - 1
mu = rd_cont - rf_cont
mulTv = (mu - lambda * Egamma0 - implied_vol * implied_vol * 0.5) * time
sum = 0
prod = 1
Do While sum <= time
U = Rnd()
Pois = -Log(U) / lambda
sum = sum + Pois
Q = Application.NormInv(Rnd(), muY, sigmaY)
gamma = Exp(Q) - 1
prod = prod * (1 + gamma)
Loop
prod = prod / (1 + gamma)
Z = Sqr(time) * Application.NormSInv(Rnd())
spotnext = spotnext * Exp(mulTv + implied_vol * Z) * prod
sum1 = sum1 + spotnext
Next i
mean = sum1 / 6
payoff = Application.Max(mean - K, 0)
payoff_sum = payoff_sum + payoff
secondmoment = secondmoment + payoff * payoff
Next j
Following up on the community wiki answer I posted, a possible solution is this:
Function RndExcludingZero()
Do
RndExcludingZero = Rnd()
Loop While RndExcludingZero = 0
End Function
Usage:
Z = Sqr(time) * Application.NormSInv(RndExcludingZero())
Rnd() returns values >=0 and <1.
At some point it is bound to return 0. When given 0 as input in Excel, NormSInv returns the #NUM!
Excel error.* When called in VBA via Application.NormSInv(0), it returns a Variant of subtype Error with value "Error 2036" (equivalent to the #NUM! Excel error).
Such Variant/Errors cannot be implicitly coerced to a numerical value (which is what the * operator expects) and thus in this case, you will get the type mismatch error.
You will only get this error when Rnd() happens to return 0, which is consistent with your observation that the error occurs only sometimes.
* This was first remarked by user3964075 in a now defunct comment to the question.