Calculating the angle of a Line in a circle(resolved)

Hi all,

I've got some complicated physics equations I need to use in VB, and I need to figure out the angle of a line, from 0° to 180°...


Here is a screen shot of the line:

what data are you given? startpoint and end point or line?

yes? you have the starting point, ending point of the line, it's X1, X2, Y1, and Y2...

VB Code:

Private Function GetAngle(X1 As Long, Y1 As Long, X2 As Long, Y2 As Long) As Double
    Dim PI As Double
    Dim theta As Double
    PI = Atn(1) * 4
    'get angle of line in radians
    'deltay/deltax = tan(angle)
    theta = Atn((Y2 - Y1) / (X2 - X1))
    'convert to degrees
    'there are 180 degrees per PI radians
    GetAngle = theta * 180 / PI
End Function

Quote:

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Originally Posted by moeur

VB Code:

Private Function GetAngle(X1 As Long, Y1 As Long, X2 As Long, Y2 As Long) As Double
    Dim PI As Double
    Dim theta As Double
    PI = Atn(1) * 4
    'get angle of line in radians
    'deltay/deltax = tan(angle)
    theta = Atn((Y2 - Y1) / (X2 - X1))
    'convert to degrees
    'there are 180 degrees per PI radians
    GetAngle = theta * 180 / PI
End Function

---

umm, what is Atn function? sorry i forgot.

got it, inverse of TANGENT!!
:ehh: am i right??

Atn is called ArcTangent (Gupta is correct)
Atn(x) is the angle who's Tangent is x

Quote:

---

Originally Posted by moeur

Atn is called ArcTangent (Gupta is correct)
Atn(x) is the angle who's Tangent is x

---

mouer, sorry this time you confused me. is Atn different from Atn(x)??

No, they're the same thing. He was just using it in a sentence.

Quote:

---

Originally Posted by mendhak

No, they're the same thing. He was just using it in a sentence.

---

thnx moeur, mendhak

Moeur,

I dont think this is working correctly because this happens:

that should be 180 degrees

What are your X1, Y1 and X2,Y2?

here they are



they can change when the user clicks and drags the line:

VB Code:

Private Sub Form_MouseMove(Button As Integer, Shift As Integer, X As Single, Y As Single)
    If Moving Then
        Line1.X2 = X
        Line1.Y2 = Y
        angle = GetAngle(Line1.Y2, Line1.Y1, Line1.X2, Line1.X1)
        lblDegrees.Caption = angle & "°"
    End If
End Sub

Try putting the parameters in the right order ;)

VB Code:

angle = GetAngle(Line1.X1, Line1.Y1, Line1.X2, Line1.Y2)

yeah that'd probably help :wave:

ok now it works better, but on the right side of 90 degrees it goes negatively...

use the abs() ? even though it shouldnt return negative...

Or swap x2 and x1 of x2>x1, and the same for y

Quote:

---

Originally Posted by |2eM!x

use the abs() ? even though it shouldnt return negative...

---

the negative is not the only problem, when you move the line to the right of the 90 degrees it should return numbers from 91 - 180 degrees...

Quote:

---

Originally Posted by dglienna

Or swap x2 and x1 of x2>x1, and the same for y

---

I tried this also and I get the same exact thing for whichever I try :confused:

VB Code:

Private Function GetAngle(X1 As Long, Y1 As Long, X2 As Long, Y2 As Long) As Double
    Dim PI As Double
    Dim theta As Double
    Dim deltaX As Long
    Dim deltaY As Long
    'get angle of line in radians
    'deltay/deltax = tan(angle)
    deltaY = Y1 - Y2
    deltaX = X2 - X1
    PI = Atn(1) * 4
    'adjust to our coordinate system
        If deltaX = 0 Then
            'fix division by zero errors
            If Y1 > Y2 Then
                theta = 180
            Else
                theta = 90
            End If
        ElseIf deltaX > 0 And deltaY >= 0 Then '1st quadrant
            theta = Atn((Y1 - Y2) / (X2 - X1))
        ElseIf deltaX < 0 And deltaY >= 0 Then '2nd quadrant
            theta = Atn((Y1 - Y2) / (X2 - X1)) + PI
        ElseIf deltaX < 0 And deltaY < 0 Then '3rd quadrant
            theta = Atn((Y1 - Y2) / (X2 - X1)) + PI
        ElseIf deltaX > 0 And deltaY < 0 Then '4th quadrant
            theta = Atn((Y1 - Y2) / (X2 - X1)) + 2 * PI
        End If
    
    'convert to degrees
    'there are 180 degrees per PI radians
    'if theta < 0 then theta =
    GetAngle = theta * 180 / PI
End Function

great thanks moeur :thumb:

moeur, could you possible explain that? I thought that it had something to do with quadrants, but didn't want to try to guess at the answer. If not, that's ok, also. I was thinking that 0 degrees was at 9'oclock, and that was the issue. Maybe I was close.

The problem is how the ArcTangent is defined
-90° < Atn(x) < 90°
for
-infinity < x < infinity

But that is not what we wanted so I had to adjust

Well, I wish that I could have seen an explanation, but I found this link, and it didn't help much.

http://mathworld.wolfram.com/InverseTangent.html

I looked up tangent, and saw the same images.

http://mathworld.wolfram.com/Tangent.html

I guess I don't know what I'm looking at.

I used to know the formula to draw a circle with COS and SIN, so thought this wouldn't be too hard to grasp