Arc Degrees

I tried all kinds of things to calculate the Arc degrees. And did lots of research. Anyways, I made a diagram and I will post my code attempt as well. Any assistance with the getDegreesArcText function would be great :)

vb Code:

Private Shared Function getDegreesArcText(g As Graphics, text As String, font As Font, circleRadius As Double) As Double
        Dim degrees As Double
        Dim linearSize As SizeF
        linearSize = g.MeasureString(text, font)
 
        'TODO: use curved line to calculate
        'Dim thePoint As Point = clsJoystickSlim.DegreesToPoint(New Point(0, 0), circleRadius, 10)
        'Dim arcLineSize As Double = joystick.findAngle(New Point(0, 0), New Point(circleRadius, 0), New Point(circleRadius, linearSize.Width)) 'thePoint)
 
 
        degrees = 'linearSize.Width
        Return degrees
    End Function
 
    Public Function findAngle(p1 As Point, p2 As Point, p3 As Point) As Double
        Dim a_x As Double = p2.X - p1.X
        Dim a_y As Double = p2.Y - p1.Y
 
        Dim b_x As Double = 1.0
        Dim b_y As Double = 0.0
 
        Return Math.Acos((a_x * b_x + a_y * b_y) / Math.Sqrt(a_x * a_x + a_y * a_y))
    End Function

I'll answer the questions in your diagram. It shouldn't be hard to convert to code.

We're given a circle centered at (CX, CY) of radius R. There is an arc starting at (CX + R, CY) going counterclockwise a length L. The arc's angle D is related to the other parameters by L = DR, where D is measured in radians. [For instance, if L = 2*pi*R, L is the entire circumference, forcing D = 2*pi.] That is, D = L/R. The coordinates are then just

(E, F) = (CX + R*cos(D), CY + R*sin(D))

Maybe I missed something, but how do I account for curvedLength (5 in diagram) (the length of the arc in units, not degrees)?

vb Code:

Public Function findAngle(circleCenter As Point, radius As Double, curvedLength As Double)
        'L = 2*pi*R, L is the entire circumference
        Dim L As Double = 2 * Math.PI * radius
 
        'The arc's angle A is related to the other parameters by L = AR. forcing A = 2*pi. So A = L/R
        Dim A As Double = L / radius
 
        Dim arcDegrees As Double = 0 'WHAT DO I DO HERE?????
 
        Dim e As Double = circleCenter.X + radius * Math.Cos(A)
        Dim f As Double = circleCenter.Y + radius * Math.Sin(A)
 
        'We're given a circle centered at (CX, CY) of radius R. There is an arc starting at (CX + R, CY) going counterclockwise a
        'length L. The arc's angle A is related to the other parameters by L = AR, where A is measured in radians. [For instance, 
        'if L = 2*pi*R, L is the entire circumference, forcing A = 2*pi.] That is, A = L/R. The coordinates are then just
 
        '(E, F) = (CX + R*cos(A), CY + R*sin(A)) 
        Return A
    End Function

Angle(degrees):360 degrees::ArcLenth:Circumference

A/360 = L/C

jemidiah gave the formula with radians:

Angle(radians):2*pi radians::ArcLenth:Circumference

A/(2*pi) = L/C

Since the circumference, C, is 2*pi*Radius, this simplifies to

A = L/R

If you need the measure in degrees, mutliply by 180/pi

A = (180*L)/(pi*R)

I do not see how you get curvedLength included in this formula. Is L supposed to be curvedLength? And then how does curvedLength fit into this: Dim L As Double = 2 * Math.PI * radius

vb Code:

Public Function findAngle(circleCenter As Point, radius As Double, curvedLength As Double)
        'L = 2*pi*R, L is the entire circumference
        Dim L As Double = 2 * Math.PI * radius
 
        'The arc's angle A is related to the other parameters by L = AR. forcing A = 2*pi. So A = L/R
        Dim A As Double = L / radius '
 
        Dim arcDegrees As Double = (180 * L) / (Math.PI * radius)
 
        Dim e As Double = circleCenter.X + radius * Math.Cos(A)
        Dim f As Double = circleCenter.Y + radius * Math.Sin(A)
 
        'We're given a circle centered at (CX, CY) of radius R. There is an arc starting at (CX + R, CY) going counterclockwise a
        'length L. The arc's angle A is related to the other parameters by L = AR, where A is measured in radians. [For instance, 
        'if L = 2*pi*R, L is the entire circumference, forcing A = 2*pi.] That is, A = L/R. The coordinates are then just
 
        '(E, F) = (CX + R*cos(A), CY + R*sin(A)) 
        Return A
    End Function

Yes, L is the curve length.

(2 * Math.Pi * radius) is the circumference. jemidiah gave that only as an example. It is not part of the calculation.

The angle, measured in radians, is CurveLength/Radius. You need radians for the Sin and Cos functions to determine the point (E,F).