Arc Degrees

[rex64]

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

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

[jemidiah]

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))

[rex64]

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

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

[Logophobic]

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)

[rex64]

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

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

[Logophobic]

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).