• Status: Solved
• Priority: Medium
• Security: Public
• Views: 3002

# Function to find X Y coordinates of an arcs center point

Hi,
I am trying to create a function in VB.net to return to me the xy coordinates of where the center point of any arc is.
I am very poor with math so this is really over my head. I did a lot of searching but could not find anything close enough so that i could make it work.

I know the arc start point ,arc end point and the radius.

Regards,
Ben
``````Private Sub GetCenterPoints(ByVal dbArcStartX As Double, ByVal dbArStartY As Double, _
ByVal dbArcEndX As Double, ByVal dbArcEndY As Double, ByVal dbRadius As Double)

Dim dbCenterPointX As Double
Dim dbCenterPointY As Double

MessageBox.Show("The center point = X " & dbCenterPointX & " Y " & dbCenterPointY)
End Sub
``````
Arc.JPG
0
bguenther
1 Solution

Senior .Net DeveloperCommented:
0

Commented:
Note that there are two possible centers, not just one.  This page explains why:

http://mathforum.org/library/drmath/view/54490.html
0

Commented:
http://en.wikipedia.org/wiki/Versine

Find the midpoint between your start(x1,y1) and end(x2,y2) points:
midpoint(x3,y3) = (x1 + x2)/2 , (y1 + y2)/2

the slope of the versine segment will be perpendicular to the line between the start and end points
slope = -(y2-y1)/(x2-x1)

starting at the center of the circle, calculate the endpoint of a radius segment that passes through the midpoint with the same slope as the versine.
x4 = sqrt(r^2 / (1+slope^2))
y4 = x4*slope

Assumptions: a clockwise path between start and end points, circle centered at (0,0) - if not, then just apply a linear offset before and after your calculation
0

Author Commented:
Thanks for the reply's, I will play with this and see if i can get it going.

What i am trying to achieve here is by knowing start and end point and radius and whether it is clockwise or counter clockwise is to get the start angle and end angle of the arc.

As for my original question "How to get the center point of the arc" is it possible to get just one possible to get just one center with the specs that i have?

The code that i have pasted below is the function that i was going to use once i had the center point of the arc. It only works on arcs under 180 degrees. So it is useless to me because my arcs will be all over the place.
It also does not work quite right it does not figure out the end angle properly.

I am not sure if this is allowed but I would like to rephrase  my original question. How to get the start and end angle of an arc. knowing start and end point and radius and whether it is clockwise or counter clockwise

Regards,
Ben
``````Private Shared Sub GetCirlce(ByVal dbArcStartX As Double, ByVal dbArStartY As Double, _
ByVal dbArcEndX As Double, ByVal dbArcEndY As Double, _
ByVal dbArcMidX As Double, ByVal dbArcMidY As Double, _
ByRef dbArcCenterX As Double, ByRef dbArcCenterY As Double, _
ByRef dbArcStartAngle As Double, ByRef dbArcEndAngle As Double, _

Dim dbUnknown1 As Double
Dim dbUnknown2 As Double
Dim dbUnknown3 As Double
Dim dbUnknown4 As Double
Dim dbUnknown5 As Double
Dim dbUnknown6 As Double

'Unknown
dbUnknown2 = (2 * dbArcStartX * (2 * dbArcEndY - 2 * dbArcMidY) - 2 * dbArStartY * (2 * dbArcEndX - 2 * dbArcMidX) + 4 * dbArcEndX * dbArcMidY - 4 * dbArcEndY * dbArcMidX)

'Get the arc center point for X
dbArcCenterX = ((dbArcStartX ^ 2 + dbArStartY ^ 2) * (2 * dbArcEndY - 2 * dbArcMidY) - 2 * dbArStartY * (dbArcEndX ^ 2 + dbArcEndY ^ 2 - dbArcMidX ^ 2 - dbArcMidY ^ 2) + 2 * dbArcMidY * (dbArcEndX ^ 2 + dbArcEndY ^ 2) - 2 * dbArcEndY * (dbArcMidX ^ 2 + dbArcMidY ^ 2)) / dbUnknown2
'Get the arc center point for X
dbArcCenterY = (2 * dbArcStartX * (dbArcEndX ^ 2 + dbArcEndY ^ 2 - dbArcMidX ^ 2 - dbArcMidY ^ 2) - (dbArcStartX ^ 2 + dbArStartY ^ 2) * (2 * dbArcEndX - 2 * dbArcMidX) + 2 * dbArcEndX * (dbArcMidX ^ 2 + dbArcMidY ^ 2) - 2 * dbArcMidX * (dbArcEndX ^ 2 + dbArcEndY ^ 2)) / dbUnknown2
'Unknown
dbUnknown1 = (2 * dbArcStartX * (2 * dbArcEndY * (dbArcMidX ^ 2 + dbArcMidY ^ 2) - 2 * dbArcMidY * (dbArcEndX ^ 2 + dbArcEndY ^ 2)) - 2 * dbArStartY * (2 * dbArcEndX * (dbArcMidX ^ 2 + dbArcMidY ^ 2) - 2 * dbArcMidX * (dbArcEndX ^ 2 + dbArcEndY ^ 2)) + (dbArcStartX ^ 2 + dbArStartY ^ 2) * (4 * dbArcEndX * dbArcMidY - 4 * dbArcEndY * dbArcMidX)) / dbUnknown2

dbArcRadius = Math.Sqrt(dbUnknown1 + dbArcCenterX ^ 2 + dbArcCenterY ^ 2)

If dbArcCenterX < dbArcStartX And dbArcCenterY <> dbArStartY Then
dbArcStartAngle = Math.Atan((dbArStartY - dbArcCenterY) / (dbArcStartX - dbArcCenterX)) * 180 / 3.1415927
End If

If dbArcCenterX > dbArcStartX And dbArcCenterY <> dbArStartY Then
dbArcStartAngle = 180 - Math.Atan((dbArStartY - dbArcCenterY) / (dbArcCenterX - dbArcStartX)) * 180 / 3.1415927
End If

If dbArcCenterX = dbArcStartX And dbArcCenterY < dbArStartY Then
dbArcStartAngle = 90
End If

If dbArcCenterX < dbArcStartX And dbArcCenterY = dbArStartY Then
dbArcStartAngle = 0
End If

If dbArcCenterX = dbArcStartX And dbArcCenterY > dbArStartY Then
dbArcStartAngle = -90
End If

If dbArcCenterX > dbArcStartX And dbArcCenterY = dbArStartY Then
dbArcStartAngle = 180
End If

dbUnknown3 = 1 - (((dbArcStartX - dbArcMidX) ^ 2 + (dbArStartY - dbArcMidY) ^ 2) / (2 * (dbArcRadius ^ 2)))
If dbUnknown3 <> 1 And dbUnknown3 <> -1 Then
dbArcEndAngle = Math.Atan(-dbUnknown3 / Math.Sqrt(1 - (dbUnknown3 ^ 2))) * 180 / 3.1415927 + 90
End If
If dbUnknown3 = -1 Then
dbArcEndAngle = 180
End If

dbUnknown6 = (dbArcMidX - dbArcStartX)

If dbUnknown6 <> 0 Then
dbUnknown5 = (dbArcMidY - dbArStartY) / dbUnknown6
End If

dbUnknown4 = dbUnknown5 * (dbArcEndX - dbArcStartX) + dbArStartY

If dbArcStartX < dbArcMidX And dbArcEndY > dbUnknown4 Then
dbArcEndAngle = dbArcEndAngle * (-1)
End If

If dbArcStartX > dbArcMidX And dbArcEndY < dbUnknown4 Then
dbArcEndAngle = dbArcEndAngle * (-1)
End If

If dbArcStartX = dbArcMidX And dbArcMidY > dbArStartY And dbArcEndX < dbArcStartX Then
dbArcEndAngle = dbArcEndAngle * (-1)
End If

If dbArcStartX = dbArcMidX And dbArcMidY < dbArStartY And dbArcEndX > dbArcStartX Then
dbArcEndAngle = dbArcEndAngle * (-1)
End If
End Sub
``````
0

Author Commented:
Hi,

Well after more searching and crying below is what i found and am going to settle for, thanks for all your help.
Regards,
Ben
0

Author Commented:
Sorry forgot to attach code
``````Public Function DrawArc(ByVal StartPoint As PointF, ByVal EndPoint As PointF, ByVal Radius As Single, ByVal Clockwise As Boolean, ByRef sngStartAngle As Single, ByRef sngEndAngle As Single)
Try
' Find the point that corresponds to the midpoint of the line.
Dim pLineMidPoint As New PointF((StartPoint.X + EndPoint.X) / 2, (StartPoint.Y + EndPoint.Y) / 2)

' Calculate 1/2 the distance between the start and end point.
Dim sngMidpointDistance As Single = PointDistance(StartPoint, pLineMidPoint)

' If 1/2 between the two points is greater than the radius, then a
' circular segment cannot be draw. Instead, draw a straight line
' between the two points.
Return False
End If

' Find the slope of a line drawn between the start and end point.
Dim sngLineSlope As Single = GetSlope(StartPoint, EndPoint)
' Find the distance between the midpoint of the line and the midpoint
' of the arc.

Dim sngDistanceToCenterPoint As Single = CSng(Math.Sqrt(Radius ^ 2 - sngMidpointDistance ^ 2))

' Find the mid point of the arc.
Dim pOffset As New PointF(0, sngDistanceToCenterPoint)
If Not Clockwise Then pOffset.Y *= -1
Dim pArcMidpoint As PointF = OffsetPointsByAngle(pLineMidPoint, pOffset, sngLineSlope)

' Build a bounding box to hold the arc.

' Find the starting angle.
sngStartAngle = 360 - GetSlope(pArcMidpoint, StartPoint)
' Find the the ending angle.
sngEndAngle = 360 - GetSlope(pArcMidpoint, EndPoint)

If sngEndAngle < sngStartAngle Then sngEndAngle += 360
' Convert the ending angle to the number of degrees between the start
' and end angle.
sngEndAngle = Math.Abs(sngStartAngle - sngEndAngle)
If Not Clockwise Then sngEndAngle = sngEndAngle - 360

Return True
Catch ex As Exception
Return False
End Try
End Function
``````
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.