Solved

Function to find X Y coordinates of an arcs center point

Posted on 2010-08-24
6
2,040 Views
Last Modified: 2012-05-10
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.

Please help,
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

Open in new window

Arc.JPG
0
Comment
Question by:bguenther
6 Comments
 
LVL 10

Expert Comment

by:Jini Jose
Comment Utility
0
 
LVL 13

Expert Comment

by:Carl Bohman
Comment Utility
Note that there are two possible centers, not just one.  This page explains why:

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

Expert Comment

by:tyogerst
Comment Utility
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.
radiusendpoint(x4,y4):
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
IT, Stop Being Called Into Every Meeting

Highfive is so simple that setting up every meeting room takes just minutes and every employee will be able to start or join a call from any room with ease. Never be called into a meeting just to get it started again. This is how video conferencing should work!

 

Author Comment

by:bguenther
Comment Utility
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, _

                               ByRef dbArcRadius 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



        'Get the arc radius

        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

Open in new window

0
 

Author Comment

by:bguenther
Comment Utility
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
 

Accepted Solution

by:
bguenther earned 0 total points
Comment Utility
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.
            If sngMidpointDistance > Radius Then
                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.
            Dim rCircle As New RectangleF(pArcMidpoint.X - Radius, (pArcMidpoint.Y - Radius), Radius * 2, Radius * 2)

            ' 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

Open in new window

0

Featured Post

Threat Intelligence Starter Resources

Integrating threat intelligence can be challenging, and not all companies are ready. These resources can help you build awareness and prepare for defense.

Join & Write a Comment

Suggested Solutions

Introduction As chip makers focus on adding processor cores over increasing clock speed, developers need to utilize the features of modern CPUs.  One of the ways we can do this is by implementing parallel algorithms in our software.   One recent…
Foreword (May 2015) This web page has appeared at Google.  It's definitely worth considering! https://www.google.com/about/careers/students/guide-to-technical-development.html How to Know You are Making a Difference at EE In August, 2013, one …
When you create an app prototype with Adobe XD, you can insert system screens -- sharing or Control Center, for example -- with just a few clicks. This video shows you how. You can take the full course on Experts Exchange at http://bit.ly/XDcourse.
This video demonstrates how to create an example email signature rule for a department in a company using CodeTwo Exchange Rules. The signature will be inserted beneath users' latest emails in conversations and will be displayed in users' Sent Items…

744 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

17 Experts available now in Live!

Get 1:1 Help Now