C# Drawing Arc with 3 Points

Question

Hi, I need to draw an arc using GraphicsPath and having initial, median and final points. The arc has to pass on them.

I tried .DrawCurve and .DrawBezier but the result isn't exactly an arc.

What can I do?

SOLUTION:

After a couple of hours of code writing I managed to draw what I wanted with this algorithm (give 3 Point a,b,c and a GraphicsPath path):

double d = 2 * (a.X - c.X) * (c.Y - b.Y) + 2 * (b.X - c.X) * (a.Y - c.Y);
double m1 = (Math.Pow(a.X, 2) - Math.Pow(c.X, 2) + Math.Pow(a.Y, 2) - Math.Pow(c.Y, 2));
double m2 = (Math.Pow(c.X, 2) - Math.Pow(b.X, 2) + Math.Pow(c.Y, 2) - Math.Pow(b.Y, 2));
double nx = m1 * (c.Y - b.Y) + m2 * (c.Y - a.Y);
double ny = m1 * (b.X - c.X) + m2 * (a.X - c.X);
double cx = nx / d;
double cy = ny / d;
double dx = cx - a.X;
double dy = cy - a.Y;
double distance = Math.Sqrt(dx * dx + dy * dy);
Vector va = new Vector(a.X - cx, a.Y - cy);
Vector vb = new Vector(b.X - cx, b.Y - cy);
Vector vc = new Vector(c.X - cx, c.Y - cy);
Vector xaxis = new Vector(1, 0);
float startAngle = (float)Vector.AngleBetween(xaxis, va);
float sweepAngle = (float)(Vector.AngleBetween(va, vb) + Vector.AngleBetween(vb, vc));
path.AddArc(
    (float)(cx - distance), (float)(cy - distance),
    (float)(distance * 2), (float)(distance * 2), 
    startAngle, sweepAngle);

Answer 1

Have you tried the DrawArc method and seeing if u can manipulate your 3 points somehow?

maybe

Pen blackPen= new Pen(Color.Black, 3);
// Create rectangle to bound ellipse.
Rectangle rect = new Rectangle(initial x, initial y, final x, median y);
// Create start and sweep angles on ellipse.
float startAngle =  0F;
float sweepAngle = 270.0F;
// Draw arc to screen.
e.Graphics.DrawArc(blackPen, rect, startAngle, sweepAngle);

http://msdn.microsoft.com/en-us/library/system.drawing.graphics.drawarc%28VS.71%29.aspx

Answer 2

I would use DrawArc() as suggested by ANC_Michael. To find an arc that passes through 3 points you want to calculate the circumcircle of the triangle formed by the points.

Once you have the circumcircle calculate a bounding box for the circle to use with DrawArc using the min/max dimensions (center +/- radius). Now calculate your start and stop angles by translating the points so that the circumcircle is centered on the origin (translate by -circumcenter) and take the dot-product of the normalized start and end vectors with the X-axis:

double startAngle = Math.Acos(VectorToLeftPoint.Dot(XAxis));
double stopAngle = Math.Acos(VectorToRightPoint.Dot(XAxis));

Note that DrawArc expects angles clockwise from the X-axis so you should add Math.PI if the calculated vector is above the x-axis. That should be enough information to call DrawArc().

Edit: This method will find a circular arc and not necessarily the 'best fit' arc depending on your expected endpoint behavior.

Answer 3

hi keeper we know , Since .Net 2.0 vector is no available. now how can i change your above code ?

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