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Plot/Draw Circle in WindowsForms Chart

Is there any possibility to plot a circle in a WindowsForm Chart?
A method-call as follows would be really nice!
Graph.Series["circle"].Circle.Add(centerX, centerY, radius);

Well, I created myself a work around.
Maybe it helps someone
public void DrawCircle(Chart Graph, double centerX, double centerY, double radius, int amountOfEdges)
{
string name = "circle_" + centerX + centerY + radius + amountOfEdges;
// Create new data series
if (Graph.Series.IndexOf(name) == -1)
Graph.Series.Add(name);
// preferences of the line
Graph.Series[name].ChartType = SeriesChartType.Spline;
Graph.Series[name].Color = Color.FromArgb(0, 0, 0);
Graph.Series[name].BorderWidth = 1;
Graph.Series[name].IsVisibleInLegend = false;
// add line segments (first one also as last one)
for (int k = 0; k <= amountOfEdges; k++)
{
double x = centerX + radius * Math.Cos(k * 2 * Math.PI / amountOfEdges);
double y = centerY + radius * Math.Sin(k * 2 * Math.PI / amountOfEdges);
Graph.Series[name].Points.AddXY(x, y);
}
}
You can call it for example via
DrawCircle(Graph, 5, 4, 3, 30);
Around 30 points should be enough to get a nice circle instead of a polygon, but depends on the size of your chart.

draw an equilateral triangle C#

I have tried to adapt some code I came across to draw an equilateral triangle in c#
public void drawTriangle(PaintEventArgs e, int x, int y, int distance)
{
float angle = 0;
SolidBrush brs = new SolidBrush(Color.Green);
PointF[] p = new PointF[3];
p[0].X = x;
p[0].Y = y;
p[1].Y = (float)( x + distance * Math.Cos(angle + Math.PI / 3));
p[1].X = (float)( y + distance * Math.Sin(angle + Math.PI / 3));
p[2].Y = (float)( x + distance * Math.Cos(angle - Math.PI / 3));
p[2].X = (float)( y + distance * Math.Sin(angle - Math.PI / 3));
e.Graphics.FillPolygon(brs, p);
}
Unfortunately, this doesn't even come close. I have drawn equilateral triangles, but the points were always based on the centers of congruent circles. I am trying to find a simpler way. I am sure there must be an obvious problem with this code, but I am trying to learn the math needed as I go, so I don't know what it is. Thanks for your time.

Try this approach. I assume that for zero angle p[0] is left bottom vertex, p[1] is right bottom (the same horizontal).
(BTW, you have got strange mangling of Y/X)
p[0].X = x;
p[0].Y = y;
p[1].X = (float)( x + distance * Math.Cos(angle));
p[1].Y = (float)( y + distance * Math.Sin(angle));
p[2].X = (float)( x + distance * Math.Cos(angle + Math.PI / 3));
p[2].Y = (float)( y + distance * Math.Sin(angle + Math.PI / 3));

Get a parabola/arc based on vertex and a point

I'm trying to get a character to throw something in an arc at a target.
I know the vertex(x,y) and the target(x,y) and I want to get an arc from the origin(x,y) to the target with a max height of vertex.y
What I have is based off the vertex form of y = a(x-h)^2 + k
public static Vector3 parabola(Vector2 origin, Vector2 target, float height)
{
float dist = target.x - origin.x;
Vector2 vertex = new Vector2(origin.x + (dist / 2), origin.y + height);
//a = (y-k) / (x-h)^2
float a = (target.y - vertex.y) / ((target.x - vertex.x) * (target.x - vertex.x));
//b = (-h + -h) * a
float b = (-vertex.x + -vertex.x) * a;
//c = (h * h) * a + k
float c = (vertex.x * vertex.x) * a + vertex.y;
return new Vector3(a, b, c);
}
x += Time.DeltaTime;
float yPos = a * ((x - h) * (x - h)) + k;
This doesn't produce the correct arc. It's usually much too steep or much too shallow. Is my algebra wrong, or am I using the wrong approach?
Thanks

Here is a good solution: Wiki:Trajectory of a projectile.

Clockwise right hand turn angle in 2D?

If I have two points p1 and p2 where p1 is the pivot point and p2 is the original direction the user was headed and they have a number of possible directions to go p3...pn in random sequence. How do I get the angles between the choices and the segment formed by p1,p2 as clockwise(right hand) positive values between 0 and 360 so that I can sort them from least to greatest?
Also the points p1...pn will be in any quadrant, I can’t assume they will always be in the positive x,y direction. The grid is a standard Cartesian grid not screen coordinates so Y gets smaller as you go down not larger.
So in this example (sorry for the poor drawing but Paint was all I had on my laptop) I need to get the angles:
(p2-p1-p3)
( p2-p1-p4)
( p2-p1-p5)
( p2-p1-p6)
In this order(smallest right hand turn to largest right hand turn):
[( p2-p1-p4), ( p2-p1-p6), ( p2-p1-p5), (p2-p1-p3)]
The points in my case are a class called Vertex:
public class Vertex
{
public double X = 0;
public double Y = 0;
public Vertex() { }
public Vertex(double x, double y)
{
X = x;
Y = y;
}
}
And the code for getting the angles and sorting looks like this right now but has a problem:
private static IEnumerable<Vertex> SortByAngle(Vertex original, Vertex pivot, List<Vertex> choices)
{
choices.Sort((v1, v2) => GetTurnAngle(original, pivot, v1).CompareTo(GetTurnAngle(original, pivot, v2)));
return choices;
}
private static double GetTurnAngle(Vertex original, Vertex pivot, Vertex choice)
{
var a = original.X - pivot.X;
var b = original.Y - pivot.Y;
var c = choice.X - pivot.X;
var d = choice.Y - pivot.Y;
var rads = Math.Acos(((a * c) + (b * d)) / ((Math.Sqrt(a * a + b * b)) * (Math.Sqrt(c * c + d * d))));
return (180 / Math.PI * rads);
}
The problem is the above is if I check it for:
original 66,-66
pivot 280,-191
choice 200,-180
I get an angle of 22.460643124 instead of 337.539356876 which means it went counter-clockwise from the original direction to get that angle. I need it to always go clockwise to get the angle.
What am I doing wrong and how do I fix it?
Update: OK so according to what you guys are saying I can probably use some cross product like math to determine CW vs CCW so the new method would look like this:
private static double GetTurnAngle(Vertex original, Vertex pivot, Vertex choice)
{
var a = original.X - pivot.X;
var b = original.Y - pivot.Y;
var c = choice.X - pivot.X;
var d = choice.Y - pivot.Y;
var angle = Math.Acos(((a * c) + (b * d)) / ((Math.Sqrt(a * a + b * b)) * (Math.Sqrt(c * c + d * d))));
angle = (180 / Math.PI * angle);
var z = (choice.X - pivot.X) * (original.Y - pivot.Y) - (choice.Y - pivot.Y) * (original.X - pivot.X);
if (z < 0)
{
return 360 - angle;
}
return angle;
}
Update 2:
Using the accepted solution it now looks like so:
private static double GetTurnAngle(Vertex original, Vertex pivot, Vertex choice)
{
var angle1 = Math.Atan2(original.Y - pivot.Y, original.X - pivot.X);
var angle2 = Math.Atan2(choice.Y - pivot.Y, choice.X - pivot.X);
var angleDiff = (180 / Math.PI * (angle2 - angle1));
if (angleDiff > 0)//It went CCW so adjust
{
return 360 - angleDiff;
}
return -angleDiff;//I need the results to be always positive so flip sign
}
So far as I can tell that works great so far. Thank you guys for the help!

Take a look at atan2 function. It takes delta y and delta x, so can distinguish all angles.
angle1 = atan2(p1.y-p0.y, p1.x-p0.x);
angle2 = atan2(p2.y-p0.y, p2.x-p0.x);
angle = angle2 - angle1;
If angle is negative, then CW, if positive CCW (or other way around depending on your axis orientation). Note |angle| may be > 180, in which case you may want to do 360-|angle| and reverse the CW CCW conclusion if you're after the shortest route.

You find the Dn=direction from p1 to pn (x=pn.x-p1.x and y=pn.y-p1.y) by the formula:
Dn=f(x,y)=180-90*(1+sign(y))* (1-sign(x^2))-45*(2+sign(y))*sign(x)
-180/pi()*sign(x*y)*atan((abs(y)-abs(x))/(abs(y)+abs(x)))
So the angles are Angle(p2-p1-pn)=Dn-D2.

How do I draw an image based on a simple polygon?

I'd like to copy a roughly rectangular area to a rectangular area. Example:
Both areas are defined by their corner points. The general direction is kept (no flipping etc).
Simply rotating the source image does not work since opposing sides may be of different length.
So far I found no way to do this in pure C# (except manual pixel copying), so I guess I have to resort to the Windows API or some 3rd party library?

Since I could not find an answer, I wrote a naive implementation myself. It works reasonably well.
Examples
I drew all examples manually in Paint, so they are not very exact - it was just enough to test some basics.
a) Slight rotation.
Source:
Result:
b) Various sides
Source:
Result:
c) Perspective
Source:
Result:
Code
(it's specialized to my use case, but it should be easy to adapt):
// _Corners are, well, the 4 corners in the source image
// _Px is an array of pixels extracted from the source image
public void Rescale ()
{
RescaleImage (
_Corners[0],
_Corners[1],
_Corners[3],
_Corners[2],
100,
100);
}
private void RescaleImage (PointF TL, PointF TR, PointF LL, PointF LR, int sx, int sy)
{
var bmpOut = new Bitmap (sx, sy);
for (int x = 0; x < sx; x++) {
for (int y = 0; y < sy; y++) {
/*
* relative position
*/
double rx = (double) x / sx;
double ry = (double) y / sy;
/*
* get top and bottom position
*/
double topX = TL.X + rx * (TR.X - TL.X);
double topY = TL.Y + rx * (TR.Y - TL.Y);
double bottomX = LL.X + rx * (LR.X - LL.X);
double bottomY = LL.Y + rx * (LR.Y - LL.Y);
/*
* select center between top and bottom point
*/
double centerX = topX + ry * (bottomX - topX);
double centerY = topY + ry * (bottomY - topY);
/*
* store result
*/
var c = PolyColor (centerX, centerY);
bmpOut.SetPixel (x, y, c);
}
}
bmpOut.Save (_Path + "out5 rescale out.bmp");
}
private Color PolyColor (double x, double y)
{
// get fractions
double xf = x - (int) x;
double yf = y - (int) y;
// 4 colors - we're flipping sides so we can use the distance instead of inverting it later
Color cTL = _Px[(int) y + 1, (int) x + 1];
Color cTR = _Px[(int) y + 1, (int) x + 0];
Color cLL = _Px[(int) y + 0, (int) x + 1];
Color cLR = _Px[(int) y + 0, (int) x + 0];
// 4 distances
double dTL = Math.Sqrt (xf * xf + yf * yf);
double dTR = Math.Sqrt ((1 - xf) * (1 - xf) + yf * yf);
double dLL = Math.Sqrt (xf * xf + (1 - yf) * (1 - yf));
double dLR = Math.Sqrt ((1 - xf) * (1 - xf) + (1 - yf) * (1 - yf));
// 4 parts
double factor = 1.0 / (dTL + dTR + dLL + dLR);
dTL *= factor;
dTR *= factor;
dLL *= factor;
dLR *= factor;
// accumulate parts
double r = dTL * cTL.R + dTR * cTR.R + dLL * cLL.R + dLR * cLR.R;
double g = dTL * cTL.G + dTR * cTR.G + dLL * cLL.G + dLR * cLR.G;
double b = dTL * cTL.B + dTR * cTR.B + dLL * cLL.B + dLR * cLR.B;
Color c = Color.FromArgb ((int) (r + 0.5), (int) (g + 0.5), (int) (b + 0.5));
return c;
}

Generally speaking, what you want to do is map the destination coordinates to the source coordinates through a transform function:
for (int y = 0; y < destHeight; y++) {
for (x=0; x < destWidth; x++) {
Color c = Transform(x, y, sourceImage, sourceTransform);
SetPixel(destImage, x, y, c);
}
}
Let's assume that sourceTransform is an object that encapsulates a transformation from source to dest coordinates (and vice versa).
Working in dest coordinates will make it easier to avoid that curve in your retransformed source image and will allow you to better antialias, as you can map the corners of the dest pixel to the source image and sample within it and interpolate/extrapolate.
In your case you're going to have a set of linear equations that do the mapping - in this case this is known as quadrilateral warping - see this previous question.