I am relatively new to c# and i am trying to draw the quadratic curve with an X and Y graph to scale with. The i drew curve although appears at the upper left corner of the screen which is very small and barely noticeable. Is there possible way to enlarge my curve line and align it to the middle so it can be shown properly?
protected override void OnPaint(PaintEventArgs e)
{
float a = 1, b = -3, c = -4;
double x1, x2, x3, y1, y2, y3, delta;
delta = (b * b) - (4 * a * c);
x1 = ((b * (-1)) + Math.Sqrt(delta)) / (2 * a);
y1 = a * (x1 * x1) + b * (x1) + c;
x2 = x1 + 1;
y2 = a * (x2 * x2) + b * (x2) + c;
x3 = x1 - 3;
y3 = a * (x3 * x3) + b * (x3) + c;
int cx1 = Convert.ToInt32(x1);
int cx2 = Convert.ToInt32(x2);
int cx3 = Convert.ToInt32(x3);
int cy1 = Convert.ToInt32(y1);
int cy2 = Convert.ToInt32(y2);
int cy3 = Convert.ToInt32(y3);
Graphics g = e.Graphics;
Pen aPen = new Pen(Color.Blue, 1);
Point point1 = new Point(cx1, cy1);
Point point2 = new Point(cx2, cy2);
Point point3 = new Point(cx3, cy3);
Point[] Points = { point1, point2, point3 };
g.DrawCurve(aPen, Points);
Yes it is possible and even rather simple to both move (Translate) and enlarge (Scale) the Graphics results by using Graphics.TranslateTransform and Matrix and Graphics.MultiplyTransform:
using System.Drawing.Drawing2D;
//..
int deltaX = 100;
int deltaY = 100;
g.TranslateTransform(deltaX, deltaY);
float factor = 2.5f;
Matrix m = new Matrix();
m.Scale(factor, factor);
g.MultiplyTransform(m);
Note that the scaling works like a lens and will enlarge the pixels. So you may want to scale down the Pen.Width when you scale up the Graphics..
Using one before..
g.DrawEllipse(Pens.Blue, 11, 11, 55, 55);
..and two after the transformations..
g.DrawEllipse(Pens.Red, 11, 11, 55, 55);
using (Pen pen = new Pen(Color.Green, 1/factor))
g.DrawEllipse(pen, 11, 11, 44, 44);
..these calls result in this image:
(I have changed the green circle's radius to avoid complete overlaying..)
It will be up to you to find the desired numbers for the moving and scaling; this will probably involve finding the minimum and maximum values for points involved..
I would suggest that you look into Microsoft Chart controls, it has a lots of interesting features regarding how to do this kind of curves with the ability to parameterize them.
A link to a more recent version of it: here
Related
I'm working on a Android app for receipts preprocessing. My first step would be to find the edges from the receipt in the given image. Then I would like to use the edges to remove the background and perform a perspective correction.
Im trying to do it using OpenCv 4.5.3.
My first try has been to perform HoughLinesP to detect the lines in the image, but the result is not that good.
Mat sampledImage = OpenCV.ImgCodecs.Imgcodecs.Imread(imagePath);
Mat gray = new Mat();
Imgproc.CvtColor(sampledImage, gray, Imgproc.ColorRgb2gray);
//Imgproc.GaussianBlur(gray, gray, new Size(7, 7), 0);
Mat edgeImage = new Mat();
Imgproc.Canny(gray, edgeImage, 50, 200, 3);
Mat lines = new Mat();
int threshold = 80;
Imgproc.HoughLinesP(edgeImage, lines, 1.0, Math.PI / 180, threshold, 30, 10);
//OpenCV.ImgCodecs.Imgcodecs.Imwrite(imagePath, edgeImage);
for (int x = 0; x < lines.Rows(); x++)
{
double[] vec = lines.Get(x, 0);
double x1 = vec[0],
y1 = vec[1],
x2 = vec[2],
y2 = vec[3];
Point start = new Point(x1, y1);
Point end = new Point(x2, y2);
double dx = x1 - x2;
double dy = y1 - y2;
double dist = Math.Sqrt(dx * dx + dy * dy);
if (dist > 50D)
Imgproc.Line(sampledImage, start, end, new Scalar(0, 255, 0), 5);
}
OpenCV.ImgCodecs.Imgcodecs.Imwrite(imagePath, sampledImage);
Original image
So basically the edges of the receipt are not 100% detected. Any idea of how could I improve this edge detection to remove the background and perform perspective correction afterwards?
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.
The title for this post was quite hard to think of, so if you can think of a more descriptive title please tell me. Anyway, my problem is quite specific and requires some simple maths knowledge. I am writing a C# WinForms application which is a bit like the old 'xeyes' Linux application. It basically is a set of eyes which follow around your mouse cursor. This may sound easy at first, however can get rather complicated if you're a perfectionist like me :P. This is my code so far (only the paint method, that is called on an interval of 16).
int lx = 35;
int ly = 50;
int rx;
int ry;
int wx = Location.X + Width / 2;
int wy = Location.Y + Height / 2;
Rectangle bounds = Screen.FromControl(this).Bounds;
// Calculate X
float tempX = (mx - wx) / (float)(bounds.Width / 2);
// Calculate Y
float tempY = (my - wy) / (float)(bounds.Height / 2);
// Draw eyes
e.Graphics.FillEllipse(Brushes.LightGray, 10, 10, 70, 100);
e.Graphics.FillEllipse(Brushes.LightGray, 90, 10, 70, 100);
// Draw pupils (this only draws the left one)
e.Graphics.FillEllipse(Brushes.Black, lx += (int)(25 * tempX), ly += (int)(40 * tempY), 20, 20);
Now this does work at a basic level, however sometimes this can happen if the user puts the cursor at 0,0.
Now my question is how to fix this? What would the IF statement be to check where the mouse pointer is, and then reduce the pupil X depending on that?
Thanks.
Edit: This is where I get the mouse positions (my and mx):
private void timer_Tick(object sender, EventArgs e)
{
mx = Cursor.Position.X;
my = Cursor.Position.Y;
Invalidate();
}
The timer is started in the eyes_Load event and the interval is 16.
Edit 2: Final solution: http://pastebin.com/fT5HfiQR
Modelling the eyeball as the following ellipse:
Its equation is:
And that of the line joining its center and the cursor:
(don't worry about the singularity)
We can then solve to get the intersection point:
Where
Now you can calculate the distance to the eyeball's edge, by dividing the distance from the center to the cursor by sigma. What remains is just interpolating to cap the position of the pupil:
The if statement you want is then
(N.B. for math-mo's out there the above was a slight simplification, which assumes your ellipse is not too narrow; the exact solution is non-analytical)
EDIT: my tests in VB.NET:
EDIT 2: C# port
PointF Bound(double xc, double yc, double w, double h, double xm, double ym, double r)
{
double dx = xm - xc, dy = ym - yc;
if (Math.Abs(dx) > 0.001 && Math.Abs(dy) > 0.001)
{
double dx2 = dx * dx, dy2 = dy * dy;
double sig = 1.0 / Math.Sqrt(dx2 / (w * w * 0.25) + dy2 / (h * h * 0.25));
double d = Math.Sqrt(dx2 + dy2), e = d * sig;
if (d > e - r)
{
double ratio = (e - r) / d;
return new PointF((float)(xc + dx * ratio),
(float)(yc + dy * ratio));
}
}
return new PointF((float)xm, (float)ym);
}
xc, yc: Center coordinates of the ellipse
w, h: Width and height of the ellipse
xm, ym: Mouse coordinates
r: Radius of the circle you wanna constrain (the pupil)
Returns: The point where you wanna place the center of the circle
EDIT 3: Many thanks to Quinchilion for the following optimization (gawd damn this smacked me hard in the face)
PointF Bound(double xc, double yc, double w, double h, double xm, double ym, double r)
{
double x = (xm - xc) / (w - r);
double y = (ym - yc) / (h - r);
double dot = x*x + y*y;
if (dot > 1) {
double mag = 1.0 / Math.Sqrt(dot);
x *= mag; y *= mag;
}
return new PointF((float)(x * (w - r) + xc), (float)(y * (h - r) + yc));
}
I'm creating and drawing a triangle mesh in wpf c# using GeometryModel3D. I've been trying to figure out how to create a smooth shading over the triangles, like the classic openGL smooth shaded triangle.
I would like to define a colour for each vertex, and then having the colours interpolated over the face, like this, assuming the three colour where red, green and blue.
I assumed I would need to use a brush, but I haven't been able to figure out how.
So any help would be appreciated, or any pointer to a guide that shows me how to achieve this.
EDIT:
I've looked at Triangular Gradient in WPF3D, which seems to answer the question partly, just using xaml.
Unfortunatly it seems like it need equilateral triangles.
2nd EDIT
The answer above, uses the RadialGradientBrush. Is the RadiusXand RadiusY used to make it elliptic instead of circular?
3rd EDIT
Okay, I'm fairly sure I can use the RadialGradientBrush. What I think I can do is, find the center of the circumcircle of the triangles, and create a RadialGradientBrush with RadiusX and RadiusY equal to the radius if the circumcircle. I would then move the focal point of the RadialGradientBrush to the vertices with GradientOrigin.
GradientOrigin takes two doubles X,Y as the center, with both of them being in the interval [0,1]. From what I can read is X = 0.0 is the left side and X = 1.0 is the right side and Y = 0.0 is the top and Y = 1.0 is the bottom. What I can't figure out, is this mapping [0,1]x[0,1] to a circle, or is it a square? The mapping from the vertices of the triangle to [0,1]x[0,1], depends on what shape this interval represents.
Have you heard of Helix 3D Toolkit for WPF ?
I didn't go as far as you'd like but I guess it is possible by looking at the Surface Demo example :
There are surely libraries for that, but to give some simple way, searching through some google,http://www.geeksforgeeks.org/check-whether-a-given-point-lies-inside-a-triangle-or-not/
computing the distance from corners, gives info about the smooth color. Checking if point is in triangle.
float area(int x1, int y1, int x2, int y2, int x3, int y3)
{
return (float)Math.Abs((x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2)) / 2.0);
}
bool isInside(int x1, int y1, int x2, int y2, int x3, int y3, int x, int y)
{
/* Calculate area of triangle ABC */
float A = area(x1, y1, x2, y2, x3, y3);
/* Calculate area of triangle PBC */
float A1 = area(x, y, x2, y2, x3, y3);
/* Calculate area of triangle PAC */
float A2 = area(x1, y1, x, y, x3, y3);
/* Calculate area of triangle PAB */
float A3 = area(x1, y1, x2, y2, x, y);
/* Check if sum of A1, A2 and A3 is same as A */
return (A == A1 + A2 + A3);
}
for (int ii = 5; ii < 100; ii++)
{
for (int jj = 5; jj < 100; jj++)
{
int distanceRed =0, distanceGreen =0,distanceBlue =0;
if (isInside(30, 50, 30, 90, 20, 70, ii, jj))
{
distanceRed = (int)Math.Sqrt(((ii - 30) * (ii - 30) + (jj - 50) * (jj - 50)));
distanceGreen = (int)Math.Sqrt(((ii - 30) * (ii - 30) + (jj - 90) * (jj - 90)));
distanceBlue = (int)Math.Sqrt(((ii - 20) * (ii - 20) + (jj - 70) * (jj - 70)));
}
else
{
distanceRed = 0; distanceGreen = 0; distanceBlue = 0;
}
ptr[(((int)jj) * 3) + ((int)ii) * stride] = (byte)(distanceRed % 256);
ptr[(((int)jj) * 3) + ((int)ii) * stride + 1] = (byte)(distanceGreen % 256);
ptr[(((int)jj) * 3) + ((int)ii) * stride + 2] = (byte)(distanceBlue % 256);
}
}
gives the result:
Couldnt fit the red. Maybe the modulo is wrong here.
Also the sqrt is inefficient.
i am currently try to inscribe diagonals of a decagon inside a circle
like this
in c# my approach would be creating a circle
e.Graphics.DrawEllipse(myPen, 0, 0, 100, 100);
and draw lines inside using
e.Graphics.DrawLine(myPen, 20, 5, 50, 50);
after that i would draw a decagon polygon.
currently im stuck at how to divide the circle into 10 parts/ finding the correct coordiantes of the points on the circumference of the circles because im not good in math,
i want to know how would i know the next point in a circumference of the circle the size of my circle is indicated above.
and also i want also to ask a better approach for my problem.
Thank you :)
Just for grits and shins, here's a generic implementation that will inscribe an X-sided polygon into the Rectangle you pass it. Note that in this approach I'm not actually calculating any absolute points. Instead, I am translating the origin, rotating the surface, and drawing the lines only with respect to the origin using a fixed length and an angle. This is repeated in a loop to achieve the end result below, and is very similar to commanding the Turtle in Logo:
public partial class Form1 : Form
{
PictureBox pb = new PictureBox();
NumericUpDown nud = new NumericUpDown();
public Form1()
{
InitializeComponent();
this.Text = "Inscribed Polygon Demo";
TableLayoutPanel tlp = new TableLayoutPanel();
tlp.RowCount = 2;
tlp.RowStyles.Clear();
tlp.RowStyles.Add(new RowStyle(SizeType.AutoSize));
tlp.RowStyles.Add(new RowStyle(SizeType.Percent, 100));
tlp.ColumnCount = 2;
tlp.ColumnStyles.Clear();
tlp.ColumnStyles.Add(new ColumnStyle(SizeType.AutoSize));
tlp.ColumnStyles.Add(new ColumnStyle(SizeType.AutoSize));
tlp.Dock = DockStyle.Fill;
this.Controls.Add(tlp);
Label lbl = new Label();
lbl.Text = "Number of Sides:";
lbl.TextAlign = ContentAlignment.MiddleRight;
tlp.Controls.Add(lbl, 0, 0);
nud.Minimum = 3;
nud.Maximum = 20;
nud.AutoSize = true;
nud.ValueChanged += new EventHandler(nud_ValueChanged);
tlp.Controls.Add(nud, 1, 0);
pb.Dock = DockStyle.Fill;
pb.Paint += new PaintEventHandler(pb_Paint);
pb.SizeChanged += new EventHandler(pb_SizeChanged);
tlp.SetColumnSpan(pb, 2);
tlp.Controls.Add(pb, 0, 1);
}
void nud_ValueChanged(object sender, EventArgs e)
{
pb.Refresh();
}
void pb_SizeChanged(object sender, EventArgs e)
{
pb.Refresh();
}
void pb_Paint(object sender, PaintEventArgs e)
{
// make circle centered and 90% of PictureBox size:
int Radius = (int)((double)Math.Min(pb.ClientRectangle.Width, pb.ClientRectangle.Height) / (double)2.0 * (double).9);
Point Center = new Point((int)((double)pb.ClientRectangle.Width / (double)2.0), (int)((double)pb.ClientRectangle.Height / (double)2.0));
Rectangle rc = new Rectangle(Center, new Size(1, 1));
rc.Inflate(Radius, Radius);
InscribePolygon(e.Graphics, rc, (int)nud.Value);
}
private void InscribePolygon(Graphics G, Rectangle rc, int numSides)
{
if (numSides < 3)
throw new Exception("Number of sides must be greater than or equal to 3!");
float Radius = (float)((double)Math.Min(rc.Width, rc.Height) / 2.0);
PointF Center = new PointF((float)(rc.Location.X + rc.Width / 2.0), (float)(rc.Location.Y + rc.Height / 2.0));
RectangleF rcF = new RectangleF(Center, new SizeF(1, 1));
rcF.Inflate(Radius, Radius);
G.DrawEllipse(Pens.Black, rcF);
float Sides = (float)numSides;
float ExteriorAngle = (float)360 / Sides;
float InteriorAngle = (Sides - (float)2) / Sides * (float)180;
float SideLength = (float)2 * Radius * (float)Math.Sin(Math.PI / (double)Sides);
for (int i = 1; i <= Sides; i++)
{
G.ResetTransform();
G.TranslateTransform(Center.X, Center.Y);
G.RotateTransform((i - 1) * ExteriorAngle);
G.DrawLine(Pens.Black, new PointF(0, 0), new PointF(0, -Radius));
G.TranslateTransform(0, -Radius);
G.RotateTransform(180 - InteriorAngle / 2);
G.DrawLine(Pens.Black, new PointF(0, 0), new PointF(0, -SideLength));
}
}
}
I got the formula for the length of the side here at Regular Polygon Calculator.
One way of dealing with this is using trigonometric functions sin and cos. Pass them the desired angle, in radians, in a loop (you need a multiple of 2*π/10, i.e. a = i*π/5 for i between 0 and 9, inclusive). R*sin(a) will give you the vertical offset from the origin; R*cos(a) will give you the horizontal offset.
Note that sin and cos are in the range from -1 to 1, so you will see both positive and negative results. You will need to add an offset for the center of your circle to make the points appear at the right spots.
Once you've generated a list of points, connect point i to point i+1. When you reach the ninth point, connect it to the initial point to complete the polygon.
I don't test it, but i think it is ok.
#define DegreeToRadian(d) d * (Pi / 180)
float r = 1; // radius
float cX = 0; // centerX
float cY = 0; // centerY
int numSegment = 10;
float angleOffset = 360.0 / numSegment;
float currentAngle = 0;
for (int i = 0; i < numSegment; i++)
{
float startAngle = DegreeToRadian(currentAngle);
float endAngle = DegreeToRadian(fmod(currentAngle + angleOffset, 360));
float x1 = r * cos(startAngle) + cX;
float y1 = r * sin(startAngle) + cY;
float x2 = r * cos(endAngle) + cX;
float y2 = r * sin(endAngle) + cY;
currentAngle += angleOffset;
// [cX, cY][x1, y1][x2, y2]
}
(fmod is c++ function equals to floatNumber % floatNumber)