How do I determine the rotation point of a polygon (triangle)? - c#

I would like to draw a thick, transparent arrow with an arrowhead:
Here's the code that draws the arrow shaft. Notice that I have to offset the rectangle so the calculations are done from the midpoint of the rectangle.
private void DrawMovementArrow(bool color, double StartX, double StartY, double EndX, double EndY)
{
SolidColorBrush partiallyTransparentSolidColorBrush;
Rectangle myRectangle = new Rectangle();
// This will be replaced by piece size
int width = 35;
myRectangle.Width = width;
// Apparently necessary to offset the drawing of the path so that the point is in the center of the path; not the edge.
StartX -= width / 2;
EndX -= width / 2;
myRectangle.Height = Map.EuclideanDistance(StartX, StartY, EndX, EndY) ;
int angle = CalculateAngle(StartX , StartY , EndX , EndY );
// This selects the midpoint of edge of the rectangle to rotate around (weird system)
myRectangle.RenderTransformOrigin = new Point(0.5, 0);
angle = angle - 180;
RotateTransform rotateTransform1 = new RotateTransform(angle, 0 , 0 );
myRectangle.RenderTransform = rotateTransform1;
if (color)
partiallyTransparentSolidColorBrush = new SolidColorBrush(Colors.Blue);
else
partiallyTransparentSolidColorBrush = new SolidColorBrush(Colors.Red);
partiallyTransparentSolidColorBrush.Opacity = 0.4;
myRectangle.Fill = partiallyTransparentSolidColorBrush;
MovementCanvas1.Children.Clear();
MovementCanvas1.Children.Add(myRectangle);
Canvas.SetTop(myRectangle, StartY);
Canvas.SetLeft(myRectangle, StartX);
DrawArrowhead(color, EndX, EndY, angle + 90, width);
ShowUnitCenter(MovementArrowList[0]);
}
Note that this code selects a point in the middle of the edge to rotate the rectangle:
// This selects the midpoint of edge of the rectangle to rotate around (weird system)
myRectangle.RenderTransformOrigin = new Point(0.5, 0);
The problem is that I can't find that point with the arrowhead (triangle). Here's the code that draws the arrowhead:
public void DrawArrowhead(bool color, double x, double y, int angle, int width)
{
x += width /2 ;
width = width + (width / 2);
//Add the Polygon Element
Polygon myPolygon = new Polygon();
myPolygon.Opacity = 0.4;
if (color)
{
myPolygon.Fill = new SolidColorBrush(Colors.Blue);
myPolygon.Stroke = System.Windows.Media.Brushes.Blue;
}
else
{
myPolygon.Fill = new SolidColorBrush(Colors.Red);
myPolygon.Stroke = System.Windows.Media.Brushes.Red;
}
myPolygon.StrokeThickness = 0;
RotateTransform rotateTransform1 = new RotateTransform(angle, 0, 0);
myPolygon.RenderTransform = rotateTransform1;
// This selects the midpoint of edge of the triangle to rotate around (weird system)
myPolygon.RenderTransformOrigin = new Point(0.0, 0.5);
System.Windows.Point Point1 = new System.Windows.Point(0, 0);
System.Windows.Point Point2 = new System.Windows.Point(width / 2, width / 2);
System.Windows.Point Point3 = new System.Windows.Point(0,width);
PointCollection myPointCollection = new PointCollection();
myPointCollection.Add(Point1);
myPointCollection.Add(Point2);
myPointCollection.Add(Point3);
myPolygon.Points = myPointCollection;
MovementCanvas1.Children.Add(myPolygon);
Canvas.SetTop(myPolygon, y );
Canvas.SetLeft(myPolygon, x );
}
Note the myPointCollection that creates the triangle. The problem is that I've tried almost every conceivable combination of values in RenderTransformOrigin to find the point that (center bottom of triangle) to use for the rotation point. Nothing seems to be working out.
Can anybody suggest the correct value?
Edit Solved
I solved it by changing the points of the triangle. That was easier than trying to figure out the rotation point.

Changing the points that made up the triangle solved the problem. This was easier than trying to find the rotation point.

Related

How to increase the height of curve in windows forms

I want to increase the height of the curve but its left and right position should remain same. Just want to lift up from center to give it a shape like curve as height changes.
Pen blackPen = new Pen(Color.Black, 3);
// Create coordinates of rectangle to bound ellipse.
int x = 93;
int y = 136;
int width = 320;
int height = 50;
// Create start and sweep angles on ellipse.
int startAngle = 0;
int sweepAngle = -180;
// Draw arc to screen.
e.Graphics.DrawArc(blackPen, x, y, width, height, startAngle, sweepAngle);
In the most direct way your problem can be solved like this:
int change = 0;
e.Graphics.DrawArc(blackPen, x, y-change, width, height+change, startAngle, sweepAngle);
By increasing the variable change the ellipse will curve up more and more:
private void button1_Click(object sender, EventArgs e)
{
change += 10;
panel1.Invalidate();
}
But maybe you want more control over the shape? Let's have a look at the options:
Here are examples of the three 'curve' drawing methods:
Curves, Beziers & Ellipses
And here is the code to draw that image.
Please ignore the Graphics.xxxTransform calls! They only are meant to shift the curves a little bit upwards so they don't overlap too much to see them properly.
Also note that the curves in the first image are not completely convex. See the last part of the answer to see a DrawCurve call that avoids the concave segments!
The important part are the Points! And just as the comments suggest, in the third part the ellipses are being changed by making the height larger and moving the top of the Rectangle up by the same amount.
The complexity DrawArc of and DrawCurve is pretty much equal; both are controlled by four integers with a rather clear meaning: They either make one rectangle or the corners of a symmetrical triangle. (Plus one counterpoint for the convex call.)
DrawBezier is more complex, especially since the controls point(s) are not actually on the resulting curve. They can be thought of force vectors that pull the line into a curved shape and are harder to calculate.
private void panel1_Paint(object sender, PaintEventArgs e)
{
Point a = new Point(0, 200);
Point c = new Point(200, 200);
for (int i = 0; i< 10; i++)
{
e.Graphics.TranslateTransform(0, -5);
Point b = new Point(100, 50 + i * 10);
e.Graphics.DrawCurve(Pens.Maroon, new[] { a, b, c }, 0.7f);
}
e.Graphics.ResetTransform();
Point pa = new Point(250, 200);
Point pb = new Point(450, 200);
for (int i = 0; i < 10; i++)
{
e.Graphics.TranslateTransform(0, -5);
Point pc = new Point(350, 200 - i * 10);
e.Graphics.DrawBezier(Pens.ForestGreen, pa, pc, pc, pb);
}
e.Graphics.ResetTransform();
int x = 500;
int y0 = 200;
int w = 200;
for (int i = 0; i < 10; i++)
{
e.Graphics.TranslateTransform(0, -5);
Rectangle rect = new Rectangle(x, y0 - i * 10, w, 10 + i * 10);
e.Graphics.DrawArc(Pens.DarkBlue, rect, -0, -180);
}
e.Graphics.ResetTransform();
}
Notes:
The Curve (1st image) can be further controlled by the Tension parameter. The lower the tension the more pointed it gets, approaching 1f it makes the curve broader..
The Bezier curve (2nd image) is using only one control point. (Twice.) The curve gets a little pointed this way. You can make it broader and broader by using two different points the move apart little by little..
The Ellipse can't be controlled; it will always fill the bounding Rectangle.
Here is an example of varying the Curves and the Beziers:
The Curves are drawn with varying Tensions. Also I have used an overload that helps to get rid of the concave part at the start and end of the curve. The trick is to add a suitable extra point to the start and end and to tell the DrawCurve to leave out these 1st and last segments.
The simplest point to use (for both ends actually) is the counterpoint of the one at the top.
The Beziers are drawn using two control points, moving out and up a little.
Here is the code for the variations:
Point a = new Point(0, 200);
Point c = new Point(200, 200);
for (int i = 1; i < 10; i++)
{
e.Graphics.TranslateTransform(0, -5);
Point b = new Point(100, 50);
Point b0 = new Point(b.X, a.Y + (a.Y - b.Y));
e.Graphics.DrawCurve(Pens.Maroon, new[] { b0, a, b, c, b0 }, 1, 2, 0.1f * i);
}
e.Graphics.ResetTransform();
Point pa = new Point(250, 200);
Point pb = new Point(450, 200);
for (int i = 0; i < 10; i++)
{
e.Graphics.TranslateTransform(0, -5);
Point ca = new Point(350 - i * 9, 100 - i * 5);
Point cb = new Point(350 + i * 9, 100 - i * 5);
e.Graphics.DrawBezier(Pens.ForestGreen, pa, ca, cb, pb);
}
e.Graphics.ResetTransform();
here was the solution just increase value of y0 as u increase the value of
y0-i here i=20
int x = 96;
int y0 = 260;
int w = 320;
e.Graphics.TranslateTransform(0, -5);
Rectangle rect = new Rectangle(x, y0 - 20 * 10, w, 10 + 20 * 10);
e.Graphics.DrawArc(Pens.DarkBlue, rect, -0, -180);
e.Graphics.ResetTransform();

Align text to center in System.Drawing.RectangleF

I am trying to create a captcha image. I am generating a random string and rotating the text with a random angle and trying to create a byte array. Below is my code snippet:
Image img = Image.FromFile(#"C:\Images\BackGround.jpg");
RectangleF myRect = new RectangleF(0, 0, width, height);
objGraphics.DrawImage(img, myRect);
Matrix myMatrix = new Matrix();
int i = 0;
StringFormat formatter = new StringFormat();
formatter.Alignment = StringAlignment.Center;
for (i = 0; i <= myString.Length - 1; i++)
{
myMatrix.Reset();
int charLenght = myString.Length;
float x = width / (charLenght + 1) * i;
float y = height / 30F;
myMatrix.RotateAt(oRandom.Next(-40, 40), new PointF(x, y));
objGraphics.Transform = myMatrix;
objGraphics.DrawString(myString.Substring(i, 1), MyFont, MyFontEmSizes, MyFontStyles,
MySolidBrush, x, Math.Max(width, height) / 50, formatter );
objGraphics.ResetTransform();
}
Every thing is working fine, except that, the first character in my final image on the web page is crossing my left border of the rectangle. How can I align my text to the center of the rectangle?
Thanks.

Draw on MSChart control

I've got a simple line graph and I'd like to highlight some parts of this graph by drawing a rectangle around the line (ideally a filled rectangle with transparency...). I haven't any idea if this is possible with the MS chart control ?
I recommend you download the code samples from MS and checkout the section on annotations. In there you will find all the documentation you require to achieve what you described:
private void AddRectangleAnnotation()
{
RectangleAnnotation annotation = new RectangleAnnotation();
annotation.AnchorDataPoint = Chart1.Series[0].Points[2];
annotation.Text = "I am a\nRectangleAnnotation";
annotation.ForeColor = Color.Black;
annotation.Font = new Font("Arial", 12);;
annotation.LineWidth = 2;
annotation.BackColor = Color.PaleYellow;
annotation.LineDashStyle = ChartDashStyle.Dash;
Chart1.Annotations.Add(annotation);
}
Do you mean:
using (Graphics g = Graphics.FromImage(pictureBox1.Image))
{
using(Brush brush = new SolidBrush(your_color))
{
g.FillRectangle(brush , x, y, width, height);
}
}
or you can use
Brush brush = new SolidBrush(Color.FromArgb(alpha, red, green, blue))
where alpha goes from 0 to 255, so a value of 128 for your alpha will give you 50% opactity.
When you'd like to draw on a chart you can take add a LineAnnotation or RectangleAnnotation. if however you'd like more control you can use the chart's PrePaint and PostPaint events. And if you can paint, well then you can paint anything. Also using this will make the chart "printing" and "exporting" look the same as you painted it. Paining over it will look funny when the chart's location is changed on the screen, so always paint in it.
Say you have a trading chart and you need to draw a line as to where you become profitable or as square as to state where you're "To Something" tzhen just add the coordinates from where to where you'd like to be and of you go...
MS Chart sample project shows how to do this with the following code (vb.net also available):
using System.Windows.Forms.DataVisualization.Charting;
...
private void chart1_PostPaint(object sender, System.Windows.Forms.DataVisualization.Charting.ChartPaintEventArgs e)
{
if(sender is ChartArea)
{
ChartArea area = (ChartArea)sender;
if(area.Name == "Default")
{
// If Connection line is not checked return
if( !ConnectionLine.Checked )
return;
double max;
double min;
double xMax;
double xMin;
// Find Minimum and Maximum values
FindMaxMin( out max, out min, out xMax, out xMin );
// Get Graphics object from chart
Graphics graph = e.ChartGraphics.Graphics;
// Convert X and Y values to screen position
float pixelYMax = (float)e.ChartGraphics.GetPositionFromAxis("Default",AxisName.Y,max);
float pixelXMax = (float)e.ChartGraphics.GetPositionFromAxis("Default",AxisName.X,xMax);
float pixelYMin = (float)e.ChartGraphics.GetPositionFromAxis("Default",AxisName.Y,min);
float pixelXMin = (float)e.ChartGraphics.GetPositionFromAxis("Default",AxisName.X,xMin);
PointF point1 = PointF.Empty;
PointF point2 = PointF.Empty;
// Set Maximum and minimum points
point1.X = pixelXMax;
point1.Y = pixelYMax;
point2.X = pixelXMin;
point2.Y = pixelYMin;
// Convert relative coordinates to absolute coordinates.
point1 = e.ChartGraphics.GetAbsolutePoint(point1);
point2 = e.ChartGraphics.GetAbsolutePoint(point2);
// Draw connection line
graph.DrawLine(new Pen(Color.Yellow,3), point1,point2);
}
}
}
private void chart1_PrePaint(object sender, System.Windows.Forms.DataVisualization.Charting.ChartPaintEventArgs e)
{
if(sender is ChartArea){
ChartArea area = (ChartArea)sender;
if(area.Name == "Default")
{
double max;
double min;
double xMax;
double xMin;
// Find Minimum and Maximum values
FindMaxMin( out max, out min, out xMax, out xMin );
// Get Graphics object from chart
Graphics graph = e.ChartGraphics.Graphics;
// Convert X and Y values to screen position
float pixelYMax = (float)e.ChartGraphics.GetPositionFromAxis("Default",AxisName.Y,max);
float pixelXMax = (float)e.ChartGraphics.GetPositionFromAxis("Default",AxisName.X,xMax);
float pixelYMin = (float)e.ChartGraphics.GetPositionFromAxis("Default",AxisName.Y,min);
float pixelXMin = (float)e.ChartGraphics.GetPositionFromAxis("Default",AxisName.X,xMin);
// Specify width of triangle
float width = 3;
// Set Maximum points
PointF [] points = new PointF[3];
points[0].X = pixelXMax - width;
points[0].Y = pixelYMax - width - 2;
points[1].X = pixelXMax + width;
points[1].Y = pixelYMax - width - 2;
points[2].X = pixelXMax;
points[2].Y = pixelYMax - 1;
// Convert relative coordinates to absolute coordinates.
points[0] = e.ChartGraphics.GetAbsolutePoint(points[0]);
points[1] = e.ChartGraphics.GetAbsolutePoint(points[1]);
points[2] = e.ChartGraphics.GetAbsolutePoint(points[2]);
// Draw Maximum trangle
graph.FillPolygon(new SolidBrush(Color.Red), points);
// Set Minimum points
points = new PointF[3];
points[0].X = pixelXMin - width;
points[0].Y = pixelYMin + width + 2;
points[1].X = pixelXMin + width;
points[1].Y = pixelYMin + width + 2;
points[2].X = pixelXMin;
points[2].Y = pixelYMin + 1;
// Convert relative coordinates to absolute coordinates.
points[0] = e.ChartGraphics.GetAbsolutePoint(points[0]);
points[1] = e.ChartGraphics.GetAbsolutePoint(points[1]);
points[2] = e.ChartGraphics.GetAbsolutePoint(points[2]);
// Draw Minimum triangle
graph.FillPolygon(new SolidBrush(Color.Blue), points);
}
}
}

Drawing a polygon according to the input coordinates

How can i draw a polygon according to the input coordinates which are given in C#.
You didn't show any code because based on those coordinate, you are applying some form of scaling to the image.
Using the Paint event of a PictureBox, here is an example using those coordinates on the screen. It fills in the polygon, then draws the border, then it loops through all the points to draw the red circle:
void pictureBox1_Paint(object sender, PaintEventArgs e) {
e.Graphics.SmoothingMode = SmoothingMode.AntiAlias;
e.Graphics.Clear(Color.White);
// draw the shading background:
List<Point> shadePoints = new List<Point>();
shadePoints.Add(new Point(0, pictureBox1.ClientSize.Height));
shadePoints.Add(new Point(pictureBox1.ClientSize.Width, 0));
shadePoints.Add(new Point(pictureBox1.ClientSize.Width,
pictureBox1.ClientSize.Height));
e.Graphics.FillPolygon(Brushes.LightGray, shadePoints.ToArray());
// scale the drawing larger:
using (Matrix m = new Matrix()) {
m.Scale(4, 4);
e.Graphics.Transform = m;
List<Point> polyPoints = new List<Point>();
polyPoints.Add(new Point(10, 10));
polyPoints.Add(new Point(12, 35));
polyPoints.Add(new Point(22, 35));
polyPoints.Add(new Point(24, 22));
// use a semi-transparent background brush:
using (SolidBrush br = new SolidBrush(Color.FromArgb(100, Color.Yellow))) {
e.Graphics.FillPolygon(br, polyPoints.ToArray());
}
e.Graphics.DrawPolygon(Pens.DarkBlue, polyPoints.ToArray());
foreach (Point p in polyPoints) {
e.Graphics.FillEllipse(Brushes.Red,
new Rectangle(p.X - 2, p.Y - 2, 4, 4));
}
}
}
You may use Graphics.DrawPolygon. You can store the coordinates in an array of Point and then you can pass that to DrawPolygon method. You may wanna see:
Drawing with Graphics in WinForms using C#
private System.Drawing.Graphics g;
System.Drawing.Point[] p = new System.Drawing.Point[6];
p[0].X = 0;
p[0].Y = 0;
p[1].X = 53;
p[1].Y = 111;
p[2].X = 114;
p[2].Y = 86;
p[3].X = 34;
p[3].Y = 34;
p[4].X = 165;
p[4].Y = 7;
g = PictureBox1.CreateGraphics();
g.DrawPolygon(pen1, p);
This simple function is able to generate an array of PointF equal to the vertices of the regular polygon to be drawn, where "center" is the center of the polygon, "sides" is its number of sides, "sideLength" is the size of each side in pixels and "offset" is its slope.
public PointF[] GetRegularPolygonScreenVertex(Point center, int sides, int sideLength, float offset)
{
var points = new PointF[sides];
for (int i = 0; i < sides; i++)
{
points[i] = new PointF(
(float)(center.X + sideLength * Math.Cos((i * 360 / sides + offset) * Math.PI / 180f)),
(float)(center.Y + sideLength * Math.Sin((i * 360 / sides + offset) * Math.PI / 180f))
);
}
return points;
}
The result obtained can be used to draw a polygon, e.g. with the function:
GraphicsObject.DrawPolygon(new Pen(Brushes.Black, GetRegularPolygonScreenVertex(new Point(X, Y), 6, 30, 60f));
Which will generate a regular hexagon with a side of 30 pixels inclined by 30°.
hex

How do I rotate image then move to the top left 0,0 without cutting off the image

How do I rotate image then move to the top left 0,0 without cutting off the image.
Please read the comments inside the code. I got stuck at STEP 3
I think using trigonometry should be able to solve this problem.
thanks
private Bitmap RotateImage(Bitmap b, float angle)
{
//create a new empty bitmap to hold rotated image
Bitmap returnBitmap = new Bitmap(b.Width, b.Height);
//make a graphics object from the empty bitmap
Graphics g = Graphics.FromImage(returnBitmap);
//STEP 1 move rotation point to top left
g.TranslateTransform((float)0, (float)0);
//STEP 2 rotate
g.RotateTransform(angle);
//STEP 3 move image back to top left without cutting off the image
//SOME trigonometry calculation here
int newY = b.Height;
g.TranslateTransform(-(float)0, -newY);
//draw passed in image onto graphics object
g.DrawImage(b, new Point(0, 0));
return returnBitmap;
}
Does this cover the 'trigonometry'? I have made it step 0 because I think you need to do it first. That way you can calculate the size of the resulting bitmap, which will be bigger - see my comments in the code.
private Bitmap RotateImage(Bitmap b, float Angle) {
// The original bitmap needs to be drawn onto a new bitmap which will probably be bigger
// because the corners of the original will move outside the original rectangle.
// An easy way (OK slightly 'brute force') is to calculate the new bounding box is to calculate the positions of the
// corners after rotation and get the difference between the maximum and minimum x and y coordinates.
float wOver2 = b.Width / 2.0f;
float hOver2 = b.Height / 2.0f;
float radians = -(float)(Angle / 180.0 * Math.PI);
// Get the coordinates of the corners, taking the origin to be the centre of the bitmap.
PointF[] corners = new PointF[]{
new PointF(-wOver2, -hOver2),
new PointF(+wOver2, -hOver2),
new PointF(+wOver2, +hOver2),
new PointF(-wOver2, +hOver2)
};
for (int i = 0; i < 4; i++) {
PointF p = corners[i];
PointF newP = new PointF((float)(p.X * Math.Cos(radians) - p.Y * Math.Sin(radians)), (float)(p.X * Math.Sin(radians) + p.Y * Math.Cos(radians)));
corners[i] = newP;
}
// Find the min and max x and y coordinates.
float minX = corners[0].X;
float maxX = minX;
float minY = corners[0].Y;
float maxY = minY;
for (int i = 1; i < 4; i++) {
PointF p = corners[i];
minX = Math.Min(minX, p.X);
maxX = Math.Max(maxX, p.X);
minY = Math.Min(minY, p.Y);
maxY = Math.Max(maxY, p.Y);
}
// Get the size of the new bitmap.
SizeF newSize = new SizeF(maxX - minX, maxY - minY);
// ...and create it.
Bitmap returnBitmap = new Bitmap((int)Math.Ceiling(newSize.Width), (int)Math.Ceiling(newSize.Height));
// Now draw the old bitmap on it.
using (Graphics g = Graphics.FromImage(returnBitmap)) {
g.TranslateTransform(newSize.Width / 2.0f, newSize.Height / 2.0f);
g.RotateTransform(Angle);
g.TranslateTransform(-b.Width / 2.0f, -b.Height / 2.0f);
g.DrawImage(b, 0, 0);
}
return returnBitmap;
}

Categories

Resources