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Transforming Rectangle Coordinates to a Large Resolution Image

I'm using the following code to Transform a small rectangle coordinates to a larger one ie: A rectangle position on a small image to the same position on the larger resolution of the same image
Rectangle ConvertToLargeRect(Rectangle smallRect, Size largeImageSize, Size smallImageSize)
{
double xScale = (double)largeImageSize.Width / smallImageSize.Width;
double yScale = (double)largeImageSize.Height / smallImageSize.Height;
int x = (int)(smallRect.X * xScale + 0.5);
int y = (int)(smallRect.Y * yScale + 0.5);
int right = (int)(smallRect.Right * xScale + 0.5);
int bottom = (int)(smallRect.Bottom * yScale + 0.5);
return new Rectangle(x, y, right - x, bottom - y);
}
But there seems to be a problem with some images.The transformed rectangle coordinates seems to be off the image.
UPDATE:
img.Draw(rect, new Bgr(232, 3, 3), 2);
Rectangle transret= ConvertToLargeRect(rect, orgbitmap.Size, bit.Size);
target = new Bitmap(transret.Width, transret.Height);
using (Graphics g = Graphics.FromImage(target))
{
g.SmoothingMode = SmoothingMode.HighQuality;
g.DrawImage(orgbitmap, new Rectangle(0, 0, target.Width, target.Height),
transret, GraphicsUnit.Pixel);
}
Rectangle Drawn on small resolution Image
{X=190,Y=2,Width=226,Height=286}
Rectangle Transformed into Orginal Large Resolution Image {X=698,Y=7,Width=830,Height=931}
Original Image

First of all, if you resize the shape it shouldn't move position. That's not what one would expect out of enlarging a shape. This means the X,Y point of the top-left corner shouldn't be transformed.
Second, you shouldn't be adding 0.5 manually to operations, that's not a clean way to proceed. Use the ceiling function as suggested by #RezaAghaei
Third, you should not substract X/Y from the height/width, your calculations should be done as width * scale.
Please correct those mistakes, and if it doesn't work I'll update the answer with extra steps.

Reading a wedge area of Circular bitmap in c#

I am working on a program that takes a Bitmap and converts it into circular form. The code is as follows:
public static Image CropToCircle(Image srcImage, Color backGround)
{
Image dstImage = new Bitmap(srcImage.Width, srcImage.Height, srcImage.PixelFormat);
Graphics g = Graphics.FromImage(dstImage);
using (Brush br = new SolidBrush(backGround)) {
g.FillRectangle(br, 0, 0, dstImage.Width, dstImage.Height);
}
GraphicsPath path = new GraphicsPath();
path.AddEllipse(0, 0, dstImage.Width, dstImage.Height);
g.SetClip(path);
g.DrawImage(srcImage, 0, 0);
return dstImage;
}
It returns the image in circular shape; however I need to read an image wedge in the form of degrees; that is, the circle has 360 degrees and I am trying to write a function that will accept a degree (e.g. 10) and will return the pixels of the image that fall in 10th degree. Such that entire image will be readable in 1 to 360 degrees.

Since my hint was actually rather misleading, let me make up by giving you a working code:
// collect a list of colors from a bitmap with a cetner c and radius r
List<Color> getColorsByAngle(Bitmap bmp, Point c, int r, float angle)
{
List<Color> colors = new List<Color>();
for (int i = 0; i < r; i++)
{
int x = (int)(Math.Sin(angle / 180f * Math.PI) * i);
int y = (int)(Math.Cos(angle / 180f * Math.PI) * i);
colors.Add(bmp.GetPixel(c.X + x, c.Y + y));
}
return colors;
}
Here it is at work:
(The gif is rather quantized for size..)
Note that
Pixels close to the center will be read multiple time, the center itself even each time
To collect all outer pixels you need to read as many angles as the circumference of the circle has pixels, ie 2 * PI * radius. So for a circle with a radius of 300 pixels you need to step the angle in 360° / (600 * 3.14) or about 0.2°..
Also note the the coordinate systems in GDI and in geometry are not the same, neither in the direction of the axes nor the angles. Adapting this is left for you..
The original version didn't mention a 'wedge area'. To read an area or the whole image simply loop over an angle range in suitable steps!

Crop a diagonal area from an image in WPF

I want to crop from an image using user-drawn rectangles on a canvas. The rectangles can be moved, re-sized, and rotated.
When the user selects "Get Cropped Image", the area inside the rectangle should be saved in a second image location on the page, which I can do perfectly well, so long as the rectangle is not rotated. (Straight-forward use of CroppedBitmap.) However, when the rectangle is at an angle I do not know how to perform the crop.
This is what I want to do (forgive my poor MS Paint skills):
My questions are:
1) How do I correctly track or calculate the points of the rectangle?
and,
2) Once I have the points, how do I crop the rotated rectangle?
EDIT:
Thanks to user Rotem, I believe that I have the answer to the second question. Using code modified from the following answers: Answer 1, Answer 2, I am seeing good results. Unfortunately, I am still unable to track the correct location points for the rectangle, so I cannot fully test this as of yet.
public static Bitmap CropRotatedRect(Bitmap source, System.Drawing.Rectangle rect, float angle, bool HighQuality)
{
Bitmap result = new Bitmap((int)rect.Width, (int)rect.Height);
using (Graphics g = Graphics.FromImage(result))
{
g.InterpolationMode = HighQuality ? InterpolationMode.HighQualityBicubic : InterpolationMode.Default;
using (Matrix mat = new Matrix())
{
mat.Translate(-rect.Location.X, -rect.Location.Y);
mat.RotateAt(-(angle), rect.Location);
g.Transform = mat;
g.DrawImage(source, new System.Drawing.Point(0, 0));
}
}
return result;
}
EDIT:
The answer to the first point is much easier than I had originally thought. You can always get the top-left corner of the rectangle by calling—
double top = Canvas.GetTop(rect);
double left = Canvas.GetLeft(rect);
You can then calculate the rest of the points using the width and the height—
Point topLeft = new Point(left, top);
Point topRight = new Point(left + rect.Width, top);
Point bottomLeft = new Point(left, top + rect.Height);
Point bottomRight = new Point(left + rect.Width, top + rect.Height);
Point centerPoint = new Point(left + (rect.Width / 2), top + (rect.Height / 2));
If your rectangle is rotated, then you have to translate these points to determine where they truly lie on the canvas—
public Point TranslatePoint(Point center, Point p, double angle)
{
// get the point relative to (0, 0) by subtracting the center of the rotated shape.
Point relToOrig = new Point(p.X - center.X, p.Y - center.Y);
double angleInRadians = angle * Math.PI / 180;
double sinOfA = Math.Sin(angleInRadians);
double cosOfA = Math.Cos(angleInRadians);
Point translatedPoint = new Point(relToOrig.X * cosOfA - relToOrig.Y * sinOfA,
relToOrig.X * sinOfA + relToOrig.Y * cosOfA);
return new Point(translatedPoint.X + center.X, translatedPoint.Y + center.Y);
}
Once you are able to translate the top-left corner, you can use Rotem's cropping method. You can also calculate the position of the rest of the rectangle, so you are able to determine if the rectangle is within the bounds of the image, if it is touching an edge, or any other thing that you might want to do in regards to the position.

I discovered the answer to my own question(s), and made the appropriate edits along the way. Please see above for the answer.

C# Rotate a Polygon(Triangle)

I have a method which is to draw a polygon, and then rotate that polygon 90 degrees to the right so that its original top point is now pointing towards the right.
This is the code to draw the polygon(triangle) how ever I'm lost on how to rotate this.
Point[] points = new Point[3];
points[0] = new Point((int)top, (int)top);
points[1] = new Point((int)top - WIDTH / 2, (int)top + HEIGHT);
points[2] = new Point((int)top + WIDTH / 2, (int)top + HEIGHT);
paper.FillPolygon(normalBrush, points);
Thanks in advance.

http://msdn.microsoft.com/en-us/library/s0s56wcf.aspx#Y609
public void RotateExample(PaintEventArgs e)
{
Pen myPen = new Pen(Color.Blue, 1);
Pen myPen2 = new Pen(Color.Red, 1);
// Draw the rectangle to the screen before applying the transform.
e.Graphics.DrawRectangle(myPen, 150, 50, 200, 100);
// Create a matrix and rotate it 45 degrees.
Matrix myMatrix = new Matrix();
myMatrix.Rotate(45, MatrixOrder.Append);
// Draw the rectangle to the screen again after applying the
// transform.
e.Graphics.Transform = myMatrix;
e.Graphics.DrawRectangle(myPen2, 150, 50, 200, 100);
}
You can use TransformPoints method of Matrix class to just rotate the points

See this informative Wikipedia article for a great explanation of rotation matrices. When rotating 90 degrees we note that cos 90 collapses into zero yielding the following simple transformation where x' and y' are your rotated coordinates and x and y are the previous coordinates.
x' = -y
y' = x
Applying this simple replacement on your example yields the following code. I've also used a shorthand collection initializer expression for added readability.
var points = new[]
{
new Point(-(int) top, (int) top),
new Point((int) -(top + HEIGHT), (int) top - WIDTH/2),
new Point((int) -(top + HEIGHT), (int) top + WIDTH/2)
};
paper.FillPolygon(normalBrush, points);
I also recommend reading up on linear algebra using for example Anton Rorres, et al.

You can rotate your polygon if you rotate every point. You also have to find out your rotation center O. Probably you want to use your polygon center as rotation center.

Rotating graphics?

I have this code, that draws an image.
private void timer1_Tick(object sender, EventArgs e)
{
Invalidate();
}
protected override void OnPaint(PaintEventArgs e)
{
var tempRocket = new Bitmap(Properties.Resources.rocket);
using (var g = Graphics.FromImage(tempRocket))
{
e.Graphics.DrawImage(tempRocket, 150, 150);
}
}
Yet what do I do to rotate it?

public static 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
using(Graphics g = Graphics.FromImage(returnBitmap))
{
//move rotation point to center of image
g.TranslateTransform((float)b.Width / 2, (float)b.Height / 2);
//rotate
g.RotateTransform(angle);
//move image back
g.TranslateTransform(-(float)b.Width / 2, -(float)b.Height / 2);
//draw passed in image onto graphics object
g.DrawImage(b, new Point(0, 0));
}
return returnBitmap;
}

There are overloads of Graphics.DrawImage that take an array of three points used to define a parallelogram for the destination, such as:
Graphics.DrawImage Method (Image, Point[])
Remarks
The destPoints parameter specifies
three points of a parallelogram. The
three Point structures represent the
upper-left, upper-right, and
lower-left corners of the
parallelogram. The fourth point is
extrapolated from the first three to
form a parallelogram.
The image represented by the image
parameter is scaled and sheared to fit
the shape of the parallelogram
specified by the destPoints
parameters.
There is also an article on MSDN describing the use of this method: How to: Rotate, Reflect, and Skew Images, with the following code example. Unfortunately, the example complicates the issue by also skewing the image.
Point[] destinationPoints = {
new Point(200, 20), // destination for upper-left point of original
new Point(110, 100), // destination for upper-right point of original
new Point(250, 30)}; // destination for lower-left point of original
Image image = new Bitmap("Stripes.bmp");
// Draw the image unaltered with its upper-left corner at (0, 0).
e.Graphics.DrawImage(image, 0, 0);
// Draw the image mapped to the parallelogram.
e.Graphics.DrawImage(image, destinationPoints);
The main differences compared to using the Graphics.Transform property are:
This method does not allow you to specify the rotation angle in degrees -- you have to use some simple trigonometry to derive the points.
This transformation applies only to the specific image.
Good if you only need to draw one rotated image and everything else is non-rotated since you don't have to reset Graphics.Transform afterward.
Bad if you want to rotate several things together (i.e., rotate the "camera").

Use Graphics.RotateTransform to rotate the image.
protected override void OnPaint(PaintEventArgs e)
{
var tempRocket = new Bitmap(Properties.Resources.rocket);
e.Graphics.RotateTransform(30.0F);
e.Graphics.DrawImage(tempRocket, 150, 150);
}

Version without clipping:
private Bitmap RotateBitmap(Bitmap bitmap, float angle)
{
int w, h, x, y;
var dW = (double)bitmap.Width;
var dH = (double)bitmap.Height;
double degrees = Math.Abs(angle);
if (degrees <= 90)
{
double radians = 0.0174532925 * degrees;
double dSin = Math.Sin(radians);
double dCos = Math.Cos(radians);
w = (int)(dH * dSin + dW * dCos);
h = (int)(dW * dSin + dH * dCos);
x = (w - bitmap.Width) / 2;
y = (h - bitmap.Height) / 2;
}
else
{
degrees -= 90;
double radians = 0.0174532925 * degrees;
double dSin = Math.Sin(radians);
double dCos = Math.Cos(radians);
w = (int)(dW * dSin + dH * dCos);
h = (int)(dH * dSin + dW * dCos);
x = (w - bitmap.Width) / 2;
y = (h - bitmap.Height) / 2;
}
var rotateAtX = bitmap.Width / 2f;
var rotateAtY = bitmap.Height / 2f;
var bmpRet = new Bitmap(w, h);
bmpRet.SetResolution(bitmap.HorizontalResolution, bitmap.VerticalResolution);
using (var graphics = Graphics.FromImage(bmpRet))
{
graphics.Clear(Color.White);
graphics.TranslateTransform(rotateAtX + x, rotateAtY + y);
graphics.RotateTransform(angle);
graphics.TranslateTransform(-rotateAtX - x, -rotateAtY - y);
graphics.DrawImage(bitmap, new PointF(0 + x, 0 + y));
}
return bmpRet;
}
Primary source

You need to apply transformation matrix.
Here you can find good example about transformations in GDI+