Every sample code I've ever found for drawing rounded rectangles using GDI+ goes something like this (lifted and slightly modified from BobPowell.net):
Private Sub Panel1_Paint(ByVal sender As Object, ByVal e As PaintEventArgs) Handles Panel1.Paint
e.Graphics.Clear(SystemColors.Window)
e.Graphics.SmoothingMode = SmoothingMode.None
Call DrawRoundRect(e.Graphics, Pens.Red, 10, 10, 48, 24, 6)
End Sub
Public Sub DrawRoundRect(ByVal g As Graphics, ByVal p As Pen, ByVal x As Single, ByVal y As Single, ByVal width As Single, ByVal height As Single, ByVal radius As Single)
Using gp As New GraphicsPath()
gp.StartFigure()
gp.AddArc(x + width - radius, y, radius * 2, radius * 2, 270, 90)
gp.AddArc(x + width - radius, y + height - radius, radius * 2, radius * 2, 0, 90)
gp.AddArc(x, y + height - radius, radius * 2, radius * 2, 90, 90)
gp.AddArc(x, y, radius * 2, radius * 2, 180, 90)
gp.CloseFigure()
g.DrawPath(p, gp)
End Using
End Sub
This produces a rounded rectangle where only the top left corner is accurate.
AntiAliasing has to be turned off because it is going through a remote desktop connection, and I can't depend on it being available. Besides, I am looking for a crisp rounded rectangle.
I've tried resizing the other corners and changing the pen alignments, but nothing seems to produce a simple, accurate rounded rectangle.
Is there a way to draw a better rounded rectangle than this in good old winforms?
1) Resize your source image to a binary multiple of its original size. Typically, I'll resample to a width and height 4 times greater (or 8, or 16) than the original.
2) Perform all my GDI+ drawing operations (taking into account, of course, that my co-ordinates will need to be multiplied by a factor of 4). There is no need to use any fancy anti-aliasing.
3) Re-sample the image back down to the original dimensions. Shrinking the image results in a nice smoothing effect, and minimizes any rounding errors in lines, curves, etc.
private Bitmap GenerateButton(int overSampling) {
int overSampling = 8;
int width=(48 + 10 + 10 + 6) * overSampling;
int height=(24 + 10 + 10 + 6) * overSampling;
// Draw the button with the rounded corners, but do
// so at 8 times the normal size.
Bitmap bitmap=new Bitmap(width,height);
using (Graphics g = Graphics.FromImage(bitmap)) {
g.Clear(Color.White);
g.SmoothingMode = SmoothingMode.None;
DrawRoundRect(overSampling, g, new Pen(Color.Red, overSampling), 10, 10, 48, 24, 6);
}
// Shrink the image down to its intended size
Bitmap shrunkVersion=new Bitmap(bitmap.Width / overSampling, bitmap.Height / overSampling);
using (Graphics g = Graphics.FromImage(shrunkVersion)) {
// Use hi-quality resampling for a nice, smooth image.
g.InterpolationMode = InterpolationMode.HighQualityBicubic;
g.DrawImage(bitmap, 0, 0, shrunkVersion.Width, shrunkVersion.Height);
}
return shrunkVersion;
}
private void DrawRoundRect(int overSampling, Graphics g, Pen p, float x, float y, float width, float height, float radius)
{
using (GraphicsPath gp = new GraphicsPath())
{
gp.StartFigure();
gp.AddArc((x + width - radius) * overSampling, y * overSampling, (radius * 2) * overSampling, (radius * 2) * overSampling, 270, 90);
gp.AddArc((x + width - radius) * overSampling, (y + height - radius) * overSampling, (radius * 2) * overSampling, (radius * 2) * overSampling, 0, 90);
gp.AddArc(x * overSampling, (y + height - radius) * overSampling, radius * 2 * overSampling, radius * 2 * overSampling, 90, 90);
gp.AddArc(x * overSampling, y * overSampling, radius * 2 * overSampling, radius * 2 * overSampling, 180, 90);
gp.CloseFigure();
g.DrawPath(p, gp);
}
}
Without oversampling:
With 8 times oversampling:
I have found the best solution to be just old-school Windows API:
Private Sub DrawRoundRect(ByVal g As Graphics, ByVal r As Rectangle)
Dim hDC As IntPtr = g.GetHdc
Dim hPen As IntPtr = CreatePen(PS_SOLID, 0, ColorTranslator.ToWin32(Color.Red))
Dim hOldPen As IntPtr = SelectObject(hDC, hPen)
SelectObject(hDC, GetStockObject(NULL_BRUSH))
RoundRect(hDC, r.Left, r.Top, r.Right - 1, r.Bottom - 1, 12, 12)
SelectObject(hDC, hOldPen)
DeleteObject(hPen)
g.ReleaseHdc(hDC)
End Sub
This produces the symmetrical rounded rectangle I've been looking for:
Because no-one's answered you yet here is a trick I have used in the past. It works reasonably well, and definitely looks better than the classic implementation with AddArc().
It uses circles and clipping to achieve the result you want. It may show slight artefacts when using pens with a width greater than 1px, but other than that it works well.
I hope it will be good enough for your project.
private void DrawRoundedRectangle(Graphics g, Pen pen, Rectangle rect, int radius)
{
g.DrawLine(pen, rect.Left + radius, rect.Top, rect.Right - radius, rect.Top);
g.DrawLine(pen, rect.Right, rect.Top+radius, rect.Right, rect.Bottom - radius);
g.DrawLine(pen, rect.Left + radius, rect.Bottom, rect.Right - radius, rect.Bottom);
g.DrawLine(pen, rect.Left, rect.Top + radius, rect.Left, rect.Bottom - radius);
g.SetClip(new Rectangle(rect.Left, rect.Top, radius, radius));
g.DrawEllipse(pen, rect.Left, rect.Top, radius * 2, radius * 2);
g.ResetClip();
g.SetClip(new Rectangle(rect.Right-radius, rect.Top, radius+1, radius+1));
g.DrawEllipse(pen, rect.Right - radius * 2, rect.Top, radius * 2, radius * 2);
g.ResetClip();
g.SetClip(new Rectangle(rect.Right - radius, rect.Bottom-radius, radius+1, radius+1));
g.DrawEllipse(pen, rect.Right - radius * 2, rect.Bottom - (radius * 2), radius * 2, radius * 2);
g.ResetClip();
g.SetClip(new Rectangle(rect.Left, rect.Bottom - radius, radius+1, radius+1));
g.DrawEllipse(pen, rect.Left, rect.Bottom - (radius * 2), radius * 2, radius * 2);
g.ResetClip();
}
The method's interface is straightforward, but post a comment if you need assistance.
Edit: Something else that should work is to draw the same arc four times, but flipped using TranslateTransform and TranslateScale. That should mean the arc appears identical in each corner.
private void DrawRoundedRectangle(Graphics g, Pen pen, Rectangle rect, int radius)
{
g.DrawLine(pen, rect.Left + radius, rect.Top, rect.Right - radius, rect.Top);
g.DrawLine(pen, rect.Right-1, rect.Top+radius, rect.Right-1, rect.Bottom - radius);
g.DrawLine(pen, rect.Left + radius, rect.Bottom-1, rect.Right - radius, rect.Bottom-1);
g.DrawLine(pen, rect.Left, rect.Top + radius, rect.Left, rect.Bottom - radius);
g.TranslateTransform(rect.Left, rect.Top);
g.DrawArc(pen, 0, 0, radius * 2, radius * 2, 180, 90);
g.ResetTransform();
g.TranslateTransform(rect.Right, rect.Top);
g.ScaleTransform(-1, 1);
g.DrawArc(pen, 1, 0, radius * 2, radius * 2, 180, 90);
g.ResetTransform();
g.TranslateTransform(rect.Right, rect.Bottom);
g.ScaleTransform(-1, -1);
g.DrawArc(pen, 1, 1, radius * 2, radius * 2, 180, 90);
g.ResetTransform();
g.TranslateTransform(rect.Left, rect.Bottom);
g.ScaleTransform(1, -1);
g.DrawArc(pen, 0, 1, radius * 2, radius * 2, 180, 90);
g.ResetTransform();
}
This is similar to the old Computer Graphics method of drawing a circle, where you'd draw a quarter circle four times to avoid rounding errors such as the one in GDI.
Another alternative is to draw the first arc onto an image, and then draw the image four times, flipping as required. Below is a variation on the second method, using an image to draw the arcs.
private void DrawRoundedRectangle(Graphics g, Pen pen, Rectangle rect, int radius)
{
g.DrawLine(pen, rect.Left + radius, rect.Top, rect.Right - radius, rect.Top);
g.DrawLine(pen, rect.Right - 1, rect.Top + radius, rect.Right - 1, rect.Bottom - radius);
g.DrawLine(pen, rect.Left + radius, rect.Bottom - 1, rect.Right - radius, rect.Bottom - 1);
g.DrawLine(pen, rect.Left, rect.Top + radius, rect.Left, rect.Bottom - radius);
Bitmap arc = new Bitmap(radius, radius, g);
Graphics.FromImage(arc).DrawArc(pen, 0, 0, radius * 2, radius * 2, 180, 90);
g.TranslateTransform(rect.Left, rect.Top);
g.DrawImage(arc, 0, 0);
g.ResetTransform();
g.TranslateTransform(rect.Right, rect.Top);
g.ScaleTransform(-1, 1);
g.DrawImage(arc, 0, 0);
g.ResetTransform();
g.TranslateTransform(rect.Right, rect.Bottom);
g.ScaleTransform(-1, -1);
g.DrawImage(arc, 0, 0);
g.ResetTransform();
g.TranslateTransform(rect.Left, rect.Bottom);
g.ScaleTransform(1, -1);
g.DrawImage(arc, 0, 0);
g.ResetTransform();
arc.Dispose();
}
On occasion, I've used a "low tech" approach to deal with the rounding errors in GDI+
1) Resize your source image to a binary multiple of its original size. Typically, I'll resample to a width and height 4 times greater (or 8, or 16) than the original.
2) Perform all my GDI+ drawing operations (taking into account, of course, that my co-ordinates will need to be multiplied by a factor of 4). There is no need to use any fancy anti-aliasing.
3) Re-sample the image back down to the original dimensions. Shrinking the image results in a nice smoothing effect, and minimizes any rounding errors in lines, curves, etc.
Related
I'm creating a game in LibGDX and am using Tiled as my map system.
I'm trying to contain an OrthographicCamera within the bounds of my TiledMap. I use MathUtils.clamp to achieve this. When the camera is at a normal zoom of 1.0f, it works perfectly. However when the camera is zoomed in further, to lets say .75f, the camera is clamped to the wrong location because it has no information of the zoom value.
position.x = MathUtils.clamp(position.x * (gameScreen.gameCamera.camera.zoom), gameScreen.gameCamera.camera.viewportWidth / 2, gameScreen.mapHandler.mapPixelWidth - (gameScreen.gameCamera.camera.viewportWidth / 2));
position.y = MathUtils.clamp(position.y * (gameScreen.gameCamera.camera.zoom), (gameScreen.gameCamera.camera.viewportHeight / 2), gameScreen.mapHandler.mapPixelHeight - (gameScreen.gameCamera.camera.viewportHeight / 2));
My question: How do I include the zoom value in my clamp code so the camera is correctly clamped? Any ideas?
Thank you!
- Jake
You should multiply by zoom the world size, not the camera position:
float worldWidth = gameScreen.mapHandler.mapPixelWidth;
float worldHeight = gameScreen.mapHandler.mapPixelHeight;
float zoom = gameScreen.gameCamera.camera.zoom;
float zoomedHalfWorldWidth = zoom * gameScreen.gameCamera.camera.viewportWidth / 2;
float zoomedHalfWorldHeight = zoom * gameScreen.gameCamera.camera.viewportHeight / 2;
//min and max values for camera's x coordinate
float minX = zoomedHalfWorldWidth;
float maxX = worldWidth - zoomedHalfWorldWidth;
//min and max values for camera's y coordinate
float minY = zoomedHalfWorldHeight;
float maxY = worldHeight - zoomedHalfWorldHeight;
position.x = MathUtils.clamp(position.x, minX, maxX);
position.y = MathUtils.clamp(position.y, minY, maxY);
Note, that if a visible area can be smaller than the world size, then you must handle such situations differently:
if (maxX <= minX) {
//visible area width is bigger than the worldWidth -> set the camera at the world centerX
position.x = worldWidth / 2;
} else {
position.x = MathUtils.clamp(position.x, minX, maxX);
}
if (maxY <= minY) {
//visible area height is bigger than the worldHeight -> set the camera at the world centerY
position.y = worldHeight / 2;
} else {
position.y = MathUtils.clamp(position.y, minY, maxY);
}
I am developing a app that use picking selection using color depth in OpenGL for Mac. The selection works great, but I observed that OpenGL not draw perfect circles (circunferences). I use GL_LINE_LOOP for render 360 segments for complete a close circunference. But if, I see the render in SELECTION mode I observe some kind of bites (triangles in white) over the circunference, 4 in total for each circle.
Is a OpenGL problem or I am doing something wrong?
This is the code for draw each circunference.
glLineWidth(10.0f); // Exaggerated for image.
glBegin(GL_LINE_LOOP);
for (int i = 0; i <= 360; i++) {
calcX = radio * cos((i * M_PI) / 180) + offset_x;
calcY = radio * sin((i * M_PI) / 180) + offset_y;
glVertex3f(calcX, calcY, offset_z);
}
glEnd();
Thanks "radical7, for the the suggest to use Triangles instead lines. I was able to obtain a perfect RING, works as smooth .
This is the code:
color = 0.00277778;
glBegin(GL_TRIANGLE_STRIP);
for (int i = 0; i <= 360; i++) {
glColor4f(color, 0, 0, 1.0f);
calcX = radio * cos((i * M_PI) / 180) + offset_x;
calcY = radio * sin((i * M_PI) / 180) + offset_y;
glVertex3f(calcX, calcY, offset_z);
radio +=20;
calcX = radio * cos((i * M_PI) / 180) + offset_x;
calcY = radio * sin((i * M_PI) / 180) + offset_y;
glVertex3f(calcX, calcY, offset_z);
radio -=20;
color += 0.00277778;
}
glEnd();
Result:
Currently, I am working with a ray tracer that takes an iterative approach towards developing the scenes. My goal is to turn it into a recursive ray tracer.
At the moment, I have a ray tracer defined to do the following operation to create the bitmap it is stored in:
int WIDTH = 640;
int HEIGHT = 640;
BMP Image(WIDTH, HEIGHT); // create new bitmap
// Slightly shoot rays left of right camera direction
double xAMT, yAMT;
*/
Color blue(0.1, 0.61, 0.76, 0);
for (int x = 0; x < WIDTH; x++) {
for (int y = 0; y < HEIGHT; y++) {
if (WIDTH > HEIGHT) {
xAMT = ((x + 0.5) / WIDTH) * aspectRatio - (((WIDTH - HEIGHT) / (double)HEIGHT) / 2);
yAMT = ((HEIGHT - y) + 0.5) / HEIGHT;
}
else if (HEIGHT > WIDTH) {
xAMT = (x + 0.5) / WIDTH;
yAMT = (((HEIGHT - y) + 0.5) / HEIGHT) / aspectRatio - (((HEIGHT - WIDTH) / (double)WIDTH) / 2);
}
else {
xAMT = (x + 0.5) / WIDTH;
yAMT = ((HEIGHT - y) + 0.5) / HEIGHT;
}
..... // calculate intersections, shading, reflectiveness.... etc
Image.setPixel(x, y, blue); // this is here just as an example
}
}
Is there another approach to calculating the reflective and refractive child rays outside the double for-loop?
Are the for-loops necessary? // yes because of the bitmap?
What approaches can be taken to minimize/optimize an iterative ray tracer?
I am writing an objective-c method that draws a series of triangles on a slope. In order to complete this, I need to calculate the vertex point of each triangle (C,D). The position starting and ending points are variable.
This seems like it should be an easy math problem. But so far I haven't been able to work it out on paper. Can anyone point me in the right direction?
No trigonometry involved.
Let D= Sqrt(X12^2+Y12^2) the Euclidean distance between P1 and P2 (X12 = X2-X1 and Y12 = Y2-Y1), and let p= P/D, a= A/D.
If P1P2 was the line segment (0, 0)-(1, 0), the vertices would be at (0, 0), (a, p/2), (0, p), (a, 3p/2), (0, 2p)...
The transform below scales and rotates (0, 0)-(1, 0) to P1P2:
X = X1 + X12.x - Y12.y
Y = Y1 + Y12.x + X12.y
Set triangle at origin horizontally:
(0, 0), (p, 0), (p/2, a)
Rotate to get needed slope alpha:
(0, 0), (p*cos(alpha), p*sin(alpha)), (p/2 * cos(alpha) - a * sin(alpha), p/2 * sin(alpha) + a*sin(alpha))
Shift by adding (x1, y1) to all of the coordinates.
The third coordinate is your vertex:
(Cx, Cy) = (p/2 * cos(alpha) - a * sin(alpha) + x1, p/2 * sin(alpha) + a*sin(alpha) + y1)
To find other vertices use the fact that they are shifted by p from each other, under the angle alpha:
(Cx_i, Cy_i) = (Cx, Cy) + i*(p * cos(alpha), p * sin(alpha))
I have a rectangle that has to be rotated always the same amount of degrees. Lets call this angle alpha (𝜶).
The width (w) and height (h) of this rectangle can vary. The rectangle has always to fit rotated inside the big rectangle. It must be scaled up or down to fit inside the gray rectangle.
NOTE: Alpha is the angle between w and the horizontal line.
So, there are 3 kinds of rectangles where
w > h
w < h or
w = h
See the picture below.
What I know:
The big rectangle has width of R and height of K and I know both values;
w and h are unknown;
the rectangle is always rotated 𝜶 degrees;
I know the value of w/h. I call this "ratioWH";
red rectangle is always centered horizontally and vertically on the gray rectangle
what I need to know:
the maximum values of w and h that will fit the gray rectangle for each case of w and h.
the coordinates of point P, assuming that 0,0 is at the upper left of the gray rectangle.
This is what I did so far, but this is not giving the correct values:
CGPoint P = CGPointZero;
if (ratioWH > 0) { // means w > h
maxH = R / (ratioWH * fabsf(cosf(theta)) + fabsf(sinf(theta)));
maxW = maxH * ratioWH;
// P.x = 0.0f; // P.x is already zero
CGFloat marginY = (K - maxW * fabsf(sinf(theta)) - maxH * fabsf(cosf(theta))) / 2.0f;
P.y = marginY + maxW * fabsf(sinf(theta));
} else { // w <= h
maxW = K / (fabsf(cosf(theta) / ratioImagemXY) + fabsf(sinf(theta)));
maxH = maxW / ratioWH;
P.x = (R - maxW * fabsf(cosf(theta)) - maxH * fabsf(sinf(theta))) / 2.0f;
P.y = maxW * fabsf(sinf(theta));
}
any clues? Thanks.
The way I see it is like this... You work out the total width and total height of the rectangle. For that, you simply walk along two edges. Like this:
dx = w * cos(theta) + h * sin(theta)
dy = h * cos(theta) + w * sin(theta)
These could be negative, so special handling would apply if the rectangle is rotated into other quadrants. This will happen later.
You now just need the ratio between the width and height. This is where you decide whether to scale by the vertical amount or the horizontal amount. It's nothing to do with w and h -- it's actually about where the rectangle ends up as a result of rotation. That's what dx and dy are for.
rectratio = abs( dx / dy )
viewratio = R / K
If rectratio comes out larger than viewratio that means the rotated rectangle's horizontal footprint needs to be scaled. Otherwise you scale by the vertical footprint.
if rectratio > viewratio
scale = R / abs(dx)
else
scale = K / abs(dy)
end
And the scale itself is applied to the original width and height
sw = scale * w
sh = scale * h
So now you can compute the corners of your rectangle. It doesn't matter where you start.
x[0] = 0
x[1] = x[0] + sw * cos(theta)
x[2] = x[1] + sh * sin(theta)
x[3] = x[2] - sw * cos(theta)
y[0] = 0
y[1] = y[0] - sw * sin(theta)
y[2] = y[1] + sh * cos(theta)
y[3] = y[2] + sw * sin(theta)
I've assumed image co-ordinates given that (0,0) is top-left, so increasing y moves down. So, if I haven't made a mistake in my math, the above gives you the rectangle vertices (in clockwise order).
The last thing to do is normalise them... This means finding the min value of px and py. Call them pxmin and pymin. I don't need to show code for that. The idea is to calculate an offset for your rectangle such that the view area is defined by the rectangle (0,0) to (R,K).
First we need to find the left and right value of the subview that completely contains our rotated rectangle... Remember the ratio before:
if( rectratio > viewratio )
// view is too tall, so centre vertically:
left = 0
top = (K - scale * abs(dy)) / 2.0
else
// view is too wide, so centre horizontally:
left = (R - scale * abs(dx)) / 2.0
top = 0
end
left and top are now the 'minimum' co-ordinate of our subview that exactly contains the rectangle (floating point rounding errors exempted). So:
left += pxmin
top += pymin
Now they are the offset required to shift the rectangle to where it's wanted. All you do is add left and top to all your rectangle co-ordinates, and you are done. The position of P is px[0] and py[0]. If you rotated by 90 degrees or more, it won't be the top-left vertex.