Circles are solid and there's no way to create an empty space inside them, is there?
It is possible to create it using a combination of the basic shapes (Chains/Edges) but I was wondering if there was an easier way?
The easiest way to make a ring would be draw a larger circle of the primary color, then draw a smaller circle inside of the background color (if it's a solid color.) Otherwise, use a texture with a transparent center.
No, there is no way to create a space inside a circle in box2d. ChainShape and EdgeShape will work, but only if you are creating a ring that doesn't move. If you want a dynamic body your only option is compound polygon shape (Body with few convex polygon fixtures). Compound polygon shape is a good idea for static ring too.
I have an hourglass like vector shape and I'd like to use it to mask an image. I'd like to feather the edges - have a soft falloff in transparency that follows the contours of the hour glass. Any ideas how I can do this?
I tried using a gradient fill on a closed shape (using beginGradientFill() and curveTo() functions) but that falloff doesn't follow the contour of the vector shape, it can only go one direction.
Maybe there is a better solution but until somebody comes up with it... I assume you could do the following:
Draw whatever shape you want to use as mask into a transparent bitmap.
Scale a bit the bitmap down (or use a matrix while drawing its bitmapdata).
Apply a blur filter to it.
Put the bitmap's center to the masked clip's center so they are aligned.
Set the masked clip's cacheAsBitmap property to true.
I'm developing a simple a graphical editor for my flash-based app. In my editor there's a posibility of scaling, range of scaling is big (maximum scale is 16.0, minimum scale is 0.001 and default scale is 0.2). So it's quite possible that a user can draw a line with thickness 0.1 or 300.0, and it looks that line possible thickness (in Graphics.lineStyle()) has upper border. As I found out from livedocs maximum value is 255. So if thickness is greater then 255.0 there'is drawn a line of thickness 255.0. Whether mentioned upper border exists and how big is it. Here're my questions:
Right now I'm drawing lines with drawPath() or lineTo() methods. Natural walkarround if thickness is greater then 255.0 is to draw a rectange instead of segment and two circles on the ends of segment (instead of lineTo()). Or even to draw two thin segments and two half-circles and fill interior. Maybe there's more elegant/quick solution?
Another question is if the thickness of line is big but less then 255.0 (e.g. 100.0), what is faster drawing a line with lineTo() or drawing two thin segments and two half-circles and fill interior?
And finally, maybe someone knows a good article/book where I can read what's inside all methods of flash.display.Graphics class (or even not flash specific article/book on graphics)?
Any thoughts are appreciated. Thank you in advance!
I agree with f-a that putting the line in a container would probably be better and more efficient than drawing a rectangle and extra circles.
I don't think that the math would be too difficult to work out. For efficiency you should probably only do this if the line style is going to be over 255.
To setup the display object to hold your line I would start by halving the width of your line (the length can stay the same). Then create a new sprite and draw the line in the sprite at half size (e.g. if you wanted 300, just draw it at 150). It would be most simple to just start at (0,0) and draw the segment straight so that all of your transformations can be applied to the new sprite.
From here you can just double the scaleY of the sprite to get the desired line weight. It should keep the same length and the ends should also be rounded correctly.
Hope this helped out!
A cool resource for working with the graphics class is Flash and Math. This site has several cool effects and working examples and source code.
http://www.flashandmath.com/
DisplayObject.getBounds in actionscript returns the bounds of the object with the strokes included. The left, top, width, height properties of a SymbolInstance in JSFL don't seem to include the strokes. That's the only way I've found to get the bounds of a symbol from JSFL. Is there another way?
You are looking for the Edge object on a Shape. The Edge has a Stroke object that has a thickness property.
// This will show the selected shape's first edge's thickness:
fl.trace(fl.getDocumentDOM().selection[0].edges[0].stroke.thickness );
You will have to loop over all the shapes and all of their edges to determine final bounds (if you are confident that all the edges have the same thickness, just check one).
Strokes have 0 width to JSFL, when it comes to getting the bounds of an object.
The only method I can think of is to edit the symbol, select the shape, and either
1.) get the stroke size and add 1/2 of its value to your calculation, or
2.) convert the stroke to a fill (unreliable for complex outlines)
If you only wish to include the strokes but exact sizing is not crucial, you can just arbitrarily add some pixels to the result of getBounds.
I am trying to understand the method transition that falls in the Matrix Class. I am using it to copy pieces of a bitMapData. But I need to better understand what transitions do.
I have a tilesheet that has 3 images on it. all 30x30 pixels. the width of the total bitmap is 90pxs.
The first tile is green, the second is brown, and the third is yellow. If I move over 30pxs using the matrix that transitions, instead of getting brown, I get yellow, if I move over 60px, I get brown.
If I move -30 pixels, then the order is correct. I am confused on what is going on.
tileNum -= (tileNumber * tWidth);
theMatrix = new Matrix();
theMatrix.translate(tileNum,0);
this.graphics.beginBitmapFill(tileImage,theMatrix);
this.graphics.drawRect(0, 0,tWidth ,tHeight );
this.graphics.endFill();
Can someone tell me how transitions work, or some resources that show how they work. I ultimately want to know a good way to switch back and forth between each tile.
First of all, don't confuse translation with transition. The latter is a general English word for "change", whereas to translate in geometry and general math is to "move" or "offset" something.
A transformation matrix defines how to transform, i.e. scale, rotate and translate, an object, usually in a visual manner. By applying a transformation matrix to an object, all pixels of that object are rotated, moved and scaled/interpolated according to the values stored inside the matrix. If you'd rather not think about matrix math, just think of the matrix as a black box which contains a sequence of rotation, scaling, and translation commands.
The translate() method simply offsets the bitmap that you are about to draw a number of pixels in the X and Y dimensions. If you use the default ("identity") matrix, which contains no translation, the top left corner of your object/bitmap will be in the (0,0) position, known as the origin or registration point.
Consider the following matrix:
var mtx : Matrix = new Matrix; // No translation, no scale, no rotation
mtx.translate(100, 0); // translated 100px on X axis
If you use the above matrix with a BitmapData.draw() or Graphics.beginBitmapFill(), that means that the top left corner of the original bitmap should be at (x=100; y=0) in the target coordinate system. Sticking to your Graphics example, lets first consider drawing a rectangle without a matrix transformation.
var shape : Shape = new Shape;
shape.graphics.beginBitmapFill(myBitmap);
shape.graphics.drawRect(0, 0, 200, 200);
This will draw a 200x200 pixels rectangle. Since there is no transformation involved in the drawing method (we're not supplying a transformation matrix), the top left corner of the bitmap is in (x=0; y=0) of the shape coordinate system, i.e. aligned with the top left corner of the rectangle.
Lets look at a similar example using the matrix.
var shape : Shape = new Shape;
shape.graphics.beginBitmapFill(myBitmap, mtx);
shape.graphics.drawRect(0, 0, 200, 200);
This again draws a rectangle that is 200px wide and 200px high. But where inside this rectangle will the top left corner of myBitmap be? The answer is at (x=100, y=0) of the shape coordinate system. This is because the matrix defines such a translation.
But what then will be to the left of (x=100; y=0)? With the above code, the answer is that the bitmap repeats to fill the entire rectangle, and hence you will see the rightmost side of the bitmap, to the left of the leftmost side, as if there was another instance of the bitmap right next to it. If you want to disable the repeating image, set the third attribute of beginBitmapFill() to false:
shape.graphics.beginBitmpFill(myBitmap, mtx, false);
Lets take a look at one last example that might help your understanding. Remember that the translation matrix defines the position of the top left corner of an image, in the coordinate system of the shape. With this in mind, consider the following code, using the same matrix as before.
var shape : Shape = new Shape;
shape.graphics.beginBitmapFill(myBitmap, mtx);
shape.graphics.drawRect(100, 0, 100, 100);
Notice that this will draw the rectangle 100px in on the X axis. Not coincidentally, this is the same translation that we defined in our matrix, and hence the position of the top left corner of the bitmap. So even though repeating is enabled, we will not see a repeating image to the left of our rectangle, because we only start drawing at the point where the bitmap starts.
So the bottom line is, I guess, that you could think of the transform matrix as a series of transformation commands that you apply to your image as you draw it. This will offset, scale and rotate the image as it's drawn.
If you are curious about the inner workings of the matrix, Google transformation matrices, or read up on Linear Algebra!