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!
Related
I split my canvas into two. I need to draw on each side, defining a clipping region for both. Each side has to do similar things, like drawing text in colours, drawing circles etc.
I have read that you should keep your fillStyle and strokeStyle changes to a minimum. However I have also read you must keep your save and restores to a minimum as well.
So what is faster?
Save the canvas, clip the left hand side, do ALL the drawing for that side for the multiple colours, then restore and repeat for the right hand side?
Or ... Set the first text colour. Clip the left hand side. Draw all text for this colour. Then clip the right hand side (without restting fillStyle) and draw all the same colour text for the right hand side. Then set the next fillStyle and then clip each side and draw text for this colour? Etc
Anyone know?
Also if I set one clip region, then set another clip region without saving and restoring, what actually happens?
As always, you must performance test with your own project's code.
The context maintains internal variables relating to its current state (colors, transforms, current path, compositing applied, etc).
You can save a copy of the current context state using context.save. Then after altering the context state, you can restore the original context state with context.restore. Note: you can nest multiple save/restore if needed.
fillStyle is a context state so changing fillStyle='green' followed later by fillStyle='blue' is faster than save/restore because just 1 state variable is being reset rather than every state variable that's done during save/restore.
The point is that you often gain performance by saving+resetting individual state values rather than doing a full context.save/context.restore.
Minimizing state changes will maximize performance, so for example, batching all green drawings will help performance.
Clipping is a complex operation and is more expensive than simple state changes like changing fillStyle. With GPU acceleration the clipping is accomplished much more efficiently and is therefore considerably less costly than clipping without a GPU.
About changing clipping regions: Clipping is always done only on the last path defined. So setting another clipping region will undo the previous clipping region (unless the previous and current path are identical).
About left and right clipping regions: If your design permits, you might want to define both the left and right clipping regions at once into 1 combined clipping region. Since paths can be disconnected, defining 1 clipping path with 2 non-intersecting parts is allowed.
As the car companies warn: "Your mileage may vary". But you might perf test this method:
Define your left and right clipping regions in 1 clipping path
Set fillStyle="red" and do all your red drawings at once (both right and left)
Reset fillStyle="blue" and do all your blue drawings at once, etc, etc.
One final thought: All projects have their own distinct requirements so you must perf test your actual code rather than relying exclusively on general rules. Don't skip the perf tests--especially if you will be deploying on mobile where canvas is slow by nature.
Good luck with your project!
[ Additional note about clipping ]
A clipping region is semi-permanent. Once it is set it remains even if the context.clip command is issued again. The new clip command will further restrict all previous clipping regions.
To clear a clipping path you must wrap your clips inside context.save & context.restore(). Alternatively, resizing the canvas will force the context state to be restored to defaults--but resizing will also clear all canvas content: canvas.width=canvas.width
Here's and example of a single path consisting of a left and right square:
// left square
ctx.beginPath();
ctx.moveTo(50,50);
ctx.lineTo(150,50);
ctx.lineTo(150,150);
ctx.lineTo(50,150);
ctx.closePath();
// right square
ctx.moveTo(200,50);
ctx.lineTo(300,50);
ctx.lineTo(300,150);
ctx.lineTo(200,150);
ctx.closePath();
ctx.strokeStyle="red";
ctx.stroke();
Create a clipping region from this left+right path.
// create a clipping region from the left+right path
ctx.clip();
Fill the entire canvas with green. The green will only be drawn inside the clipping region (inside the left and right squares).
// draw a green rect over the entire canvas
// the green will only be drawn inside the clipping region
ctx.fillStyle="green";
ctx.fillRect(0,0,canvas.width,canvas.height);
If you add a second clipping region, the result is that drawing will only be visible in the union of all clipping paths.
So if the second clipping region slightly overlaps the first, all drawings will only be visible inside the union of the first and second clipping regions. The blue section in the following illustration is the union of the 2 clipping paths.
ctx.beginPath();
ctx.moveTo(0,115);
ctx.lineTo(canvas.width,115);
ctx.lineTo(canvas.width,135);
ctx.lineTo(0,135);
ctx.clip();
ctx.closePath();
ctx.fillStyle="blue";
ctx.fillRect(0,0,canvas.width,canvas.height);
A few ideas :
To handle your two screens, you might use two canvases side by side.
Even if you keep only one canvas, you can reduce the use of clipping by doing :
- erase left half of screen
- draw left part
- erase right part
- clip on right half
- draw right part
Thus you clip only once.
For your color vs clip concern, clipping is much costly than changing color.
And if one keeps on clipping with no save/restore, clipping zones will add-up.
For your strokeStyle/fillStyle, yes there's a cost, especially if using color names ('blue'), or rgb 'rgb()' strings or worse 'hsl()' strings.
So here's a trick : pre-compute and store the colors whenever you can. Simply use the context to convert !
var blue = 'hsl(23, 75%, 75%)';
context.fillStyle = blue ;
blue = context.fillStyle ;
Using the first photo below, let's say:
The red outline is the stage bounds
The gray box is a Sprite on the stage.
The green box is a child of the gray box and has a rotation set.
both display object are anchored at the top-left corner (0,0).
I'd like to rotate, scale, and position the gray box, so the green box fills the stage bounds (the green box and stage have the same aspect ratio).
I can negate the rotation easily enough
parent.rotation = -child.rotation
But the scale and position are proving tricky (because of the rotation). I could use some assistance with the Math involved to calculate the scale and position.
This is what I had tried but didn't produce the results I expected:
gray.scaleX = stage.stageWidth / green.width;
gray.scaleY = gray.scaleX;
gray.x = -green.x;
gray.y = -green.y;
gray.rotation = -green.rotation;
I'm not terribly experienced with Transformation matrices but assume I will need to go that route.
Here is an .fla sample what I'm working with:
SampleFile
You can use this answer: https://stackoverflow.com/a/15789937/1627055 to get some basics. First, you are in need to rotate around the top left corner of the green rectangle, so you use green.x and green.y as center point coordinates. But in between you also need to scale the gray rectangle so that the green rectangle's dimensions get equal to stage. With uniform scaling you don't have to worry about distortion, because if a gray rectangle is scaled uniformly, then a green rectangle will remain a rectangle. If the green rectangle's aspect ratio will be different than what you want it to be, you'd better scale the green rectangle prior to performing this trick. So, you need to first transpose the matrix to offset the center point, then you need to add rotation and scale, then you need to transpose it away. Try this set of code:
var green:Sprite; // your green rect. The code is executed within gray rect
var gr:Number=green.rotation*Math.PI/180; // radians
var gs:Number=stage.stageWidth/green.width; // get scale ratio
var alreadyTurned:Boolean; // if we have already applied the rotation+scale
function turn():void {
if (alreadyTurned) return;
var mat:flash.geom.Matrix=this.transform.matrix;
mat.scale(gs,gs);
mat.translate(-gs*green.x,-gs*green.y);
mat.rotate(-1*gr);
this.transform.matrix=mat;
alreadyTurned=true;
}
Sorry, didn't have time to test, so errors might exist. If yes, try swapping scale, translate and rotate, you pretty much need this set of operations to make it work.
For posterity, here is what I ended up using. I create a sprite/movieClip inside the child (green) box and gave it an instance name of "innerObj" (making it the actually content).
var tmpRectangle:Rectangle = new Rectangle(greenChild.x, greenChild.y, greenChild.innerObj.width * greenChild.scaleX, greenChild.innerObj.height * greenChild.scaleY);
//temporary reset
grayParent.transform.matrix = new Matrix();
var gs:Number=stage.stageHeight/(tmpRectangle.height); // get scale ratio
var mat:Matrix=grayParent.transform.matrix;
mat.scale(gs,gs);
mat.translate(-gs * tmpRectangle.x, -gs * tmpRectangle.y);
mat.rotate( -greenChild.rotation * Math.PI / 180);
grayParent.transform.matrix = mat;
If the registration point of the green box is at one of it's corners (let's say top left), and in order to be displayed this way it has a rotation increased, then the solution is very simple: apply this rotation with negative sign to the parent (if it's 56, add -56 to parent's). This way the child will be with rotation 0 and parent -> -56;
But if there is no rotation applied to the green box, there is almost no solution to your problem, because of wrong registration point. There is no true way to actually determine if the box is somehow rotated or not. And this is why - imagine you have rotated the green box at 90 degrees, but changed it's registration point and thus it has no property for rotation. How could the script understand that this is not it's normal position, but it's flipped? Even if you get the bounds, you will see that it's a regular rectangle, but nobody know which side is it's regular positioned one.
So the short answer is - make the registration point properly, and use rotation in order to display it like in the first image. Then add negative rotation to the parent, and its all good :)
Edit:
I'm uploading an image so I can explain my idea better:
As you can see, I've created a green object inside the grey one, and the graphics INSIDE are rotated. The green object itself, has rotation of 0, and origin point - top left.
#Vesper - I don't think that the matrix will fix anything in this situation (remember that the green object has rotation of 0).
Otherwise I agree, that the matrix will do a pretty job, but there are many ways to do it :)
The problem
I'm trying to create a brush tool with opacity jitter (like in Photoshop). The specific problem is:
Draw a stroke on an HTML canvas with different levels of opacity. Pixels with higher opacity should replace pixels with lower opacity; otherwise, pixels are left unchanged.
Transparency should not be lost in the process. The stroke is drawn on a separate canvas and merged with a background canvas afterwards.
The result should look like this. All code and the corresponding output can be found here (JSFiddle).
Because you can't stroke a single path with different levels of opacity (please correct me if I'm wrong) my code creates a path for each segment with different opacity.
Non-solution 1, Using the 'darken' blend mode
The darken blend mode yields the desired result when using opaque pixels but doesn't seem to work with transparency. Loosing transparency is a dealbreaker.
With opaque pixels:
With transparent pixels:
Non-solution 2, Using the 'destination-out' compositing operator
Before drawing a new stroke segment, subtract its opacity from subjacent pixels by using the 'destination-out' compositing operator. Then add the new stroke segment with 'source-over'. This works almost but it's a little bit off.
Looking for a solution
I want to avoid manipulating each pixel by hand (which I have done in the past). Am I missing something obvious? Is there a simple solution to this problem?
"Links to jsfiddle.net must be accompanied by code."
Because you can't stroke a single path with different levels of opacity (please correct me if I'm wrong)
You're wrong =)
When you use globalCompositeOperation = 'destination-out' (which you are in lineDestinationOut) you need to set the strokeStyle opacity to 1 to remove everything.
However, simply changing that in your fiddle doesn't have the required effect due to the order of your path build. Build the 10% transparent one first, the whole length, then delete and draw the two 40% transparent bits.
Here's a jsfiddle of the code below
var canvas = document.getElementById('canvas');
var cx = canvas.getContext('2d');
cx.lineCap = 'round';
cx.lineJoin = 'round';
cx.lineWidth = 40;
// Create the first line, 10% transparency, the whole length of the shape.
cx.strokeStyle = 'rgba(0,0,255,0.1)';
cx.beginPath();
cx.moveTo(20,20);
cx.lineTo(260,20);
cx.lineTo(220,60);
cx.stroke();
cx.closePath();
// Create the first part of the second line, first by clearing the first
// line, then 40% transparency.
cx.strokeStyle = 'black';
cx.globalCompositeOperation = 'destination-out';
cx.beginPath();
cx.moveTo(20,20);
cx.lineTo(100,20);
cx.stroke();
cx.strokeStyle = 'rgba(0,0,255,0.4)';
cx.globalCompositeOperation = 'source-over';
cx.stroke();
cx.closePath();
// Create the second part of the second line, same as above.
cx.strokeStyle = 'black';
cx.globalCompositeOperation = 'destination-out';
cx.beginPath();
cx.moveTo(180,20);
cx.lineTo(260,20);
cx.stroke();
cx.strokeStyle = 'rgba(0,0,255,0.4)';
cx.globalCompositeOperation = 'source-over';
cx.stroke();
cx.closePath();
Use two layers to draw to:
First calculate the top layer opacity 40% - 10% and set this as alpha on top layer
Set bottom layer to 10%
Set top layer with dashed lines (lineDash) (calculate the dash-pattern size based on size requirements)
Draw lines to both layers and the bottom layer will be a single long line, the top layer will draw a dashed line on top when stroked.
Copy both layers to main canvas when done.
#HenryBlyth's answer is probably the best you're going to get; there's no native API to do what you're being asked to do (which, in my opinion, is kinda weird anyways... opacity isn't really supposed to replace pixels).
To spell out the solution in one paragraph: Split up your "stroke" into individual paths with different opacities. Draw the lowest opacity paths as normal. Then, draw the higher opacities with "desitination-out" to remove the low-opacity paths that overlap. Then, draw the high opacity paths as usual, with "source-over", to create the effect desired.
As suggested in the comments to that answer, #markE's comment about making each path an object that is pre-sorted before drawing is a great suggestion. Since you want to perform manual drawing logic that the native API can't do, turning each path into an object and dealing with them that way will be far easier than manually manipulating each pixel (though that solution would work, it could also drive you mad.)
You mention that each stroke is being done on another canvas, which is great, because you can record the mouseevents that fire as that line is being drawn, create an object to represent that path, and then use that object and others in your "merged" canvas without having to worry about pixel manipulation or anything else. I highly recommend switching to an object-oriented approach like #markE suggested, if possible.
I'm currently struggling on a problem that seems far beyond my maths capacities (been a long time since I've made some proper maths...) and I would appreciate some help on that.
Here's my setting :
I got some simple shapes (rectangles), and I "project" their bottom points on a line, coming from an Origin point.
Up to this point everything is fine.
But now I'd like to draw the original shape distorted as if it was projected with some perspective on a plane.
Please consider that I have nothing related to any rotation, isometric or any 3D or fake 2D perspective in my code, I'm only trying to draw some shapes using the graphics library to only have a feeling of something real.
Here's a quick drawing of what I'm trying to do :
What I know :
Origin point coordinates
the rect position & sizes
the red line position
the A & B points coordinates
What I want to determine is the coordinates of the C & D points, thing that could be easy if I wasn't struggling to find the "Origin bis" coordinates.
What I'm trying to do is to fake the projection of my rectangle on something that can be considered as a "floor" (related to the plane where my original rectangle is that can be seen as a wall).
Maybe I'm over-complicating the problem or maybe I fail to see any other easier way to do it, but I'm really not good anymore in any geometry or maths thing... :-(
Thanks a lot for your answers !
hmm i don't know if I undestood it correctly but I think you have too few input parameters:
you said the following information is given:
Origin point coordinates
the rect position & sizes
the red line position
the A & B points coordinates
I don't think it is possible to get your projected rectangle with this information alone.
Additionally, I think your green lines and the 'origin Bis' aren't helpful as well.
Perhaps, try this:
Supose, a blue line going through the points C & D is given as well.
Then you could find your projected rectangle by projecting the top of the rectangle onto that blue line.
So in summary:
You define an origin + two parallel lines, a red and a blue one.
Then you can project the top of the rect onto the blue line and the bottom of the rect onto the red line, yielding the points A,B,C,D
I hope this helps.
If I'm right, this code will show what you wanted to see.
First of all, I've ignored your initial setup of objects and information, and focused on the example situation itself; fake-projecting shadow for a "monolith" (any object is possible with the example below, even textured)
My reason was that it's really quite easy with the Matrix class of ActionScript, a handy tool worth learning.
Solution:
You can use the built-in Matrix class to do skew transform on DisplayObjects.
Try this example:
(The "useful" part lies in the _EF EnterFrame handler ;) )
import flash.display.MovieClip;
import flash.geom.Matrix;
import flash.events.Event;
import flash.display.BitmapData;
const PIP180:Number = Math.PI / 180;
const MAX_SHADOW_HEIGHT_MULTIPLIER:Number = 0.25; // you can also calculate this from an angle, like ... = Math.sin(angle * PIP180);
const ANIM_DEG_PER_FRAME:Number = 1.0 * PIP180; // the shadow creeps at a +1 degree per frame rate
var tx:BitmapData = new MonolithTexture(); // define this BitmapData in the library
var skew:Number = -10 * PIP180; // initial
var mono:MovieClip = new MovieClip();
mono.graphics.beginBitmapFill(tx);
// drawn that way the registration point is 0,0, so it's standing on the ground
mono.graphics.drawRect(0, -tx.height, tx.width, tx.height);
mono.graphics.endFill();
// align monolith to the "ground"
mono.x = stage.stageWidth / 2;
mono.y = stage.stageHeight - 100;
// make it be 100x300 pixel
mono.width = 100;
mono.height = 300;
var shad:MovieClip = new MovieClip();
// colored:
shad.graphics.beginFill(0x000000);
// or textured:
//shad.graphics.beginBitmapFill(tx);
shad.graphics.drawRect(0, -tx.height, tx.width, tx.height);
shad.graphics.endFill();
addChild(shad); // shadow first
addChild(mono); // then the caster object
addEventListener(Event.ENTER_FRAME, _EF);
function _EF(e:Event):void {
// animate skew on the positive half circle
skew = (skew + ANIM_DEG_PER_FRAME) % Math.PI;
// Matrix takes 6 parameters: a, b, c, d, x, y
// for this shadow trick, use them as follows:
// a = width scaling (as mono and shad are drawn in the same way, copy mono.scaleX for a perfect fit
// b = 0, because we don't want to project the vertical axis of transformation to the horizontal
// c = horizontal skew
// d = height scaling * skew * making it a bit flat using the constant
// x = mono.x, ...
// y = mono.y since originally mono and shad look alike, only the Matrix makes shad render differently
var mtx:Matrix = new Matrix(mono.scaleX, 0, Math.cos(skew), mono.scaleY * Math.sin(skew) * MAX_SHADOW_HEIGHT_MULTIPLIER, mono.x, mono.y);
shad.transform.matrix = mtx;
}
Now all you got to know to utilize this in your case, is the following N factors:
Q1: from what angle you want to project the shadow?
A1: horizontal factor is the skew variable itself, while vertical angle is stored as constant here, called MAX_SHADOW_HEIGHT_MULTIPLIER
Q2: do you want to project shadow only "upwards", or freely?
A2: if "upwards" is fine, keep skew in the positive range, otherwise let it take negative values as well for a "downward" shadow
P.S.: if you render the internals of the objects that they don't snap to 0 y as a base point, you can make them seem float/sink, or offset both objects vertically with a predefined value, with the opposite sign.
You face 1 very simple problem, as you said:
'What I want to determine is the coordinates of the C & D points, thing that could be easy if I wasn't struggling to find the "Origin bis" coordinates.'
But these co-ordinates relate to each other, so without one (or another value such as an angle) you cannot have the other. If you are to try this in 3D you are simply allowing the 3D engine to define 'Origin bis' and do your calculating for C and D itself.
So regardless you will need an 'Original bis', another value relating to the redline or your Rect for which to calculate the placement of C and D.
I remember making stuff like this and sometimes it's better to just stick with simple, you either make an 'Original bis' defines by yourself (it can be either stationary or move with the player/background) and get C and D the way you got A and B only that you use a lower line than the red line, or as I would of done, once you have A and B, simple skew/rotate your projection from those points down a bit further, and you get something the same as an 'Original bis' that follows the player. This works fine at simulating 'feeling of something real' but sadly as has been said, it looking real depends on what you are portraying. We do not know what the areas above or below the red line are (sky/ground, ground/water) and whether 'Origin' and 'Origin bis' is your light source, vanishing point, etc.
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.