HTML5 Canvas - Zooming into a Point - html

So I know there are threads about it already here, like that one.
I followed the idea that was proposed in the thread above, and it works. However, I don't understand WHY it works.
Here is an example:
Let's say that i have a square centered at (100, 100), and its width/height is 100. So its top-left corner will be at (50, 50).
Now let's say that i want to zoom X2 into the square's center, that is, to zoom into (100, 100). So i will write the following transformation sequence:
translate(100, 100);
scale(2, 2);
translate(-100, -100);
So because the canvas apply the transformation in reverse order, my transformed square's top-left corner will be now at (0, 0), and its height/width will be 200.
Ok, let's say that now i want to zoom X2 into the right-bottom corner of the already transformed square. So intuitively, i would like to perform the following transformation sequence:
translate(200, 200);
scale(2, 2);
translate(-200, -200);
But it wont work, because again, the canvas apply transfomations in reverse order. That is to say, that if i sum up my two transformation sequences, i'll get:
// First Sequence
translate(100, 100);
scale(2, 2);
translate(-100, -100);
// Second Sequence
translate(200, 200);
scale(2, 2);
translate(-200, -200);
This means that the second sequence will be applied to each point before the first sequence (because the canvas will apply the transformation from bottom to top), and this is wrong. So as the thread in the link above suggest the following:
Because sequence 2 will be applied first, i should transform the point (200, 200) to its original coordinates, by applying to it the inverse of the first sequence. that is to say, if T1 is the matrix that represents the first sequence, then it will look like that:
// First Sequence
translate(100, 100);
scale(2, 2);
translate(-100, -100);
// Second Sequence
var point = SVGPoint(200, 200);
var transformedPoint = point.matrixTransform(T1.inverse());
translate(-transformedPoint.x, -transformedPoint.y);
scale(2, 2);
translate(transformedPoint.x, transformedPoint.y);
But why it works? I really don't understand why it should work like that... can anyone elaborate about it?
Thanks!

The HTML5 canvas transformations happen top-down, not bottom-up as you believe. The reason for the distinction is because the transformations applied to the canvas affect the coordinate system, not your logical coordinates.
Translating by translate(100, 100) will move your coordinate system right and down, which appears hauntingly similar to moving your logical coordinate up and left.
Let's take the first sequence (I have changed your use of transform to translate):
translate(100, 100);
scale(2, 2);
translate(-100, -100);
Naturally, when we think to scale an object from it's center, we translate the object to (0,0), scale the object, then move the object back. The above code, when read in reverse, would appear to do that. However, that's not the case.
When we read the above code from top-down, it says (assume we start with an identity transform):
Move the context's (0,0) right 100 units and down 100 units. This takes it to the canvas's (100,100) location.
Make the coordinate system 2x bigger.
Move the context's (0,0) left 100 units and up 100 units, essentially returning it to it's original location (in context coordinate space, not canvas space).
The scaling happens relative to the context's (0,0) point, which is at (100,100) on the canvas.
If we were to now add your second sequence:
translate(200, 200);
scale(2, 2);
translate(-200, -200);
This will:
Move the context's (0,0) to the coordinate system's (200,200) location.
Make the coordinate system 2x bigger than it already was.
Return the context's (0,0) back to where it was previously (in context coordinate space, not canvas space).
As you've found out, that does not give you what you are expecting because (200,200) is not the point about which you want to scale. Remember, all units are relative to the context coordinate system. So we need to convert the canvas location of (200,200) to the context coordinate location of (150,150) which is the original bottom-right corner of our rectangle.
So we change sequence #2 to be:
translate(150, 150);
scale(2, 2);
translate(-150, -150);
This gives us what we are expecting (to zoom in on the bottom-right corner of the rectangle). That's why we do the inverse-transform.
In the demo application, when the app zoom's in, it's taking the coordinate in canvas units where the user's mouse was, inverse-transforming that using the context transformation thus-far, to get the location in context coordinate space that was clicked on. The context origin is moved to that location, zoomed, then returned to it's previous location.
References:
Safari HTML5 Canvas Guide: Translation, Rotation, and Scaling

You seem to be way overthinking transforms!
Here’s the simple rule:
If you apply any set of transforms, then you must undo all of them if you want to get back to your untransformed state.
Period !!!!
So let say you do these 4 transforms:
Do #1. context.translate(100,100);
Do #2. context.scale(2,2);
Do #3. context.translate(-20,50);
Do #4. context.scale(10,10);
To get back to your original untransformed state, you must undo in exactly reverse order:
Undo #4: context.scale( 0.10, 0.10 ); // since we scaled 10x, we must unscale by 0.10
Undo #3: context.translate(20,-50);
Undo #2: context.scale( 0.50, 0.50 ); // since we scaled 2x, we must unscale by 0.50
Undo #1: context.translate(-100,-100);
Think of it like walking to your friends house.
You turn Right + Left + Right.
Then to go home you must reverse that: Left + Right + Left
You must undo your walking path in exactly the reverse of your original walk.
That’s how transforms work too.
That’s Why !!

Related

Scale, Position & Rotate Parent object to make child object take up entire stage

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 :)

Random bezier blob?

I have some ActionScript3 code I'm using to create liquid-like "droplets", and when they're first generated they look like a curved square (that's as close as I can get them to being a circle). I've tried and failed a lot here but my goal is to make these droplets look more organic and free-form, as if you were looking closely at rain drops on your windshield before they start dripping.
Here's what I have:
var size:int = (100 - asset.width) / 4,
droplet:Shape = new Shape();
droplet.graphics.beginFill(0xCC0000);
droplet.graphics.moveTo(size / 2, 0);
droplet.graphics.curveTo(size, 0, size, size / 2);
droplet.graphics.curveTo(size, size, size / 2, size);
droplet.graphics.curveTo(0, size, 0, size / 2);
droplet.graphics.curveTo(0, 0, size / 2, 0);
// Apply some bevel filters and such...
Which yields a droplet shaped like this:
When I try adding some randomness to the size or the integers or add more curves in the code above, I end up getting jagged points and some line overlap/inversion.
I'm really hoping someone who is good at math or bezier logic can see something obvious that I need to do to make my consistently rounded-corner square achieve shape randomness similar to this:
First off, you can get actual circle-looking cirles using beziers by using 0.55228 * size rather than half-size (in relation to bezier curves, this is sometimes called kappa). It only applies if you're using four segments, and that's where the other hint comes in: the more points you have, the more you can make your shape "creep", so you might actually want more segments, in which case it becomes easier to simply generate a number of points on a circle (fairly straight forward using good old sine and cosine functions and a regularly spaced angle), and then come up with the multi-segment Catmul-Rom curve through those points instead. Catmul-Rom curves and Bezier curves are actually different representations of the same curvatures, so you can pretty much trivially convert from one to the other, explained over at http://pomax.github.io/bezierinfo/#catmullconv (last item in the section gives the translation if you don't care about the maths). You can then introduce as much random travel as you want (make the upper points a little stickier and "jerk" them down when they get too far from the bottom points to get that sticky rain look)

What's the purpose of Canvas.Context Save and Restore in this example?

This page shows some animations in HTML5 canvas. If you look at the source of the scroller, there's a statement to save the context after clearing the rectangle and restoring it after the animation. If I substitute the restore statement with another ctx.clearRect(0, 0, can.width, can.height statement, nothing works. I thought the restore is restoring the cleared rectangle but it seems its restoring more info. What's that extra info that's needed for the next frame?
I am not looking for HTML5 textbook definitions of Save and Restore but I want to understand why they are needed in this specific example.
UPDATE
It's frustrating to get an answer where I specifically already mentioned in the question I don't want to get the definitions of save() and restore(). I already know Save() saves the state of the context and Restor()e restores it. My question is very specific. Why is restore() used in the manner in the example when all the Save did is saved an empty canvas. Why is restoring an empty canvas not the same as clearing it?
Canvas state isn't what's drawn on it. It's a stack of properties which define the current state of the tools which are used to draw the next thing.
Canvas is an immediate-mode bitmap.
Like MS Paint. Once it's there, it's there, so there's no point "saving" the current image data, because that would be like saving the whole JPEG, every time you make a change, every frame...
...no, the state you save is the state which will dictate what coordinate-orientation, dimension-scale, colour, etc, you use to draw the NEXT thing (and all things thereafter, until you change those values by hand).
var canvas = document.createElement("canvas"),
easel = canvas.getContext("2d");
easel.fillStyle = "rgb(80, 80, 120)";
easel.strokeStyle = "rgb(120, 120, 200)";
easel.fillRect(x, y, width, height);
easel.strokeRect(x, y, width, height);
easel.save(); // stores ALL current status properties in the stack
easel.rotate(degrees * Math.PI / 180); // radians
easel.scale(scale_X, scale_Y); // any new coordinates/dimensions will now be multiplied by these
easel.translate(new_X, new_Y); // new origin coordinates, based on rotated orientation, multiplied by the scale-factor
easel.fillStyle = "gold";
easel.fillRect(x, y, width, height); // completely new rectangle
// origin is different, and the rotation is different, because you're in a new coordinate space
easel.clearRect(0, 0, width, height); // not even guaranteed to clear the actual canvas, anymore
easel.strokeRect(width/2, height/2, width, height); // still in the new coordinate space, still with the new colour
easel.restore(); // reassign all of the previous status properties
easel.clearRect(0, 0, width, height);
Assuming that you were only one state-change deep on the stack, that last line, now that your canvas' previous state was restored, should have successfully cleared itself (subpixel shenanigans notwithstanding).
So as you can see, it has very, VERY little to do with erasing the canvas.
In fact, it has nothing to do with erasing it, at all.
It has to do with wanting to draw something, and doing the basic outlining and sweeping colours/styles, and then manually writing in the colours for the smaller details on top, and then manually writing all of the styles back the way they were before, to go back to sweeping strokes for the next object, and on and on...
Instead, save general states that will be reused, create a new state for smaller details, and return to the general state, without having to hard-code it, every time, or write setter functions to set frequently-used values on the canvas over and over (resetting scale/rotation/affine-transforms/colours/fonts/line-widths/baseline-alignment/etc).
In your exact example, then, if you're paying attention, you'll see that the only thing that's changing is the value of step.
They've set the state of a bunch of values for the canvas (colour/font/etc).
And then they save. Well, what did they save?
You're not looking deep enough. They actually saved the default translation (ie: origin=0,0 in original world-space).
But you didn't see them define it?
That's because it's defined as default.
They then increase the step 1 pixel (actually, they do this first, but it doesn't matter after the first loop -- stay with me here).
Then they set a new origin point for 0,0 (ie: from now on, when they type 0,0 that new origin will point to a completely different place on the canvas).
That origin point is equal to x being the exact middle of the canvas, and y being equal to the current step (ie: pixel 1 or pixel 2, etc... and why the difference between starting at 0 and starting at 1 really doesn't matter).
Then what do they do?
They restore.
Well, what have they restored?
...well, what have they changed?
They're restoring the point of origin to 0,0
Why?
Well, what would happen if they didn't?
If the canvas is 500px x 200px, and it starts at 0,0 in our current screen space... ...that's great...
Then they translate it to width/2, 1
Okay, so now when they ask to draw text at 0,0 they'll actually be drawing at 250, 1
Wonderful. But what happens next time?
Now they're translating by width/2, 2
You think, well, that's fine... ...the draw call for 0,0 is going to happen at 250, 2, because they've set it to clear numbers: canvas.width/2, 2
Nope. Because current 0,0 is actually 250,1 according to our screen. And one translation is relative to its previous translation...
...so now you're telling the canvas to start at it's current-coordinates' 0,0 and go left 250, and down 2.
According to the screen (which is like a window, looking at the map, and not the map, itself) we're now 500px to the right, and 3 pixels down from where we started... And only one frame has gone by.
So they restore the map's coordinates to be the same origin as the screen's coordinates (and the rotation to be the same, and the scale, and the skew, etc...), before setting the new one.
And as you might guess, by looking at it, now, you can see that the text should actually move top to bottom. Not right to left, like the page says...
Why do this?
Why go to the trouble of changing the coordinate-system of the drawing-context, when the draw commands give you an x and y right there in the function?
If you want to draw a picture on the canvas, and you know how high and wide it is, and where you'd like the top-left corner to be, why can't you just do this:
easel.drawImage(myImg, x, y, myImg.width, myImg.height);
Well, you can.
You can totally do that. There's nothing stopping you.
In fact, if you wanted to make it zoom around the screen, you could just update the x and y on a timer, and call it a day.
But what about if you were drawing a game character? What if the character had a hat, and had gloved hands, and big boots, and all of those things were drawn separate from the character?
So first you'd say "well, he's standing at x and y in the world, so x plus where his hand is in relation to his body would be x + body.x - hand.x...or was that plus..."
...and now you have draw calls for all of his parts that are all looking like a notebook full of Grade 5 math homework.
Instead, you can say: "He's here. Set my coordinates so that 0,0 is right in the middle of my guy". Now your draw calls are as simple as "My right hand is 6 pixels to the right of the body, my left hand is 3 pixels to the left".
And when you're done drawing your character, you can set your origin back to 0,0 and then the next character can be drawn. Or, if you want to attempt it, you can then translate from there to the origin of the next character, based on the delta from one to the other (this will save you a function call per translation). And then, if you only saved state once the whole time (the original state), at the end, you can return to 0,0 by calling .restore.
The context save() saves stuff like transformation color among other stuff. Then you can change the context and restore it to have the same as when you saved it. It works like a stack so you can push multiple canvas states onto the stack and recover them.
http://html5.litten.com/understanding-save-and-restore-for-the-canvas-context/

AS3 smooth rotation direction

I'm not very good with radial calculations, I can't imagine thus I can't be sure. I need some explanation of Math.atan2() thing, please.
Usual task - to make an object rotate after the mouse. I get the differences, get the angle, I see angles in the text areas and DIRECTLY the object does follow the mouse. What I need now is everything to be smooth. I need angles to be 0-360 but after 180 object rotation becomes -180 and counts backwards, and mouse rotation becomes -90 after 270 and also counts back to 0.
More deeply, I want a smooth rotation, it means a set speed of say 2 per frame, to meet the mouse angle the shortest way. It takes to set conditions and I can't do that cause I don't even understand the logic of these values. They are almost random! I don't need it to be done or copied, I need to understand to move on so if you could please explain how does it work and what I do wrong...
Code is simple:
angle = Math.atan2(deltaY,deltaX)/(Math.PI/180) + 90; //+90 cause it lacks it to look at the mouse//
Object01.rotation = angle;
So the problem is I don't even get how it works... if 2 values are different the object can't point at the mouse but it does. Numbers lie and if I need something based on these numbers it will be wrong. Very wrong... Need organization. Meaning I want everything to be ready for further coding that will be based on the rotations to not jump up and down cause of misfit ends.
Add: Explanation of how does it happen, what I described above. Why such a chaos of the values? And an advice on how could I arrange it for further coding, just as I said. Animation alone wont work if I want to make rotation an element of important events such as shooting direction and aiming speed. Or changes of speed rotation of a lockpicked lock. Or anything much more complicated that wont work if I don't make straight and clear values: from A to Z, from 1 to 10, no 8s between 2 and 3, no R before B, no mouse angle 270 when object facing it -90 when they both started from 0 and reached 180 together.
Oh, and as I said, mouse facing works but when I try to make a certain speed of chasing mouse the shortest way it turns the object wrong directions in all 4 quarters. I assume it's also about this arctangens thing that has issues with delta values becoming negative in different quarters. And when I change it, some other value goes wrong... So I need to know exactly what I'm doing to know what's wrong and how to fix it. So yep. Need explanation. Please.
Add2: angleZ = Math.atan2(oppSide,adjSide)/(Math.PI/180);
So I divided rotation to 4 quarters, for each I count atan as opp. side to adj. side, then add 90, 180 and 270 respectively. My mouse rotation does 360, but the object that follow through simple object.rotation = angleZ; still goes to 180, then from -180 to 0 on the left side. Why does it ignore the simple command? The rotation fits but I need it to be equal, no surprises! Why is it happening? How can a number I directly set to be equal to another number as a base of the action change itself to the one of same rotation but completely different number? It doesn't even know it's degrees! It's as simple as "object.rotation, please be equal to the number I choose!"
It's just different coordinate systems. Like how x starts at 0 at the left of the stage, goes +x to the right, and -x to the left, object rotation starts at 0˚ pointing up, and goes +180˚ clockwise and -180˚ anti-clockwise.
Math.atan2 happens to start at 0 pointing left (-x), and go +270˚ clockwise and -90˚ anti-clockwise, which is annoying, but it just means you have to convert between coordinate systems by adding 90˚.
You can spin something around over and over of course, so the numbers jump so that they always stay within the same range, because 361˚ is the same as 1˚, and -270˚ is the same as 90˚. You can tell an object to rotate outside of the -180˚ to 180˚ range, and it will normalise the rotation to within those values.
As mitim described, to smoothly animate rotation you'll either need to use Event.ENTER_FRAME, a Timer, or a tweening library like TweenLite. Most tweening libraries can work out the shortest rotation direction for you, but otherwise you can simply calculate both and see which is smaller.
As an idea, since it seems like you know the angle you need to rotate towards and it's direction, would it be easier to just animate towards that angle to get your smooth rotation? Basically treat it like any other animatable property and just add on your rotation speed (2 degrees it looks like) per frame tick, until it reaches the desired rotation.
Find angle amount needed to rotate by
Figure out if clockwise or counter clockwise direction and set the rotation amount. This can be figured out by checking if the angle is great then 180 / positive or negative
Add the rotation amount * direction every frame tick, until the desired rotation is less then or equal to the rotation amount per frame
Set rotation to desired rotation

understanding matrix.transition(); as3

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!