I have 2 DisplayObject, one containing the other like the code below:
var sprite1:Sprite = new Sprite();
sprite1.addChild(loader1); // assume that I have load picture on to loader1
loader1.rotationZ = 30;
sprite1.rotationZ = -30;
If I run the code above, the image on loader1 will look just like it hasn't been rotate at all because its container rotate in opposite direction. Which is correct, as what I expected.
But, if I had multiple axis rotation to the code:
loader1.rotationZ = 30;
loader1.rotationY = 50;
sprite1.rotationZ = -30;
sprite1.rotationY = -50;
Now, loader1 will rotate in different angle. My question is, why it doesn't offset each other?
Note that, both loader1 and sprite1's x,y,z position are all at 0,0,0. And this problem occurs regardless of rotationX,Y or Z. If you have 1 axis, it works fine. 2 axis, it won't.
I post this sample code because I'm trying to understand how Flash rotate in 3D. In AS3 document, it claims that it rotate around its "3D parent container" What is that?
Rotation is relative to the parent, and when you rotate the parent object it will change the frame of reference for the child. This doesn't matter when you are only rotating around one axis, but makes a difference when you rotate around two. To undo the rotations, you have to undo all of them, in the reverse order that they were applied.
Try it with a book, or your phone, rotating by 90 degrees on one axis, then a second. You can't get back to where you started and you will always have involved the 3rd axis.
Related
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 :)
I want to set the rotateX property of a movie clip (so it tilts backwards in 3d), but whenever I attempt it with
myMC.rotationX = -90;
it tilts back at an odd angle. How can I rotate the movieclip so that it leans backwards perfectly straight? How do I set a center point for this transform?
A screenshot would be helpful to explain "...an odd angle" but it sounds like your 'vanishing point' is in the wrong place. Take a look at http://help.adobe.com/en_US/FlashPlatform/reference/actionscript/3/flash/geom/PerspectiveProjection.html
You set the vanishing point on the parent (only needs to be done once, unless you wish to move it or stage resizes), eg:
//set vanishing point to center of current DisplayObject
this.projectionCenter = new Point(this.width/2, this.height/2);
Then rotate:
myMC.rotationX = -90;
Please this fiddle I have copied my complete project in it
here if you open the fiddle in the output you can see an image, scribble on the image selecting pen,add text etc(like this perform some functions).then rotate the group using rotate button and then try to scribble you will understand what is the problem exactly.
In me view Problem is I am having a stage and a layer is added to the stage and a group is added to the layer and different elements like lines text etc are added to the group. then group is rotated the i am trying to draw the line based on the mouse position of the stage.But it is not coming correctly because the group got rotated the x and y what we are taking to draw a line is from stage.I need to take the x and y from the group not from the stage is their any way.If hav't understand please ask me or post a reply.
This should get you fairly close: http://jsfiddle.net/k4qB8/24/
// This rotates that added active line along with your group.
// This makes the draw direction correct
activeline.setRotationDeg(0-rootGroup.getRotationDeg());
// Here you'll have to figure a way to calculate how much to move the
// line over so the draw is on the correct spot
// This is as close as I got it
if(Math.abs(rootGroup.getRotationDeg()%360)==0)
activeline.move(rootGroup.getX()-375, rootGroup.getY()-175);
if(Math.abs(rootGroup.getRotationDeg()%360)==90)
activeline.move(rootGroup.getX()-175, rootGroup.getY()+375);
if(Math.abs(rootGroup.getRotationDeg()%360)==180)
activeline.move(rootGroup.getX()+375, rootGroup.getY()+175);
if(Math.abs(rootGroup.getRotationDeg()%360)==270)
activeline.move(rootGroup.getX()+175, rootGroup.getY()-375);
Also, add some more logic for counter-clockwise rotation, as this doesn't work 100%.
I think the real solution would be to just draw on separate layers for each rotation, kind of like this:
if (rotation is 90) : draw on lineLayer1;
if (rotation is 180) : draw on lineLayer2;
if (rotation is 270) : draw on lineLayer3;
if (rotation is 360 || 0) : draw on lineLayer4;
This way you could just rotate the layers which are not drawn on to simulate the feel of rotation.
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.
I have a task:
I need to place about 100 sprites on one canvas (with prepared grid on it). I need to place them as invisible (circles) stones, on the board, and make visible only on mouseover.
The problem I come across is following, I can't place those objects accurately into the nodes on the grid.
E.g.
if I define stones (it's just a sprite, as I said earlier) this way:
var stone:StoneSprite = new StoneSprite();
stone.x = this.x + 2*cellWidth;
stone.graphics.beginFill( 0x000000 );
stone.graphics.drawCircle(stone.x , this.y + cellWidth, cellWidth/3 );
stone.graphics.endFill();
rawChildren.addChild(stone);
They don't sit on the node...
See image:
http://img.skitch.com/20091014-kuhfyjeg1g5qmrbyxbcerp4aya.png
And if I do it this way:
var stone:StoneSprite = new StoneSprite();
stone.graphics.beginFill( 0x000000 );
stone.graphics.drawCircle(this.x + 2*cellWidth , this.y + cellWidth, cellWidth/3 );
stone.graphics.endFill();
rawChildren.addChild(stone);
The stone is displayed correctly in the grid node... See image 2:
http://img.skitch.com/20091014-f595tksjxramt98s7yfye591bh.png
So I wonder what is the difference between these 2 approaches.
Also, I think I need to pass correct coordinates to the stone class... In case I would like to change some properties of the stone object. E.g. visibility, or radius.
Could you please suggest, what's wrong in defining coordinates as stone.x, stone.y
How to fix the problem with incorrect positioning.
Would really appreciate ideas about the problem, I am trying to solve for so long :(
Assume x & y are 30 and cellWidth is 30.
First Example:
stone.x = 30 + 60; //90
drawCircle(90, 60, 10);
This means if you were to draw a rectangle around your circle, it would be at [170,50]. (x,y).
Second Example:
stone.x = 0;
drawCircle(90, 60, 10)
This means the rectangle around your circle is at [80,50];
In the first example, you are moving the sprite to position x==90. Then drawing a circle whose center is at x==90 inside the sprite. So relative to this, you're at x==180. But because a circle's x,y coords are the center, subtract 10 for the radius to get the boundary x position.
In the second example, the sprite defaults to position x==0 relative to this and you're drawing the circle inside the sprite at position x==90. (therefore it begins at x==80).
I am not sure what's causing the issue - might be some padding induced by the container - can't say without testing. But I believe that adding a Sprite (say board) to canvas.rawChildren and using it as the parent for the grid and stones would fix the issue.