I'm trying to tint a circle around the player in my 2D side scroller but I keep getting an oval! Here's the part of the code I'm using that matters:
if(length(abs(coords - playerCoords)) < .1)
{
color = color *float4(1,0,1,1);
}
return color;
My screen size is 1280 wide x 720 tall. I know that this is the reason for the distortion, but I don't know enough about my issue in order to come up with or find a solution. Can someone explain to me how to compensate for the screen stretch?
Thanks!
-ATD
multiply the "abs()" term by "float2((720./1280.),1.0)" -- or whatever your y/x aspect ration might be
The coords you are using are normalized in 0-1 space, so just correct them
Related
I realized that a sprite which texture is set to “setAliasTexParameters” must be placed on integer positions to avoid pixel bleeting when moved.
That’s fine but if I use a RenderTexture with screen size and scale it by a non-integer value (for example 1.5f) I get again pixel bleeding although the position of the sprite is rounded.
I guess that I have to take the scalefactor of the rendertexture into account when I calculate the position but I did not find the correct formula yet.
Really appreciate if someone can help here!
I would like to know how to compute rotation components of a rectangle in space according to four given points in a projection plane.
Hard to depict in a single sentence, thus I explain my needs.
I have a 3D world viewed from a static camera (located in <0,0,0>).
I have a known rectangular shape (an picture, actually) That I want to place in that space.
I only can define points (up to four) in a spherical/rectangular referencial (camera looking at <0°,0°> (sph) or <0,0,1000> (rect)).
I considere the given polygon to be my rectangle shape rotated (rX,rY,rZ). 3 points are supposed to be enough, 4 points should be too constraintfull. I'm not sure for now.
I want to determine rX, rY and rZ, the rectangle rotation about its center.
--- My first attempt at solving this constrint problem was to fix the first point: given spherical coordinates, I "project" this point onto a camera-facing plane at z=1000. Quite easy, this give me a point.
Then, the second point is considered to be on the <0,0,0>- segment, which is about an infinity of solution ; but I fix this by knowing the width(w) and height(h) of my rectangle: I then get two solutions for my second point ; one is "in front" of the first point, and the other is "far away"... I now have a edge of my rectangle. Two, in fact.
And from there, I don't know what to do. If in the end I have my four points, I don't have a clue about how to calculate the rotation equivalency...
It's hard to be lost in Mathematics...
To get an idea of the goal of all this: I make photospheres and I want to "insert" in them images. For instance, I got on my photo a TV screen, and I want to place a picture in the screen. I know my screen size (or I can guess it), I know the size of the image I want to place in (actually, it has the same aspect ratio), and I know the four screen corner positions in my space (spherical or euclidian). My software allow my to place an image in the scene and to rotate it as I want. I can zoom it (to give the feeling of depth)... I then can do all this manually, but it is a long try-fail process and never exact. I would like then to be able to type in the screen corner positions, and get the final image place and rotation attributes in a click...
The question in pictures:
Images presenting steps of the problem
Note that on the page, I present actual images of my app. I mean I had to manually rotate and scale the picture to get it fits the screen but it is not a photoshop. The parameters found are:
Scale: 0.86362
rX = 18.9375
rY = -12.5875
rZ = -0.105881
center position: <-9.55, 18.76, 1000>
Note: Rotation is not enought to set the picture up: we need scale and translation. I assume the scale can be found once a first edge is fixed (first two points help determining two solutions as initial constraints, and because I then know edge length and picture width and height, I can deduce scale. But the software is kind and allow me to modify picture width and height: thus the constraint is just to be sure the four points are descripbing a rectangle in space, with is simple to check with vectors. Here, the problem seems to place the fourth point as a valid rectangle corner, and then deduce rotation from that rectangle. About translation, it is the center (diagonal cross) of the points once fixed.
I am working with OpenGL and I wanted to invert the image. So I went here, asked a question and finally I had the following code:
glMatrixMode(GL_PROJECTION);
glScalef(-1,1,1);
glTranslatef(-width(),0,0);
From what I understand from this, the position of every pixel gets inverted, so the pixels that were on the right of the image are now on the same absolute position, but are the left of the image, so I have to move the entire thing back exactly as many pixels as its wide: 360 (which is the size of the "canvas", so in the snippents the function width() is being used)! So to undo this process I would invert the image again and then move it back to where it came from:
glMatrixMode(GL_PROJECTION);
glScalef(-1,1,1);
glTranslatef(width(),0,0);
Nope, blackscreen. I have to do exactly the same thing twice to undo the flipping: I have to move with -360 every time I flip the image. Why?
It's exactly as Daniel Fischer mentioned in the comment. Here is an illustration of the process.
What you must have in mind is that the transformations operate on the transformed coordinate systems.
We start with the image (grey) on the screen (green):
Then we scale the image. So the origin is preserved, but the x-axis is mirrored.
No we have to move the image onto the screen again. Because the x-axis points to the left (but we want to move the image to the right), we have to use a negative offset for the translation:
If we flip the image again, the following happens. The origin is preserved and the x-axis is mirrored:
So we must translate the image by a negative offset:
Another way of undoing the flip is undoing the operations (but in the opposite order):
glTranslatef(width, 0, 0);
glScalef(-1,1,1);
The mathematical reason for that is that inversion reverses the oder. If we have Matrix A = B * C then A^-1 = (C^-1 * B^-1).
I want to repeat a background image that is rotated. Trying to make it seamless is destroying my soul.
Starting with something simple, consider each image is laid out like bricks. Creating a seamless repeating background image is pretty simple:
(the red area is the crop). You can see this working as expected at http://jsfiddle.net/mPqfB.
Now let's say I want to rotate the image by 45 degrees:
Unfortunately, the same crop no longer works, as you can see on http://jsfiddle.net/mPqfB/1.
I'm trying to figure out how to crop the image correctly so that we have a seamless repeat. There's probably some fairly trivial maths involved to do this but I can't for the life of me figure it out.
[Update]
I'm attempting to follow #oezi's calculations so to make things easier have created an image of dimensions: 100px x 50px.
Therefore:
Least Common Multiple = 100
Hypotenuse = 1002 + 1002 = 20000
Now I'm assuming this means we don't have to create an image of 20000px x 20000px. Am hoping that #oezi can clarify how he performs his resizing??
If this is a2 + b2 = c2 is equal to c = square root of (a2 + b2)
Then we can concur that our crop should be 141px?
Finally, this doesn't actually explain where we take the crop from?
[Update 2]
It does look like this is how the resize should be created. Taking a 141px x 141px crop of the image yielded the correct results - http://jsfiddle.net/EfuV2/
As far as where to crop from, it doesn't actually matter!
is the rotation is exactly 45 degrees, you'll have to find out the least common multiple of the width and height of your unrotated pattern.
in your case, that's 15100 (width 100 and height 151)
it would be much better to scale your pattern to width 100 and height 150, so the least common multiple is only 300
Take that number and some math (pythagorean theorem). Assume your number is the length of the two short arms and calculate the length of the hypotenuse - that's our result (make a square image of that size to get your pattern).
in your case, that's 21355
with resizing, it's ~ 424
Note that this is just typed straight from my head because i can't try it out practically at the moment - but i'm really sure it's correct.
edit: a fast (and messy) test got me to this:
http://i.imgur.com/rZuu9.jpg
http://jsfiddle.net/mPqfB/2/ (click the image-link first, otherwise jsfiddle doesn't show the image)
accidentally i made the pattern only be 423 in height and the rotation isn't perfect (don't have photoshop here), but it's good enough to prove that my math is correct.
The trick is to crop the pattern at points where the section being cut off matches the section remaining on the opposite side of the crop area (see example cuts in blue). It'll probably take some trial and error to get it right but you should be able to do it easily enough.
I need to know what the visible height of a display object will be after I change it's rotationX value.
I have an application that allows users to lay out a floor in 3D space. I want the size of the floor to automatically stretch after a 3D rotation so that it always covers a certain area.
Anyone know a formula for working this out?
EDIT: I guess what I am really trying to do is convert degrees to pixels.
On a 2D plane say 100 x 100 pixels, a -10 degree change on rotationX means that the plane has a gap at the top where it is no longer visible. I want to know how many pixels this gap will be so that I can stretch the plane.
In Flex, the value for the display objects height property remains the same both before and after applying the rotation, which may in fact be a bug.
EDIT 2: There must be a general math formula to work this out rather than something Flash/Flex specific. When viewing an object in 3D space, if the object rotates backwards (top of object somersaults away from the viewer), what would the new visible height be based on degrees of rotation? This could be in pixels, metres, cubits or whatever.
I don't have a test case, but off the top of my head I'd guess something like:
var d:DisplayObject;
var rotationRadians:Number = d.rotationX * Math.PI / 180;
var visibleHeight:Number = d.height * Math.cos(rotationRadians);
This doesn't take any other transformations into account, though.
Have you tried using the object's bounding rectangle and testing that?
var dO:DisplayObject = new DisplayObject();
dO.rotation = 10;
var rect:Rectangle = dO.getRect();
// rect.topLeft.y is now the new top point.
// rect.width is the new width.
// rect.height is the new height.
As to the floor, I would need more information, but have you tried setting floor.percentWidth = 100? That might work.
Have you checked DisplayObject.transform.pixelBounds? I haven't tried it, but it might be more likely to take the rotation into account.
Rotation actually changes DisplayObject's axis's (i.e. x and y axes are rotated). That is why you are not seeing the difference in height. So for getting the visual height and y you might try this.var dO:DisplayObject = new DisplayObject();
addChild();
var rect1:Rectangle = dO.getRect(dO.parent);
dO.rotation = 10;
var rect2:Rectangle = dO.getRect(dO.parent);
rect1 and rect2 should be different in this case. If you want to check the visual coordinates of the dO then just change dO.parent with root.