Note:
The question is specific to PHP GD library only
This question is NOT about how to crop image to a target aspect ratio, rather it is about how to draw overlay extending outside the image
I want to create custom graphic by putting together a background image with a polygon and some texts.
Input background images are of varied dimensions and aspect-ratios, but the final graphic has to be of a fixed (2:1) aspect ratio (it also has to be of pre-defined dimensions, but resizing of image is trivial, so correct aspect ratio is my only target).
Presently I'm cropping-to-fit my input image to target aspect ratio (2:1) by performing max-center area cropping using imagecrop function. Thereafter I draw red polygon on it as shown below (ignore the texts drawn on red band) using imagefilledpolygon method [cropping screenshot below is for demonstration purpose only, it is actually being done programmatically via imagecrop function]
Here's my function that draws the overlay (this function is called after cropping of image to 2:1 aspect ratio is done)
/**
* adds overlay (colored band) on the image
* for better output, overlay must be added before resizing
*/
public function withOverlay(): NotifAdsCreativeGenerator {
// Prepare custom red-color Hex to RGB https://stackoverflow.com/a/15202130/3679900
list($r, $g, $b) = sscanf(self::OVERLAY_COLOR, "#%02x%02x%02x");
$custom_red_color = imagecolorallocate($this->getCrrImage(), $r, $g, $b);
// prepare coordinates for polygon
$coords = [
[0, 0],
[(int) ($this->getCrrWidth() * self::OVERLAY_BEGIN_X_RATIO), 0],
[(int) ($this->getCrrWidth() * self::OVERLAY_END_X_RATIO), $this->getCrrHeight()],
[0, $this->getCrrHeight()]
];
$flattened_coords = array_merge(...$coords);
// draw polygon on image
imagefilledpolygon($this->getCrrImage(), $flattened_coords, count($flattened_coords) / 2, $custom_red_color);
return $this;
}
But what I want is to crop the image to ~ 1.28:1 aspect ratio (the approx ratio of right part of graphic without the red band) and then draw the polygon (extending) outside the image so as to obtain the final graphic in the same same 2:1 aspect ratio as shown below
I'm able to crop image to my desired aspect ratio (1.28:1) but I can't figure out a way to draw the polygon outside the image bounds (effectively expanding the image in the process). Is there a way to do this using PHP-GD library?
It was just a lack of understanding (about working of PHP-GD, available methods) on my part, but the solution is pretty simple
create an empty 'canvas' image of desired dimensions (and the 2:1 target aspect ratio) using imagecreatetruecolor function
(after cropping), copy the image on right side of canvas using imagecopy method (some basic maths has to be done to determine the offset where the image has to be placed on canvas)
now as before, the red polygon can be drawn on the left side on canvas to obtain the final graphic
/**
* adds overlay (colored band) on the image
* for better output, overlay must be added before resizing
*
* This method tries to preserve maximum center region of input image by performing minCenterCrop(.) on it
* before drawing an overlay that extends beyond left border of the cropped image
*
* (since this method incorporates call to 'withMinCenterCrop', calling that method before this is not required
* (and is redundant). For benefits of this method over 'withOverlay', read docstring comment of
* 'withMinCenterCrop' method
* #return NotifAdsCreativeGenerator
*/
public function withExtendedOverlay(): NotifAdsCreativeGenerator {
// perform min center crop to the 1.28:1 aspect ratio (preserve max central portion of image)
$this->withMinCenterCrop();
// this $required_canvas_aspect_ratio calculates to 2.0 (2:1 aspect ratio)
// calculate aspect ratio & dimensions of empty 'canvas' image
// since canvas is wider than min center-cropped image (as space on the left will be occupied by red overlay)
// therefore it's height is matched with cropped image and width is calculated
$required_canvas_aspect_ratio = self::IMAGE_WIDTH / self::IMAGE_HEIGHT;
// height of cropped image
$canvas_height = $this->getCrrHeight();
$canvas_width = $required_canvas_aspect_ratio * $canvas_height;
// create a new 'canvas' (empty image) on which we will
// 1. draw the existing input 'min-cropped' image on the right
// 2. draw the red overlay on the left
$canvas_image = imagecreatetruecolor($canvas_width, $canvas_height);
// copy contents of image on right side of canvas
imagecopy(
$canvas_image,
// cropped image
$this->getCrrImage(),
self::OVERLAY_BEGIN_X_RATIO * $canvas_width,
0,
0,
0,
// dimensions of cropped image
$this->getCrrWidth(),
$this->getCrrHeight()
);
// draw red band overlay on left side of canvas
$this->crr_image = $canvas_image;
return $this->withOverlay();
}
Related
(More info at end)----->
I am trying to render a small picture-in-picture display over my scene. The PiP is just a smaller texture, but it is intended to reveal secret objects in the scene when it is placed over them.
To do this, I want to render my scene, then render the SAME scene on the smaller texture, but with the exact same positioning as the main scene. The intended result would be something like this:
My problem is... I cannot get the scene on the smaller texture to match up 1:1. I keep trying various kludges, but ultimately I suspect that I need to do something to the projection matrix to pan it over to the location of the frame. I can get it to zoom correctly...just can't get it to pan.
Can anyone suggest what I need to do to my projection matrix to render my scene 1:1 (but panned by x,y) onto a smaller texture?
The data I have:
Resolution of the full-screen framebuffer
Resolution of the smaller texture
XY coordinate where I want to draw the smaller texture as an overlay sprite
The world/view/projection matrices from the original full-screen scene
The viewport from the original full-screen scene
(Edit)
Here is the function I use to produce the 3D camera:
void Make3DCamera(Vector theCameraPos, Vector theLookAt, Vector theUpVector, float theFOV, Point theRez, Matrix& theViewMatrix,Matrix& theProjectionMatrix)
{
Matrix aCombinedViewMatrix;
Matrix aViewMatrix;
aCombinedViewMatrix.Scale(1,1,-1);
theCameraPos.mZ*=-1;
theLookAt.mZ*=-1;
theUpVector.mZ*=-1;
aCombinedViewMatrix.Translate(-theCameraPos);
Vector aLookAtVector=theLookAt-theCameraPos;
Vector aSideVector=theUpVector.Cross(aLookAtVector);
theUpVector=aLookAtVector.Cross(aSideVector);
aLookAtVector.Normalize();
aSideVector.Normalize();
theUpVector.Normalize();
aViewMatrix.mData.m[0][0] = -aSideVector.mX;
aViewMatrix.mData.m[1][0] = -aSideVector.mY;
aViewMatrix.mData.m[2][0] = -aSideVector.mZ;
aViewMatrix.mData.m[3][0] = 0;
aViewMatrix.mData.m[0][1] = -theUpVector.mX;
aViewMatrix.mData.m[1][1] = -theUpVector.mY;
aViewMatrix.mData.m[2][1] = -theUpVector.mZ;
aViewMatrix.mData.m[3][1] = 0;
aViewMatrix.mData.m[0][2] = aLookAtVector.mX;
aViewMatrix.mData.m[1][2] = aLookAtVector.mY;
aViewMatrix.mData.m[2][2] = aLookAtVector.mZ;
aViewMatrix.mData.m[3][2] = 0;
aViewMatrix.mData.m[0][3] = 0;
aViewMatrix.mData.m[1][3] = 0;
aViewMatrix.mData.m[2][3] = 0;
aViewMatrix.mData.m[3][3] = 1;
if (gG.mRenderToSprite) aViewMatrix.Scale(1,-1,1);
aCombinedViewMatrix*=aViewMatrix;
// Projection Matrix
float aAspect = (float) theRez.mX / (float) theRez.mY;
float aNear = gG.mZRange.mData1;
float aFar = gG.mZRange.mData2;
float aWidth = gMath.Cos(theFOV / 2.0f);
float aHeight = gMath.Cos(theFOV / 2.0f);
if (aAspect > 1.0) aWidth /= aAspect;
else aHeight *= aAspect;
float s = gMath.Sin(theFOV / 2.0f);
float d = 1.0f - aNear / aFar;
Matrix aPerspectiveMatrix;
aPerspectiveMatrix.mData.m[0][0] = aWidth;
aPerspectiveMatrix.mData.m[1][0] = 0;
aPerspectiveMatrix.mData.m[2][0] = gG.m3DOffset.mX/theRez.mX/2;
aPerspectiveMatrix.mData.m[3][0] = 0;
aPerspectiveMatrix.mData.m[0][1] = 0;
aPerspectiveMatrix.mData.m[1][1] = aHeight;
aPerspectiveMatrix.mData.m[2][1] = gG.m3DOffset.mY/theRez.mY/2;
aPerspectiveMatrix.mData.m[3][1] = 0;
aPerspectiveMatrix.mData.m[0][2] = 0;
aPerspectiveMatrix.mData.m[1][2] = 0;
aPerspectiveMatrix.mData.m[2][2] = s / d;
aPerspectiveMatrix.mData.m[3][2] = -(s * aNear / d);
aPerspectiveMatrix.mData.m[0][3] = 0;
aPerspectiveMatrix.mData.m[1][3] = 0;
aPerspectiveMatrix.mData.m[2][3] = s;
aPerspectiveMatrix.mData.m[3][3] = 0;
theViewMatrix=aCombinedViewMatrix;
theProjectionMatrix=aPerspectiveMatrix;
}
Edit to add more information:
Just playing and tweaking numbers, I have come to a "close" result. However the "close" result requires a multiplication by some kludge numbers, that I don't understand.
Here's what I'm doing to to perspective matrix to produce my close result:
//Before calling Make3DCamera, adjusting FOV:
aFOV*=smallerTexture.HeightF()/normalRenderSize.HeightF(); // Zoom it
aFOV*=1.02f // <- WTH is this?
//Then, to pan the camera over to the x/y position I want, I do:
Matrix aPM=GetCurrentProjectionMatrix();
float aX=(screenX-normalRenderSize.WidthF()/2.0f)/2.0f;
float aY=(screenY-normalRenderSize.HeightF()/2.0f)/2.0f;
aX*=1.07f; // <- WTH is this?
aY*=1.07f; // <- WTH is this?
aPM.mData.m[2][0]=-aX/normalRenderSize.HeightF();
aPM.mData.m[2][1]=-aY/normalRenderSize.HeightF();
SetCurrentProjectionMatrix(aPM);
When I do this, my new picture is VERY close... but not exactly perfect-- the small render tends to drift away from "center" the further the "magic window" is from the center. Without the kludge number, the drift away from center with the magic window is very pronounced.
The kludge numbers 1.02f for zoom and 1.07 for pan reduce the inaccuracies and drift to a fraction of a pixel, but those numbers must be a ratio from somewhere, right? They work at ANY RESOLUTION, though-- so I have have a 1280x800 screen and a 256,256 magic window texture... if I change the screen to 1024x768, it all still works.
Where the heck are these numbers coming from?
If you don't care about sub-optimal performance (i.e., drawing the whole scene twice) and if you don't need the smaller scene in a texture, an easy way to obtain the overlay with pixel perfect precision is:
Set up main scene (model/view/projection matrices, etc.) and draw it as you are now.
Use glScissor to set the rectangle for the overlay. glScissor takes the screen-space x, y, width, and height and discards anything outside that rectangle. It looks like you have those four data items already, so you should be good to go.
Call glEnable(GL_SCISSOR_TEST) to actually turn on the test.
Set the shader variables (if you're using shaders) for drawing the greyscale scene/hidden objects/etc. You still use the same view and projection matrices that you used for the main scene.
Draw the greyscale scene/hidden objects/etc.
Call glDisable(GL_SCISSOR_TEST) so you won't be scissoring at the start of the next frame.
Draw the red overlay border, if desired.
Now, if you actually need the overlay in its own texture for some reason, this probably won't be adequate...it could be made to work either with framebuffer objects and/or pixel readback, but this would be less efficient.
Most people completely overcomplicate such issues. There is absolutely no magic to applying transformations after applying the projection matrix.
If you have a projection matrix P (and I'm assuming default OpenGL conventions here where P is constructed in a way that the vector is post-multiplied to the matrix, so for an eye space vector v_eye, we get v_clip = P * v_eye), you can simply pre-multiply some other translate and scale transforms to cut out any region of interest.
Assume you have a viewport of size w_view * h_view pixels, and you want to find a projection matrix which renders only a tile w_tile * h_tile pixels , beginning at pixel location (x_tile, y_tile) (again, assuming default GL conventions here, window space origin is bottom left, so y_tile is measured from the bottom). Also note that the _tile coordinates are to be interpreted relative to the viewport, in the typical case, that would start at (0,0) and have the size of your full framebuffer, but this is by no means required nor assumed here.
Since after applying the projection matrix we are in clip space, we need to transform our coordinates from window space pixels to clip space. Note that clip space is a 4D homogeneous space, but we can use any w value we like (except 0) to represent any point (as a point in the 3D space we care about forms a line in the 4D space we work in), so let's just use w=1 for simplicity's sake.
The view volume in clip space is denoted by the [-w,w] range, so in the w=1 hyperplane, it is [-1,1]. Converting our tile into this space yields:
x_clip = 2 * (x_tile / w_view) -1
y_clip = 2 * (y_tile / h_view) -1
w_clip = 2 * (w_tile / w_view) -1
h_clip = 2 * (h_tile / h_view) -1
We now just need to translate the objects such that the center of the tile is moved to the center of the view volume, which by definition is the origin, and scale the w_clip * h_clip sized region to the full [-1,1] extent in each dimension.
That means:
T = translate(-(x_clip + 0.5*w_clip), -(y_clip + 0.5 *h_clip), 0)
S = scale(2.0/w_clip, 2.0/h_clip, 1.0)
We can now create the modified projection matrix P' as P' = S * T * P, and that's all there is. Rendering with P' instead of P will render exactly the region of your tile to whatever viewport you are using, so for it to be pixel-exact with respect to your original viewport, you must now render with a viewport which is also w_tile * h_tile pixels big.
Note that there is also another approach: The viewport is not clamped against the framebuffer you're rendering to. It is actually valid to provide negative values for x and y. If your framebuffer for rendering your tile into is exactly w_tile * h_tile pixels, you simply could set glViewport(-x_tile, -y_tile, x_tile + w_tile, y_tile + h_tile) and render with the unmodified projection matrix P instead.
Laravel has Intervention Image, with which you can resize images
// create instance
$img = Image::make('public/foo.jpg');
// resize image to fixed size
$img->resize(300, 200);
Everything is fine if the uploaded image has a size of 3000 x 2000 and we will make 300 x 200 from it.
But if we load images with different heights and widths, and do a 300 x 200 resize, then the images will stretch and look awful.
I need them to be cut a little at the edges.
I have tried using $img->fit(300, 200); but it cuts too much.
Is it possible to do two steps in Intervention Image for resizing, for example, defining the size and removing a few unnecessary parts to make the picture look more or less normal?
To remove a few unnecessary parts you can use the crop() function first and then resize the image maintaining its aspect ration.
$img->crop(width, height, x, y);
This function cuts out a rectangular part of the current image with given width and height. Define optional x,y coordinates to move the top-left corner of the cutout to a certain position. By default the rectangular part will be centered on the current image if you do not provide the x,y co-ordinates.
After cropping the image, you can resize the image maintaining its aspect ration, so the image will not look stretchy or abnormal.
// resize the image to a width of 300 and constrain aspect ratio (auto height)
$img->resize(300, null, function ($constraint) {
$constraint->aspectRatio();
});
// resize the image to a height of 200 and constrain aspect ratio (auto width)
$img->resize(null, 200, function ($constraint) {
$constraint->aspectRatio();
});
However, if you always want the image to be a specific dimension, which is in your case 300 X 200 , You can try the fit function with the callback functionality to retain maximal original image size.
// add callback functionality to retain maximal original image size
$img->fit(300, 200, function ($constraint) {
$constraint->upsize();
});
I have a background pixmap, basically a canvas, which I draw a bunch of
rectangles on and I need to rotate the pixmap and rectangles.
However rotating the background pixmap and the rectangles needs to be done
seperately, that is the rotation of the background pixmap gets handled via an
external library routine and I need to rotate and redraw the rectangles
on top manually.
So far I am actually able to rotate the rectangles by applying a
transformation matrix I got from Wikipedia
to each vertex. What I don't know is how to translate them that each rectangle retains its position relative to the canvas.
Here is a quick drawing for illustration of what I want to achieve:
I need to do this with C and Xlib, but I'm not necessarily looking for code but would appreciate some general hints/algorithms.
To get the translated position for the child object, you need to rotate the relative position vector for the child object, and then add it to the origin:
Pseudocode would be:
public static Vector2 OffsetByRotation(Vector2 childPos, Vector2 parentPos, float angle)
{
var relativeVector = childPos - parentPos;
relativeVector = Rotate(relativeVector, angle);
return parentPos + relativeVector;
}
Note that your example image not only rotates the parent object, but also translates it: your left image is rotated around (0, 300), but this point is then translated to (0, 0).
The requested transformation is
X' = 300 - Y
Y' = X
Is it possible to add image/text on the 2d image so that it gives a real view.For example as present in : http://www.zazzle.com/make_your_own_iphone_5_case-179092402149274498.
These views are orthographic and isometric views and they can be reproduced using affine transformations in canvas, as they are also parallelograms.
First you will need to make masks for the different cases. These needs to be drawn in the same orientation as the case in the "photo". Use solid pixels (any color will do, it won't show in later step) where you want the custom graphics to show, transparent pixels anywhere else (anti-aliased pixels are fine).
Then draw in the mask in the canvas, select composite mode "source-in" to replace non-transparent pixels and finally, select blending mode "multiply" and draw the case "photo" on top to mix in shadows and highlights. The latter step is what will give the illusion of the image having depth.
For the isometric views, calculate the skew angle (or use trial and error if you're not sure if the image is accurate - this is as a rule-of-thumb usually tan(60°), ie. transform(1, 0, Math.tan(60/180*Math.PI), 1, 0, 0)), then do the same process as above. Just remember only apply transformation when drawing the custom image, mask and top layer must be drawn without transformations.
The orthographic side views can be generated using scaling for the x-axis. Depending on which angle, add a stripe for the side of the case.
Example of steps
var img = new Image(),
cust = new Image(),
count = 2,
ctx = document.querySelector("canvas").getContext("2d");
img.onload = cust.onload = comp;
img.src = "http://i.stack.imgur.com/je0Jh.png";
cust.src = "http://i.stack.imgur.com/uRPDt.png";
function comp() {
if (--count) return;
// draw in mask
ctx.drawImage(img, 0, 0);
// comp. mode source-in
ctx.globalCompositeOperation = "source-in";
// draw in custom graphics
ctx.drawImage(cust, 0, 0, ctx.canvas.width, ctx.canvas.height);
// blend mode multiply
ctx.globalCompositeOperation = "multiply";
// draw in original case multiplied (does not work in IE)
ctx.drawImage(img, 0, 0);
}
<canvas with=263 height=505></canvas>
The quality largely depends on the quality of the mask - I made a very quick-n-dirty version here as you can see (your case image can also act as the mask btw).
The steps are the same for the isometric view with the exception of the skew transform. Multiply does not work in IE, you can use alpha here instead or make a separate mask containing only shadows etc.
That being said: remember that this is not the image sent to production. This will just show a representation of the final result. What is used is the image, image position and size. These data is then used to build an unmasked flat print-template which is used to make the phone-case.
PIX* returnRotatedImage(PIX* image, float theta)
{
PIX* rotated = pixRotate(image, -theta, L_ROTATE_AREA_MAP, L_BRING_IN_BLACK, image->w, image->h);
return rotated;
}
When I execute the above code on an image, the resulting image has the edges cut off.
Example: the original scan, followed by the image after being run through the above function to rotate it by ~89 degrees.
I don't have 10 reputation yet, so I can't embed the images, but here's a link to the two pictures: http://imgur.com/a/y7wAn
I need it to work for arbitrary angles as well (not just angles close to 90 degrees), so unfortunately the solution presented here won't work.
The description for the pixRotate function says:
* (6) The dest can be expanded so that no image pixels
* are lost. To invoke expansion, input the original
* width and height. For repeated rotation, use of the
* original width and height allows the expansion to
* stop at the maximum required size, which is a square
* with side = sqrt(w*w + h*h).
however it seems to be expanding the destination after rotation, and thus the pixels are lost, even if the final image size is correct. If I use pixRotate(..., 0, 0) instead of pixRotate(..., w, h), I end up with the image rotated within the original dimensions: http://i.imgur.com/YZSETl5.jpg.
Am I interpreting the pixRotate function description incorrectly? Is what I want to do even possible? Or is this possibly a bug?
Thanks in advance.