I'm trying to make 2D graphics for my Android app that consists of six thin rectangles that each take up about 1/6th of the screen in width and equal the screen's height. I'm not sure the right way to determine the bounds of the x and y OpenGL coordinate plane on screen. Eventually I will need to write logic that tests which of the 6 rectangles a touch event occurs in, so I have been trying to solve this problem by remapping OpenGL's coordinate plane into the device's screen coordinate plane (where the origin (0,0) is at the top left of the screen instead of the middle.
I declare one of my six rectangles like so:
private float vertices1[] = {
2.0f, 10.0f, 0.0f, // 0, Top Left
2.0f, -1.0f, 0.0f, // 1, Bottom Left
4.0f, -1.0f, 0.0f, // 2, Bottom Right
4.0f, 10.0f, 0.0f, // 3, Top Right
};
but since i'm not sure what the visible limits are on the x and y planes (in the OpenGL coordinate system) I have no concrete way of knowing what vertices my rectangle needs to be instantiated with to occupy 1/6th of the display. Whats the ideal way to do this?
I've tried approaches such as using glOrthoof() to remap OpenGL's coordinates into easy to work with device screen coordinates:
gl.glViewport(0, 0, width, height);
// Select the projection matrix
gl.glMatrixMode(GL10.GL_PROJECTION);
// Reset the projection matrix
gl.glLoadIdentity();
// Calculate the aspect ratio of the window
GLU.gluPerspective(gl, 45.0f,(float) width / (float) height,0.1f, 100.0f);
gl.glOrthof(0.0f,width,height, 0.0f, -1.0f, 5.0f);
// Select the modelview matrix
gl.glMatrixMode(GL10.GL_MODELVIEW);
// Reset the modelview matrix
gl.glLoadIdentity();
but when I do my rectangle dissapears completely.
You certainly don't want to use a perspective projection for 2D graphics. That just doesn't make much sense. A perspective projection is for... well, creating a perspective projection, which is only useful if your objects are actually placed in 3D space.
Even worse, you have two calls to set up a perspective matrix:
GLU.gluPerspective(gl, 45.0f,(float) width / (float) height,0.1f, 100.0f);
gl.glOrthof(0.0f,width,height, 0.0f, -1.0f, 5.0f);
While that's legal, it rarely makes sense. What essentially happens if you do this is that both projections are applied in succession. So the first thing to do is get rid of the gluPerspective() call.
To place your 6 rectangles, you have a few options. Almost the easiest one is to not apply any transformations at all. This means that you will specify your input coordinates in normalized device coordinates (aka NDC), which is a range of [-1.0, 1.0] in both the x- and y-direction. So for 6 rectangles rendered side by side, you would use a y-range of [-1.0, 1.0] for all the rectangles, and an x-range of [-1.0, -2.0/3.0] for the first, [-2.0/3.0, -1.0/3.0] for the second, etc.
Another option is that you use an orthographic projection that makes specifying the rectangles even more convenient. For example, a range of [0.0, 6.0] for x and [0.0, 1.0] for y would make it particularly easy:
gl.glOrthof(0.0f, 6.0f, 0.0f, 1.0f, -1.0f, 1.0f);
Then all rectangles have a y-range of [0.0, 1.0], the first rectangle has a x-range of [0.0, 1.0], the second rectangle [1.0, 2.0], etc.
BTW, if you're just starting with OpenGL, I would pass on ES 1.x, and directly learn ES 2.0. ES 1.x is a legacy API at this point, and I wouldn't use it for any new development.
Related
I have a texture that I can render in OpenGL-ES with an orthogonal identity matrix:
gst_gl_shader_set_uniform_matrix_4fv(shader, "u_transformation", 1, FALSE, identity_matrix);
I want to move "the pixels around": half of the top is going to the left, half of the bottom is going to the right as shown on the image below. Is there an "easy" way to do that? I'm on Android.
On this related answered question How to crop/clip in OpenGL using u_transformation, I was able to keep the top part 'a' or the bottom part 'e'. Would there be a way to do a "double gst_gl_shader_set_uniform_matrix_4fv" after "cutting" the scene in two?
The transformation that you want here cannot be represented by a transformation matrix. Matrices can only represent certain classes of transformations. In 3D space:
A 3x3 matrix represents a linear transformation. Typical examples include rotation, scaling, mirroring, shearing.
A 4x3 matrix represents an affine transformation. On top of the above, this includes translations.
If you extend the 3D space to homogenous coordinates with 4 components, a 4x4 matrix can represent additional classes of transformations, like projections.
The transformation in your sketch is none of the above. So applying a matrix to your vertices will not be able to do this.
So what can you do? Two options come to mind:
If you can easily draw the two parts (top/bottom, left/right) separately, you can obviously do that, and simply change the transformation between rendering the two parts.
Apply the logic in your shader code.
For option 2, you could do this either in the vertex or fragment shader. If you have no primitives that cross the boundary between the two parts, handling it in the vertex shader would be more efficient. Otherwise, similar logic can be used in the fragment shader.
Sketching the critical parts for the vertex shader case, let's say you currently have the following that gives you the arrangement in the left side of your sketch:
// Calculate output position and store it in variable "pos".
gl_Position = pos;
To get the second arrangement, the logic could look like this (completely untested...):
if (pos.y > 0.0) {
gl_Position = vec4(0.5 * pos.x - 0.5, 2.0 * pos.y - 1.0, pos.zw)
} else {
gl_Position = vec4(0.5 * pos.x + 0.5, 2.0 * pos.y + 1.0, pos.zw);
}
The idea is that you check whether the vertex is in the top or bottom half, and scale/shift it accordingly to map the top half of the coordinate space into the left half, and the bottom half of the coordinate space into the right half.
This could be streamlined some more by replacing the conditional with a sign operation:
float s = sign(pos.y);
gl_Position = vec4(0.5 * pos.x - sign * 0.5, 2.0 * pos.y - sign, pos.zw);
Some more care will be needed if pos.w is not 1.0, which happens if you e.g. applied a perspective projection to your vertices. In that case, you'll have to incorporate the division by w in the calculations above.
The formula described in Reto answers 'semi' work as they only produce the "a" on the left or the "e" on the right but not both at the same time.
The solution I found is to double the number of vertices and indices and play around with the vertices coordinates like this:
static const GLfloat vertices[] = {
1.0f, 1.0f, 0.0f, 1.0f, 0.0f,
0.0f, 1.0f, 0.0f, 0.0f, 0.0f,
0.0f, -1.0f, 0.0f, 0.0f, 0.5f,
1.0f, -1.0f, 0.0f, 1.0f, 0.5f,
0.0f, 1.0f, 0.0f, 1.0f, 0.5f,
-1.0f, 1.0f, 0.0f, 0.0f, 0.5f,
-1.0f, -1.0f, 0.0f, 0.0f, 1.0f,
0.0f, -1.0f, 0.0f, 1.0f, 1.0f
};
static const GLushort indices[] = { 0, 1, 2, 0, 2, 3, 4, 5, 6, 4, 6, 7 };
I'm trying to draw rectangles with a certain distance in x-axis to each other.
The distance on each rectangle should be
(screen width - rectangle width)
so that only one rectangle can be fully displayed on the screen at a time. And if I drag the first rectangle to left, the portion of that rectangle not displayed (or cut) on the screen would be the amount displayed on second rectangle .
The problem is that I could not figure out how to calculate the ratio of my rectangle's width to screen width.
My rectangle vertices are
final float[] rectangleVerticesData = {
// X, Y, Z
-1.0f, 2.0f, 0.0f,
-1.0f, -2.0f, 0.0f,
1.0f, 2.0f, 0.0f,
-1.0f, -2.0f, 0.0f,
1.0f, -2.0f, 0.0f,
1.0f, 2.0f, 0.0f
};
Based above, my rectangle width is 2. I am using a tablet with 1280 pixels width.
Using the formula above, I got
1280 - 2 = 1278
When I ran the program, the expected output is not as what I am trying to achieve. When I drag the first rectangle, the next rectangle is not displayed even if the first rectangle is fully dragged to left off the screen.
When I tried hard-coding distance to 10, the rectangles are displayed with that distance. I guess 1278 is so long to be the distance between them. How can I calculate the correct distance between the triangles based on screen width so that only one rectangle can be fully displayed at a time on the screen?
Any help would be greatly appreciated. Thank you.
In OpenGL, the screen is considered as a quad of normalised size ie, it extends from vertex (-1,-1) to vertex (+1,+1). The vertex input given to OpenGL for drawing does not range according to the screen pixel size, but is converted internally for drawing to actual screen size. Since there are many devices with varying screen sizes, to make the GL code run across all devices without hardcoding screen sizes, this is required.
So you will have to scale your calculations accordingly.
I am trying to understand the basic concepts around the co-ordinates system in OpenGL so I have been making a test application from guides online.
Presently I have drawn a simple Square to the screen, using simple Co-ordinates of:
-1.0f, 1.0f, 0.0f, // 0, Top Left
-1.0f, -1.0f, 0.0f, // 1, Bottom Left
1.0f, -1.0f, 0.0f, // 2, Bottom Right
1.0f, 1.0f, 0.0f, // 3, Top Right
In my application I run the following code:
GLU.gluPerspective(gl, 45.0f, (float) width / (float) height, 0.1f, 100.0f);
My basic understanding here is that the code is setting the viewing port angle to 45 degrees and the width to height ratio of the window size.
Another thing I am doing is setting the viewing position as -4 units on the Z axis:gl.glTranslatef(0, 0, -4);
This is what the result looks like in Landscape...
And in Portrait...
My questions are:
How does the co-ordinate system work, how many Pixels does one unit represent? How does changing the orientation and width to height ratio effect the equation?
If I wanted to draw a square the size of the screen, with a View Port of 45 degrees and a Viewing position of z-4... how does one figure out the required width and height in units?
I'll try to answer the best I can.
There wouldn't be any reason to change the width to height ratio or 45 degree angle. Doing it the way you have it keeps the things from being stretched horizontally or vertically in an unusual way. Because you are using a perspective view, you have 3D space with depth as apposed to an Orthographic view where there is no depth. In doing glTranslatef(0,0,-4) what you've actually done is changed the MODELVIEW Matix, moving it 4 in the negative z direction, presumably before actually drawing the square. By default, the "camera" is sitting at 0,0,0 with Y (up) as the upward direction.
You may be able to translate 3D space to pixels, but with a Perspective view type, I'm not at all sure you'd really want or need to. A 2D Orthographic view would be a different story, though, as many people use OpenGL for 2D games as well. Wanting a square exactly the size of the screen, Orthographic is probably the way to go, and you should be able to with a few Google searches be able to figure out your pixel density to 2d space comparison.
I want to draw opengl texture at full screen.
(texture : 128x128 ===> device screen : 320x480)
Below code works good, but texture is small.
I have to use only glFrustumf function(not glOrthof function).
How can I draw texture in full screen size?
// this is android source code
float ratio = (float) screenWidth / screenHeight;
gl.glMatrixMode(GL10.GL_PROJECTION);
gl.glLoadIdentity();
gl.glFrustumf(-ratio, ratio, -1, 1, 1, 10);
GLU.gluLookAt(gl, 0.0f, 0.0f, -2.5f, // eye
0.0f, 0.0f, 0.0f, // center
0.0f, 1.0f, 0.0f); // up
// draw blah blah
Why do you have to use glFrustum only? Switching to glOrtho for drawing the background, then switching to glFrustum for regular drawing would be the canonical solution.
BTW: gluLookAt must happen in the modelview matrix, not in the projection matrix like you do right now. As it stands your code is broken and if you were a student in one of my OpenGL classes I'd give you negative points for this cardinal error.
I'm trying to draw a square in OpenGL ES (Android), 2D and covering the whole screen.
At the moment I'm just using trial and error but am sure there has got to be a better way to get the size of the screen. Below is how I'm currently initializing square:
float[] square = new float[] { -0.1f, -0.1f, 0.0f,
0.1f, -0.1f, 0.0f,
-0.1f, 0.1f, 0.0f,
0.1f, 0.1f, 0.0f };
Ideally the 0.1f in the x axis would be be the width and 0.1 in y the height of the window. Any help would be greatly appreciated.
Cheers
I think the size of the screen depends on your projection. For 2D graphics, most people use glOrtho to define the parallel projection. It's up to you to specify the size here.
Also, you can specify a larger size and have multiple 2D textures positioned within the clipping bounds, or you can have a single texture with the vertices mapped to the corners of your specified projection. This single texture would contain your entire display contents.
The following site explains this a little better.
http://www.scottlu.com/2008/04/fast-2d-graphics-wopengl-es.html
. . .
WindowManager w = getWindowManager();
Display d = w.getDefaultDisplay();
int width = d.getWidth();
int height = d.getHeight();
. . .
see http://groups.google.com/group/android-developers/browse_thread/thread/229c677ef0c5ae97