I'm using OpenGL ES 2.0 on Android and I and I initialise my display like so:
float ratio = (float) width / height;
Matrix.orthoM(mProjMatrix, 0, -ratio, ratio, -1, 1, 3, 7); //Using Orthographic as developing 2d
What I'm having trouble understanding is this:
Let's say my app is a 'fixed screen' game (like Pac-Man ie, no scrolling, just the whole game visible on the screen).
Now at the moment, if I draw a quad at -1 to +1 on both x and y I get something like this:
Obviously, this is because I am setting -ratio, ratio as seen above. So this is correct.
But am I supposed to use this as my 'whole' screen? With rather massive letterboxing on the left and right?
I want a rectangular display that is the whole height of the physical display (and as much of the width as possible), but this would mean drawing at less that -1 and more than +1, is this a problem?
I realise the option may be to use clipping if this was a scrolling game, but for this particular scenario I want the whole 'game board' on the screen and to be static (And to use as much of the available screen real estate as possible without 'stretching' thus causing elongation of my sprites).
As I like to work with 0,0 as the top of the screen, basically what I do is pass my draw method something like so:
quad1.drawQuad (10,0);
When the drawQuad method get's this, it basically takes the range from left to right as expressed my openGL and divide the the screen width (so, in my case -1.7 through +1.7 so 3.4/2560 = 0.001328125). And say I specify 10 as my X (as above), it will say something like:
-1.7 + (10*0.001328125) = -1.68671875
It then plots the quad at -1.68671875.
Doing this I am able to work with normal co-ords (and I just subtract rather than add for y axis so I can have 0 at the top).
Is this a good way to do things?
Because with this method, at the moment, if I specify a 100,100 square, it isn't a square, it's rectangle. However, on the plus side, I can fill the whole physical screen by scaling the quad by width x height.
You are drawing a 1x1 quad, so that is why you see a 1x1 quad. Try translating the quad 0.25 to the right or left and you will see that you can draw in that space too.
In graphics, you create an object, like a quad, in your case you made it 1x1. Then you position it wherever you want. If you do not position it, then it will be at the origin, which is what you see.
If you draw a wider shape, you will also see you can draw outside this area on the screen.
By the way, with your ortho matrix function, you aren't just specifying the screen aspect ratio, you are also specifying the coordinate unit size you have to work with. This is why a 1x1 is filling the height the of the screen, because your upper and lower boundaries are set to 1 and -1. Your ratio is a little more than one, since your width is longer than your height, so your left and right boundaries are essentially something like -1.5 and 1.5 (whatever your ratio happens to be).
But you can also do something like this;
Matrix.orthoM(mProjMatrix, 0, -width/2, width/2, -height/2, height/2, 3, 7);
Here, your ratio is the same, but you are sending it to your ortho projection with screen coordinates. (Disclaimer: I don't use the same math library you do, but these appears to be a conventional ortho matrix function based on the arguments you are passing to it).
So lets say you have a 1000x500 pixel resolution. In OpenGL your origin of 0,0 is in the middle. So now your left edge is at (-500,y), right edge at (500,y) and your top is (x,250). So if you draw your 1x1 quad, it will be very tiny, but if you draw a 250x250 square, it will look like your 1x1 quad in your previous ortho projection.
So you can specify the coordinates you want, the ratio, the unit size, etc for how you want to work. Personally, I dont't like specifying coordinates as fractions between 0 and 1, I like to think about them in the same sense as the screen pixels.
But whether or not you choose to do this, hopefully you understand what you are actually passing to these matrix functions.
One of the best ways to learn is draw an object to the screen and just play around with different numbers you send to your modelview and projection matrices so you can see what it is they are actually doing.
Related
I have a question regarding transformations in OpenGL ES 2. I'm currently drawing a rectangle using triangle fans as depicted in the image below. The origin is located in its center, while its width and height are 0.6 and 2 respectively. I assume that these sizes are related to the model space. However, in order to maintain the ratio of height and width on a tablet or phone one has to do a projection that considers the proportion of the device lengths (again width and height). This is why I call orthoM(projectionMatrix, 0, -aspectRatio, aspectRatio, -1f, 1f, -1f, 1f);and the aspectRatio is given by float aspectRatio = (float) width / (float) height. This finally leads to the rectangle shown in the image below. Now, I would like to move the rectangle along the x-axis to the border of the screen. However, I was not able to come up with the correct calculation to do so, either I moved it too little or too much. So how would the calculation look like? Furtermore, I'm a little bit confused about the sizes given in the model space. What are the max and min values that can be achieved there?
Thanks a lot!
Vertex position of the rectangle are in world space. A way to do this it could be get the screen coordinates you want to move to and then transform them into world space.
For example:
If the screen is 300 x 200 and you are in the center 0,0 in world space (or 150, 100) in screen space). You want to translate to 300.
So the transformation should be screen_position to normalized device coordiantes and then multiply by inverseOf(projection matrix * view matrix) and divided by the w component.
Here it is explained for mouse that it is finally the same, just that you know the z because it is the one you used for your rectangle already (if it is on the plane x,y): OpenGL Math - Projecting Screen space to World space coords.
I am using OpenGl to draw an image. Now when i try to move the image, it moves by too much. So if i say the following:
gl.glTranslatef(0, 1, -5.0f);
squirrel.draw(gl);
If i out one as a parameter, the image is now located half way of screen. How do i make it so i can say things like:
gl.glTranslatef(screen_width - image_width , 0);
Is there an alternative method for drawing images in OpenGl?
I previously used canvas to draw images, and i had no problem positioning images on the screen. However with openGl i'm experiencing issues.
All you need to remember is, the screen space in OpenGL ranges from -1,-1 (top left), and 1,1 (bottom right). So you need to provide normalized values to OpenGL. To move a point along x direction from one end of the screen (-1.0) to another (1.0), left to right, you will have to Translate by 2.0 by using glTranslatef(2.0, 0, 0). This point is on the border, so you will have to adjust depending on the actual size of your object and its location.
I'm writing a little cross platform game engine for iOS, Android and BADA. I have a question about setting the perspective to be consistent regardless of screen resolution and ratio.
I have the following set up for my flipped normalized orthographic projection which works fine:
glViewport(0, 0, mWidth, mHeight);
glMatrixMode(GL_PROJECTION);
glOrthof(-1.0, //LEFT
1.0, //RIGHT
-1.0 * mHeight / mWidth, //BOTTOM
1.0 * mHeight / mWidth, //TOP
-2.0, //NEAR
100.0); //FAR
On my iPhone's this is fine and I get the desired position of world objects but on some of the Android devices and iPads the position when retaining the correct ratio.
All of the meshes are the right proportions but the position obviously alters so that if something is aligned to the bottom of the screen, when rendered on the iPad the objects will be drawen partly off the screen.
So the question:
Is this correct and I need to place objects relative to the width and height of the viewport?
Or
Is there a method to setting up the orthogonal perspective so that regardless of screen ratio the positioning will remain constant without damaging the perspective of world objects?
I am thinking from what I know and the math I did the second isn't an option because perspective is defined based on ratio.
The iPad and iPhone screens have different proportions. The iPhone is 3:2 and the iPad is 4:3. Android phones have too wide range of different proportions to list and I wouldn't really like to comment on Bada. So unless you're going to show your image in a letterbox or pillarbox, or stretch it so that the aspect ratio changes from device to device, the amount of your internal world that's visible is going to change between devices.
At the minute you've fixed the left and right parameters while calculating the top and bottom based on the screen proportions. So you'll get exactly the same amount of your world across the screen on every device but the amount on the vertical will change.
If your game involves a camera moving in 3d then there's really not much you can do about it. But since you're talking about things being aligned to sides of the screen I guess the camera moves in 2d?
As a general rule, if the camera moves along the vertical then you probably want to keep what you have. Your level layouts that are exactly the width of the screen will be the width of everybody's screen. Wider devices will be able to see further ahead or behind, but there you go.
If the camera moves along the horizontal then you probably want to switch to supplying fixed values for top and bottom, and calculating left and right as per the aspect ratio. So I guess that'd be:
glOrthof(-1.0 * mWidth / mHeight, //LEFT
1.0 * mWidth / mHeight, //RIGHT
-1.0 , //BOTTOM
1.0 , //TOP
-2.0, //NEAR
100.0); //FAR
In terms of being able to close this all off inside a library, you'll probably just need to receive a flag as to whether logical viewport width should be fixed and height adapted to the screen or height fixed and width adapted.
Android opengl-es view question. So in openGL, the default position of the camera and view is at 0,0. How do you set the view and camera so that it acts basically the same as computer screen coordinates with 0,0 at the top. I've called gl.glOrthof(-screenWidth/2, ScreenWidth/2, -ScreenHeight/2, ScreenHeight/2). But I think this is wrong. I also need to set the camera to view the entire field. I'm not sure how to use gl.glFrustumf to accomplish this task.
To use your vertex coordinates as screen coordinates, just use glOrtho(0, width, height, 0, -1, 1) on the projection matrix and keep your modelview matrix identity (which is the default). Note that I flipped bottom and top, as in GL (0,0) is at the lower left and you want it at the top (but keep in mind that this also flips every object and therefore the triangle ordering). You also forgot to set the near and far planes (everything with a z out of this interval won't get displayed). But when you now draw all your objects with z=0 (which is the default, when drawing only 2d vertices), all should be fine.
glFrustum is just an alternative to glOrtho. Where glOrtho constrcuts an orthographic (parallel) view, glFrustum constructs a perspective view. So you don't need glFrustum.
I am trying to learn opengl stuff on Android. In the gl.gltranslatef(x,y,z) call, I am shifting my texture by some units in the +ve x direction. But I am unable to find the number of pixels does 1 unit of x belong to?
Here is what I am doing:
I call gl.glviewport(0,0,width,height); // This will set my rectangle with 0,0 as lowerleft corner and then extend it to accommodate width and height.
Then
I call to gl.glfrustrum(-5,5,-7,7,3,7); // I am little confused how this call is using the dimensions I set in gl.glviewport.
How will -5 to 5 units from left to right in the above call, translate to pixels on the screen of android?
I mean if width = 320 and height = 533 pixels, then what will be the number of pixels occupied on the screen due to the gl.glfrustrum call?
I am experimenting in the gl.gltranslatef call by specifying xshift as 5.0, but it does not translate the bitmap at the right or left corner of the screen, when I increase it to 6, part of it is still visible on the screen.
Thanks
Siddhesh
In short, I am searching for the maximum number of units (in terms of X) which will represent extreme corners of my android phone screen.
glViewpoint tells it what rectangle (in pixels) your OpenGL output should be displayed in.
glFrustum tells it what coordinates in your "world" units should be mapped to that viewport.
An important point: your glFrustum call includes not only a height and width, but also a depth. Since you are specifying a Frustum, not a cube, that means anything with a Z coordinate anywhere but the very front of your frustum will be scaled down appropriately for its distance from the viewer.
As such, when you to a glTranslatef, the distance by which a particular object will move (in terms of pixels) will depend on its distance from the viewer. The further away it is from the viewer, the fewer pixels a particular sideways or up/down will translate to.
Depending on what else you're doing, one easy way to deal with this might be to use glOrtho instead of glFrustum. glOrtho gives orthographic mode, which means no perspective scaling is done, so a given X or Y distance will translate to the same number of pixels, regardless of distance from the viewer.