Orthographic projection - What is the process converting 3d point to 2d - actionscript-3

I'm implementing a simple penalty shootout game using actionscript 3.0. The view of the game is similar to view of the old "Sensible World of Soccer". I want to use 3d game logic by using dimension z as I think that it could help me in order to achieve better collision detection - response results. However, I would like to keep the graphics style and view equivalent to old 2d soccers'. Hence, I assume that orthographic projection is suitable for this implementation. Although there is plenty of information in the internet regarding orthographic projection, I'm a little bit confused about how someone can apply it in his/her code.
So my questions are:
Which is the procedure step by step in order for someone to convert a 3d (x, y, z) point to 2d (x', y') point in orthographic projection?
Can we avoid using matrices? If yes, what are the equations that associate coordinates x', y' with x, y, z?
Do we have to define a camera position and angle before applying the conversion? In my case, camera will be in a fixed position and angle.

DisplayObjects and their descendants (ie MovieClip and Sprite) have a z property you can use to do this without the headaches - they also have rotationX/Y/Z and scaleX/Y/Z properties too!
Using 'z' will adjust the position and scale of an object accordingly (though it will convert vectors to bitmaps), there's no depth sorting, so it will stay on top of objects even if its z co-ord suggests it should be behind them, but for the project you have in mind I can't see this being a problem - it's pretty easy to fix anyway, have an array of objects in the scene, sort it according to z-position and reset the depth index of each/re-add to stage in sorted order.
You can use the perspectiveProjection member of a clip to adjust the FOV, origin etc -
Perspective Tutorial
..but you don't need to get any more sophisticated than that. Certainly you don't need to dabble with matrices with a fixed camera view, even if you wanted to calculate this manually as an experiment.
Hope this helps

Related

When drawing on a canvas, should calculations be done relative to cartesian plane coordinates?

I've been seeing a lot of canvas-graphics-related javascript projects and libraries lately and was wondering how they handle the coordinate system. When drawing shapes and vectors on the canvas, are the points calculated based on a cartesian plane and converted for the canvas, or is everything calculated directly for the canvas?
I tried playing around with drawing a circle by graphing all its tangent lines until the line intersections start to resemble a curve and found the difference between the cartesian planes I'm familiar with and the coordinate system used by web browsers very confusing. The function for a circle, for example, "y^2 + x^2 = r^2" would need to be translated to "(y-1)^2 + (x-1)^2 = r^2" to be seen on the canvas. And then negative slopes were positive slopes on the canvas and everything would be upside down :/ .
The easiest way for me to think about it was to pretend the origin of a cartesian plane was in the center of the canvas and adjust my coordinates accordingly. On a 500 x 500 canvas, the center would be 250,250, so if I ended up with a point at 50,50, it would be drawn at (250 + 50, 250 - 50) = (300, 200).
I get the feeling I'm over-complicating this, but I can't wrap my mind around the clean way to work with a canvas.
Current practice can perhaps be exemplified by a quote from David Flanagan's book "JavaScript : The Definitive Guide", which says that
Certain canvas operations and attributes (such as extracting raw
pixel values and setting shadow offsets) always use this default
coordinate system
(the default coordinate system is that of the canvas). And it continues with
In most canvas operations, when you specify the coordinates
of a point, it is taken to be a point in the current coordinate system
[that's for example the cartesian plane you mentioned, #Walkerneo],
not in the default coordinate system.
Why is using a "current coordinate system" more useful than using directly the canvas c.s. ?
First and foremost, I believe, because it is independent of the canvas itself, which is tied to the screen (more specifically, the default coordinate system dimensions are expressed in pixels). Using for instance a Cartesian (orthogonal) coordinate system makes it easy for you (well, for me too, obviously :-D ) to specify your drawing in terms of what you want to draw, leaving the task of how to draw it to the transformations offered by the Canvas API. In particular, you can express dimensions in the natural units of your drawing, and perform a scale and a translation to fit (or not, as the case may be...) your drawing to the canvas.
Furthermore, using transformations is often a clearer way to build your drawing since it allows you to get "farther" from the underlying coord system and specify your drawing in terms of higher level operations ('scale', 'rotate', 'translate' and the more general 'transform'). The abovementioned book gives a very nice exemple of the power of this approach, drawing a Koch (fractal) snowflake in many fewer lines that would be possible (if at all) using canvas coordinates.
The HTML5 canvas, like most graphics systems, uses a coordinate system where (0,0) is in the top left and the x-axis and y-axis go from left to right and top down respectively. This makes sense if you think about how you would create a graphics system with nothing but a block of memory: the simplest way to map coordinates (x,y) to a memory slot is to take x+w*y, where w is the width of a line.
This means that the canvas coordinate system differs from what you use in mathematics in two ways: (0,0) is not the center like it usually is, and y grows down rather than up. The last part is what makes your figures upside down.
You can set transformations on the canvas that make the coordinate system more like what you are used to:
var ctx = document.getElementById('canvas').getContext('2d');
ctx.translate(250,250); // Move (0,0) to (250, 250)
ctx.scale(1,-1); // Make y grow up rather than down

How do I pass barycentric coordinates to an AGAL shader? (AGAL wireframe shader)

I would like to create a wire frame effect using a shader program written in AGAL for Stage3D.
I have been Googling and I understand that I can determine how close a pixel is to the edge of a triangle using barycentric coordinates (BC) passed into the fragment program via the vertex program, then colour it accordingly if it is close enough.
My confusion is in what method I would use to pass this information into the shader program. I have a simple example set up with a cube, 8 vertices and an index buffer to draw triangles between using them.
If I was to place the BC's into the vertex buffer then that wouldn't make sense as they would need to be different depending on which triangle was being rendered; e.g. Vetex1 might need (1,0,0) when rendered with Vetex2 and Vetex3, but another value when rendered with Vetex5 and Vetex6. Perhaps I am not understanding the method completely.
Do I need to duplicate vertex positions and add the aditional data into the vertex buffer, essentially making 3 vertices per triangle and tripling my vertex count?
Do I always give the vertex a (1,0,0), (0,1,0) or (0,0,1) value or is this just an example?
Am I over complicating this and is there an easier way to do wire-frame with shaders and Stage3d?
Hope that fully explains my problems. Answers are much appreciated, thanks!
It all depends on your geomtery, and this problem is in fact a problem of graph vertex coloring: you need your geometry graph to be 3-colorable. The good starting point is the Wikipedia article.
Just for example, let's assume that (1, 0, 0) basis vector is red, (0, 1, 0) is green and (0, 0, 1) is blue. It's obvious that if you build your geometry using the following basic element
then you can avoid duplicating vertices, because such graph will be 3-colorable (i.e. each edge, and thus each triangle, will have differently colored vertices). You can tile this basic element in any direction, and the graph will remain 3-colorable:
You've stumbled upon the thing that drives me nuts about AGAL/Stage3D. Limitations in the API prevent you from using shared vertices in many circumstances. Wireframe rendering is one example where things break down...but simple flat shading is another example as well.
What you need to do is create three unique vertices for each triangle in your mesh. For each vertex, add an extra param (or design your engine to accept vertex normals and reuse those, since you wont likely be shading your wireframe).
Assign each triangle a unit vector A[1,0,0], B[0,1,0], or C[0,0,1] respectively. This will get you started. Note, the obvious solution (thresholding in the fragment shader and conditionally drawing pixels) produces pretty ugly aliased results. Check out this page for some insight in techniques to anti-alias your fragment program rendered wireframes:
http://cgg-journal.com/2008-2/06/index.html
As I mentioned, you need to employ a similar technique (unique vertices for each triangle) if you wish to implement flat shading. Since there is no equivalent to GL_FLAT and no way to make the varying registers return an average, the only way to implement flat shading is for each vertex pass for a given triangle to calculate the same lighting...which implies that each vertex needs the same vertex normal.

Calculate 3D coordinates from 2D Image plane accounting for perspective without direct access to view/projection matrix

First time asking a question on the stack exchange, hopefully this is the right place.
I can't seem to develop a close enough approximation algorithm for my situation as I'm not exactly the best in terms of 3D math.
I have a 3d environment in which I can access the position and rotation of any object, including my camera, as well as run trace lines from any two points to get distances between a point and a point of collision. I also have my camera's field of view. I do not have any form of access to the world/view/projection matrices however.
I also have a collection of 2d images that are basically a set of screenshots of the 3d environment from the camera, each collection is from the same point and angle and the average set is taken at about an average of a 60 degree angle down from the horizon.
I have been able to get to the point of using "registration point entities" that can be placed in the 3d world that represent the corners of the 2d image, and then when a point is picked on the 2d image it is read as a coordinate with range 0-1, which is then interpolated between the 3d positions of the registration points. This seems to work well, but only if the image is a perfect top down angle. When the camera is tilted and another dimension of perspective is introduced, the results become more grossly inaccurate as there no compensation for this perspective.
I don't need to be able to calculate the height of a point, say a window on a sky scraper, but at least the coordinate at the base of the image plane, or which if I extend a line out from my image from a specified image space point I need at least the point that the line will intersect with the ground if there was nothing in the way.
All of the material I found about this says to just deproject the point using the world/view/projection matrices, which I find straightforward in itself except I don't have access to these matrices, just data I can collect at screenshot time and other algorithms use complex maths I simply don't grasp yet.
One end goal of this would be able to place markers in the 3d environment where a user clicks in the image, while not being able to run a simple deprojection from the user's view.
Any help would be appreciated, thanks.
Edit: Herp derp, while my implementation for doing so is a bit odd due to the limitations of my situation, the solution essentially boiled down to ananthonline's answer about simply recalculating the view/projection matrices.
Between position, rotation and FOV of the camera, could you not calculate the View/Projection matrices of the camera (songho.ca/opengl/gl_projectionmatrix.html) - thus allowing you to unproject known 3D points?

Calculate the area a camera can see on a plane

I have a camera with the coordinates x,y at height h that is looking onto the x-y-plane at a specific angle, with a specific field of view. I want to calculate the 4 corners the camera can see on the plane.
There is probably some kind of formula for that, but I can't seem to find it on google.
Edit: I should probably mention that I mean a camera in the 3D-Graphics sense. Specifically I'm using XNA.
I've had to do similar things for debugging graphics code in 3D games. I found the easiest way of thinking about it was creating the vectors representing the corners of the screen and then calculating the intersection with whatever relevant objects (in this case, a plane).
Take a look at your view-projection (or whatever your Camera's matrix stack looks like multiplied together) and realize the screen space they output to has corners with homogonized coordinates of (-1, -1), (-1, 1), (1, -1), (1, 1). Knowing this, you're left with one free variable and can solve for the vector representing the corner of the camera.
Although, this is a pain. It's much easier to construct a vector for each corner as if the camera isn't rotated or translated and then transform them by the view matrix to give you world space vectors. Then you can calculate the intersection between the vectors and the plane to get your four corners.
I have a day job, so I leave the math to you. Some links that may help with that, however:
Angle/Field of view calculations
Line plane intersection
ignoring lens distortions and assuming the lens is almost at the focal point then you just have a triangle formed by the sensor size and the lens, then the lens to the subject - similar trianlges gives you the size of the subject plane.
If you want a tilted object plane that's just a projection onto the perpendicular object plane

Vectors for Game Programming

Im not sure how to use vectors correctly in game programming. I have been reading advanced game design with flash which shows you how to create a vector with a start point and endpoint and how to use that for games, for example the start point would be used for a characters position in a game and the x and y length would be used for velocity. But since I have started looking online I have found that vectors are usually just x and y with no start point or end point and a character would be moved by having a position vector and a velocity vector and acceleration vector. I have started creating my own vector class. And wondered what the reasons for and against each method are. Or is it completely not important?
Initially a vector means direction. Classical vector is used in physics to present a velocity so that the vector direction stands for the heading and the vector length is a speed.But in graphics vectors are used also to present position. So if you have some point, let's say, in 2d space noted by x ,y it remains point if you don't want to know in what direction it set relating to the origin which is usually a center of the coordinate system. In 2d graphics we deal with Cartesian coordinate system which has an origin in top left corner of the screen. But you can also have a direction of some vector relative to any other point in the space.That is why you have also vector operations like addition, subtracting ,dot product, cross product. All those help you to measure distances and angles between vectors. I would suggest you to buy a book on graphics programming. Most of them introduce an easy to grasp primer to vector math.And you don't need to write a vector class in AS 3.0 You have a generic one - Vector3d