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• Screen space to world space

  - by user13414

  I am writing a 2D game where my game world has x axis running left to right, y axis running top to bottom, and z axis out of the screen: Whilst my game world is top-down, the game is rendered on a slight tilt: I'm working on projecting from world space to screen space, and vice-versa. I have the former working as follows: var viewport = new Viewport(0, 0, this.ScreenWidth, this.ScreenHeight); var screenPoint = viewport.Project(worldPoint.NegateY(), this.ProjectionMatrix, this.ViewMatrix, this.WorldMatrix); The NegateY() extension method does exactly what it sounds like, since XNA's y axis runs bottom to top instead of top to bottom. The screenshot above shows this all working. Basically, I have a bunch of points in 3D space that I then render in screen space. I can modify camera properties in real time and see it animate to the new position. Obviously my actual game will use sprites rather than points and the camera position will be fixed, but I'm just trying to get all the math in place before getting to that. Now, I am trying to convert back the other way. That is, given an x and y point in screen space above, determine the corresponding point in world space. So if I point the cursor at, say, the bottom-left of the green trapezoid, I want to get a world space reading of (0, 480). The z coordinate is irrelevant. Or, rather, the z coordinate will always be zero when mapping back to world space. Essentially, I want to implement this method signature: public Vector2 ScreenPointToWorld(Vector2 point) I've tried several things to get this working but am just having no luck. My latest thinking is that I need to call Viewport.Unproject twice with differing near/far z values, calculate the resultant Ray, normalize it, then calculate the intersection of the Ray with a Plane that basically represents ground-level of my world. However, I got stuck on the last step and wasn't sure whether I was over-complicating things. Can anyone point me in the right direction on how to achieve this?

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• Finding furthermost point in game world

  - by user13414

  I am attempting to find the furthermost point in my game world given the player's current location and a normalized direction vector in screen space. My current algorithm is: convert player world location to screen space multiply the direction vector by a large number (2000) and add it to the player's screen location to get the distant screen location convert the distant screen location to world space create a line running from the player's world location to the distant world location loop over the bounding "walls" (of which there are always 4) of my game world check whether the wall and the line intersect if so, where they intersect is the furthermost point of my game world in the direction of the vector Here it is, more or less, in code: public Vector2 GetFurthermostWorldPoint(Vector2 directionVector) { var screenLocation = entity.WorldPointToScreen(entity.Location); var distantScreenLocation = screenLocation + (directionVector * 2000); var distantWorldLocation = entity.ScreenPointToWorld(distantScreenLocation); var line = new Line(entity.Center, distantWorldLocation); float intersectionDistance; Vector2 intersectionPoint; foreach (var boundingWall in entity.Level.BoundingWalls) { if (boundingWall.Intersects(line, out intersectionDistance, out intersectionPoint)) { return intersectionPoint; } } Debug.Assert(false, "No intersection found!"); return Vector2.Zero; } Now this works, for some definition of "works". I've found that the further out my distant screen location is, the less chance it has of working. When digging into the reasons why, I noticed that calls to Viewport.Unproject could result in wildly varying return values for points that are "far away". I wrote this stupid little "test" to try and understand what was going on: [Fact] public void wtf() { var screenPositions = new Vector2[] { new Vector2(400, 240), new Vector2(400, -2000), }; var viewport = new Viewport(0, 0, 800, 480); var projectionMatrix = Matrix.CreatePerspectiveFieldOfView(MathHelper.PiOver4, viewport.Width / viewport.Height, 1, 200000); var viewMatrix = Matrix.CreateLookAt(new Vector3(400, 630, 600), new Vector3(400, 345, 0), new Vector3(0, 0, 1)); var worldMatrix = Matrix.Identity; foreach (var screenPosition in screenPositions) { var nearPoint = viewport.Unproject(new Vector3(screenPosition, 0), projectionMatrix, viewMatrix, worldMatrix); var farPoint = viewport.Unproject(new Vector3(screenPosition, 1), projectionMatrix, viewMatrix, worldMatrix); Console.WriteLine("For screen position {0}:", screenPosition); Console.WriteLine(" Projected Near Point = {0}", nearPoint.TruncateZ()); Console.WriteLine(" Projected Far Point = {0}", farPoint.TruncateZ()); Console.WriteLine(); } } The output I get on the console is: For screen position {X:400 Y:240}: Projected Near Point = {X:400 Y:629.571 Z:599.0967} Projected Far Point = {X:392.9302 Y:-83074.98 Z:-175627.9} For screen position {X:400 Y:-2000}: Projected Near Point = {X:400 Y:626.079 Z:600.7554} Projected Far Point = {X:390.2068 Y:-767438.6 Z:148564.2} My question is really twofold: what am I doing wrong with the unprojection such that it varies so wildly and, thus, does not allow me to determine the corresponding world point for my distant screen point? is there a better way altogether to determine the furthermost point in world space given a current world space location, and a directional vector in screen space?

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