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We are trying to align our camera in order to see as closest as possible a rectangle(a face of a parallelepiped since we are in 3D). The aspect ratio of the viewport considered here and the rectangle to align do not match.

What we do:

In our scene is this rectangle. The camera is position in front of it. Here is the configuration of the camera:

projection = Matrix.CreatePerspectiveFieldOfView(
    Fov, AspectRatio, NearPlane, FarPlane
);

view = Matrix.CreateLookAt(Position, Target, this.Up);

float Fov = PI/4;

float AspectRatio = device.Viewport.AspectRatio;

Vector3 Target is set to the center of the rectangle.

Vector3 Position is the value we are looking for.

And then to initialize the camera: oat width = (float)GraphicsDevice.Viewport.Width; float height = (float)GraphicsDevice.Viewport.Height;

float dist = (width / 2) + (height / 2);

float alpha = 0;
float beta = 0;
float gamma = 0;

alpha = (float)Math.Sqrt(Math.Pow(width / 2, 2) + Math.Pow(height / 2, 2));
beta = height / ((float)Math.Cos(MathHelper.ToRadians(70.3f)) * 2);
gamma = (float)Math.Sqrt(beta * beta - alpha * alpha);


position = new Vector3(width / 2, -height / 2, gamma);
target = new Vector3(width / 2, -height / 2, dist / 2);

camera = new Camera(GraphicsDevice.Viewport)
{
    Position = position,
    Target = target,
    Near = 1f,
    Far = 10000f
};

(VS solution)

One more precision: to create the view, we are using Matrix.CreateLookAt and for the projection Matrix.CreatePerspectiveFieldOfView. We are not sure if these are the perfect choices either but sounds like it.

What we get:

After some basic trigonometry we got a camera Position which works when the Height > Width. As: enter image description here enter image description here

But it does not when the Width > Height: enter image description here

Do you...

...have any idea what are we doing wrong? Do you have any other or better way to achieve this?

Mr.Pe
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  • If I understand you correctly, you want to place the camera as close as possible to the rectangle that allows the entire rectangle to be visible in the field of view, right? Have you considered posting the trigonometry that you're currently using that almost work? – Dervall Aug 31 '12 at 14:53
  • If you can post your code in a zip file me and other people can do a rapid test and get your solution more faster. – EdgarT Aug 31 '12 at 16:25
  • Thanks for our advices, more code is in the post and the full solution is linked. Dervall: You perfectly understand my goal. @EdgarT: That would be wonderful if you can take a minute to spot our mistake! Thanks – Mr.Pe Sep 03 '12 at 05:38

1 Answers1

0

Finding the height of the pyramid does not work. Just use the Tan and since the Fov = PI/4 the code should look like this:

float tanGamma = (height / 2) * ((float)Math.Tan(MathHelper.ToRadians(67.5f)));
alecnash
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