[HelloSupport]

I have a diagram that I wish to start in the upper left corner.  I can do this setting the camera with FitIntoView but I don't want the entire diagram to be displayed.   

My understanding is that position (0, 0, 0)  is the center of the screen and if I make the X coordinate of an objet negative, the object location moves to the left.  

In your sample code Lines3DSample, the first line is drawn with:

              <visuals:LineVisual3D StartPosition="-50,0,220" EndPosition="50,0,220" LineColor="Silver" LineThickness="1"/>
 
How do I calculate the StartPosition so that that line starts in the upper left hand corner?

[abenedik]

If you are using TargetPositionCamera and TragetPosition is set to (0,0,0), Heading = 0 and Attitude = 0, then (0,0,0) is in the center of the view.

If you are using an orthographic camera (targetPositionCamera.ProjectionType = ProjectionTypes.Orthographic), then this is much easier to achieve. In this case you set the ViewWidth that is defined in 3D space coordinates. The ViewHeight is defined as (ViewWidth * SceneView.Height) / SceneView.Width.

For example, if  ViewWidth is 400 (showing positions from -200 to +200) and SceneView.Width == SceneView.Height and TragetPosition is set to (0,0,0), Heading = 0 and Attitude = 0, then the top left position is (-200, -200, z is any value).

If you are using a perspective projection type (by default), then this is much more complicated, because the X and Y coordinates are affected by the distance to the camera. In this case you need to use Math.Tan to calculate the offset:

   
In this case fi angle is camera's FieldOfView angle (this needs to be converted into radians("/ 180 * MathPI")):
MathF.Tan((targetPositionCamera.FieldOfView / 2) / 180 * MathF.PI) = a / b.

b is the camera's Distance.
This way you can get the value a that that tells you how much you need to move the camera.