Re: [Closed REQ 5720]: Geomview external module

[daemon]

Hi -- yes, we did get your earlier note, sorry not to have responded.

Though you can't directly get a "scale factor" from geomview, you can
get information that lets you do what you're asking: specifically, you can 
have geomview send you a transformation matrix giving the current position
of the object.

You could get geomview to notify you whenever the transformation changes;
however, it's probably easier to write it as a polling loop, which periodically
asks geomview for the current transformation.  In that case, suppose your
object's name (in the geomview object browser) is "mesh".  You'd send
a command to geomview like:

    (write transform - mesh focus)

This means:
   Send me the transformation matrix which transform from coordinates of "mesh"
   to the coordinates of "focus" (which is the currently-active camera window).

Your module will receive, on its standard input, something like

transform {
    0.74639952   -0.00292574    0.05561925             0
   -0.00632918     0.7381438     0.1237648             0
   -0.05533536   -0.12389196     0.7360726             0
  6.4132237e-09  -7.4794553e-07   -4.45694399             1
}

Call this 4x4 matrix T.  Let p = (x,y,z,1) be the coordinates of a typical
point in your object.  Appending the "1" represents it properly in homogeneous
coordinates, which are what geomview uses.  To get the corresponding
coordinates in the camera's coordinate system, multiply p . T  (note that p
appears as a row vector multiplied on the left, not a column vector on the
right; that's a common convention in computer graphics).  
Let p . T = (X, Y, Z, 1), so X,Y,Z are camera coordinates.  Assuming it's
a perspective camera (the default), that point projects on the screen
to something related to (X/Z, Y/Z), multiplied by the window size in pixels,
and divided by roughly the camera's field of view in radians (actually,
it's twice the tangent of half the angular field of view).
GCL commands "(write window - focus)" and "(write camera - focus)" would
return information which would let you compute that.  However, if you typically
adjust the view by scaling the object and/or moving the camera (as opposed to
zooming, which changes the angular field of view), it might be easiest to
just assume a typical value for window size and field of view: say,
500*(X/Z, Y/Z).  That's about right for the default 350-pixel window and
40-degree field of view.

Now you can estimate the screen position of a given point on the mesh.
The first three elements of the first column of T (.746, -.006, -.055 in this
example) are a vector parallel to the screen X-axis; call this vector Sx.
So, the screen radius of a small ball of radius r, lying near p in the mesh, is
 500 * (magnitude(Sx) * r / Z)

This is probably the kind of expression you need to compute optimal feature
sizes in the objects you're drawing.