"Bald Eagle" <cre### [at] netscapenet> wrote:
> "Kenneth" <kdw### [at] gmailcom> wrote:
>
> > The object rotational problems I see when using the Point_At_Trans and
> > Reorient_Trans macros are specific to those tools, I agree. I've tried mightily
> > to 'correct' the resulting 'unexpected' object rotations (which DO result from
> > the normal's direction into space), but it's all quite mathematically
> > complicated.
>
> It's always complicated when we're using black-box macros and shooting in the
> dark.
>
> > Try tracing a simple sphere from random directions with thousands of greeble
> > objects applied, while using the found normals and those alignment macros to
> > place them, and you'll see what I mean.
>
> Well, I tend to do things like that "by hand" using macros that I write deriving
> everything from first-principles.
>
> > Here's an example using the Point_At_Trans macro, applying typical pre-made
> > objects. Note the delineated 'diamond' rotational patches.
>
> They're not "diamonds" - they're triangles.
>
> So let's peek into tranforms.inc:
>
> // Similar to Reorient_Trans(), points y axis along Axis.
> #macro Point_At_Trans (YAxis)
>      // this is just a simple vector normalization.  No mystery.
>      #local Y = vnormalize (YAxis);
>      // this is a macro - I can't tell if it's a source-code thing, or a
> math.inc thing. But here's where I think is what causes what you see. (* see
> below)
>      #local X = VPerp_To_Vector (Y);
>      // vcross returns 0 when vectors are (anti-)parallel, and +/-1 when
> perpendicular
>      #local Z = vcross (X, Y);
>      // As far as I can see, shear trans just applies a straightforward
> transformation matrix, using x, y, and z.  And all those three vectors are
> perpendicular, which is merely what all of the above code does.  And so the
> whole object gets oriented to a "new y vector"
>      Shear_Trans (X, Y, Z)
> #end
>
> * So, let's take a look at VPerp_To_Vector()
> It takes a single argument.  Given a vector, find a vector perpendicular to it.
> Which is ridiculous, given that I can spin a T-square around, and come up with
> literally an infinite number of parallel vectors.
>
> So let's take a look at what the macro specifically DOES.
>
> Peeking into math.inc:
>
> // Returns a vector perpendicular to V
> // Author: Tor Olav Kristensen
> #macro VPerp_To_Vector(v0)
>    #if (vlength(v0) = 0)
>       #local vN = <0, 0, 0>;
> // Sanity check.   If you give it a/the zero vector, send it right back.
>
>    #else
>       #local Dm = min(abs(v0.x), abs(v0.y), abs(v0.z));
>
> // And here's where we get the three-fold symmetry.  Find the smallest component
> of the supplied vector.  This drives the rest of the macro results, which
> computes the vector cross-product between your "new y vector" (the surface
> normal), and the cardinal axis that it is "farthest away from".  What's the
> center point of that 3-pronged choice tree?  The center of the triangle(s) in
> your render.
> Take a cardinal axis and start generating results from this macro as you rotate
> it.  Eventually you will rotate it so far that a DIFFERENT cardinal axis will be
> chosen by the macro to calculate the vector cross-product with.  And there are
> only 3 axes, so you get triangular wedges of each octant of your sphere.
>
>       #if (abs(v0.z) = Dm)
>          #local vN = vnormalize(vcross(v0, z));
>       #else
>          #if (abs(v0.y) = Dm)
>             #local vN = vnormalize(vcross(v0, y));
>          #else
>             #local vN = vnormalize(vcross(v0, x));
>          #end
>       #end
>    #end
>    vN // <--- returns one of three results
> #end
>
>
> So, now that we see where the problem is and why, we can formulate any number of
> solutions.
>
> The first and easiest is just write a new macro where you either use only one
> dedicated axis as your second vector cross-product vector, or one where you can
> choose which vector you want.  Or generate a random vector.
>
> Second, you could take a look at which cardinal vector was used, and further
> rotate your greeble based on that, to "correct" its orientation.
>
> Maybe there are other options as well.
>
> Maybe the POV-Ray light_source will shine its benevolent rays upon us, and TOK
> will chime in with something elegant and brilliant.   :D


Hehe. I did not notice this post before long after you posted it...

You probably meant to write that there are an infinite number of vectors that
are orthogonal to a given vector.

I was a bit frustrated when somebody started to make a VPerp_To_Vector macro,
because how can one know which of these inifinite number of orthogonal vectors
the user needs or wants?

One can not know that. (More information is needed to calculate one specific
vector.)

But if the user doesn't care about anything else than that the returned vector
is orthogonal to the given vector, then a simple VPerp_To_Vector macro would be
feasible.

So I thought that if we were going to have such a macro, it should be a
numerical stable one. And my suggestion ended up as the VPerp_To_Vector macro in
math.inc. (Btw.: I don't like the name that it got.)

To be numerical stable, it should not try to calculate the cross product with a
vector that is parallel or almost parallel to the given vector. The simplest
candidate is the zero vector, which is orthogonal to all vectors. (The zero
vector can also be considered to be parallel to all vectors). But that would
certainly not be a useful candidate for the user. The second simplest candidate
for a vector to cross it with is the basis vector for the axis that is
"farthest" from it.

If you have a vector vA where the x component is the smallest, then the returned
vector is <0, -A.z, +A.y>.
If you have a vector vA where the y component is the smallest, then the returned
vector is <+A.z, 0, -A.x>.
If you have a vector vA where the z component is the smallest, then the returned
vector is <-A.y, +A.x, 0>.

For a vector to be orthogonal to another their dot product must be zero. I.e.:

<0, -A.z, +A.y> * <A.x, A.y, A.z> = 0
(0*A.x) + (-A.z*A.y) + (A.y*A.z) = 0
-A.y*A.z +A.y*A.z = 0

<+A.z, 0, -A.x> * <A.x, A.y, A.z> = 0
(+A.z*A.x) + (0*A.y) + (-A.x*A.z) = 0
+A.x*A.z -A.x*A.z = 0

<-A.y, +A.x, 0> * <A.x, A.y, A.z> = 0
(-A.y*A.x) + (A.x*A.y) + (0*A.z) = 0
-A.x*A.y +A.x*A.y = 0

Also note that in two dimensions there are only two 2 vectors that are
orthogonal to a given vector vA:

Those vecors are <+vA.v, -vA.u> and <-vA.v, +vA.u>.

Now notice the simlariy with the results from the VPerp_To_Vector macro.

--
Tor Olav
http://subcube.com
https://github.com/t-o-k

Post a reply to this message