Ok this is attempt number 2 at getting an answer to this question.
I would therefore please request that you DO NOT answer unless you are
sure you can provide a full working solution.

Please do comment on the question if the parameters are not clear.

This diagram represents a plane in 3d space, it is sitatuted at a
distance S from the origin (i.e. distance O to fC is S). A, B, C and D
are corners of a square, the distance A to B, A to C, C to D, B to D
are all a distance 2S apart. A,B,C and D are all equidistant from O.


                   A ________________B
                    /               /
                   /           .X  /
                  /       .fC     /
                 /               /
                /_______________/
               C                D

                          .O

corner A = (Ax,Ay,Az)
corner B = (Bx,By,Bz)
corner C = (Cx,Cy,Cz)
corner D is not actually defined, but obviously can be derived from
the other information
point fC (face centre) = (fCx, fCy, fCz)
point 0 (origin) = (0,0,0)
point X = (Xx,Xy,Xz) -- this is the point on the plane.

The point X actually represents an intersection of the plane And I
need to convert the 3d plane into a separate 2D representation of the
plane. (like a map)
i.e.

      A?_______B?
       |    .X?|
       |       |
       |   .fC?|
       |       |
     C?|_______|

point X? can either be a coord, or be represented in terms of A?,B?
and C?. Or it can be represented as a vector (although obviously in XY
rather than XYZ)

The only thing that needs to also be considered is that sometimes a
face will be rotated so that the point of intersection will be exactly
the same in 3d space, but the points A,B,C and D will all have rotated
about point fC. This will obviously have an effect on the 2d
representation: rotating X? about fC? by the same amount.

I would like a full breakdown of how to convert the point in 3d space
X to a point in 2d space X?.

---

Clarification of Question by charltonian-ga on 24 May 2005 00:22 PDT

Ok, it seems that by trying to simplify the question by only refering
to one plane I have actually made the question more difficult.
I will cancel this question, and replace it with a new reworded
question, and a working example of what I mean (I just need to add
some colour coding to the 3d model.)
Please do not now answer this question.

I'm probably missing something non-obvious... X is by definition on
the 2D plane defined by ABC. Unless my geometry fails my mind, there
can be no such thing as a 3D plane -- it is defined by three points.
("I need to convert the 3d plane into a separate 2D representation of
the plane. (like a map)") ... O is not relevant, etc.

X is the same distance from A as it is from A', and from B as it is
from B', and on the same plane, etc...

If X (a point) was not on ABC, then you could use pythagorean theorem
to get your visual representation of X on ABC. A right triangle can be
gathered with known information of the delta between A and X , and B
and X. The imaginary perpendicular to the ABC plane, intersecting at
X' on ABC, is at this point rather trivial to obtain, once Az', Bz',
and Cz' are adjusted to z'=0

This is a free comment.

Yes, sorry, probably a bad explanation.
X is not the same distance from A, as X' is to A' although they would
be proportional. It is important to point out that they are not the
same thing, the 2d version is a representation of the plane in 3d
space. O probably isn't relevant to any calculation, but is given as
background info.
The plane exists in 3d space (as opposed to being 3 dimensional) and
can be at any angle. I have the data for all the points listed
(A,B,C,fC,X), but my 'camera' view could be as the first diagram, it
also could be as below.

                fC   X
       .________.____.__.
      A/C              B/D

essentially the only difference in the points now is in the Z axis.

What I need is way to convert it to the 2d map version however the
points are oriented in 3d space.

I really hope it is, or there is, a trivial solution, as I would love
to see it (wit h a full working though).

May I ask again if X is on plane ABC?
and A'B'C' is simply an orientation of ABC?
Um. wait a sec. You already know what ABC is, and can transform it,
right? How do you do that? Why is that any different than transforming
X?

Yes, I transform the plane, unfortunately the transformations are
handled incrementally (and there are lots of them), and then they are
adjusted for floating point errors, so I never have one reverse
transform to move the plane back to perpendicular to the camera.
It is conceivable that I could calculate a reverse transform from some
other data, but it seemed to become fiendishly complex very quickly I
was hoping for some kind of pythagorean solution.

ALSO...
X isn't on the plane, it is an intersection of the plane. So it
doesn't transform when the plane is moved. However (referring to my
last post), I could take the current position of X, and then transform
it to perpendicualr to the 'camera' position, and then convert the
scale. (Like I say tho, this becomes complex quite quickly -  for me
at any rate.)

I'm so confused. How in the world can an intersection not be a part of
the plane? That is its definition.

The intersection of a cylinder and a plane is an elipse *on*/in? the
plane. The intersection of a line and a plane is a point on the plane.
The intersection of two planes is a line segment on the plane. If you
can transform ABC, then certainly you can transform ABX. It's the same
plane.

If you're trying to transform the "rest" of whatever X is... 

 A_______B
 |    X  |
 |Y___.--|---Z
 |       |
 |_______|
 C
                        __
(Let's say line segment YZ passes through plane ABC at point X), then
what we need to do is render visible stuff outside of plane ABC, which
is not what you asked.

Also, you said both "point X = (Xx,Xy,Xz) -- this is the point on the
plane." and "The point X actually represents an intersection of the
plane"

Let's go to some fundamental questions ... what is intersecting with
the plane? a point, a line(segment), a plane, a cylinder, what? Is the
problem that we don't know (Xx, Xy, Xz)? or is it that we can
calculate where it is relative to (Ax, Ay, Az) and (Bx, By, Bz) ... do
we know or need to know distance or location? Is there anything that
is known absolutely about X relative to A and B? Well, sure ... X
intersects ABC. Do we know if X is merely a point or a point on a
geometric object that we might know a trajectory or something? Do we
need to render anything but this point on the plane?

Ok, it seems I'm trying your patience, so I'll try to explain.
You actually have grasped completely what the problem is, X is the
point of intersection of a line (that goes from an arbitrary point P
to the origin, I do have the coordinates of P)
I have calculated the point Xx,Xy,Xz, which is on the plane. And from
this using any of the the other points we know (A,B,C,fC) I need to
create some function which I can use to calculate where X' is in
relation ot A',B'...etc
In answer to your questions: the plane is a face of a cube, but if I
can calculate point X' for one plane I can do it for the rest of them
(I would just use different values of A,B,C etc.)
Not sure about the trajectory part.
And no, we don't need to render anything but this point on the plane.

I should have added the line from PO does not move, it is fixed.

I should have added the line PO does not move, it is fixed.

ok, but ... let me clarify a few more things. and, I'm still not sure
why X, a point, cannot itself be transformed ... we'll get back to
that in a sec.
    <-->
Line OP passes through plane ABC at point X, which is able to be
calculated. This calculation gives a point in space at (Xx,Xy,Xz).
Line OP is fixed. By definition, all points A, B, C are equidistant
from O. I'd gather that perhaps ABC is fixed as well. We certainly
don't want X to move relative to ABC, do we?

So, that brings us to a camera view. I still am having trouble
understanding how you have the math to transform A and not X. It
appears that either by rounding or not trusting the results, X no
longer appears on plane ABC. So, let's start with an orthogonal camera
view of the square, effectively making your first diagram:

      A?_______B?
       |    .X?|
       |       |
       |   .fC?|
       |       |
     C?|_______|

Now, apparently, I think I understand that A, B, C aren't really
orthogonal to the x and y axes, so to get the A', B', C' drawing
requires rotating ABC.
Assuming that z=0, X is still on the same plane as ABC...
http://mathworld.wolfram.com/Distance.html tells us the distance
between A and X is d=sqrt((Ax-Xx)^2+(Ay-Xy)^2))... This is sorta
interesting. I suppose what'd be more interesting is the distance from
X to AB and X to AC ...
http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html
This would give us the approximate (rotated) position of X, assuming
A'B' and A'C' are orthogonal to the axes. (then, you can use this
distance to

But still, if you can transform ABC, you can transform ABX, and
frankly you should be able to transform the point easier than the
plane. I assume you know how to transform Fc as well? What formula do
you use to do that? (Well, I take distance AB and distance CD and make
it 1/2 the way ...) ok. So if we can determine the X-AB distance and
divide it by 2s, and then X-AC distance and divide by 2S, we have a
ratio ... if we learn the new ratio of the distance (A'B' to C'D')/2S
, we can apply that distance to X's location, relative to AB and AC.

oh, but what if we have a parallelogram type thing? ... Well, somehow
you know how to place Fc in the right place. I'm certain you can
figure out how to shift  X. It's simply the relative position, shifted
the direction and amount that the parallelogram is shifted

This is more painful than simply figuring out that if you know how the
camera looks at point A, it'll look at point X the same way it looks
at point Fc, and it really doesn't care that X is a point on ABC. The
only reason that it might is if you're considering ABC to be opaque,
and rounding errors make X *under* the opaqueness.

BTW... did I mention this is a free comment? :) This just seemed
interesting to me, is all.

OH, btw :) I think you might be interested in www.blender.org ... Oh,
no reason, really :).

Ah I can see where the confusion lies, because I feel confident that
you understand what I am talking about.
The first problem is: the cube/face is not rotated about the global
axes but its internal axes. I worked out a method of rotating a point
about an arbitrary axis (which essentially tends to be rotating the
points about fC - bear in mind this is a cube in the real project not
just one plane, and through some fairly simple calculations I decide
which axis is which.)
However, I really don't think it's possible for me to transform it
back (I have tried, the results were not very impressive, but the main
problem is, it's hard to see where some of the strange results come
from, so I have no way to improve from them.)
Yes, I was looking at some kind of proportional transform, and got a
little stuck at the parallelogram stage, but it doeas appear that
there should be enough information there to calculate it (well except
that I couldn't of course.)
To be honest if there were rounding errors, they probably wouldn't be
too noticable - so I would be ok with that.
Yes, I am aware that this is free, but I'm hoping that maybe between
us we will get to a solution and then you can take the reward.
I had a quick look at www.blender.org, is it something you work
on/for? (and I will have a proper look tomorrow)
I must got to bed now it's late here, and tomorrow is another day at
the coal face. But beleive me when I say that the fact you even
understand my little puzzle is quite reassuring...

Yes! That is why I was confused! I assumed that you were *only*
talking about a square plane with points equidistant from O, and you
were panning a camera around this plane and wanted to view the dot X
on the square ... Well, you didn't say anything about a cube...

Now I think about a laser pointer centered at O and pointing at P
shining through a rotating? clear cube that needs to show a point on a
plane that varies as the cube rotates. Which, again, isn't the same
thing as the question, but let's go with it.

I think the question is now: Where is the red laser dot on the
external plane (furthest from O) as the cube rotates? Indeed, how do
you describe the shape of the plane and the location of the dot in
that plane given the location of A, B, C and X ... The camera, I hope,
is centered viewing the x and y axes in orthogonal viewpoints.

Hold on. The cube can't rotate about its own axes *AND* the surface
being equidistant from O. Unless of course we're talking about a
camera view observing the cube.

Well, that's where we need to talk. Is the cube centered on O? That helps. ...

The cube, centered on O, rotating strangely, has a laser pointer
pointing at point P. As the cube rotates, its front view changes from
a square to a trapezoid and other geometric formations. The laser
pointer, fixed, intersects this floating plane at a point, and you're
trying to describe in 2D space the position of the intersection as
well as the shape of the plane that it intersects. How am I doing?

And, of course, the intersection makes (could make?) the dot vary a
bit in location, even though OP is fixed.

Can I fix the camera to look at x and y axes being east-west and
north-south respectively, or is the camera position still random as
well? I don't necessarily care, but the math is easier...

Somehow, if z'=0, and we also assume that there is no refraction
because of the glass we're using in our cube ... if we've been able to
calculate (Xx,Xy,Xz), then I'd have to say that Xz doesn't matter ...
I'm still back to if you can transform A, you can transform X. What am
I missing?

PS: No, I don't work on blender. It's a 3D modeling software that's
Open Source (read: free to use and free to modify) and certainly can
help with observing this as you want.