Mtg 7/26: Tue-28-Jan-2025

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Geometry and Transformations

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How's everyone today?
Good?
So I've tried To take the feedback. I
so I've specified chapter four, section one, and I have a link
to that for next class reading, as well as the PBR key version
for User's Guide and PBR key version for file format the
so a word about assignments. Since I said I'd talk about them
this Week, I
so there were 30 marks and
I think that adds up to 30.
Oh, I didn't.
So the first one will be to familiar, familiarize yourselves
with pbrt Making some,
some simple, some straightforward changes to the
example input file that generated the sphere floating
over the checkerboard and
is there any interest? So I see the second and third assignments
about Gen dealing with generating more complex images.
Is there any interest in providing an option to instead
of two and Three, do
Some implementation And
it. So I'm thinking option C,
so we're not putting all your eggs in one basket.
I'll make a choice on your courses for this. Does that any
discussion about this to begin with? Yeah, explain what that
option C is. A
bit i I'm thinking to give you a bit more variety of
opportunities to get marks so so this isn't weighted as much,
yeah, so do a variation of assignment of one of the bigger
assignments, two or three, And then the implementation is worth
a bit less. I
okay So I Will. I
I thought I had a bigger version of this. I
it. Usually, if you open it in a new tab, you can zoom in on it.
So this was done before 3d printers were really a thing. So
this was done with a program called Ray shade, written by
Craig Cole and
he was once considering doing his PhD at the University of
Regina,
thinking he went to Stanford instead. I
anyway, so one of
one of the assignments will be to take that shape and
reinterpret it. Take this image and reinterpret it. I
I don't know how many hours it took to compute this. I don't
recall, but it would not be long. Would be a few seconds
These days, I'm sure you
Does that seem okay. I'm
okay, let's Talk about The quiz. I
spherical ordinance and
yeah, octahedral encoding an equal
area mapping,
an equal area. So what about spherical coordinates. So it's a
nice representation, but if we're going to there are some
drawbacks, because we there's distortion of the poles. And if
we want to think about set uniform sampling using that
parameterization, we'll get A we'll get a
very equal distribution of points I
So the reasons are talking about octahedral encoding as a way to
be more efficient in terms of storage and
I've Never tried to draw an octahedron on the board.
And How does that look? I
so the idea is, we're going From the top, we're going To do
so this I'm
this is the upper part, and then we Fold out The bottom and
and the equal Area mapping
is a way to the
Instead
of I'll bring up the picture
here. So I'm just curious, is anyone able to click on the
links to the pictures and actually have something show up?
What are you using for a browser? Yep, fireproof. I have
them on Chrome. You're
just talking about big in Chapter pictures, right?
Yeah. So there, let me see,
there was An open image In your tab. I
I wonder
what the problem is. Anyway,
It's good to know you're still not numbered. The way they
advertise. I
So the third prioritization. So the idea is we're Taking the
unit square and we're
so we create the mapping from unit square to the unit disc,
and then we can
so we do the mapping so that we're preserving area, and the
points on the disk are mapped to the hemisphere for
so here's a link in the text to an appendix talks about sampling
a unit disc.
So this is about the spherical prioritization of spherical
coordinates. So we get a using that parameterization, we get a
collection of points at the pole, at the center, so we're
not getting uniform distribution over the whole disk. We're
so a correct map, it gives uniform distribution of light.
So I
So what about the Class, direction, code?
Oh, God. That direction from that's the it's similar to what
we talked about yesterday, when, when light rays are popping out,
it calls them right away, so you don't have any calculations on
them. So it keeps them sort of like in A spotty direction or In
the Correction direction. Yeah.
So then we can
bound the directions so that we know that we're going to do.
Using to reduce complexity. So here's something about the
constructor for the direction cone I
here for finding an angle then, so we want to create the
directions we so we can create a cone. So we have a point, p and
a bounding, in this case, a sphere. So we want to determine
the size of the cone that includes the directions here, so
we have point P and then the center of The sphere. And
so this is u of R equals one. So
so the idea is, we get, we can find the cone, which will
include the directions from point P and including that
bounding sphere and
so then we can look at whether these bonding cones, these
direction cones, are included in One in the other. So we can
simplify our calculations that way.
And similarly, we could take cones that are not inside the
other but we can Find a new cone that bounds both of them and
so what are two interpretations of the matrix transformation
in Frame and Frame too frank, yeah, That's right. You
so within frame means we're describing how points are points
and geometric objects are positioned within the same
frame. In that tetrahedron, example, had a bunch of spheres
that were transformed to create that shape. And then so that's
within frame and between frame
that's like when we're setting up the camera. So we specify an
arbitrary position for the camera, an orientation, so then
we can set up.
We can set up a matrix to go from the world coordinate frame
to the camera coordinate frame. We
and we'll talk about that.
So homogeneous coordinates and so we have a fourth element to
our vector. Instead of x, y and z, we have x, y, z and w.
So that w is generally one and so you can see in the textbook
to talk about making things more efficient
by omitting the division by W in some cases. So it's really the
projection, the projective transformations, that we'll be
talking about chapter five, that's the case where we need to
be concerned with W and do that division and
so what's the benefit? What's a practical benefit of using
homogeneous coordinates for Lyd geometry? I
You're asking. What's the point of keeping that last bit? Is
that what
you're asking? Well, I has
to do with multiplication. For instance, if you want to, if
everything is on zero, then you won't be able to save whether or
not it's scaled to a certain degree, and it will just kind of
ignore that bit. It's just a way of returning everything back to
its origin. I can
you explain that a bit more?
If you take a shape and you perform a bunch of
multiplications on its dimensions, so it's like three
times bigger. That last bit will give you the ability to divide
it by three. It sort of is. It will remember that was sort of
multiplied by three, and it allows you to say, Oh, this
thing is scaled by three. You could tell, by the way, the
weight variable is three, for instance.
Ok. So if you want to scale by 3x, Y and Z, I,
so if, if this last entry, it's a three as well, then It's, it's
really, this is Really the identity matrix.
You so we're talking about three dimensions here, so x, y and z,
and this last column is the so we can think of these as as the
basis vectors for our frame, and this is the origin i
Okay, so I
so we're representing three dimensions. Why is it a four by
four matrix and
well, because we have
x, y, z and w, I, I
so let's do some multiplication on the board. This is shout out
when I make a mistake, although this is pretty straightforward,
so we have three times x and nothing else i
three times y and nothing else, three Times said and nothing
else that we just Want. Nothing
so let's say w equals one. I'm
so that's the scaling.
What if we want to
what if we wanted to move the point as well, instead of just
scaling it, I
it. That's like the last column on
the right. Yes, so if we just have,
if we have three by three matrix, we have 3x
reduce the multiplication We
so if you want to translate to 123, and
so we're doing not just the multiplication, But we're doing
an addition here. So if we If
here We get i
so we get the addition just by the matrix multiplication. Does
that make sense?
Any questions or concerns about that? I
Okay, so there's a translation matrix that we just talked
About.
So there's a distinction here, applying transformations to
points and to vectors. So w equals one is a point and a
vector w is zero because we can think of a vector being defined
as The difference between two points.
So x1, y1, z1, one. And
so x1, minus x2, y1, minus y2, z1, minus z2, and then one minus
One, if equals zero and
so if vector, we have the W set term equal to zero, then we
don't get in
the get in these terms, because the plus one is because of The
one here.
So 3x plus zero, y plus zero, z plus one times zero, so that's
zero.
So the translation affects points, not vectors. Does that
make sense? I
it. So we can do a scaling where it's uniform, like we did here,
or we can assign different values for each term so they
have a function as scale to determine whether there's a
scaling component in the transformation. And so they're
testing the the length if it's greater, if the difference in
length after the transformation has been applied is greater than
A tolerance.
So then we have rotations as well.
So if we rotate around the x axis. So we can see the first
column is just what we'd expect. It's it doesn't do anything to
the x coordinate, it just multiplies the x coordinate
value by one, so it leaves it unaffected.
So Y and Z axes are changed. So base take the cosine of theta,
the sine of theta, that's how the y axis is transformed, and
then minus sine theta, cos theta is what the that access becomes
so.
So we're combining X and Y or Y and Z values with cos and sine
functions and
so the x coordinate stays the same, and we have Y And Z here,
and then we see the rotation I, a theta.
So here, for rotating around the Y axis, the Y column is the
identity, so it's unchanged. And for the Z axis, Z for rotating
about the z axis, the z coordinate is unchanged, but
then the X and Y are modified by those trig functions.
So we can rotate around an arbitrary axis and
so a is the vector we want to rotate about, and V is the
vector We'd like to rotate about that
so we find
V c by projecting v onto a, and that's the dot product. So
so one vector we can determine by subtracting V from
subtracting V C from V so that's v1 and then v2 you can take the
cross product v1 and a, a is normalized, so v1 and v2 have
the same length. And and then we can compute the angle and
so we can convert that into a matrix. Here's part of that,
that transformation in matrix form we
I can also rotate one vector to another. And then I just want to
talk about the lookout transformation. I
so we can specify Camera Orient, position and orientation and
so sometimes we get, we have one vector here, defined by the
difference between the look at and the i point, and we have the
vector from the up, and we do a cross product, and then we get a
third vector.
Then here they re compute up by doing a cross product so that If
well. So to get an ortho orthonormal basis, I
it. So I just wanted to point out in the last minute here that
we have that we were transforming normals. So if we
scaled the circle to be half as tall in the y direction, then we
think we're doing the right thing in B to get the normal to
transform the normal, but we've lost the property that normal is
perpendicular to the surface, so we need to do a bit of work to
handle the transformation of normal vectors.
So the takeaway is that normals must be transformed by the
inverse transpose of the transformation matrix, And
that's our time for today.
Any questions or concerns.
Thank you for today, and if you're doing Wiki pages, please
post the ID of your wiki page or your just the link to your Ricky
page to the class discussion so that I can add it to the
meeting, please. And thank you. Great. Rest of your day and see
you Thursday.
How did you meet? Meeting. Thanks.
How did you meet? Meeting. Thanks.

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