Mtg 11/23: Mon-21-Oct-2024

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How's that i
Happy Monday you
This is 11, number, 23,
so suddenly realized There was no quiz. It occurred to me last
night that I had not done that on Friday. I
the assignment who is posted? I posted that before break, but
now I've actually linked it to you our courses. So if you're
looking in your courses for it, you can click here on the
description.
Click here for the description, and it will take you to the
description of the assignment.
You so they're both programming and non programming questions.
There any questions about the questions? Yeah, pretty exactly
what it says. But like, how do you determine
the highest quality image without doing extra work. What
exactly are you asking for there? I don't know. I even
asked to say anything on the first question mark. Saw for
that. I thought I answered it well, but terrific, sure.
I didn't repeat myself. Did I miss it? Does have that like
the bottom of the questions thereafter I
can be defined transformations.
No, I said just not
supposed to do. Yeah.
So thanks for that. Heads up. I will
clean that up.
So the intent of discussion about how to create a high
quality image without doing extra work, is it
supposed to be related to the gasket? Or was that just like
its own separate question?
No, it was related to the gasket. So if we have a stream
that's defined against 512 by 512 do if we have US
coordinates that go from minus one to one and
so
we're dividing the
two into 512 we're dividing two by 512 so the sample, each
sample, is assigned that size of space. So if we make, if we
don't generate, so we
can say this whole space, what happens to it under
transformations?
So we start out with
two and then we shrink by one half and we apply Another
transformation, which tricks it by a half.
So is this? So we're getting
two to one quarter
and so on. So it's was asking for a discussion about when we
could stop how many, how many levels of recursion do we need
to get the size where we
Yeah, so the sample relates To the resolution of the canvas,
and Then further random or chaos game i
Well, that's,
that's more than 250,000 points.
So that's probably a little bit of overkill. We don't need to
generate that many points to generate a high resolution
representation of the gasket and
So we can estimate I
so 2500 is too Few, but it's up discussion about how we might I
how we might decide what reasonable number. So the issue
here, because it's random, you don't have the same idea of the
precision with the recur recursive subdivision. So it's
it's a simple algorithm to apply, but it's frustrating
because it's random and it's hard to get precision, precise
answers about it. So anyway, that was the kind of discussion
that I was thinking of. Does that make sense? I
so I wanted to talk about major scenes and
I'm viewing
less. I also want to say, if you have some questions about
assignment two programming we I'll have a sound a bit of code
to illustrate some of the ideas I'm Looking for, and we can
discuss that on Wednesday. You
it so I promised no more Matt on the board. So I did
it with Blake tack. I didn't plug in the cable. That's why
it's not working. You
okay, you see that all right, should I zoom in a bit? I
i can see the cursor on the big monitor that isn't connected to
my computer. I
so 50th anniversary celebration, we tried to use the owl camera.
But do you know what that is? It's it looks a bit like an owl.
It turns its head 360 degrees to track who's speaking. Then I got
two cursors on my laptop, and anyway, it was not so great.
Sometimes less technology is better, I think. Anyway, so here
we have matrix multiplication. So we're going to deal with four
by four matrices, so I've laid them out here, and I've labeled
the entries in column major format. So a, one A, one one
through a, one four is column one A, two, one through two,
four is column two, column three, column four and
so
multiplying The A times B will give a matrix C and
and where we do the multiplication is we
pick a row, not just pick A row, take the row correspondence. So
a one, so c1, one, take row one from a in column b1,
let me say that again. So to compute the value at the
position c1, one,
we take row
one from matrix A, column one from matrix B, and we sum them,
a, one, One times b1, one A, two, one times b1, two, A, 231,
times b1, three, a, four, one times b1, four. So here are the
rotation matrices around different axes and
so we have rotation around x, y and z. So here is our example
from the other day.
Rotate around x by 90 degrees, rotate around y by minus 90 and
rotate around z by 90. So this is not too complicated, but it's
to keep track of the ones and servers because because cosine
of 90 is zero and sine of 90 is one, because It's all in the y
component.
So the first step was to combine and multiply these y and x
matrices together.
That gives this matrix. And then this is the rotation around z.
So multiply those two together,
you get now, that matrix, so there's no zero columns that I
came up with on the board way back when.
Thanks to Corbin for pointing out my mistake.
So then I
did something a little different. Here we took the
point we came up with in class seven to five. So then I
I'm going to rotate about that point. So I
so the first step is to translate that point to the
origin. So we write minus seven, minus two, minus five. That's
the translation matrix to bring our new point for the rotation
to the origin, then we apply our rotation matrix, and then we
translate back and it. So this on the right is the
multiplication of these two matrices, and then we multiply
it by the third one to translate it back. And then you get the
final matrix.
So you notice that the rotation is still the same.
The upper three by three, the left upper three by three
portion is the same, but the but the translations are different.
So how can I check that I've gotten to write that I haven't
made another mistake in my late tech I
I'm not talking about writing latex, just to be clear, does
anyone use latex For document processing? Yeah, it's quite
nice, I think I
so if we're rotating about the point 725, and
what Should that point map to under the transformation?
Yes, I'm going to resist doing math on The board. Let's try if
so okay. I sun, Okay, I'll
stop trying to understand why I can't see a curse on my laptop.
So, 7250,
times, seven, zero times two, five times one and
And two times, one let
me try that again.
00, five times one plus two times one. So the first entry is
zero is seven part of me,
then we have
zero minus two, zero plus four. So that gives us plus two, and
then 700
minus two. That gives us five, so 725, and
is a fixed point under that transformation. So wherever we
center the rotation and scaling and cheer, that'll be the fixed
point of The transformation you
Does that seem okay? I
so the nice thing is that we can do everything by multiplying
matrices together And
sorry, I'm having the same problem. I did
the october 11 event and
so when we're Talking about viewing with the computer,
we have to position the camera, select a lens for the camera,
which means a projection matrix, and we need to set the view
volume for clipping.
So that just means that
we we
we determine what's visible. We
so the default projection is orthogonal, and that's what
we're using in our examples for 2d and for early ones in 3d
if we specify objects outside of the view volume, they'll be
clipped out so we won't see them.
So we can reposition the camera. It's more it's originally at z
equals zero.
We can
move it back so we can see More objects and so
we
translate by some distance and
so here's the effect of the translation,
and we can move the Camera to any desired position by sequence
of rotations and translations. So if we were in the side view,
we could rotate by 90 degrees and Then apply the translation
to move the camera
away. I
so it says, Remember the last transformation specified is the
first to be applied. I
so that means so we're Multiplying the matrices
together to get The
So
multiplying them.
So remember, we did a multiplication like this first
so we could start with identity, and then
apply rotation matrix X, about X, rotation matrix For the
rotation about y, and then rotation about said.
So the way To do This Is The
that makes sense. So
we talked a little bit about the look at the look At function or
approach to specifying the counter position i,
so we have the camera and a position that We look at.
We have a camera at a position that's where I
looks out of the scene, so that we need something to look at. So
so that gives us one dimension of the vehicle. Coordinates i
right? So what else do we need to specify our other two
dimensions, other two.
Other two vectors are coordinate basis you
the camera could be We
have a suspected orientation for the Camera. How could we do
that? I
so what if I'm taking a picture like this of
the landscape orientation,
if I want to do I want To do a
core object or something in between, I
we need to pick a direction. Here you
so we specify the view out right here, Like
and how can we Get
the third axis and
when should the third axis be in relation to These two axes? Why
can
that's take a
minute today. What property does the third axis have in relation
to the other two that are here? Yeah. Does that make sense?
Yeah.
It so we can find a perpendicular vector by doing
the cross product of the vector from the eye point to the look
point and the left vector does that.
Makes sense? Maybe?
So when we're doing
by default, we give them
orthogonal projections. So that means that the ones that are in
the view volume are clipping coordinates x and y are
preserved, but the projected value of z is zero, because
that's we're removing the Z, the depth information.
So this is a matrix that we could use to do that. Notice
that it's the identity matrix less than one for z,
so
that force is said to be zero. So he makes a note. In practice,
you could use that any matrix of n, set z equal to zero,
center of projection for a simple perspective projection,
the center is at the origin, and the projection plane is at z
equal to D, with D greater d less than zero, because positive
z is coming out of the screen in our right hand, fourth system.
So negative z will be into The screen, into our model and
it. So here we have a top and a side view. So this we're looking
down the y axis here, so we have x and z. So the point at x, said
is going to be mapped to the point
XP, at distance d,
so looking down the x axis, you see that the y projected point
is going to be At that this is D. So we can calculate x, p is
equal to x over z divided by d and y, p is y divided by z over
d and z p equals d and
so the projection is,
we have The one over d term in the bottom row the matrix. So we
get,
we get that perspective division and
so here's the view volume for an orthogonal viewing. Specify the
left, the right, at the bottom, The top and the near and far and
so then we can also, here's an example with the Perspective
view volume. That's a few frustum, frustum and
so it comes to a plane at the eye point, and then we have near
and far planes as well.
So we can specify left, bottom, left,
right, bottom, top, near and far. We can also specify a field
of view and An aspect ratio and and an aspect ratio. I
Ah, so this is a perspective view volume.
So we can change The field of view and
and if we modify the Near
So, you can see if we
don't set the near and far to include the models that we want
to see will get
clipped and
and then we can also change the zip position of The model. We
so this is the one Daryl point out
so we can Change the near The
so just to show you the shaders. The vertex shader is
we're taking the model view projection, model view matrix
and a projection matrix from that are specified in the
JavaScript file. So we're taking the attribute position,
multiplying it pre multiplying it by the model V matrix, and
then pre multiplying that by the projection matrix, so that these
are uniform variables and so they're not changing with the
attributes, and then The fragment shader is just Passing
through the color and
so here we get we set up the Matrix Model view matrix
location and the projection Matrix location.
We have the callbacks we're
so these are functions in the MV new.js file. So we specify we
get the i vector and
from the radius and the theta and the phi angles and
I'm just looking for where they specified. Oh, okay, so here the
app, it's looking at the origin and and the up is the y
direction, so not To pretty straightforward there. And
so here's the function to generate the viewing coordinates
Based on the eye position the at Looking at and the up vector.
So v is normalized vector that goes from the look at position
to the eye position that's v n is
the view direction vector cross with the up vector. And then u
is the cross between the
so n is the normal vector to the plane defined by the up and the
view vectors, And then u is the cross between N and V. So we
have the view vector the normal to the plane that's
perpendicular
to the
viewing surface. And
so that gives us our three vectors, and then so to return
and n. So this is the column major format. So n is the first
column and Then last entry, Not column Is I
so the look at function creates the transformation matrix that
takes us From the object coordinates To the camera
coordinates and
so we'll get into more details the matrices for viewing next
day. I
and we'll also have a discussion about what the midterm should
look like, and I'll solicit Your input for that. I
any questions or concerns I
so I think I have a feeling the software that I downloaded has
messed up my because the cursor on the big screen is a different
location than it is on my laptop, so maybe I'll have to
anyway, not that you care about that. Thank you for your time
today. Have a good rest of your day and see you on Wednesday.
Take care.
Take care.

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