Practice English Speaking&Listening with: Lecture - 8 3D Viewing

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So we have been talking about 3D viewing, the transformations pertaining to 3D viewing.

Today we will continue on it. Earlier we basically looked at 3D viewing as a process where there

is an object AB here and we have a projection plane and what we get are these projectors

emanating from the object on to the projection plane meeting at the center of projection

and this is the process of projection. And in this way we get what we call as perspective

projection. So A prime B prime is basically the projected image of the object AB. And

you notice here that these projectors are not parallel to each other and they are basically

meeting at this point which is center of projection.

Now the other type of viewing projection is the parallel projection where we have this

projectors coming out the object AB parallel to each other and this is again the projection

plane. So, in effect what we get is the center of projection now moves to infinity. It does

not meet at the point which we will call it as the center of projection. So this is parallel

projection and we basically looked at the various types of parallel projections.

For instance, we looked at the orthographic projection where we basically have an object

and the projectors are parallel to each other and they are perpendicular to the principle

planes which are actually planes perpendicular to the principle or the coordinate axis. So

this is for instance gives us the top view the plane which is perpendicular to the y

axis and similarly the front view and the side view we get.

Mathematically we can get this projection again using as a matrix transformation. Therefore

this matrix P basically gives as an orthographic projection on Z is equal to 0 plane. What

we get basically is this third column and all the values are 0. So we can represent

this orthographic projection again as a matrix transformation. So the property here we have

basically preserves the true silent shape for the object. So these are the examples.

Here we notice that these are the projectors, this is the projection plane and these projectors

are parallel to each other and they are also perpendicular to the projection plane and

these are the various views we can get of this object such as the front view, the top

view and the side view.

The next category we looked at was the axonometric projections. These are also parallel projection

meaning that the projectors are parallel to each other. So the way we can look at these

axonometric projections are basically as additional transformation before we apply the conventional

orthographic projection. For instance, we can have additional rotations or translation

and then we have the projection on Z is equal to 0 plane. Now as a consequence if I consider

the unit vectors in the three coordinate axis 1 0 0 1 represented in homogeneous coordinates

0 1 0 as Y axis or unit vector in Y direction and 0 0 1 as unit vector in Z direction and

if I apply a transformation T to it what I observe is and then this T is also consisting

of the projection of Z is equal to 0 plane which basically means that the third column

is going to be all 0.

So what I will get is basically this meaning that this particular unit vector which I considered

will get transformed to this Xx Yx star 0 1 which says that there is some sort of a

foreshortening which has happened to the unit vector which I had considered in X direction.

And the amount of foreshortening is nothing but this distance in X direction. Basically

I just measure these and that gives me fx which is foreshortening in X. Similarly I

get foreshortening in Y and foreshortening in Z. Now depending on the relationship of

these foreshortenings I have various kinds of axonometric projections because these foreshortenings

could be equal or may not be equal.

Therefore for instance I can basically define three types of axonometric projection; trimetric

where no foreshortening is the same. So fx is not equal to fy is not equal to fz. In

diametric two foreshortenings are the same out of these three and in isometric all the

foreshortening are the same.

As an example here we have the trimetric projection where fx is not equal fy is not equal to fz.

This is diametric so you that this basically equals to this if I had taken the object to

start with as a queue. So when we observe the foreshortening these two directions are

the same. And in isometric I have all the foreshortening. What basically happens here

is that these projectors are parallel, these projectors are also perpendicular to the projection

plane. It is just the projection plane which is placed differently as opposed to in the

case of orthographic projection where it was kept actually at the principle planes. So

this basically adds the visual realism to the 3 dimensional object. When you view the

object it looks more like a 3 dimensional object then you would see in orthographic

projection. Orthographic projection in isolation for a particular view does not give you any

3 dimensional information whereas these actually help you viewing in 3 dimension. That is the

basic advantage we have here.

If we are dealing with isometric projection the basic thing is that the projecting plane

undergoes some kind of a transformation before we perform the conventional orthographic projection

on Z is equal to 0 plane. These transformations are of the kind rotations. So, if I use two

rotations the first one is about Y axis by an angle phi and the other one is about X

axis by an angle theta. The corresponding transformations I have given, this is the

rotation about the Y axis and this is the rotation about X axis and then I apply the

projection on Z is equal to 0. So this whole matrix now would give me a transformation

which I want to use as an isometric projection.

This is the resulting transformation I get after concatenating these individual transformations.

It is just a multiplication of those matrices. Now if you go back and see how we obtain those

foreshortenings for the unit vectors let us again try to see these foreshortenings with

this transformation matrix.

Again I have the three unit vectors and this is the transformation. Once I apply the transformation

on to these three unit vectors the modified or the transformed unit vectors are dmos.

So the first row gives me the transform unit vector in X, the second row gives me the transform

unit vector in Y and the third one gives me the transform unit vector in Z. So these are

the three unit vectors transform. Now, from these I try to compute the foreshortening

just in the way I had done earlier. So I get the square of foreshortening in X given by

this which are the square of these two entries I have here.

Similarly I have the foreshortening in Y the square of it given here and foreshortening

in Z. Now I impose the constraint of the transformation to be isometric which in turn means that these

three foreshortenings should be the same which basically means now I have fx is equal to

fy is equal to fz is equal to some f which gives me these two equations.

Now I have these three equations where I can solve it for pi, theta and f. So basically

it can give me the amount of rotations I should be performing in order to get isometric projection

for a given value of foreshortening. Now we move on to the other type of parallel projections.

So what were the observations we had in the axonometric projection? It was that the projectors

were parallel and they were perpendicular to the projection plane. Now we relax one

of the restrictions.

Since these are in parallel projections we will still retain the projectors to be parallel

because we want the center of projections to lie at infinity whereas we relax this condition

that they need not be perpendicular to the projection. So what this gives us we call

it as oblique projection. Here the projectors are not perpendicular to the projection. For

instance, if this is the object and this is the projector I have this is not perpendicular

to the projection. Therefore now with this property the true shape and size will be preserved

only for the faces which are parallel to the projection.

This is again an example here, here these projectors are not perpendicular to the projection

plane. And again the motivation of having oblique projection is to add the element of

3D viewing, to make it look like a 3 dimensional object. So this is the example of oblique

projection.

And mathematically what we can observe here is let us say we have a point P(x, y, z) and

this is my projection plane the (x, y) plane and this is the projector I have from P to

P double prime and now I want to get the value of Xp Yp which are the coordinates on (x, y) plane of the

point P so what I observe is that the projector which is parallel to the Z axis goes to P

prime which gives me (x, y) basically so I can represent Xp Yp in terms of this (x, y).

Hence xp is nothing but X plus L cos phi where L is this length from p prime to p double

prime and similarly Yp is y plus L sin pi using this triangle. So phi is basically the

angle between P prime and P double prime and X axis on this plane. Now the angle which

this projector is making with the projection plane is the alpha. Now for a given alpha

we can get many projectors for the same value of alpha so I have to restrict the location

of the projector and I am doing that using this phi because it basically gives me a cone.

The set of all projectors is basically a cone. Now this alpha is actually related through

Z and L so I have this tan alpha given by Z by L or L is equal to Z cot alpha from this

triangle. Depending on what alpha we choose we get a particular type of oblique projection

for some value of phi by fixing the value of phi.

Then we choose alpha to be 45 degree, the type of public projection we get is called

a cavalier and what we are observe here is that the lines which are perpendicular to

the projection plane are not foreshorten. Remember how L and Z are related. Basically

cot alpha is 1. The lines which are perpendicular to the projection plane will basically have

the same length whereas if I choose alpha in a manner that cot alpha is half the projection

which we obtain we call it as cabinet and here we observe that the lines perpendicular

to the projection plane are foreshorten by half for the same reason what we observe in

this case because they are related by cot alpha. And the five is actually typically

30 degree or 45 degree. So this is basically the oblique projection. So here we observe

that the projectors are at an angle other than 90 degree to the projection.

Now we move on to the more general kind of a projection which also matches to the visual

system we have which is the perspective projection. And in the beginning we observed that these

projectors are not parallel and they actually meet at a point which is the center of projection.

So here parallel lines seem to converge and the foreshortening could be non uniform. And

in fact this perspective projection helps us in the 3D viewing because it gives you

some sort of a depth perception. The front objects seem to look bigger there is a depth

queue and the shape is not preserved although the lines map to lines but the parallelism

is not preserved and the foreshortening is non uniform therefore the shape will not be

different. This is the projection plane and this is the object here and the center of projection is sitting

here. Now let us try to see it mathematically.

Just as we have seen axonometric and orthographic projections even the oblique projection can

be represented in the matrix form because the way you had seen the transformation of

obtaining Xp using Xy you can actually write that in the matrix form. So that can also

be obtained using a matrix form. So now we try to represent this perspective transformation

using a matrix form. Here we have the input given as x y z 1 in homogeneous coordinate

system and this is the matrix we have. In 4 into 4 general matrix of transformation

we have not been considering the last column.

Now let us consider the last column where we put a non zero value in the third row of

this column this r. Now using this transformation the point x y z 1 becomes this in homogeneous

coordinates. So, to get the Cartesian coordinates the normal coordinates which we use for our

purposes we divide it by the homogeneous coordinates all through and we get this X by rz plus 1Y

by rz plus 1, Z by rz plus 1 so this is still 3D. We are basically making a transformation

from 3D to 3D.

Now if I take a projection on Z is equal to 0 plane which basically means I multiply by

this matrix where the third column again is all 0s this would give me this, basically

this value becomes 0. Therefore this is the projected point now after having done the

perspective transformation.

Using a matrix form of transformation we know what we got. Now let us try to see geometrically.

So here is your center of projection along the Z direction Z axis located at Zc and this

is your point P(x, y, z) and we are looking at projection on Z is equal to 0 plane meaning

(x, y) plane. So I basically project this point through the center of projection on

to this plane and I get T star which has the coordinate x star y star and that what is

happening as a process when I say that the point p is projected on Z is equal to 0 plane

from the center of projection located at ZC along the Z direction.

So now consider the properties from similar triangles and try to obtain the values of

x star and y star. Now I consider the similar triangle first this where this is y star which

is already present and this is y which is here. And these lengths are given as l1 and

l2. So using this triangle I can now get the value of y star. Basically y star by l2 is

equal to y by l2 minus l1 and also now using these similar triangles on the Xz plane I

have Zc by Zc minus Z from this triangle on the base equals to l2 by l2 minus l1. So first

I consider this triangle which is there and then I consider these triangles from the base.

So this gives me y star is equal to y by 1 minus z by Zc. Similarly I can obtain x star.

I can obtain X star in a similar way. So X star is given as X by 1 minus Z by Zc.

If you go back and see what we have obtained as x star and y star are these values and

here what we got is X star as X by 1 minus Zc and Y star as Y by 1 minus Zc. So, if you

substitute r as minus 1 by Zc they are equivalent. So this establishes the correspondence between

the two representations what we do geometrically and what we obtain from the matrix. Therefore

if I have r is equal to minus 1 by Zc I basically get the same transformation which I derive

here geometrically. And Zc is nothing but the location of center of projection along

Z axis. Now there is another property which we observe in the perspective transformation.

The fact that the parallel lines seem to converge is being observed. Even in the real life scenario

we observe that the two parallel lines need to converge, there will be tracks and so on.

But in reality they do not converge. So what we get are these vanishing points which are

basically the points where the set of parallel lines not parallel to the projection plane

converge.

Here is an example where if I look at an object like a cube or parallelepiped I see that the

lines which are perpendicular to the projection plane in this case the lines which are parallel

to Z axis seem to converge somewhere else. This is what we call as the vanishing point

in Z direction. In the world this point is located at infinity. It is just that the transformation

which is the perspective transformation making it appear there. So what does it mean? If

I want to know this point I can actually take a point at infinity and apply my perspective

transformation. That will locate the vanishing point corresponding to the point at infinity.

That is what is happening now.

If I consider a point at infinity on positive Z axis, remember the homogenous coordinates

enable us to represent the point at infinity just by using the homogenous part of the coordinate

as 0. So this is the point at infinity on positive Z axis. Now if I take this point

and apply the perspective transformation using this r here in the last column what we obtain is this value 0 0 1 by r1.

This is after homogenizing it after dividing by the last homogenous coordinates I get this.

And recall that this r is nothing but minus 1 by Zc. This is the scenario I am taking

where the center of projection is laying along the Z axis at Zc. So r is equal to minus 1

by Zc which means that the vanishing point which I am trying to obtain here is at Zc.

So it is an image of center of projection on

the other side.

This is after the perspective transformation. But you are trying to locate what image would

it have after the projection then it is going to lie on the origin. This is after the perspective

transformation I am not projecting it on a 2D plane. The moment I do the projection on

2D plane this point will get there because this is along the Z axis. So this is the before

the projection has taken place. In this way I can obtain the vanishing point.

Similarly, this was the example where I had considered the single perspective so I call

the single perspective transformation when one of the three entries here and the last

column of the matrix is non zero. I had earlier considered the third row where the center

of projection was along Z axis similarly I can consider the other axis. Here if I consider

center of projection on X axis then I am taking about a non zero value here p and I have a

very similar treatment which can give me the transformed point x y z 1 as this and correspondingly

the center of projection is located at minus 1 by p 0 0 1 and the vanishing point along

X axis is located at 1 by p 0 0 1. So I call the type of perspective transformation as

a single point perspective transformation when one of the three entries here is non

zero. So this is again when I would consider the second row now I get center of projection

as this and this is VPy.

Now the idea can be extended of single point perspective to have a two point perspective

just using the same framework of values in the matrix. So instead of having one of the

values in the last column to be non zero let us say I have two values P and Q which can

also be obtained by concatenating the individual matrices of one point perspective like Pp

and Pq. If I have just multiplied them I would have obtain this again. This is to get a two

point perspective. So this would give me the two centers of projections and correspondingly

two principle vanishing points.

Now I can further extend this idea where all the three entries here are non zero p q r.

And once again this can be obtained as concatenation of the three one point perspective transformation.

And this gives me three centers of projections and the corresponding vanishing points.

Here this is the example where I have as one point or single point perspective. So these

lines of the columns are still parallel but these lines if you consider this to be as

Z axis then they seem to meet somewhere. Therefore this is a one point or a single point perspective,

this is a two point perspective. So here I still have the lines which are along the y

axis that is still parallel whereas the other ones seem to converge this one and these ones.

So this is a two point perspective and this is a three point perspective and none of the

parallel lines now seem to remain parallel but they seem to converge at some point. Therefore

again the idea of using one point two point three point perspective is to add wherever

required the 3 dimensional viewing in

the object.

What do you mean by two point perspective relating to what happens in the reality because

even when you take a camera there is only one camera from the image perspective. This

is just with respect to the matrix we have. So I basically looked at the entries in the

last column and defined the various types of perspective there and correspondingly what

we get is the principle vanishing points. But we can generate these even using single

maybe a camera. If you actually apply certain transformations before you take a single perspective

transformation you can create two or three point perspective.

So all it is saying is that if you apply a certain transformation to the object and then

you take a single point perspective you may get two point perspective or you may get even

three point perspective. That is the process of generating two point or three point perspective.

There I had considered the location of the position of the object in such a way that

they were aligned to the principle axis and the vanishing points which I am referring

to here are basically the principle vanishing points because so I am looking at the lines

which are parallel to either of the coordinate axis. But you can actually get vanishing points

for the lines which are not parallel to one of these coordinate axis and they still converge

to a point. Hence there might be just vanishing points or sometimes they are also referred

as tracing points.

If you take a point in (x, y) plane, treating it from the representation of a matrix was

to define these single point, two point and three point perspective. When you are generating

them that is when you have to generate such a kind of a projection you may still be using

one camera and that is where your viewer or the camera is located. But you can create

the effect of multiple vanishing points. So let us look at how to generate these perspective

views.

Even when we are talking about the aspect of enhancing the 3 dimensionality in viewing

of the objects a single perspective may not be adequate and that is the idea. For instance,

if I put an object or a cube in such a way that the cube is basically having all the

edges parallel to the coordinate axis and I am viewing from the Z axis then what will

happen is I will just see the front face considering there is a hidden line or hidden surface elimination process going on. Therefore

all I will see is the first front face and nothing else and that would not give me any

3 dimensional viewing so I need to enhance that in some ways. So the way we can do is

we can actually have additional transformation before we apply the single perspective transformation.

Hence let us consider a simple transformation like translation. If I apply a translation

by the offset l m n to the object and again I consider the center of projection located

at Z is equal to c and the projection plane Z is equal to 0 then I just concatenate these

two matrices the translation matrix and the projection matrix and this is what I get.

So the example of this, if I consider the translation along y is equal to x line so

an object initially located here when translated along y is equal to x line may actually look

like this or this which adds to the 3D viewing. So a simple translation is adding to the process

of 3D image.

Similarly, if I have a translation in Z that is another aspect of translation would actually

give me the notion of scaling and that we absorb as zooming in zooming out. This is

one of the things we often absorb. The object here looks big and the object here will look

small and this is when everything is in front of the projection plane if it goes behind

the projection plane then the object becomes smaller than its original shape. So, just

a simple addition of transformation of like translation can change the process of viewing.

Further I add rotation.

I do a rotation before I take the perspective transformation and now I am always doing perspective

transformation which is a single point that is one point. Now I do a rotation about y

axis by an angle phi. So the total transformation is basically the rotational transformation

and the projection transformation which is this multiplied by this.

And if I see the resulting matrix this is what I get. In the last column we considered

single point perspective. The only thing is I applied a transformation of rotation before

I gave the perspective transformation. So now this gives me a two point perspective

transformation. Therefore in this way I can have a two point perspective transformation

using single camera.

Similarly if I use two rotations about Y axis by an angel phi and about X axis about an

angle theta so the total transformation is the rotation about Y, rotation about X and

the projection transformation. So if I multiply these three matrices this is what I get.

Now I have all those three entries non zero which basically means that I get a three point

perspective transformation. So now using two rotations but still using a same single point

perspective transformation I can give the effect of three point perspective transformation.

This is how I can generate the various types of perspective transformation.

The Description of Lecture - 8 3D Viewing