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# Assignment No. 03 Semester: Fall 2012 CS602-Computer Graphics

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is it correct?

points of intersection (10,-2,-3)

kindly confirm it.

mja ya bta da k -3 k bad is value ko t ma put krna ha kia?

According 2 mine calculation

t = 3

Here is proof:

2(-t + 7) – 3(2t + 4) + 6(t) – 8 = 0

-2t + 14 -6t -12 + 6t – 8 = 0

-2t = -6

t = 3

no..its wrong.

2(-t + 7) – 3(2t + 4) + 6(t) – 8 = 0

-2t + 14 -6t -12 + 6t – 8 = 0

-2t  -6 =0

2t=-6

t=-3

yup u r write....Thnx a ton

wud u plz solve the following 4 me. i get hung in the middle of que 1

=(P1-P2) x (P3-P2)

={(4.0-9.0),(5.0-9.0),(6.0-8.0)} x {(2.0-9.0),(3.0-9.0),(4.0-8.0)}

={-5.0,-4.0,-2.0} x {-7.0,-6.0, -4.0}

see cross product from handouts

it is not like this in the way you are going.

hmmm.......got it....!!!!

Assignment No. 03
Semester: Fall 2012
CS602-Computer Graphics

Total Marks: 25

Due Date: 14/01/2013

Objective

To build concepts concerning Line, plane, lighting and shading in Computer Graphics.

■Your assignment must be in .doc format.(Any other formats like scan images, PDF,
Zip, rar, bmp etc will not be accepted).
■No assignment will be accepted through email.

Rules for Marking:

It should be clear that your assignment will not get any credit if:

■The assignment is submitted after due date.
■The submitted assignment does not open or file is corrupted.
■Your assignment is copied from internet, handouts or from any other student
(Strict disciplinary action will be taken in this case).

Assignment

QNo. 1 [5]
Given are the points that lie on a plane P1,

P1<4.0,5.0,6.0>
P2<9.0,9.0,8.0>
P3<2.0,3.0,4.0>

You are required to find the equation of a normal to the plane P1.

Q No. 2 [5]

Find the point of intersection between the plane 2x−3y+6z−8 = 0 and the line

Q No. 3 [15]

Differentiate between Ambient, Diffuse and Specular lighting with the help of an example.

You can send email at cs602@vu.edu.pk for any assignment related problem.

Your work must be original. No marks will be given in case of cheating or copying
from the internet, handouts or from any other student and strict action will be taken
against that student.

CS602 Computer Graphics Assignment No. 03 Solutions Question No. 03

Re: Whats the difference in diffuse, ambient and specular colors?
Specular color sets the color for highlights.
Diffuse color is the color of the mesh when it is illuminated.
Ambiant color is the color of the mesh when it's not illuminated.
Note that the ambient color works also when the mesh is illuminated, therefore, on illuminated parts of the mesh, you get the add of both diffuse and ambiant colors.
To change the color of shadows, you can play with the ambient color but that requires some fine adjusting on every material group. It would be more simple to use a light dedicated to add a colored ambiance, such as an IBL.
Whats the difference in diffuse, ambient and specular colors?
In the material room, what is the difference among diffuse, ambient and specular colors? Also, how can you change the color of shadows in poser?
Re: Whats the difference in diffuse, ambient and specular colors?
So if I wanted the shadows of my figure to be an orangy or any warm color, I would make the ambient color orange or make the color of the IBL orange?
Yes, that's the idea.

................................
This is the diffuse float_color value for the Light. This parameter is 3 floating point numbers (red, green, blue). Color component values (red, green and blue) should be between 0.0 and 1.0, though using values greater than 1 can make the objects brighter. When calculating the color and shading of an object, we take each light source within range and cumulatively multiply their diffuse and ambient colors by the diffuse and ambient colors of the object respectively. The diffuse light is adjusted for attenuation (distance to light) and the angle between the surface and the light (shading on a per vertex basis using vertex normals) while ambient light is NOT adjusted for distance or angle. Emissive is then added to the diffuse and ambient values. If there is a texture, it will be adjusted by the sum of the lit diffuse, ambient and emissive values. So setting diffuse, ambient and/or emissive color values to (1, 0, 0) will eliminate all greens and blues from the displayed texture. Then the calculated lit specular value is added to produce the final shading for the 3D picture part. Black is (0.0, 0.0, 0.0). White is (1.0, 1.0, 1.0).
ambient This is the ambient float_color value for the Light. This parameter is 3 floating point numbers (red, green, blue). Color component values (red, green and blue) should be between 0.0 and 1.0, though using values greater than 1 can make the objects brighter. When calculating the color and shading of an object, we take each light source within range and cumulatively multiply their diffuse and ambient colors by the diffuse and ambient colors of the object respectively. The diffuse light is adjusted for attenuation (distance to light) and the angle between the surface and the light (shading on a per vertex basis using vertex normals) while ambient light is NOT adjusted for distance or angle. Emissive is then added to the diffuse and ambient values. If there is a texture, it will be adjusted by the sum of the lit diffuse, ambient and emissive values. So setting diffuse, ambient and/or emissive color values to (1, 0, 0) will eliminate all greens and blues from the displayed texture. Then the calculated lit specular value is added to produce the final shading for the 3D picture part. Black is (0.0, 0.0, 0.0). White is (1.0, 1.0, 1.0).
specular This is the specular float_color value for the Light. This parameter is 3 floating point numbers (red, green, blue). Color component values (red, green and blue) should be between 0.0 and 1.0, though using values greater than 1 can make the specular highlight brighter. Specular highlights are affected by attenuation. The specular highlights are added to the lit diffuse, ambient and emissive values to get the final displayed color. Setting specular to anything other than black will incur a slight performance hit for each time the picture is drawn.

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