tnm046: datorgrafik texture mapping - linköping universitysasgo26/tnm046/lectures/lecture_5.pdf5...
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TNM046: Datorgrafik
Texture mapping
Sasan Gooran VT 2014
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Texture Mapping The process of mapping an image onto an object (/triangle) in order to increase the detail of the rendering. We can get fine details without needing to render millions of tiny triangles. The image that gets mapped onto the object is called a texture map and is usually a regular color image.
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Texture Mapping
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The problem
Source: Teaching Texture Mapping Visually, Rosalee Wolfe, DePaul University
For each pixel on an object, the question is: where to look in the texture map to find the color?
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Planar/Linear mapping
Source: Teaching Texture Mapping Visually, Rosalee Wolfe, DePaul University
Take the (x, y, z) values from the object and drop one of the components (here z). This gives us a 2D coordinate (planar), which can be used to look up the color in the texture map.
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Linear mapping
Source: Teaching Texture Mapping Visually, Rosalee Wolfe, DePaul University
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Linear mapping
Source: Teaching Texture Mapping Visually, Rosalee Wolfe, DePaul University
The color remains constant when x changes
The color remains constant when y changes
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Cylindrical mapping
Source: Teaching Texture Mapping Visually, Rosalee Wolfe, DePaul University
The (x, y, z) values are converted to cylindrical coordinates (r, φ, h). For the texture mapping only φ and h are used (φ is converted to an x and h to a y-coordinate)
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Cylindrical mapping
Source: Teaching Texture Mapping Visually, Rosalee Wolfe, DePaul University
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Cylindrical mapping
Source: Teaching Texture Mapping Visually, Rosalee Wolfe, DePaul University
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Spherical mapping
Source: Teaching Texture Mapping Visually, Rosalee Wolfe, DePaul University
The (x, y, z) values are converted to spherical coordinates (r, θ, φ). For the texture mapping only θ and φ are used (θ is converted to an x and φ to a y-coordinate)
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Spherical mapping
Source: Teaching Texture Mapping Visually, Rosalee Wolfe, DePaul University
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Spherical mapping
Source: Teaching Texture Mapping Visually, Rosalee Wolfe, DePaul University
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Basic Idea
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Basic Idea
Source: http://blog.tartiflop.com/2008/11/first-steps-in-away3d-part-3-texture-mapping/
First: Map a point on the object to a point on a unit square (usually denoted by st or uv), second: map the unit square to the texture.
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Basic Idea 1- For each coordinate (x, y, z) find its corresponding coordinate (s, t) in the unit square. 2- For each (s, t) coordinate in the unit square find the corresponding coordinate in the texture map. Let’s start with number 2, which means how to map unit square to Texture map. After that, we will see how to map (x, y, z) to the unit square (s, t). Note: The coordinate in the unit map (s, t) are denoted by (u, v) in some literature.
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Map unit square to Texture
Texture map
(0,0)
(128,128)
s
t
(0,0)
(1,1)
Unit texture plane
Above example: (0,0)à (0,0), (1,1)à (128,128), (0.25,0.5)à (32,64)
In general: For any point (s, t) on unit plane, corresponding point in texture plane is: (s*texture width, t*texture height)
Once we have the coordinates, we just need to look up the color of the texture at that location.
T(m, n)
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Map unit square to Texture Above example: (0.2, 0.3)à (25.6, 38.4), how do we look up the color of the texture at this location? The simplest way: (nearest neighbor interpolation), just look up the color at the closest possible neighbor, in this example (26, 38)
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Map unit square to Texture Above example: (0.2, 0.3)à (25.6, 38.4), how do we look up the color of the texture at this location? Or use bilinear interpolation,
T(i, j) T(i+1, j)
T(i, j+1) T(i+1, j+1)
(a, b) Δx
Δy
T (a,b) = (1−Δx)(1−Δy)T (i, j)+Δx(1−Δy)T (i+1, j)+ (1−Δx)ΔyT (i, j +1)+ΔxΔyT (i+1, j +1)
For the above example:
T (25.6,38.4) = 0.4 ⋅0.6 ⋅T (25,38)+ 0.6 ⋅0.6 ⋅T (26,38)+0.4 ⋅0.4 ⋅T (25,39)+ 0.6 ⋅0.4 ⋅T (26,39)
Texture map
a
b
Close up
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Map unit square to Texture
Nearest Neighbor Bilinear
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Map from point object to unit square Plane
Given (x, y, z), you can drop for example z component (or x or y, depending on which plane you are using) The plane can also be arbitrary, in this case you just need to do some transformations (for example rotation)
Then: (x, y, z) à (x, y) à (s, t) = (x/w, y/h)
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Map from point object to unit square Plane
What if you get a texture coordinate outside [0, 1]? Tiling • Loop around and start at the other side (wrapping) • Reflect the image (mirroring) • Repeat the edge pixels (clamping) • Default to some other color (bordering)
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Map from point object to unit square Cylinder
x = rcosϕy = rsinϕ0 ≤ z ≤ h
"
#$
%$
0 ≤ϕ ≤ 2π
rϕ
x
y
z
OR −π / 2 ≤ϕ ≤ 3π / 2
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Map from point object to unit square Cylinder
x = rcosϕy = rsinϕ0 ≤ z ≤ h
"
#$
%$
−π / 2 ≤ϕ ≤ 3π / 2
rϕ
x
y
z
tanϕ = yx
ϕ = arctan(x, y)
Note: φ is between –π/2 and 3π/2, but the result of tan-1 is between – π/2 and π/2.
ϕ = arctan(x, y) =
tan−1(y / x), if x > 0tan−1(y / x)+π, if x < 0π / 2, if x = 0 and y > 0−π / 2, if x = 0 and y < 0
"
#
$$
%
$$
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Map from point object to unit square Cylinder
ϕ = arctan(x, y)Given (x, y, z) on the object, you find:
s = ϕ +π / 22π
t = zh
!
"##
$##
Note: φ is between –π/2 and 3π/2, and z between 0 and h. The unit square coordinates (s, t) are between 0 and 1, therefore:
Ex: 39 a and e
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Map from point object to unit square Sphere
x = rsinθ cosϕy = rsinθ sinϕz = rcosθ
!
"#
$#
ϕ
0 ≤ϕ ≤ 2π
x
y
z
OR −π / 2 ≤ϕ ≤ 3π / 2
rθ
0 ≤θ ≤ π
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Map from point object to unit square Sphere
x = rsinθ cosϕy = rsinθ sinϕz = rcosθ
!
"#
$#
ϕ
x
y
z
−π / 2 ≤ϕ ≤ 3π / 2
rθ0 ≤θ ≤ π
tanϕ = yx
ϕ = arctan(x, y)See page 24
cosθ = zr=
zx2 + y2 + z2
θ = cos−1( zx2 + y2 + z2
)
Note: θ is between 0 and π, and the result of cos-1 is also between 0 and π.
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Map from point object to unit square Sphere
ϕ = arctan(x, y)
Given (x, y, z) on the object, you find:
s = ϕ +π / 22π
t = θπ
!
"##
$##
Note: φ is between –π/2 and 3π/2, and θ between 0 and π. The unit square coordinates (s, t) are between 0 and 1, therefore:
and θ = cos−1( zx2 + y2 + z2
)For arctan see page 24
Ex: 40 a and e
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Mapping triangles to Texture
Interpolate linearly across triangles. For each (x, y, z) on the triangle find a (s, t) coordinate by interpolation and the corresponding location in the texture map (as explained above) and look up its color.
(xA, yA,, zA)
(xB, yB, zB)
(xC, yC, zC)
(xp , yp, zP)
Pre-calculate texture coordinates (or (s, t) coordinates) for each vertex and store them.
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Texture mapping, OpenGL and Lab
x
y
z
p0
p1
p2
p3
Vertex table: Nr coordinate normal texture (s, t) 0: (0,0,0) (0, 0, -1) (s, t) 1: (0,0,0) (0, -1, 0) (s, t) 2: (0,0,0) (-1, 0, 0) (s, t) 3: (1,0,0) (0, 0, -1) (s, t) 4: (1,0,0) (0, -1, 0) (s, t) 5: (1,0,0) (1,1,1)/sqrt(3) (s, t)
As we saw previously, for each vertex we define its coordinates and the normal vectors associated with the vertex. In OpenGL we also add the (s, t) texture coordinates to the vertex list.
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Texture mapping, OpenGL and Lab
x
y
z
p0
p1
p2
p3
Vertex table: Nr coordinate normal texture (s, t) 0: (0,0,0) (0, 0, -1) (s, t) 1: (0,0,0) (0, -1, 0) (s, t) 2: (0,0,0) (-1, 0, 0) (s, t) 3: (1,0,0) (0, 0, -1) (s, t) 4: (1,0,0) (0, -1, 0) (s, t) 5: (1,0,0) (1,1,1)/sqrt(3) (s, t)
Observe that both s and t are between 0 and 1 and can be calculated according to one of the three approaches that have been described (plane, cylinder or sphere)
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Texture mapping, OpenGL and Lab Lets have a look at a simple example, the unit box.
x
y
p1 p0
p2 p3 Front face
z
y
p0 p4
p5 p2 Left face
y
x
z
p0 p1
p2 p3 p5
p4
P0: (0, 0, 1)
P1: (1, 0, 1)
P2: (0, 1, 1)
P3: (1, 1, 1)
P4: (0, 0, 0)
P5: (0, 1, 0)
P6: (1, 0, 0) P7: (1, 1, 0)
p7
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Texture mapping, OpenGL and Lab Assume that the left figure shows the front face of the box, consisting of two triangles. Assume also that we want the whole texture to be mapped to the front face of the box.
s
t
(0,0)
(1,1)
Unit texture plane
This means for the front face, p0 has to be associated with (s, t)=(0, 0). p1 has to be associated with (s, t)=(1, 0). p2 has to be associated with (s, t)=(0, 1) and finally p3 has to be associated with (s, t)=(1, 1).
x
y
p1 p0
p2 p3 Front face
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Texture mapping, OpenGL and Lab
Here we just show the vertex table for p0. In the box we can see that p0 is associated with three faces of the box (front, left and bottom). Therefore there are three normal vectors and (s, t) coordinates for p0. So far we just know the (s, t) coordinates for the front face. Vertex table for p0: Nr coordinate normal texture (s, t) Front face: 0 (0,0,1) (0, 0, 1) (0, 0) Left face: 1 (0,0,1) (-1, 0, 0) (?, ?) Bottom face: 2 (0,0,1) (0, -1, 0) (?, ?)
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Texture mapping, OpenGL and Lab Lets now have a look at the left face of the box. Assume that we want the whole texture to be mapped to the left face of the box as well (meaning the front face and the left face get exactly the same texture map).
s
t
(0,0)
(1,1)
Unit texture plane
This means for the left face, p4 has to be associated with (s, t)=(0, 0). p0 has to be associated with (s, t)=(1, 0). p5 has to be associated with (s, t)=(0, 1) and finally p2 has to be associated with (s, t)=(1, 1).
z
y
p0 p4
p5 p2 Left face
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Texture mapping, OpenGL and Lab
Here we just show the vertex table for p0. In the box we can see that p0 is associated with three faces of the box (front, left and bottom). Now we also know the (s, t) coordinates for the left face. Vertex table for p0: Nr coordinate normal texture (s, t) Front face: 0 (0,0,1) (0, 0, 1) (0, 0) Left face: 1 (0,0,1) (-1, 0, 0) (1, 0) Bottom face: 2 (0,0,1) (0, -1, 0) (?, ?) The rest will be filled during “lektion 3”.
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Texture mapping
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Texture mapping
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