HemisphereOrthoSolidAngle - Revision history

Patapom at 17:43, 7 June 2017

2017-06-07T17:43:43Z

Patapom at 19:33, 2 June 2017

2017-06-02T19:33:01Z

Patapom at 19:32, 2 June 2017

2017-06-02T19:32:23Z

Patapom at 19:31, 2 June 2017

2017-06-02T19:31:39Z

Patapom at 19:31, 2 June 2017

2017-06-02T19:31:06Z

Patapom at 19:29, 2 June 2017

2017-06-02T19:29:57Z

Patapom at 19:28, 2 June 2017

2017-06-02T19:28:23Z

Patapom at 19:27, 2 June 2017

2017-06-02T19:27:45Z

Patapom at 19:26, 2 June 2017

2017-06-02T19:26:23Z

Patapom at 19:24, 2 June 2017

2017-06-02T19:24:54Z

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 Except the primitive of this expression is a bit of nightmare after all, but this is our final result:
 {| class="wikitable" align="center"
-| style="background: Periwinkle;"      | <math>A(x,y) = \int_0^y \int_0^x \frac{1}{\sqrt{1-a^2-b^2}} da \, db = \tan ^{-1}\left(\frac{x}{\sqrt{-x^2-y^2+1}}\right)+x \tan ^{-1}\left(\frac{y}{\sqrt{-x^2-y^2+1}}\right)+\frac{1}{2} \left(\tan ^{-1}\left(\frac{-x-y^2+1}{y \sqrt{-x^2-y^2+1}}\right)-\tan ^{-1}\left(\frac{x-y^2+1}{y \sqrt{-x^2-y^2+1}}\right)\right) \quad \quad \quad (2)</math>
+| style="background: Periwinkle;"      | <math>A(x,y) = \int_0^y \int_0^x \frac{1}{\sqrt{1-a^2-b^2}} da \, db = y \tan ^{-1}\left(\frac{x}{\sqrt{-x^2-y^2+1}}\right) + x \tan ^{-1}\left(\frac{y}{\sqrt{-x^2-y^2+1}}\right)+\frac{1}{2} \left(\tan ^{-1}\left(\frac{-x-y^2+1}{y \sqrt{-x^2-y^2+1}}\right)-\tan ^{-1}\left(\frac{x-y^2+1}{y \sqrt{-x^2-y^2+1}}\right)\right) \quad \quad \quad (2)</math>
 |}

← Older revision		Revision as of 19:33, 2 June 2017
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−	It's easy to notice that <math>A(1,1) = \frac{\pi}{6}</math> which is the solid angle covered by a quarter of our cube map face, the solid angle of an ~~entier~~ face would be <math>\frac{2\pi}{3}</math> and all 6 faces of the cube map are then covering a proper solid angle of <math>4\pi</math> and all is well!	+	It's easy to notice that <math>A(1,1) = \frac{\pi}{6}</math> which is the solid angle covered by a quarter of our cube map face, the solid angle of an entire face would be <math>\frac{2\pi}{3}</math> and all 6 faces of the cube map are then covering a proper solid angle of <math>4\pi</math> and all is well!

@@ Line 27: / Line 27: @@
 <math>A(x,y) = \int_0^y \int_0^x \left | \frac{\partial \mathbf{p}}{\partial x} \times \frac{\partial \mathbf{p}}{\partial y} \right | \, da \, db \\
-= \int_0^y{ \int_0^x { \left (1+a^2+b^2 \right )^{-\frac{3}{2}} \, da } \, db } \\
+= \int_0^y \int_0^x \left (1+a^2+b^2 \right )^{-\frac{3}{2}} \, da \, db \\
 = \left | atan( \frac{xy}{\sqrt{1+x^2+y^2}} ) \right | \quad \quad \quad (1)
 </math>

@@ Line 27: / Line 27: @@
 <math>A(x,y) = \int_0^y \int_0^x \left | \frac{\partial \mathbf{p}}{\partial x} \times \frac{\partial \mathbf{p}}{\partial y} \right | \, da \, db \\
-= \int_0^y \int_0^x \left (1+a^2+b^2 \right )^{-\frac{3}{2}} \, da \, db \\
+= \int_0^y{ \int_0^x { \left (1+a^2+b^2 \right )^{-\frac{3}{2}} \, da } \, db } \\
 = \left | atan( \frac{xy}{\sqrt{1+x^2+y^2}} ) \right | \quad \quad \quad (1)
 </math>

@@ Line 27: / Line 27: @@
 <math>A(x,y) = \int_0^y \int_0^x \left | \frac{\partial \mathbf{p}}{\partial x} \times \frac{\partial \mathbf{p}}{\partial y} \right | \, da \, db \\
-= \int_0^y \int_0^x (1+a^2+b^2)^{-\frac{3}{2}} \, da \, db \\
+= \int_0^y \int_0^x \left (1+a^2+b^2 \right )^{-\frac{3}{2}} \, da \, db \\
 = \left | atan( \frac{xy}{\sqrt{1+x^2+y^2}} ) \right | \quad \quad \quad (1)
 </math>

@@ Line 22: / Line 22: @@
 The idea is to compute the area of a small element of surface on the hemisphere as we make it vary on the plane, we do that by computing the partial derivatives of <math>\mathbf{p}</math> along x and y that give us the vectors <math>\frac{\partial \mathbf{p}}{\partial x}</math> and <math>\frac{\partial \mathbf{p}}{\partial y}</math>.
-As it is well-known to shader programmers, computing the cross product of these vectors gives us the normal to the sphere at this position <math>x, y</math> on the plane, and the length of that normal is the tiny area element <math>dA</math> on the hemisphere.
+It is well-known to graphics programmers that computing the cross product of these vectors gives us the normal to the sphere at position <math>\mathbf{p}</math>, and the length of that normal is the tiny area element <math>dA</math> on the hemisphere.
 Integrating this operation (cross product and norm computation) over an interval <math>[a,b]\in\mathbb{R}^2</math> yields:

@@ Line 3: / Line 3: @@
 Maybe you already bumped into the problem of projecting a cubemap into spherical harmonics and found this page to help you out: http://www.rorydriscoll.com/2012/01/15/cubemap-texel-solid-angle/ ?
-[[File:PerspectiveProjection.png|500px|Perspective projection of an area element on plane z=1 onto the hemisphere]]
+[[File:PerspectiveProjection.png|500px|Figure 1. : Perspective projection of an area element on plane z=1 onto the hemisphere]]
 But maybe you need the solution to another, similar problem that consists in finding the solid angle of a pixel lying in the z=0 plane and orthogonally projected onto the hemisphere, but couldn't find a page with that computation?
-[[File:OrthoProjection.png|500px|Orthographic projection of an area element on plane z=0 onto the hemisphere]]
+[[File:OrthoProjection.png|500px|Figure 2. : Orthographic projection of an area element on plane z=0 onto the hemisphere]]
@@ Line 18: / Line 18: @@
 ==Cubemap Projection Configuration==
-The configuration for a cube map is that pixels are lying on a plane z=1 such as <math>\mathbf{p'}=(x,y,1), -1<x<1, -1<y<1</math> and we project back onto the unit hemisphere by normalizing the vector <math>\mathbf{p}=\frac{\mathbf{p'}}{|\mathbf{p'}|}=\frac{(x,y,1)}{\sqrt{1+x^2+y^2}}</math> as shown on the figure below:
+The configuration for a cube map is that pixels are lying on a plane z=1 such as <math>\mathbf{p'}=(x,y,1), -1<x<1, -1<y<1</math> and we project back onto the unit hemisphere by normalizing the vector <math>\mathbf{p}=\frac{\mathbf{p'}}{|\mathbf{p'}|}=\frac{(x,y,1)}{\sqrt{1+x^2+y^2}}</math> as shown on figure 1 above.
 The idea is to compute the area of a small element of surface on the hemisphere as we make it vary on the plane, we do that by computing the partial derivatives of <math>\mathbf{p}</math> along x and y that give us the vectors <math>\frac{\partial \mathbf{p}}{\partial x}</math> and <math>\frac{\partial \mathbf{p}}{\partial y}</math>.

← Older revision		Revision as of 19:26, 2 June 2017
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	[[File:Fabric_weave_NDF_normalized.png\|300px]]		[[File:Fabric_weave_NDF_normalized.png\|300px]]
		+

	And you want to compute the shadowing/masking term as given by [https://hal.inria.fr/hal-01168516v2/document Dupuy et al] eq. 3:		And you want to compute the shadowing/masking term as given by [https://hal.inria.fr/hal-01168516v2/document Dupuy et al] eq. 3:

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 Say you have the normalized expression of a Normal Distribution Function D(x,y) defined in an image of the flattened hemisphere:
-[[File:Plastic_blue_NDF_normalized.png|300px]]
+[[File:Fabric_weave_NDF_normalized.png|300px]]
 And you want to compute the shadowing/masking term as given by [https://hal.inria.fr/hal-01168516v2/document Dupuy et al] eq. 3:
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 All you have to do is to write a convolution for each pixel of the image with a code similar to this:
-[[File:Plastic_blue_G.png|300px]]
+This yields the G term as an image that I'm showing here with a strong contrast applied, otherwise it's mostly white: