3D Rendering and Ray Casting

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3D Rendering and Ray Casting

Michael Kazhdan (601.457/657)

HB Ch. 13.7, 14.6 FvDFH 15.5, 15.10

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Rendering

• Generate an image from geometric primitives

Geometric Primitives (3D)

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Rendering

• Generate an image from geometric primitives

Rendering

Geometric

Primitives (3D)

Raster Image (2D)

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3D Rendering Example

What issues must be addressed by a 3D rendering system?

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Overview

• 3D scene representation • 3D viewer representation • What do we see?

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Overview

• How does it look?

How is the 3D scene described in a computer?

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3D Scene Representation

• Scene is usually approximated by 3D primitives

 Point  Line segment  Triangles  Polygon  Polyhedron  Curved surface  Solid object  etc.

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3D Point

• Specifies a location

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3D Point

• Specifies a location

 Represented by three coordinates

 Infinitely small struct Point3D { float x , y , z; }; (𝑥𝑥,𝑦𝑦,𝑧𝑧) Origin

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3D Vector

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3D Vector

• Specifies a direction and a magnitude

 Magnitude ⃗𝑣𝑣 = 𝑑𝑑𝑥𝑥2 + 𝑑𝑑𝑦𝑦2 + 𝑑𝑑𝑧𝑧2  Has no location ⃗𝑣𝑣 = (𝑑𝑑𝑥𝑥,𝑑𝑑𝑦𝑦,𝑑𝑑𝑧𝑧) struct Vector3D { float dx , dy , dz; };

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• Specifies a direction and a magnitude

 Magnitude ⃗𝑣𝑣 = 𝑑𝑑𝑥𝑥2 + 𝑑𝑑𝑦𝑦2 + 𝑑𝑑𝑧𝑧2

 Has no location

• Dot product of two 3D vectors

 ⃗𝑣𝑣₁, ⃗𝑣𝑣₂ = 𝑑𝑑𝑥𝑥₁ ⋅ 𝑑𝑑𝑥𝑥₂ + 𝑑𝑑𝑦𝑦₁ ⋅ 𝑑𝑑𝑦𝑦₂ + 𝑑𝑑𝑧𝑧₁ ⋅ 𝑑𝑑𝑧𝑧₂  ⃗𝑣𝑣₁, ⃗𝑣𝑣₂ = ⃗𝑣𝑣₁ ⋅ ⃗𝑣𝑣₂ ⋅ cos 𝜃𝜃

• Cross product of two 3D vectors

 ⃗𝑣𝑣₁ × ⃗𝑣𝑣₂ = Vector normal to 𝑣𝑣₁ and 𝑣𝑣₂

 ⃗𝑣𝑣₁ × ⃗𝑣𝑣₂ = ⃗𝑣𝑣₁ ⋅ ⃗𝑣𝑣₂ ⋅ sin 𝜃𝜃

3D Vector

⃗𝑣𝑣₁ = (𝑑𝑑𝑥𝑥₁, 𝑑𝑑𝑦𝑦₁,𝑑𝑑𝑧𝑧₁) ⃗𝑣𝑣₂ = (𝑑𝑑𝑥𝑥₂,𝑑𝑑𝑦𝑦₂,𝑑𝑑𝑧𝑧₂) 𝜃𝜃

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Cross Product: Review

• Let ⃗𝑣𝑣₁ = ⃗𝑣𝑣₂ × ⃗𝑣𝑣₃:

 𝑑𝑑𝑥𝑥₁ = 𝑑𝑑𝑦𝑦₂ ⋅ 𝑑𝑑𝑧𝑧₃ − 𝑑𝑑𝑧𝑧₂ ⋅ 𝑑𝑑𝑦𝑦₃  𝑑𝑑𝑦𝑦₁ = 𝑑𝑑𝑧𝑧₂ ⋅ 𝑑𝑑𝑥𝑥₃ − 𝑑𝑑𝑥𝑥₂ ⋅ 𝑑𝑑𝑧𝑧₃  𝑑𝑑𝑧𝑧₁ = 𝑑𝑑𝑥𝑥₂ ⋅ 𝑑𝑑𝑦𝑦₃ − 𝑑𝑑𝑦𝑦₂ ⋅ 𝑑𝑑𝑥𝑥₃

• ⃗𝑣𝑣 × 𝑤𝑤 = −𝑤𝑤 × ⃗𝑣𝑣 (remember “right-hand” rule) • We can show:

 ⃗𝑣𝑣 × 𝑤𝑤 = ⃗𝑣𝑣 ⋅ 𝑤𝑤 ⋅ sin 𝜃𝜃 ⋅ 𝑛𝑛,

where 𝑛𝑛 is the unit vector normal to ⃗𝑣𝑣 and 𝑤𝑤

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3D Line Segment

• Linear path between two points

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3D Line Segment

• Use a linear combination of two points

 Parametric representation: » 𝑝𝑝 𝑡𝑡 = 𝑝𝑝₁ + 𝑡𝑡 ⋅ (𝑝𝑝₂ − 𝑝𝑝₁), (0 ≤ 𝑡𝑡 ≤ 1) struct Segment3D { Point3D p1 , p2; }; 𝑝𝑝₁ 𝑝𝑝₂ Origin

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3D Ray

• Line segment with one endpoint at infinity

 Parametric representation: » 𝑝𝑝 𝑡𝑡 = 𝑝𝑝₁ + 𝑡𝑡 ⋅ ⃗𝑣𝑣, (0 ≤ 𝑡𝑡 < ∞) 𝑝𝑝₁ ⃗𝑣𝑣 Origin struct Ray3D { Point3D p1; Vector3D v; };

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3D Line

• Line segment with both endpoints at infinity

 Parametric representation: » 𝑝𝑝 𝑡𝑡 = 𝑝𝑝₁ + 𝑡𝑡 ⋅ ⃗𝑣𝑣, (−∞ < 𝑡𝑡 < ∞) 𝑝𝑝₁ struct Line3D { Point3D p1; Vector3D v; }; ⃗𝑣𝑣 Origin

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Origin

3D Plane

• A linear combination of three points

𝑝𝑝₁

𝑝𝑝₃ 𝑝𝑝₂

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Origin

3D Plane

• A linear combination of three points

 Implicit representation:

» Φ 𝑝𝑝 = 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑦𝑦 + 𝑐𝑐𝑧𝑧 − 𝑑𝑑 = 0

» Φ 𝑝𝑝 = 𝑝𝑝, 𝑛𝑛 − 𝑑𝑑 = 0

 𝑛𝑛 is the plane normal

» (May be) unit-length vector » Perpendicular to plane

 𝑑𝑑 is the signed (weighted) distance

of the plane from the origin.

struct Plane3D { Vector3D n; float d; }; 𝑛𝑛 = (𝑎𝑎,𝑏𝑏,𝑐𝑐) 𝑑𝑑 𝑝𝑝₁ 𝑝𝑝₃ 𝑝𝑝₂

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3D Polygon

• Area “inside” a sequence of coplanar points

 Triangle  Quadrilateral  Convex  Star-shaped  Concave  Self-intersecting

 Holes (use > 1 polygon struct)

struct Polygon3D {

Point3D *points; int npoints;

};

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3D Sphere

• All points at distance 𝑟𝑟 from center point 𝑐𝑐 = (𝑐𝑐_𝑥𝑥, 𝑐𝑐_𝑦𝑦, 𝑐𝑐_𝑧𝑧)

 Implicit representation: » Φ 𝑝𝑝 = 𝑝𝑝 − 𝑐𝑐 2 − 𝑟𝑟2 = 0  Parametric representation: » 𝑥𝑥 𝜙𝜙, 𝜃𝜃 = 𝑟𝑟 ⋅ cos 𝜙𝜙 ⋅ sin𝜃𝜃 + 𝑐𝑐𝑥𝑥 » 𝑦𝑦 𝜙𝜙, 𝜃𝜃 = 𝑟𝑟 ⋅ cos 𝜙𝜙 ⋅ sin 𝜃𝜃 + 𝑐𝑐_𝑦𝑦 » 𝑧𝑧 𝜃𝜃, 𝜙𝜙 = 𝑟𝑟 ⋅ sin𝜙𝜙 + 𝑐𝑐𝑧𝑧 struct Sphere3D { Point3D center; float radius; }; 𝑟𝑟 Origin 𝑐𝑐

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Other 3D primitives

• Cone • Cylinder • Ellipsoid • Box • Etc.

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3D Geometric Primitives

• More detail on 3D modeling later in course

 Point  Line segment  Triangle  Polygon  Polyhedron  Curved surface  Solid object  etc.

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Overview

• How does it look?

How is the viewing device described in a computer?

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Camera Models

• The most common model is pin-hole camera

 All captured light rays arrive along paths toward focal point without lens distortion (everything is in focus)

Other models consider ... Depth of field Motion blur Lens distortion View plane Eye position (focal point)

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Camera Parameters

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Camera Parameters

right back Up direction Eye Position View Plane “Look at” Point • Position

 Eye position: Point3D eye

• Orientation

 View direction: Vector3D view

 Up direction: Vector3D up

• Aperture

 Field of view angle: float xFov , yFov

 Resolution of film plane: int width , height

 Distance of film plane

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Other Models: Depth of Field

Close Focused Distance Focused

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Other Models: Motion Blur

Brostow & Essa

• Mimics effect of open camera shutter

• Gives perceptual effect of high-speed motion • Generally involves temporal super-sampling

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Traditional Pinhole Camera

• The film sits behind the pinhole of the camera.

• Rays come in from the outside, pass through the pinhole, and hit the film plane.

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Traditional Pinhole Camera

• The film sits behind the pinhole of the camera.

• Rays come in from the outside, pass through the pinhole, and hit the film plane.

Film Plane Pinhole

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Virtual Camera

• The film sits in front of the pinhole of the camera. • Rays come in from the outside, pass through the

film plane, and hit the pinhole.

Film Plane Pinhole

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Virtual Camera

• The film sits in front of the pinhole of the camera. • Rays come in from the outside, pass through the

film plane, and hit the pinhole.

Film Plane Pinhole

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Overview

• 3D scene representation • 3D viewer representation • Ray Casting

 Where are we looking?

 What do we see?

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Ray Casting

• For each sample …

 Where: Construct ray from eye through view plane

 What: Find first surface intersected by ray through pixel

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Ray Casting

• Simple implementation:

Image RayCast( Camera camera , Scene scene , int width , int height) {

Image image = new Image( width , height );

for( int i=0 ; i<width ; i++ ) for( int j=0 ; j<height ; j++ ) {

Ray ray = ConstructRayThroughPixel( camera , i , j );

Intersection hit = FindIntersection( ray , scene ); image[i][j] = GetColor( hit );

}

return image; }

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Ray Casting

Where?

}

return image; }

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Constructing a Ray Through a Pixel

right back Up direction 𝒑𝒑_𝟎𝟎 View Plane 𝒗𝒗 𝒑𝒑 𝒊𝒊 [𝒋𝒋]

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right back

Up direction _View

Plane

The ray originates at 𝑝𝑝₀ (the position of the camera).

So the equation for the ray is:

Ray 𝑡𝑡 = 𝑝𝑝0 + 𝑡𝑡 ⋅ ⃗𝑣𝑣

𝒑𝒑_𝟎𝟎

𝒗𝒗

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Constructing a Ray Through a Pixel

right back

Up direction _View

Plane

If the ray passes through the point

𝑝𝑝 𝑖𝑖 [𝑗𝑗], then the solution for ⃗𝑣𝑣 is:

⃗𝑣𝑣 = 𝑝𝑝 𝑖𝑖 [𝑗𝑗]−𝑝𝑝0

𝑝𝑝 𝑖𝑖 [𝑗𝑗]−𝑝𝑝₀

𝒑𝒑_𝟎𝟎

𝒗𝒗

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right back

Up direction _View

Plane

If 𝑝𝑝 𝑖𝑖 [𝑗𝑗] represents the (𝑖𝑖, 𝑗𝑗)-th pixel of the image, what is its position?

𝒑𝒑_𝟎𝟎

𝒗𝒗

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Constructing Ray Through a Pixel

• 2D Example: Side view of camera at 𝑝𝑝₀

 Where is the 𝑖𝑖-th pixel, 𝑝𝑝[𝑖𝑖]? (𝑖𝑖 ∈ [0, height))

𝑑𝑑

towards 𝑝𝑝₀

up 𝜃𝜃 = field of view angle (given)

𝑑𝑑 = distance to view plane (arbitrary = you pick)

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Constructing Ray Through a Pixel

𝑝𝑝₂ 𝑝𝑝₁ 2 ⋅𝑑𝑑 ⋅ tan 𝜃𝜃 /2

𝜃𝜃 = field of view angle (given)

𝑑𝑑 = distance to view plane (arbitrary = you pick) 𝑝𝑝₁ = 𝑝𝑝₀ + 𝑑𝑑 ⋅ towards − 𝑑𝑑 ⋅ tan𝜃𝜃₂ ⋅ up 𝑝𝑝₂ = 𝑝𝑝₀ + 𝑑𝑑 ⋅ towards + 𝑑𝑑 ⋅ tan𝜃𝜃₂ ⋅ up 𝑑𝑑 towards 𝑝𝑝₀ up 𝜃𝜃

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𝑝𝑝 𝑖𝑖 = 𝑝𝑝₁ + 𝑖𝑖_height+ 0.5 ⋅ 𝑝𝑝₂ − 𝑝𝑝₁ 𝜃𝜃 = field of view angle (given)

𝑑𝑑 = distance to view plane (arbitrary = you pick) 𝑝𝑝₁ = 𝑝𝑝₀ + 𝑑𝑑 ⋅ towards − 𝑑𝑑 ⋅ tan𝜃𝜃₂ ⋅ up 𝑝𝑝₂ = 𝑝𝑝₀ + 𝑑𝑑 ⋅ towards + 𝑑𝑑 ⋅ tan𝜃𝜃₂ ⋅ up 𝑝𝑝₂ 𝑝𝑝₁ 𝑑𝑑 towards 𝑝𝑝₀ up 𝑝𝑝[𝑖𝑖] 𝜃𝜃 2 ⋅𝑑𝑑 ⋅ tan 𝜃𝜃 /2

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Constructing Ray Through a Pixel

• 2D Example:

The ray passing through the 𝑖𝑖-th pixel is defined by:

Ray 𝑡𝑡 = 𝑝𝑝₀ + 𝑡𝑡 ⋅ ⃗𝑣𝑣

 𝑝𝑝₀: camera position

 ⃗𝑣𝑣: direction to the 𝑖𝑖-th pixel:

⃗𝑣𝑣 = 𝑝𝑝 𝑖𝑖 − 𝑝𝑝_{𝑝𝑝 𝑖𝑖 − 𝑝𝑝}0

0  𝑝𝑝[𝑖𝑖]: 𝑖𝑖-th pixel location:

𝑝𝑝 𝑖𝑖 = 𝑝𝑝1 + 𝑖𝑖_height+ 0.5 ⋅ 𝑝𝑝2 − 𝑝𝑝1

 𝑝𝑝₁ and 𝑝𝑝₂ are the endpoints of the view plane:

𝑝𝑝₁ = 𝑝𝑝₀ + 𝑑𝑑 ⋅ towards − 𝑑𝑑 ⋅ tan𝜃𝜃/2 ⋅ up 𝑝𝑝₂ = 𝑝𝑝₀ + 𝑑𝑑 ⋅ towards + 𝑑𝑑 ⋅ tan𝜃𝜃/2 ⋅ up 𝑝𝑝₂ 𝑝𝑝₁ 𝑑𝑑 towards 𝑝𝑝₀ up 𝑝𝑝[𝑖𝑖] 𝜃𝜃 2 ⋅𝑑𝑑 ⋅ tan 𝜃𝜃 /2

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Figuring out how to do this in 3D is assignment 2 Note:

 Given the vertical field of view angle, 𝜃𝜃_𝑣𝑣

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Constructing Ray Through a Pixel

 And the aspect ratio, 𝑎𝑎𝑟𝑟 = ℎ𝑒𝑒𝑖𝑖𝑒𝑒ℎ𝑒𝑒

𝑤𝑤𝑖𝑖𝑤𝑤𝑒𝑒ℎ

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 And the aspect ratio, 𝑎𝑎𝑟𝑟 = ℎ𝑒𝑒𝑖𝑖𝑒𝑒ℎ𝑒𝑒

𝑤𝑤𝑖𝑖𝑤𝑤𝑒𝑒ℎ

The horizontal field of view angle, 𝜃𝜃_ℎ, satisfies:

tan 𝜃𝜃𝑣𝑣/2

tan 𝜃𝜃_ℎ/2 = 𝑎𝑎𝑟𝑟

𝜽𝜽_𝒉𝒉 𝜽𝜽_𝒗𝒗

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Ray Casting

Where?

}

return image; }

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Ray Casting

What?

Intersection hit = FindIntersection( ray , scene ); image[i][j] = GetColor( hit );

}

return image; }

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Ray-Scene Intersection

• Intersections with geometric primitives

 Sphere

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Ray-Sphere Intersection

Ray: 𝑝𝑝 𝑡𝑡 = 𝑝𝑝₀ + 𝑡𝑡 ⋅ ⃗𝑣𝑣, (0 ≤ 𝑡𝑡 < ∞) Sphere: Φ 𝑝𝑝 = 𝑝𝑝 − 𝑐𝑐 2 − 𝑟𝑟2 = 0 𝑝𝑝₀ 𝒗𝒗 𝑐𝑐 𝑝𝑝 𝑟𝑟 𝑝𝑝′

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Ray-Sphere Intersection

Ray: 𝑝𝑝 𝑡𝑡 = 𝑝𝑝₀ + 𝑡𝑡 ⋅ ⃗𝑣𝑣, (0 ≤ 𝑡𝑡 < ∞)

Sphere: Φ 𝑝𝑝 = 𝑝𝑝 − 𝑐𝑐 2 − 𝑟𝑟2 = 0

Substituting for 𝑝𝑝, we get:

Φ 𝑡𝑡 = 𝑝𝑝₀ + 𝑡𝑡 ⋅ ⃗𝑣𝑣 − 𝑐𝑐 2 − 𝑟𝑟2 = 0

Solve quadratic equation:

𝑎𝑎 ⋅ 𝑡𝑡2 + 𝑏𝑏 ⋅ 𝑡𝑡 + 𝑐𝑐 = 0 where: 𝑎𝑎 = 1 𝑏𝑏 = 2 ⃗𝑣𝑣, 𝑝𝑝₀ − 𝑐𝑐 𝑐𝑐 = 𝑝𝑝₀ − 𝑐𝑐 2 − 𝑟𝑟2 𝑝𝑝0 𝒗𝒗 𝑐𝑐 𝑝𝑝 𝑟𝑟 𝑝𝑝𝑝

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Sphere: Φ 𝑝𝑝 = 𝑝𝑝 − 𝑐𝑐 2 − 𝑟𝑟2 = 0

Φ 𝑡𝑡 = 𝑝𝑝₀ + 𝑡𝑡 ⋅ ⃗𝑣𝑣 − 𝑐𝑐 2 − 𝑟𝑟2 = 0

Solve quadratic equation:

𝑎𝑎 ⋅ 𝑡𝑡2 + 𝑏𝑏 ⋅ 𝑡𝑡 + 𝑐𝑐 = 0 where: 𝑎𝑎 = 1 𝑏𝑏 = 2 ⃗𝑣𝑣, 𝑝𝑝₀ − 𝑐𝑐 𝑐𝑐 = 𝑝𝑝₀ − 𝑐𝑐 2 − 𝑟𝑟2 𝑝𝑝0 𝒗𝒗 𝑐𝑐 𝑝𝑝 𝑟𝑟 𝑝𝑝𝑝

Generally, there are two solutions to the quadratic equation, giving two points of intersection, 𝑝𝑝 and 𝑝𝑝𝑝.

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Ray-Sphere Intersection

𝑝𝑝₀ ⃗𝑣𝑣 𝑐𝑐 𝑝𝑝 𝑟𝑟 𝑛𝑛

• Need normal vector at intersection for lighting calculations:

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• More generally, if the shape is given as the set of points 𝑝𝑝 satisfying:

Φ 𝑝𝑝 = 0

for some function Φ: ℝ3 → ℝ, then the normal of the surface will be parallel to the gradient.

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Ray-Scene Intersection

• Intersections with geometric primitives

 Sphere

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Ray-Triangle Intersection

1. Intersect ray with plane

2. Check if the point is inside the triangle

𝑝𝑝

𝑝𝑝₀ ⃗𝑣𝑣

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Ray-Plane Intersection

Plane: Φ 𝑝𝑝 = 𝑝𝑝, 𝑛𝑛 − 𝑑𝑑 = 0

Φ 𝑡𝑡 = 𝑝𝑝₀ + 𝑡𝑡 ⋅ ⃗𝑣𝑣, 𝑛𝑛 − 𝑑𝑑 = 0 Solution: 𝑡𝑡 = − 𝑝𝑝0,_⃗𝑣𝑣𝑛𝑛 − 𝑑𝑑_, _𝑛𝑛 𝑛𝑛 𝑝𝑝 𝑝𝑝0 ⃗𝑣𝑣 Algebraic Method

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Ray-Triangle Intersection I

• Check for point-triangle intersection algebraically:

 Generate planes through the ray

source and each edge

 Check if the point of intersection is

above each of these planes

𝑝𝑝₀

𝑣𝑣₂ 𝑣𝑣₃

For each edge:

𝑛𝑛_𝑖𝑖 = (𝑣𝑣_𝑖𝑖+1−𝑝𝑝₀) × (𝑣𝑣_𝑖𝑖 − 𝑝𝑝₀) if ( 𝑝𝑝 − 𝑝𝑝₀, 𝑛𝑛_𝑖𝑖 < 0 ) return FALSE; 𝑛𝑛₁ 𝑣𝑣₁ 𝑝𝑝

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Ray-Triangle Intersection II

• Check for point-triangle intersection parametrically

In general, given 𝑝𝑝 ∈ ℝ3 and given

three points 𝑣𝑣₁, 𝑣𝑣₂, 𝑣𝑣₃ ⊂ ℝ3

(in general position) we can

solve for 𝛼𝛼, 𝛽𝛽, 𝛾𝛾 ∈ ℝ such that:

𝑝𝑝 = 𝛼𝛼𝑣𝑣₁ + 𝛽𝛽𝑣𝑣₂ + 𝛾𝛾𝑣𝑣₃

𝑝𝑝 is in the plane spanned by 𝑣𝑣₁, 𝑣𝑣₂, 𝑣𝑣₃ iff.:

𝛼𝛼 + 𝛽𝛽 + 𝛾𝛾 = 1

𝑝𝑝 is inside the triangle with vertices {𝑣𝑣₁, 𝑣𝑣₂, 𝑣𝑣₃} iff.:

𝛼𝛼, 𝛽𝛽, 𝛾𝛾 ≥ 0 𝑝𝑝 𝑣𝑣1 𝑣𝑣₂ 𝑣𝑣₃ 𝛼𝛼 _𝛽𝛽 𝛾𝛾

(63)

Ray-Triangle Intersection II

To get 𝛼𝛼, 𝛽𝛽, 𝛾𝛾, solve the system:

𝑣𝑣₁𝑥𝑥 𝑣𝑣₂𝑥𝑥 𝑣𝑣₃𝑥𝑥 𝑣𝑣₁𝑦𝑦 𝑣𝑣₂𝑦𝑦 𝑣𝑣₃𝑦𝑦 𝑣𝑣₁𝑧𝑧 𝑣𝑣₂𝑧𝑧 𝑣𝑣₃𝑧𝑧 𝛼𝛼 𝛽𝛽 𝛾𝛾 = 𝑝𝑝𝑥𝑥 𝑝𝑝𝑦𝑦 𝑝𝑝𝑧𝑧 ⇔ 𝛼𝛼 𝛽𝛽 𝛾𝛾 = 𝑣𝑣₁𝑥𝑥 𝑣𝑣₂𝑥𝑥 𝑣𝑣₃𝑥𝑥 𝑣𝑣₁𝑦𝑦 𝑣𝑣₂𝑦𝑦 𝑣𝑣₃𝑦𝑦 𝑣𝑣₁𝑧𝑧 𝑣𝑣₂𝑧𝑧 𝑣𝑣₁1 −1 𝑝𝑝𝑥𝑥 𝑝𝑝𝑦𝑦 𝑝𝑝𝑧𝑧 𝑝𝑝 𝛼𝛼 _𝛽𝛽 𝛾𝛾 𝑣𝑣1 𝑣𝑣₂ 𝑣𝑣₃

(64)

Ray-Triangle Intersection II

𝑣𝑣₁𝑥𝑥 𝑣𝑣₂𝑥𝑥 𝑣𝑣₃𝑥𝑥 𝑣𝑣₁𝑦𝑦 𝑣𝑣₂𝑦𝑦 𝑣𝑣₃𝑦𝑦 𝑣𝑣₁𝑧𝑧 𝑣𝑣₂𝑧𝑧 𝑣𝑣₃𝑧𝑧 𝛼𝛼 𝛽𝛽 𝛾𝛾 = 𝑝𝑝𝑥𝑥 𝑝𝑝𝑦𝑦 𝑝𝑝𝑧𝑧 ⇔ 𝛼𝛼 𝛽𝛽 𝛾𝛾 = 𝑣𝑣₁𝑥𝑥 𝑣𝑣₂𝑥𝑥 𝑣𝑣₃𝑥𝑥 𝑣𝑣₁𝑦𝑦 𝑣𝑣₂𝑦𝑦 𝑣𝑣₃𝑦𝑦 𝑣𝑣₁𝑧𝑧 𝑣𝑣₂𝑧𝑧 𝑣𝑣₃1 −1 𝑝𝑝𝑥𝑥 𝑝𝑝𝑦𝑦 𝑝𝑝𝑧𝑧 𝑝𝑝 𝛼𝛼 _𝛽𝛽 𝛾𝛾 𝑣𝑣1 𝑣𝑣₂ 𝑣𝑣₃

(65)

𝑣𝑣₁𝑥𝑥 𝑣𝑣₂𝑥𝑥 𝑣𝑣₃𝑥𝑥 𝑣𝑣₁𝑦𝑦 𝑣𝑣₂𝑦𝑦 𝑣𝑣₃𝑦𝑦 𝑣𝑣₁𝑧𝑧 𝑣𝑣₂𝑧𝑧 𝑣𝑣₃𝑧𝑧 𝛼𝛼 𝛽𝛽 𝛾𝛾 = 𝑝𝑝𝑥𝑥 𝑝𝑝𝑦𝑦 𝑝𝑝𝑧𝑧 ⇔ 𝛼𝛼 𝛽𝛽 𝛾𝛾 = 𝑣𝑣₁𝑥𝑥 𝑣𝑣₂𝑥𝑥 𝑣𝑣₃𝑥𝑥 𝑣𝑣₁𝑦𝑦 𝑣𝑣₂𝑦𝑦 𝑣𝑣₃𝑦𝑦 𝑣𝑣₁𝑧𝑧 𝑣𝑣₂𝑧𝑧 𝑣𝑣₃𝑧𝑧 −1 𝑝𝑝𝑥𝑥 𝑝𝑝𝑦𝑦 𝑝𝑝𝑧𝑧 𝑝𝑝 𝛼𝛼 _𝛽𝛽 𝛾𝛾 𝑣𝑣1 𝑣𝑣₂ 𝑣𝑣₃

This will fail if the vertices 𝑣𝑣₁, 𝑣𝑣₂, 𝑣𝑣₃ lie in a plane through the origin.

(66)

Ray-Triangle Intersection II

Embrace the problem case by translating the whole system to the origin:

𝑝𝑝 𝑣𝑣₁ _𝑣𝑣 2 𝑣𝑣₃ 𝛼𝛼 𝛾𝛾_𝛽𝛽 𝑝𝑝 − 𝑣𝑣1 𝑣𝑣₂ − 𝑣𝑣₁ 𝑣𝑣₃ − 𝑣𝑣₁ 𝛼𝛼 𝛾𝛾_𝛽𝛽 𝑣𝑣₁𝑥𝑥 𝑣𝑣₂𝑥𝑥 𝑣𝑣₃𝑥𝑥 𝑣𝑣₁𝑦𝑦 𝑣𝑣₂𝑦𝑦 𝑣𝑣₃𝑦𝑦 𝑣𝑣₁𝑧𝑧 𝑣𝑣₂𝑧𝑧 𝑣𝑣₃𝑧𝑧 𝛼𝛼 𝛽𝛽 𝛾𝛾 = 𝑝𝑝𝑥𝑥 𝑝𝑝𝑦𝑦 𝑝𝑝𝑧𝑧 0 𝑣𝑣₂𝑥𝑥 − 𝑣𝑣₁𝑥𝑥 𝑣𝑣₃𝑥𝑥 − 𝑣𝑣₁𝑥𝑥 0 𝑣𝑣₂𝑦𝑦 − 𝑣𝑣₁𝑦𝑦 𝑣𝑣₃𝑦𝑦 − 𝑣𝑣₁𝑦𝑦 0 𝑣𝑣₂𝑧𝑧 − 𝑣𝑣₁𝑧𝑧 𝑣𝑣₃𝑧𝑧 − 𝑣𝑣₁𝑧𝑧 𝛼𝛼 𝛽𝛽 𝛾𝛾 = 𝑝𝑝𝑥𝑥 − 𝑣𝑣₁𝑥𝑥 𝑝𝑝𝑦𝑦 − 𝑣𝑣₁𝑦𝑦 𝑝𝑝𝑧𝑧 − 𝑣𝑣₁𝑧𝑧

(67)

𝑝𝑝 𝑣𝑣₁ _𝑣𝑣 2 𝑣𝑣₃ 𝛼𝛼 𝛾𝛾_𝛽𝛽 𝑝𝑝 − 𝑣𝑣1 𝑣𝑣₂ − 𝑣𝑣₁ 𝑣𝑣₃ − 𝑣𝑣₁ 𝛼𝛼 𝛾𝛾_𝛽𝛽 𝑣𝑣₁𝑥𝑥 𝑣𝑣₂𝑥𝑥 𝑣𝑣₃𝑥𝑥 𝑣𝑣₁𝑦𝑦 𝑣𝑣₂𝑦𝑦 𝑣𝑣₃𝑦𝑦 𝑣𝑣₁𝑧𝑧 𝑣𝑣₂𝑧𝑧 𝑣𝑣₃𝑧𝑧 𝛼𝛼 𝛽𝛽 𝛾𝛾 = 𝑝𝑝𝑥𝑥 𝑝𝑝𝑦𝑦 𝑝𝑝𝑧𝑧 𝑣𝑣₂𝑥𝑥 − 𝑣𝑣₁𝑥𝑥 𝑣𝑣₃𝑥𝑥 − 𝑣𝑣₁𝑥𝑥 𝑣𝑣₂𝑦𝑦 − 𝑣𝑣₁𝑦𝑦 𝑣𝑣₃𝑦𝑦 − 𝑣𝑣₁𝑦𝑦 𝑣𝑣₂𝑧𝑧 − 𝑣𝑣₁𝑧𝑧 𝑣𝑣₃𝑧𝑧 − 𝑣𝑣₁𝑧𝑧 𝛽𝛽 𝛾𝛾 = 𝑝𝑝𝑥𝑥 − 𝑣𝑣₁𝑥𝑥 𝑝𝑝𝑦𝑦 − 𝑣𝑣₁𝑦𝑦 𝑝𝑝𝑧𝑧 − 𝑣𝑣₁𝑧𝑧

(68)

Ray-Triangle Intersection II

𝑝𝑝 𝑣𝑣₁ _𝑣𝑣 2 𝑣𝑣₃ 𝛼𝛼 𝛾𝛾_𝛽𝛽 𝑝𝑝 − 𝑣𝑣1 𝑣𝑣₂ − 𝑣𝑣₁ 𝑣𝑣₃ − 𝑣𝑣₁ 𝛼𝛼 𝛾𝛾_𝛽𝛽 𝑣𝑣₁𝑥𝑥 𝑣𝑣₂𝑥𝑥 𝑣𝑣₃𝑥𝑥 𝑣𝑣₁𝑦𝑦 𝑣𝑣₂𝑦𝑦 𝑣𝑣₃𝑦𝑦 𝑣𝑣₁𝑧𝑧 𝑣𝑣₂𝑧𝑧 𝑣𝑣₃𝑧𝑧 𝛼𝛼 𝛽𝛽 𝛾𝛾 = 𝑝𝑝𝑥𝑥 𝑝𝑝𝑦𝑦 𝑝𝑝𝑧𝑧 This is an over-constrained system!

This is an over-constrained system!

In general, we can’t express a 3D point as the linear combination of two 3D points.

This is not the general case!

A solution exists since 𝑝𝑝 is in the plane spanned by {𝑣𝑣₁, 𝑣𝑣₂, 𝑣𝑣₃} After solving for 𝛽𝛽 and 𝛾𝛾, we can set:

𝛼𝛼 = 1 − 𝛽𝛽 − 𝛾𝛾 𝑣𝑣₂𝑥𝑥 − 𝑣𝑣₁𝑥𝑥 𝑣𝑣₃𝑥𝑥 − 𝑣𝑣₁𝑥𝑥 𝑣𝑣₂𝑦𝑦 − 𝑣𝑣₁𝑦𝑦 𝑣𝑣₃𝑦𝑦 − 𝑣𝑣₁𝑦𝑦 𝑣𝑣₂𝑧𝑧 − 𝑣𝑣₁𝑧𝑧 𝑣𝑣₃𝑧𝑧 − 𝑣𝑣₁𝑧𝑧 𝛽𝛽 𝛾𝛾 = 𝑝𝑝𝑥𝑥 − 𝑣𝑣₁𝑥𝑥 𝑝𝑝𝑦𝑦 − 𝑣𝑣₁𝑦𝑦 𝑝𝑝𝑧𝑧 − 𝑣𝑣₁𝑧𝑧

(69)

Other Ray-Primitive Intersections

• Cone, cylinder, ellipsoid:

 Similar to sphere

• Box

 Intersect 3 front-facing planes, return closest

• Convex polygon

» Find the intersection of the ray with the plane

» Check that the intersection is above every triangle generated by the ray source and polygon edge. • Concave polygon

 Same plane intersection