2D Computer Graphics Diego Nehab Summer 2020 IMPA 1 Differential - - PowerPoint PPT Presentation

2D Computer Graphics

Diego Nehab Summer 2020

IMPA 1

Differential geometry

Inside-outside test

Ideal model is implicit Given a region Ω ⊂ R2, defjne the indicator function 1Ω : R2 → {0, 1} 1Ω(p) =    1, p ∈ Ω 0, p ∈ Ω A.k.a. characteristic function p 1 p Alternatively, using the Iverson bracket, p 1 p , where true 1 and false

2

Inside-outside test

Ideal model is implicit Given a region Ω ⊂ R2, defjne the indicator function 1Ω : R2 → {0, 1} 1Ω(p) =    1, p ∈ Ω 0, p ∈ Ω A.k.a. characteristic function χΩ(p) = 1Ω(p) Alternatively, using the Iverson bracket, p 1 p , where true 1 and false

2

Inside-outside test

Ideal model is implicit Given a region Ω ⊂ R2, defjne the indicator function 1Ω : R2 → {0, 1} 1Ω(p) =    1, p ∈ Ω 0, p ∈ Ω A.k.a. characteristic function χΩ(p) = 1Ω(p) Alternatively, using the Iverson bracket, [p ∈ Ω] = 1Ω(p), where [true] = 1 and [false] = 0

2

Inside-outside test

Simple shapes can be defjned as primitives

• p, r < 0
• ,

p, r ∈ RP2 f x y f R x y p c r p c R2 p s p1 t p2 1 s t p3 s t 1 s t R pi R2 Complex shapes can be defjned by logical expressions p r p c r Basis of CSG (constructive solid geometry)

3

Inside-outside test

Simple shapes can be defjned as primitives

• p, r < 0
• ,

p, r ∈ RP2

• f(x, y) < 0
• ,

f ∈ R[x, y] p c r p c R2 p s p1 t p2 1 s t p3 s t 1 s t R pi R2 Complex shapes can be defjned by logical expressions p r p c r Basis of CSG (constructive solid geometry)

3

Inside-outside test

Simple shapes can be defjned as primitives

• p, r < 0
• ,

p, r ∈ RP2

• f(x, y) < 0
• ,

f ∈ R[x, y]

• |p − c| < r
• ,

p, c ∈ R2 p s p1 t p2 1 s t p3 s t 1 s t R pi R2 Complex shapes can be defjned by logical expressions p r p c r Basis of CSG (constructive solid geometry)

3

Inside-outside test

Simple shapes can be defjned as primitives

• p, r < 0
• ,

p, r ∈ RP2

• f(x, y) < 0
• ,

f ∈ R[x, y]

• |p − c| < r
• ,

p, c ∈ R2

• p = s p1 + t p2 + (1 − s − t) p3 ∧ 0 ≤ s, t ≤ 1
• ,

s, t ∈ R, pi ∈ R2 Complex shapes can be defjned by logical expressions p r p c r Basis of CSG (constructive solid geometry)

3

Inside-outside test

Simple shapes can be defjned as primitives

• p, r < 0
• ,

p, r ∈ RP2

• f(x, y) < 0
• ,

f ∈ R[x, y]

• |p − c| < r
• ,

p, c ∈ R2

• p = s p1 + t p2 + (1 − s − t) p3 ∧ 0 ≤ s, t ≤ 1
• ,

s, t ∈ R, pi ∈ R2 Complex shapes can be defjned by logical expressions

• p, r < 0
• ∧ ¬
• |p − c| < r
• Basis of CSG (constructive solid geometry)

3

Inside-outside test

Simple shapes can be defjned as primitives

• p, r < 0
• ,

p, r ∈ RP2

• f(x, y) < 0
• ,

f ∈ R[x, y]

• |p − c| < r
• ,

p, c ∈ R2

• p = s p1 + t p2 + (1 − s − t) p3 ∧ 0 ≤ s, t ≤ 1
• ,

s, t ∈ R, pi ∈ R2 Complex shapes can be defjned by logical expressions

• p, r < 0
• ∧ ¬
• |p − c| < r
• Basis of CSG (constructive solid geometry)

3

Inside-outside test

More common to defjne interior by boundary (Jordan curve theorem) Let denote the boundary of region Let w p count the number of signed intersections with boundary when we move from p to infjnity in any direction w p is the winding number of the boundary around point p Defjne interior by odd or non-zero winding numbers w p 1 2

• r

w p In a sense, the defjnition is still implicit But the “function” involves a complicated decision procedure How do we defjne the boundary?

4

Inside-outside test

More common to defjne interior by boundary (Jordan curve theorem) Let ∂Ω denote the boundary of region Ω Let w p count the number of signed intersections with boundary when we move from p to infjnity in any direction w p is the winding number of the boundary around point p Defjne interior by odd or non-zero winding numbers w p 1 2

• r

w p In a sense, the defjnition is still implicit But the “function” involves a complicated decision procedure How do we defjne the boundary?

4

Inside-outside test

More common to defjne interior by boundary (Jordan curve theorem) Let ∂Ω denote the boundary of region Ω Let w(∂Ω, p) count the number of signed intersections with boundary ∂Ω when we move from p to infjnity in any direction w p is the winding number of the boundary around point p Defjne interior by odd or non-zero winding numbers w p 1 2

• r

w p In a sense, the defjnition is still implicit But the “function” involves a complicated decision procedure How do we defjne the boundary?

4

Inside-outside test

More common to defjne interior by boundary (Jordan curve theorem) Let ∂Ω denote the boundary of region Ω Let w(∂Ω, p) count the number of signed intersections with boundary ∂Ω when we move from p to infjnity in any direction w(∂Ω, p) is the winding number of the boundary ∂Ω around point p Defjne interior by odd or non-zero winding numbers w p 1 2

• r

w p In a sense, the defjnition is still implicit But the “function” involves a complicated decision procedure How do we defjne the boundary?

4

Inside-outside test

More common to defjne interior by boundary (Jordan curve theorem) Let ∂Ω denote the boundary of region Ω Let w(∂Ω, p) count the number of signed intersections with boundary ∂Ω when we move from p to infjnity in any direction w(∂Ω, p) is the winding number of the boundary ∂Ω around point p Defjne interior by odd or non-zero winding numbers

• w(∂Ω, p) = 1 (mod 2)
• r
• w(∂Ω, p) = 0]

In a sense, the defjnition is still implicit But the “function” involves a complicated decision procedure How do we defjne the boundary?

4

Inside-outside test

More common to defjne interior by boundary (Jordan curve theorem) Let ∂Ω denote the boundary of region Ω Let w(∂Ω, p) count the number of signed intersections with boundary ∂Ω when we move from p to infjnity in any direction w(∂Ω, p) is the winding number of the boundary ∂Ω around point p Defjne interior by odd or non-zero winding numbers

• w(∂Ω, p) = 1 (mod 2)
• r
• w(∂Ω, p) = 0]

In a sense, the defjnition is still implicit But the “function” involves a complicated decision procedure How do we defjne the boundary?

4

Inside-outside test

More common to defjne interior by boundary (Jordan curve theorem) Let ∂Ω denote the boundary of region Ω Let w(∂Ω, p) count the number of signed intersections with boundary ∂Ω when we move from p to infjnity in any direction w(∂Ω, p) is the winding number of the boundary ∂Ω around point p Defjne interior by odd or non-zero winding numbers

• w(∂Ω, p) = 1 (mod 2)
• r
• w(∂Ω, p) = 0]

In a sense, the defjnition is still implicit But the “function” involves a complicated decision procedure How do we defjne the boundary?

4

Inside-outside test

More common to defjne interior by boundary (Jordan curve theorem) Let ∂Ω denote the boundary of region Ω Let w(∂Ω, p) count the number of signed intersections with boundary ∂Ω when we move from p to infjnity in any direction w(∂Ω, p) is the winding number of the boundary ∂Ω around point p Defjne interior by odd or non-zero winding numbers

• w(∂Ω, p) = 1 (mod 2)
• r
• w(∂Ω, p) = 0]

In a sense, the defjnition is still implicit But the “function” involves a complicated decision procedure How do we defjne the boundary?

4

Planar parametric curve

Piecewise differentiable function α : I ⊂ R → R2 from an interval I to R2 t → α(t) =

• x(t), y(t)
• x y are the coordinate functions of

When I a b , we say the curve is closed if a b 0 2 R2 t r t r t The trace I is image of I through . It is the trace that we care about A subset S R2 is parametrized by if there is I R such that I S A subset S R2 can be parametrized in many different ways a b R2 t r t r t b a 2

5

Planar parametric curve

Piecewise differentiable function α : I ⊂ R → R2 from an interval I to R2 t → α(t) =

• x(t), y(t)
• x, y are the coordinate functions of α

When I a b , we say the curve is closed if a b 0 2 R2 t r t r t The trace I is image of I through . It is the trace that we care about A subset S R2 is parametrized by if there is I R such that I S A subset S R2 can be parametrized in many different ways a b R2 t r t r t b a 2

5

Planar parametric curve

Piecewise differentiable function α : I ⊂ R → R2 from an interval I to R2 t → α(t) =

• x(t), y(t)
• x, y are the coordinate functions of α

When I = [a, b], we say the curve α is closed if α(a) = α(b) α: [0, 2π] → R2 α(t) = (r cos t, r sin t) The trace I is image of I through . It is the trace that we care about A subset S R2 is parametrized by if there is I R such that I S A subset S R2 can be parametrized in many different ways a b R2 t r t r t b a 2

5

Planar parametric curve

Piecewise differentiable function α : I ⊂ R → R2 from an interval I to R2 t → α(t) =

• x(t), y(t)
• x, y are the coordinate functions of α

When I = [a, b], we say the curve α is closed if α(a) = α(b) α: [0, 2π] → R2 α(t) = (r cos t, r sin t) The trace α(I) is image of I through α. It is the trace that we care about A subset S R2 is parametrized by if there is I R such that I S A subset S R2 can be parametrized in many different ways a b R2 t r t r t b a 2

5

Planar parametric curve

Piecewise differentiable function α : I ⊂ R → R2 from an interval I to R2 t → α(t) =

• x(t), y(t)
• x, y are the coordinate functions of α

When I = [a, b], we say the curve α is closed if α(a) = α(b) α: [0, 2π] → R2 α(t) = (r cos t, r sin t) The trace α(I) is image of I through α. It is the trace that we care about A subset S ⊂ R2 is parametrized by α if there is I ⊂ R such that α(I) = S A subset S R2 can be parametrized in many different ways a b R2 t r t r t b a 2

5

Planar parametric curve

Piecewise differentiable function α : I ⊂ R → R2 from an interval I to R2 t → α(t) =

• x(t), y(t)
• x, y are the coordinate functions of α

When I = [a, b], we say the curve α is closed if α(a) = α(b) α: [0, 2π] → R2 α(t) = (r cos t, r sin t) The trace α(I) is image of I through α. It is the trace that we care about A subset S ⊂ R2 is parametrized by α if there is I ⊂ R such that α(I) = S A subset S ⊂ R2 can be parametrized in many different ways β : [a, b] → R2 β(t) =

• r cos(ωt + φ), r sin(ωt + φ)
• ,

b − a ≥ 2π ω

5

Regularity, velocity, and tangent

Let α : I → R2 be a parameterization of S ⊂ R2. The velocity vector is t x t y t A regular point t is a point where t exists and t

• What if

t 0? A is curve is regular in J I if all points in J are regular The speed is given by v t t If t is regular, T t t v t is the unit tangent to Are SVG paths regular? Are individual SVG segments regular?

6

Regularity, velocity, and tangent

Let α : I → R2 be a parameterization of S ⊂ R2. The velocity vector is α′(t) =

• x′(t), y′(t)
• A regular point

t is a point where t exists and t

• What if

t 0? A is curve is regular in J I if all points in J are regular The speed is given by v t t If t is regular, T t t v t is the unit tangent to Are SVG paths regular? Are individual SVG segments regular?

6

Regularity, velocity, and tangent

Let α : I → R2 be a parameterization of S ⊂ R2. The velocity vector is α′(t) =

• x′(t), y′(t)
• A regular point α(t) is a point where α′(t) exists and α′(t) = 0
• What if α′(t) = 0?

A is curve is regular in J I if all points in J are regular The speed is given by v t t If t is regular, T t t v t is the unit tangent to Are SVG paths regular? Are individual SVG segments regular?

6

Regularity, velocity, and tangent

Let α : I → R2 be a parameterization of S ⊂ R2. The velocity vector is α′(t) =

• x′(t), y′(t)
• A regular point α(t) is a point where α′(t) exists and α′(t) = 0
• What if α′(t) = 0?

A is curve α is regular in J ⊂ I if all points in α(J) are regular The speed is given by v t t If t is regular, T t t v t is the unit tangent to Are SVG paths regular? Are individual SVG segments regular?

6

Regularity, velocity, and tangent

Let α : I → R2 be a parameterization of S ⊂ R2. The velocity vector is α′(t) =

• x′(t), y′(t)
• A regular point α(t) is a point where α′(t) exists and α′(t) = 0
• What if α′(t) = 0?

A is curve α is regular in J ⊂ I if all points in α(J) are regular The speed is given by v(t) = α′(t) If t is regular, T t t v t is the unit tangent to Are SVG paths regular? Are individual SVG segments regular?

6

Regularity, velocity, and tangent

Let α : I → R2 be a parameterization of S ⊂ R2. The velocity vector is α′(t) =

• x′(t), y′(t)
• A regular point α(t) is a point where α′(t) exists and α′(t) = 0
• What if α′(t) = 0?

A is curve α is regular in J ⊂ I if all points in α(J) are regular The speed is given by v(t) = α′(t) If α(t) is regular, T(t) = α′(t)/v(t) is the unit tangent to α Are SVG paths regular? Are individual SVG segments regular?

6

Regularity, velocity, and tangent

Let α : I → R2 be a parameterization of S ⊂ R2. The velocity vector is α′(t) =

• x′(t), y′(t)
• A regular point α(t) is a point where α′(t) exists and α′(t) = 0
• What if α′(t) = 0?

A is curve α is regular in J ⊂ I if all points in α(J) are regular The speed is given by v(t) = α′(t) If α(t) is regular, T(t) = α′(t)/v(t) is the unit tangent to α Are SVG paths regular? Are individual SVG segments regular?

6

Regularity, velocity, and tangent

Let α : I → R2 be a parameterization of S ⊂ R2. The velocity vector is α′(t) =

• x′(t), y′(t)
• A regular point α(t) is a point where α′(t) exists and α′(t) = 0
• What if α′(t) = 0?

A is curve α is regular in J ⊂ I if all points in α(J) are regular The speed is given by v(t) = α′(t) If α(t) is regular, T(t) = α′(t)/v(t) is the unit tangent to α Are SVG paths regular? Are individual SVG segments regular?

6

Arc length

The arc-length of a curve segment α : [a, b] → R2 is s = b

a

• α′(t)
• dt
• Makes sense from physics’ time integral of speed
• Also makes sense from rectifjcation

Not all curves have a length

• E.g.

t t 1 t t 0 1

• Koch snowfmake

7

Arc length

The arc-length of a curve segment α : [a, b] → R2 is s = b

a

• α′(t)
• dt
• Makes sense from physics’ time integral of speed
• Also makes sense from rectifjcation

Not all curves have a length

• E.g. α(t) = t sin(1/t),

t ∈ [0, 1]

• Koch snowfmake

7

Arc length

The arc-length of a curve segment α : [a, b] → R2 is s = b

a

• α′(t)
• dt
• Makes sense from physics’ time integral of speed
• Also makes sense from rectifjcation

Not all curves have a length

• E.g. α(t) = t sin(1/t),

t ∈ [0, 1]

• Koch snowfmake

7

Reparameterization

A curve β : J → R2 is a reparameterization of α if there is a monotonic differentiable function h : J → I such that β = α ◦ h

• A positive reparameterization has h′(J) ⊂ R>0
• A negative reparameterization has h′(J) ⊂ R<0

The arc-length is invariant to reparameterizations c d R2

d c

t dt

d c

h t h t dt

b a

u du u h t

8

Reparameterization

A curve β : J → R2 is a reparameterization of α if there is a monotonic differentiable function h : J → I such that β = α ◦ h

• A positive reparameterization has h′(J) ⊂ R>0
• A negative reparameterization has h′(J) ⊂ R<0

The arc-length is invariant to reparameterizations β : [c, d] → R2 d

c

• β′(t)
• dt

d c

h t h t dt

b a

u du u h t

8

Reparameterization

A curve β : J → R2 is a reparameterization of α if there is a monotonic differentiable function h : J → I such that β = α ◦ h

• A positive reparameterization has h′(J) ⊂ R>0
• A negative reparameterization has h′(J) ⊂ R<0

The arc-length is invariant to reparameterizations β : [c, d] → R2 d

c

• β′(t)
• dt =

d

c

|α′(h(t))| h′(t) dt

b a

u du u h t

8

Reparameterization

A curve β : J → R2 is a reparameterization of α if there is a monotonic differentiable function h : J → I such that β = α ◦ h

• A positive reparameterization has h′(J) ⊂ R>0
• A negative reparameterization has h′(J) ⊂ R<0

The arc-length is invariant to reparameterizations β : [c, d] → R2 d

c

• β′(t)
• dt =

d

c

|α′(h(t))| h′(t) dt = b

a

|α′(u)| du

• u = h(t)
• 8

Arc-length reparameterization

Let α : (a, b) → R2 be a curve. The arc-length function of α is defjned by s(t) = t

a

• α′(t)
• dt

Every regular curve admits an arc-length reparameterization Regularity means s t t 0, which means s t is strictly increasing, which means s has a differentiable inverse u with u t 1 s u t 1 u t Consider u t u t u t u t u t 1 We get s t

t c

t dt t c

9

Arc-length reparameterization

Let α : (a, b) → R2 be a curve. The arc-length function of α is defjned by s(t) = t

a

• α′(t)
• dt

Every regular curve admits an arc-length reparameterization Regularity means s t t 0, which means s t is strictly increasing, which means s has a differentiable inverse u with u t 1 s u t 1 u t Consider u t u t u t u t u t 1 We get s t

t c

t dt t c

9

Arc-length reparameterization

Let α : (a, b) → R2 be a curve. The arc-length function of α is defjned by s(t) = t

a

• α′(t)
• dt

Every regular curve admits an arc-length reparameterization Regularity means s′(t) = |α′(t)| > 0, which means s(t) is strictly increasing, which means s has a differentiable inverse u with u′(t) = 1 s′ u(t) = 1

• α′

u(t)

• Consider

u t u t u t u t u t 1 We get s t

t c

t dt t c

9

Arc-length reparameterization

Let α : (a, b) → R2 be a curve. The arc-length function of α is defjned by s(t) = t

a

• α′(t)
• dt

Every regular curve admits an arc-length reparameterization Regularity means s′(t) = |α′(t)| > 0, which means s(t) is strictly increasing, which means s has a differentiable inverse u with u′(t) = 1 s′ u(t) = 1

• α′

u(t)

• Consider β = α ◦ u

|β′(t)| u t u t u t u t 1 We get s t

t c

t dt t c

9

Arc-length reparameterization

Let α : (a, b) → R2 be a curve. The arc-length function of α is defjned by s(t) = t

a

• α′(t)
• dt

Every regular curve admits an arc-length reparameterization Regularity means s′(t) = |α′(t)| > 0, which means s(t) is strictly increasing, which means s has a differentiable inverse u with u′(t) = 1 s′ u(t) = 1

• α′

u(t)

• Consider β = α ◦ u

|β′(t)| =

• α′

u(t)

• u′(t)
• u t

u t 1 We get s t

t c

t dt t c

9

Arc-length reparameterization

Let α : (a, b) → R2 be a curve. The arc-length function of α is defjned by s(t) = t

a

• α′(t)
• dt

Every regular curve admits an arc-length reparameterization Regularity means s′(t) = |α′(t)| > 0, which means s(t) is strictly increasing, which means s has a differentiable inverse u with u′(t) = 1 s′ u(t) = 1

• α′

u(t)

• Consider β = α ◦ u

|β′(t)| =

• α′

u(t)

• u′(t)
• =
• α′

u(t)

• α′

u(t)

• = 1

We get s t

t c

t dt t c

9

Arc-length reparameterization

Let α : (a, b) → R2 be a curve. The arc-length function of α is defjned by s(t) = t

a

• α′(t)
• dt

Every regular curve admits an arc-length reparameterization Regularity means s′(t) = |α′(t)| > 0, which means s(t) is strictly increasing, which means s has a differentiable inverse u with u′(t) = 1 s′ u(t) = 1

• α′

u(t)

• Consider β = α ◦ u

|β′(t)| =

• α′

u(t)

• u′(t)
• =
• α′

u(t)

• α′

u(t)

• = 1

We get s(t) = t

c

• β′(t)
• dt = t − c

9

Arc-length reparameterization

This is easier said than done Canonic parabola y2 4ax with focus at a 0 and directrix x a t at2 2at t 2a 1 t2

t

t dt at t2 1 a t2 1 t Standard ellipse t a t b t t b 1 m

2 t

m 1

a2 b2 t

t dt Elliptic integral of the second kind

10

Arc-length reparameterization

This is easier said than done Canonic parabola y2 = 4ax with focus at (a, 0) and directrix x = −a α(t) = (at2, 2at) ⇒

• α′(t)
• = 2a
• 1 + t2

t

• α′(t)
• dt = at
• t2 + 1 + a log
• t2 + 1 + t
• Standard ellipse

t a t b t t b 1 m

2 t

m 1

a2 b2 t

t dt Elliptic integral of the second kind

10

Arc-length reparameterization

This is easier said than done Canonic parabola y2 = 4ax with focus at (a, 0) and directrix x = −a α(t) = (at2, 2at) ⇒

• α′(t)
• = 2a
• 1 + t2

t

• α′(t)
• dt = at
• t2 + 1 + a log
• t2 + 1 + t
• Standard ellipse

β(t) = (a cos t, b sin t) ⇒

• β′(t)
• = b
• 1 − m sin2(t),

m = 1 − a2

b2

t

• β′(t)
• dt = Elliptic integral of the second kind

10

Curvature

Let β(t) be an arc-length parameterization T(t) = β′(t) is the unit tangent to β T t t is normal to T t T t 1 T t T t t T t is the curvature of t 1 t is the radius of curvature of t and t measure the way curve is turning N t T t T t is the unit normal to

11

Curvature

Let β(t) be an arc-length parameterization T(t) = β′(t) is the unit tangent to β T′(t) = β′′(t) is normal to β

• T(t), T(t)
• = 1

⇒

• T′(t), T(t)
• = 0

t T t is the curvature of t 1 t is the radius of curvature of t and t measure the way curve is turning N t T t T t is the unit normal to

11

Curvature

Let β(t) be an arc-length parameterization T(t) = β′(t) is the unit tangent to β T′(t) = β′′(t) is normal to β

• T(t), T(t)
• = 1

⇒

• T′(t), T(t)
• = 0

κ(t) =

• T′(t)
• is the curvature of β

t 1 t is the radius of curvature of t and t measure the way curve is turning N t T t T t is the unit normal to

11

Curvature

Let β(t) be an arc-length parameterization T(t) = β′(t) is the unit tangent to β T′(t) = β′′(t) is normal to β

• T(t), T(t)
• = 1

⇒

• T′(t), T(t)
• = 0

κ(t) =

• T′(t)
• is the curvature of β

ρ(t) = 1/κ(t) is the radius of curvature of β t and t measure the way curve is turning N t T t T t is the unit normal to

11

Curvature

Let β(t) be an arc-length parameterization T(t) = β′(t) is the unit tangent to β T′(t) = β′′(t) is normal to β

• T(t), T(t)
• = 1

⇒

• T′(t), T(t)
• = 0

κ(t) =

• T′(t)
• is the curvature of β

ρ(t) = 1/κ(t) is the radius of curvature of β κ(t) and ρ(t) measure the way curve β is turning N t T t T t is the unit normal to

11

Curvature

Let β(t) be an arc-length parameterization T(t) = β′(t) is the unit tangent to β T′(t) = β′′(t) is normal to β

• T(t), T(t)
• = 1

⇒

• T′(t), T(t)
• = 0

κ(t) =

• T′(t)
• is the curvature of β

ρ(t) = 1/κ(t) is the radius of curvature of β κ(t) and ρ(t) measure the way curve β is turning N(t) = T′(t)/

• T′(t)
• is the unit normal to β

11

Curvature

If β is an arc-length parameterization,

• β′(t) × β′′(t)
• =
• T(t) × κ(t)N(t)
• = κ(t)

If is a general parameterization with u, we obtain t u u u u u u

2

u u u u u u

2

u u u 3 u u u

3 12

Curvature

If β is an arc-length parameterization,

• β′(t) × β′′(t)
• =
• T(t) × κ(t)N(t)
• = κ(t)

If α is a general parameterization with β = α ◦ u, we obtain κ(t) =

• α(u)′ × α(u)′′
• u u

u u

2

u u u u u u

2

u u u 3 u u u

3 12

Curvature

If β is an arc-length parameterization,

• β′(t) × β′′(t)
• =
• T(t) × κ(t)N(t)
• = κ(t)

If α is a general parameterization with β = α ◦ u, we obtain κ(t) =

• α(u)′ × α(u)′′
• =
• α′(u) u′ ×
• α′′(u)(u′)2 + α′(u) u′′
• u u

u u

2

u u u 3 u u u

3 12

Curvature

If β is an arc-length parameterization,

• β′(t) × β′′(t)
• =
• T(t) × κ(t)N(t)
• = κ(t)

If α is a general parameterization with β = α ◦ u, we obtain κ(t) =

• α(u)′ × α(u)′′
• =
• α′(u) u′ ×
• α′′(u)(u′)2 + α′(u) u′′
• =
• α′(u) u′ × α′′(u) (u′)2
• u

u u 3 u u u

3 12

Curvature

If β is an arc-length parameterization,

• β′(t) × β′′(t)
• =
• T(t) × κ(t)N(t)
• = κ(t)

If α is a general parameterization with β = α ◦ u, we obtain κ(t) =

• α(u)′ × α(u)′′
• =
• α′(u) u′ ×
• α′′(u)(u′)2 + α′(u) u′′
• =
• α′(u) u′ × α′′(u) (u′)2
• =
• α′(u) × α′′(u)
• |u′|3

u u u

3 12

Curvature

If β is an arc-length parameterization,

• β′(t) × β′′(t)
• =
• T(t) × κ(t)N(t)
• = κ(t)

If α is a general parameterization with β = α ◦ u, we obtain κ(t) =

• α(u)′ × α(u)′′
• =
• α′(u) u′ ×
• α′′(u)(u′)2 + α′(u) u′′
• =
• α′(u) u′ × α′′(u) (u′)2
• =
• α′(u) × α′′(u)
• |u′|3

=

• α′(u) × α′′(u)
• α′(u)
• 3

12

Curvature

For planar curves, the normal N(t) as the right-hand rotation of T(t) Then defjne the signed curvature t u u u

3

Either way, the center of curvature for a curve is at t t N t The osculating circle has the center and radius of curvature An infmection is a point where the curvature vanishes I.e. where the 1st and 2nd derivatives are collinear

13

Curvature

For planar curves, the normal N(t) as the right-hand rotation of T(t) Then defjne the signed curvature κ(t) = α′(u) × α′′(u)

• α′(u)
• 3

Either way, the center of curvature for a curve is at t t N t The osculating circle has the center and radius of curvature An infmection is a point where the curvature vanishes I.e. where the 1st and 2nd derivatives are collinear

13

Curvature

For planar curves, the normal N(t) as the right-hand rotation of T(t) Then defjne the signed curvature κ(t) = α′(u) × α′′(u)

• α′(u)
• 3

Either way, the center of curvature for a curve α is at α(t) + ρ(t)N(t) The osculating circle has the center and radius of curvature An infmection is a point where the curvature vanishes I.e. where the 1st and 2nd derivatives are collinear

13

Curvature

For planar curves, the normal N(t) as the right-hand rotation of T(t) Then defjne the signed curvature κ(t) = α′(u) × α′′(u)

• α′(u)
• 3

Either way, the center of curvature for a curve α is at α(t) + ρ(t)N(t) The osculating circle has the center and radius of curvature An infmection is a point where the curvature vanishes I.e. where the 1st and 2nd derivatives are collinear

13

Curvature

For planar curves, the normal N(t) as the right-hand rotation of T(t) Then defjne the signed curvature κ(t) = α′(u) × α′′(u)

• α′(u)
• 3

Either way, the center of curvature for a curve α is at α(t) + ρ(t)N(t) The osculating circle has the center and radius of curvature An infmection is a point where the curvature vanishes I.e. where the 1st and 2nd derivatives are collinear

13

Stroking

A different way of specifying the interior Given a curve I R2 and a stroke width w R, we can defjne endpoints p1 p2 R2 p1 t t

w 2 N t

and p2 t t

w 2 N t

We can then defjne the line segment t 1 u p1 t u p2 t u 1 (1) The stroked region is p t t I How to decide if point p belongs to the stroked curve segment? Dashing requires the arc length

14

Stroking

A different way of specifying the interior Given a curve α : I → R2 and a stroke width w ∈ R, we can defjne endpoints p1, p2 ∈ R2 p1(t) = α(t) + w

2 N(t)

and p2(t) = α(t) − w

2 N(t)

We can then defjne the line segment t 1 u p1 t u p2 t u 1 (1) The stroked region is p t t I How to decide if point p belongs to the stroked curve segment? Dashing requires the arc length

14

Stroking

A different way of specifying the interior Given a curve α : I → R2 and a stroke width w ∈ R, we can defjne endpoints p1, p2 ∈ R2 p1(t) = α(t) + w

2 N(t)

and p2(t) = α(t) − w

2 N(t)

We can then defjne the line segment ℓ(t) =

• (1 − u) p1(t) + u p2(t), 0 < u < 1
• (1)

The stroked region is p t t I How to decide if point p belongs to the stroked curve segment? Dashing requires the arc length

14

Stroking

A different way of specifying the interior Given a curve α : I → R2 and a stroke width w ∈ R, we can defjne endpoints p1, p2 ∈ R2 p1(t) = α(t) + w

2 N(t)

and p2(t) = α(t) − w

2 N(t)

We can then defjne the line segment ℓ(t) =

• (1 − u) p1(t) + u p2(t), 0 < u < 1
• (1)

The stroked region is

• p ∈ ℓ(t), t ∈ I
• How to decide if point p belongs to the stroked curve segment?

Dashing requires the arc length

14

Stroking

A different way of specifying the interior Given a curve α : I → R2 and a stroke width w ∈ R, we can defjne endpoints p1, p2 ∈ R2 p1(t) = α(t) + w

2 N(t)

and p2(t) = α(t) − w

2 N(t)

We can then defjne the line segment ℓ(t) =

• (1 − u) p1(t) + u p2(t), 0 < u < 1
• (1)

The stroked region is

• p ∈ ℓ(t), t ∈ I
• How to decide if point p belongs to the stroked curve segment?

Dashing requires the arc length

14

Stroking

A different way of specifying the interior Given a curve α : I → R2 and a stroke width w ∈ R, we can defjne endpoints p1, p2 ∈ R2 p1(t) = α(t) + w

2 N(t)

and p2(t) = α(t) − w

2 N(t)

We can then defjne the line segment ℓ(t) =

• (1 − u) p1(t) + u p2(t), 0 < u < 1
• (1)

The stroked region is

• p ∈ ℓ(t), t ∈ I
• How to decide if point p belongs to the stroked curve segment?

Dashing requires the arc length

14

Bézier curves

Find a simple formula for the curvature at t = 0 Let curve t have endpoints p0 p1 p2 pn n p1 p0 and n 1 n p2 p1 p1 p0

3

n 1 n p1 p0 p2 p1 p1 p0 3 n 1 n h a2 How would you compute the arc length? [Jüttler, 1997] Show offset and evolute curves

15

Bézier curves

Find a simple formula for the curvature at t = 0 Let curve α(t) have endpoints p0, p1, p2, . . . , pn α′(0) = n(p1 − p0) and α′′(0) = (n − 1)n

• (p2 − p1) − (p1 − p0)
• 3

n 1 n p1 p0 p2 p1 p1 p0 3 n 1 n h a2 How would you compute the arc length? [Jüttler, 1997] Show offset and evolute curves

15

Bézier curves

Find a simple formula for the curvature at t = 0 Let curve α(t) have endpoints p0, p1, p2, . . . , pn α′(0) = n(p1 − p0) and α′′(0) = (n − 1)n

• (p2 − p1) − (p1 − p0)
• κ(0) = α′(0) × α′′(0)
• α′(0)
• 3

n 1 n p1 p0 p2 p1 p1 p0 3 n 1 n h a2 How would you compute the arc length? [Jüttler, 1997] Show offset and evolute curves

15

Bézier curves

Find a simple formula for the curvature at t = 0 Let curve α(t) have endpoints p0, p1, p2, . . . , pn α′(0) = n(p1 − p0) and α′′(0) = (n − 1)n

• (p2 − p1) − (p1 − p0)
• κ(0) = α′(0) × α′′(0)
• α′(0)
• 3

= n − 1 n (p1 − p0) × (p2 − p1) |p1 − p0|3 n 1 n h a2 How would you compute the arc length? [Jüttler, 1997] Show offset and evolute curves

15

Bézier curves

Find a simple formula for the curvature at t = 0 Let curve α(t) have endpoints p0, p1, p2, . . . , pn α′(0) = n(p1 − p0) and α′′(0) = (n − 1)n

• (p2 − p1) − (p1 − p0)
• κ(0) = α′(0) × α′′(0)
• α′(0)
• 3

= n − 1 n (p1 − p0) × (p2 − p1) |p1 − p0|3 = n − 1 n h a2 How would you compute the arc length? [Jüttler, 1997] Show offset and evolute curves

15

Bézier curves

Find a simple formula for the curvature at t = 0 Let curve α(t) have endpoints p0, p1, p2, . . . , pn α′(0) = n(p1 − p0) and α′′(0) = (n − 1)n

• (p2 − p1) − (p1 − p0)
• κ(0) = α′(0) × α′′(0)
• α′(0)
• 3

= n − 1 n (p1 − p0) × (p2 − p1) |p1 − p0|3 = n − 1 n h a2 How would you compute the arc length? [Jüttler, 1997] Show offset and evolute curves

15

Bézier curves

Find a simple formula for the curvature at t = 0 Let curve α(t) have endpoints p0, p1, p2, . . . , pn α′(0) = n(p1 − p0) and α′′(0) = (n − 1)n

• (p2 − p1) − (p1 − p0)
• κ(0) = α′(0) × α′′(0)
• α′(0)
• 3

= n − 1 n (p1 − p0) × (p2 − p1) |p1 − p0|3 = n − 1 n h a2 How would you compute the arc length? [Jüttler, 1997] Show offset and evolute curves

15

References

• B. Jüttler. A vegetarian approach to optimal parameterizations.

Computer Aided Geometric Design, 14(9):887–890, 1997.

• E. Kreyszig. Differential Geometry. Dover, 1991.
• B. O’Neill. Elementary Differential Geometry. Academic Press, revised

2nd edition, 2006.

16