Discrete Curves

Differential Geometry: Curves, Tangents, and Curvature

Differential Geometry of Curves and Surfaces, Do Carmo (Chapter 1) Differentiable Curves

Definition: A parameterized differentiable curve is a differentiable map α:I→Rn of an open interval I=(a,b) of the real line R into Rn. Differentiable Curves α(u) = (cosu,sin u,u) Examples: • α(u)=(cos u , sin u , u) • α(u)=(u , |u|) • α(u)=(u3-4u , u2-4) • α(u)=(u3 , u2)

α(u) = (u, u ) α(u) = (u3 − 4u,u 2 − 4) α(u) = (u3 ,u 2 ) Differentiable Curves

Note: If φ(u) is a differentiable map from [a’,b’] to [a,b], then the function αoφ:[a’,b’]→Rn is also a parameterized differentiable curve tracing out the same points.

α(u) = (u3 − 4u,u 2 − 4) α(u) = (u3 ,u 2 ) Regular Curves

Given a parameterized curve, α(u), the tangent of α at a point u is the vector defined by the derivative of the parameterization, α’(u).

α(u) = (u3 ,u 2 ) Regular Curves

Given a parameterized curve, α(u), the tangent of α at a point u is the vector defined by the derivative of the parameterization, α’(u). The parameterized curve α(u) is regular if α’(u)≠0 for all u∈I. This implies: –There is a “well-defined” tangent line at each point on the curve. α(u) = (u3 − 4u,u 2 − 4) –The map α(u) is “injective” Regular Curves

We say that two parameterized, differentiable, regular, curves α:[a,b]→Rn and β:[a’,b’]→Rn define the same differentiable curve if there exists a differentiable map φ:[a’,b’]→[a,b], with φ’(u)≠0, such that βoφ=α. We say that they define the same oriented differentiable curve if, in addition, φ’(u)>0. Regular Curves

Given a parameterized curve α(v), and given u∈I, the arc-length from the point u is: u 0 s(u) = ∫ α'(v) dv u0 Regular Curves

Given a parameterized curve α(v), and given u∈I, the arc-length from the point u is: u 0 s(u) = ∫ α'(v) dv u0 If we partition the interval [u0,u] into N sub- intervals, setting ∆u=(u-u0)/N and ui=u0+i∆u, we can approximate the integral as:

N −1 α(ui) α(uN) s(u) = lim α'(ui )∆u N →∞ ∑ i=0 N −1 α(u )−α(u ) = lim i+1 i ∆u N →∞ ∑ i=0 ∆u N −1 ∆u α(u0) = lim α(ui+1 )−α(ui ) N →∞ ∑ i=0 ∆u Regular Curves

Using the arc-length, we can integrate functions defined over the curve. Given a function f:Rn→R, the integral of f over the curve α can be obtained by integrating over the parameterization domain I: b ∫ f ( p)dp = ∫ f (α(u))α'(u) du p∈α a Note: If the parameterizations α and β define the same curve, they define the same integral. So integrals are properties of curves. Regular Curves

Given a regular differentiable curve α, we can define the unit-tangent vector at a point α(u) as: α'(t) t(u) = α'(t) Regular Curves

Given a regular differentiable curve α, we can define the unit-tangent vector at a point α(u) as: α'(t) t(u) = α'(t)

Note: If the parameterizations α and β define the same oriented curve, they define the same unit-tangent vectors. So unit-tangent vectors are properties of oriented curves. Regular Curves

Given a regular differentiable curve α, we can define the unit-tangent vector at a point α(u) as: α'(t) t(u) = α'(t)

Note: If the parameterizations α and β define the same oriented curve, they define the same unit-tangent vectors. So unit-tangent vectors are properties of oriented curves. Tangent lines are properties of (un-oriented) curves. Curvature

The tangent line at a point α(u0) on the curve describes the best-fitting line to the curve.

α(u0) Curvature

The tangent line at a point α(u0) on the curve describes the best-fitting line to the curve. We would like to extend this to finding the best- fit circle at a point.

α(u0) Curvature

Q: How can we compute the (reciprocal of the) radius of a circle using only local information? Curvature

Q: How can we compute the (reciprocal of the) radius of a circle using only local information? A: Look at how quickly the tangent vectors are turning as we move between adjacent points.

t1 t1 t p t p 0 1ϕ 0 1 ϕ p0 p0 Curvature

Q: How can we compute the (reciprocal of the) radius of a circle using only local information? A: Look at how quickly the tangent vectors are turning as we move between adjacent points. We use the fact that the length of the chord between p0 and p1 equals the distance t rt0 1 rt1 between the tangents, multiplied t p 0 1 ϕ p by the circle’s radius: 0

r t1 − t0 = p1 − p0 Curvature

Thus, we can express the radius as: 1 t − t = 1 0 r p1 − p0

t rt0 1 rt1 t p 0 1 ϕ p0 Curvature

Thus, we can express the radius as: 1 t − t = 1 0 r p1 − p0 In the case of the curve, we can compute the curvature by looking at the limit as u1 goes to u0:

t(u1) − t(u0 ) t'(u0 ) κ(u0 ) = lim = u →u 1 0 α(u1) −α(u0 ) α'(u0 ) Curvature (2D)

For the case of an oriented curve in 2D, we can assign a sign to the curvature, making it positive if the center of the circle is to the left of the tangent and negative if it is to the right.

κ > 0

κ < 0 Curvature (2D)

For the case of an oriented curve in 2D, we can assign a sign to the curvature, making it positive if the center of the circle is to the left of the tangent and negative if it is to the right.

Note: If parameterizations α and β define the same (oriented) curve, they κ > 0 define the same (signed) curvature. So curvature is κ < 0 a property of curves. Curvature (2D)

Theorem: For a closed oriented curve α⊂R2, the integral of the curvature over α is an integer multiple of 2π: ∫κ( p)dp = 2πl p∈α Curvature (2D)

Rough Proof: To prove this, we show that the integral of the curvature is equal to the integral of the change of angle traced out by the tangent indicatrix. Curvature (2D)

Definition: Given an oriented differentiable curve α:I→R2, we can define the tangent indicatrix, a map from the interval I to the unit-circle defined by the tangent. Curvature (2D)

Rough Proof: 1. Using the tangent indicatrix, we can assign an angle to each point on the curve, (the angle of the tangent vector). Curvature (2D)

Rough Proof: 2. With a bit of manipulation, we can show that the change in the angle, as a function of arc- length, is equal to the curvature. Curvature (2D)

Rough Proof: 2. With a bit of manipulation, we can show that the change in the angle, as a function of arc- length, is equal to the curvature. Recall that the (signed) curvature is measured as the (signed) change in tangent length, divided by the change in arc-length.

t'(u0 ) κ(u0 ) = ± α'(u0 ) Curvature (2D)

Rough Proof: 3. Since the integral is the accumulation of changes of angle, weighted by arc-length, and since we start and end with the same angle, the accumulated angle must be an integer multiple of 2π. Curvature (2D)

Theorem: For a closed curve α⊂R2, the integral of the curvature over α is an integer multiple of 2π: ∫κ( p)dp = 2πl p∈α The integer l is the winding index of the curve. Curvature (2D)

In defining the curvature, we implicitly used two different definitions: 1. Curvature is the (signed) distance between tangents divided by the distance between the corresponding points on the circle.

t1 p1 t − t 1 0 t0 κ( p0 ) = lim ϕ p → p p0 1 0 p1 − p0 Curvature (2D)

In defining the curvature, we implicitly used two different definitions: 2. Curvature is the (signed) angle between tangents divided by the arc-length between the corresponding points on the circle.

t1 ϕ p1=α(u1) ϕ κ( p0 ) = lim → u1 t u1 u0 0 ϕ p = (u ) ∫ α'(v) dv 0 α 0 u0 Curvature (2D)

In defining the curvature, we implicitly used two different definitions. In the limit, the two definitions are the same since the ratio of chord-length to arc-length goes to one as points get closer:

ϕ ϕ 3 2sin = + O(ϕ ) ϕ/2 sin(ϕ/2) 2 2

ϕ Curvature (2D)

In defining the curvature, we implicitly used two different definitions. In the limit, the two definitions are the same since the ratio of chord-length to arc-length goes to one as points get closer.

But in the discrete ϕ/2 sin(ϕ/2) setting, they gives rise to two different ϕ notions of curvature. Discrete Curves

Definition: A discrete curve is the set of points on the line segments between consecutive vertices in an n ordered set of vertices {v0,…,vN}⊂R , such that v for any index i, v ≠v . i i i+1 vi+1 Discrete Curves

The length of the discrete curve between vertex v and v is: i0 i i−1 li = ∑ v j+1 − v j j=i0

vi vi+1 Discrete Curves

The length of the discrete curve between vertex v and v is: i0 i i−1 li = ∑ v j+1 − v j j=i0 Given a function assigning a value fi to edge vi vivi+1, we can compute the vi+1 integral of the function as: N −1 F = ∑ vi+1 − vi fi i=0 Discrete Curves

The length of the discrete curve between vertex v and v is: i0 i i−1 li = ∑ v j+1 − v j j=i0 Similarly, given a function assigning a value fi to vi vertex vi, we can compute the vi+1 integral of the function as: N −1 vi+1 − vi + vi − vi−1 F = ∑ fi i=0 2 Discrete Curves

At any (point on an) edge vivi+1, we can define the tangent vector ti as the unit vector pointing from vi to vi+1: vi+1 − vi ti = vi+1 − vi vi ti vi+1

t ti-1 i+1 Discrete Curves

Q: Where and how should we define curvature?

vi ti vi+1

t ti-1 i+1 Discrete Curves

Q: Where and how should we define curvature? A: Since the tangent only changes at vertices we should define the curvature as a vertex value.

vi ti vi+1

t ti-1 i+1 Discrete Curves

Q: Where and how should we define curvature? A: Since the tangent only changes at vertices we should define the curvature as a vertex value. We should define the value of the curvature as

vi ti the change in the tangent as vi+1 we move through the vertex. t ti-1 i+1 Discrete Curves

Note that we cannot define the tangent as the result of a limiting process:

ti − ti−1 κi = lim ∆t→0 [(1− ∆t)vi + ∆tvi+1 ]−[(1− ∆t)vi + ∆tvi−1 ]

ti − ti−1 = lim vi ti ∆t→0 vi+1 ∆t vi+1 − vi−1

t ti-1 i+1 Discrete Curves

However, we can estimate it using the finite- differences using the edge centers (vi+vi+1)/2.

vi ti vi+1 vi+1 + vi 2 t ti-1 i+1 Discrete Curves

However, we can estimate it using the finite- differences using the edge centers (vi+vi+1)/2. 1. Define the curvature in terms of change in tangents divided by the distance between v edge centers: i ti vi+1 vi+1 + vi ti − ti−1 2 κi = [(vi + vi+1) / 2]− [(vi−1 + vi ) / 2] t ti-1 i+1

vi+1 + vi 2

ti-1

ti vi + vi−1 2 Discrete Curves

However, we can estimate it using the finite- differences using the edge centers (vi+vi+1)/2. 2. Define the curvature in terms of change in tangents angle divided by the arc-length v between edge centers: i ti vi+1 vi+1 + vi ∠t1t2 2 κi = (vi+1 − vi ) / 2 + (vi − vi−1) / 2 t ti-1 i+1

vi+1 + vi 2

ti-1

ti vi + vi−1 2 Discrete Curves

In the limit, as we refine the tessellation of a differentiable curve, both methods define the same curvature. Does it make a difference which one we use? Discrete Curves

Does it make a difference which one we use? Consider integrating the curvature over the closed curve: N −1 vi+1 − vi + vi − vi−1 Κ = ∑ κi i=0 2 vi ti vi+1

t ti-1 i+1 Discrete Curves

Does it make a difference which one we use? Consider integrating the curvature over the closed curve: N −1 vi+1 − vi + vi − vi−1 Κ = ∑ κi i=0 2 vi ti vi+1 Using the second definition, v + + v i 1 i we get: 2 ∠ t t1t2 ti-1 i+1 κi = (vi+1 − vi ) / 2 + (vi − vi−1) / 2

N −1 Κ = ∑∠t1t2 i=0 Discrete Curves

Does it make a difference which one we use? Consider integrating the curvature over the closed curve: N −1 vi+1 − vi + vi − vi−1 Κ = ∑ κi i=0 2 vi ti vi+1 Using the second definition, v + + v i 1 i we get the sum of angular 2 t ti+1 deficits over the vertices of i-1 a closed polygon. Discrete Curves

Does it make a difference which one we use? Consider integrating the curvature over the closed curve: N −1 vi+1 − vi + vi − vi−1 Κ = ∑ κi i=0 2 vi ti vi+1 Using the second definition, v + + v i 1 i we get the sum of angular 2 t ti+1 deficits over the vertices of i-1 a closed polygon. This has to be a multiple of 2π so this definition conforms to the winding index. Discrete Curves

Does it make a difference which one we use? Consider integrating the curvature over the closed curve: N −1 vi+1 − vi + vi − vi−1 Κ = ∑ κi i=0 2 vi ti vi+1 Using the first definition we v + + v i 1 i would get a value for the 2 t ti+1 curvature integral that is i-1 not a multiple of 2π, so this definition does not conform to the winding index.