Frenet-Serret Fram e (M oving trihedron)

Frenet-Serret Frame
(Moving trihedron)
Dr. Cengiz ERDÖNMEZ
1

Moving Trihedron
2

Parametric Curves: A Review
• A parametric curve in space has the following form:
• f: [0,1] -> ( f(u), g(u), h(u) ) where f(), g() and h() are three
real-valued functions. Thus, f(u) maps a real value u in the
closed interval [0,1] to a point in space. The domain of
these real functions and vector-valued function f() does not
have to be [0,1]. It can be any closed interval; but, for
simplicity, we restrict the domain to [0,1]. Thus, for
each u in [0,1], there corresponds to a point ( f(u),g(u),h(u))
in space. In this course, functions f(), g() and h() are always
polynomials.
• Note that if function h() is removed from the definition
of f(), f() has two coordinate components and becomes a
curve in the coordinate plane.
4

Examples
• We have seen that a straight line is defined as B+td,
where B is a base point and d is a direction vector. Thus, if f()
is defined as
f(u) = b1 + ud1 g(u) = b2 + ud2 h(u) = b3 + ud3 where B = < b1, b2, b3 >, d = < d1, d2, d3 >, and f() is a
parametric curve that maps [0,1] to the line segment
between B and B+d, inclusive.
• A circle has the following non-polynomial form:
x(u) = rcos(2*PI*u) + p
y(u) = rsin(2*PI*u) + q
It has center (p, q) and radius r. Since the parameter u is in
[0,1], the value of 2*PI*u is in [0,2*PI] (i.e., from 0 degree to
360 degree).
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• Let us eliminate u. First, change the above
equations to the following. For convenience, we
drop (u) from x(u) and y(u).
x - p = rcos(2*PI*u)
y - q = rsin(2*PI*u)
Then, squaring both equations and adding them
together yields(x - p) 2 + (y - q) 2 = r2Thus, it shows
that the given parametric curve is indeed a circle
with center at (p, q) and radius r.
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• A space cubic curve has the following form:
f(u) = u
g(u) = u2 h(u) = u3 The following figure shows this curve in the range of
[-1,1]. It is contained in the box (in white) defined by
( -1, 0, -1 ) and (1, 1, 1)
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• The circular helix is defined as follows:
f(u) = ( acos(u), asin(u), bu )
The following figure shows the curve in [0,
4*PI]. The initial point is (a, 0, 0) and the
endpoint is (a, 0, b*4*PI). Note that this curve
lies on the cylinder of radius a and axis the zaxis.
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Normal Vector and Curvature
• Consider a fixed point f(u) and two moving
points P and Q on a parametric curve. These three points
determine a plane. As P and Q moves toward f(u), this
plane approaches a limiting position. This is
the osculating plane at f(u). Obviously, the osculating
plane at f(u) contains the tangent line at f(u). It can be
shown that the osculating plane is the plane that passes
through f(u) and contains both f'(u) and f''(u). More
precisely, any point on this plane has an equation as
follows, where p and q are parameters:
f(u) + pf'(u) + qf''(u)
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• The binormal vector b(u) is the unit-length
vector of the cross-product of f'(u) and f''(u):
b(u) = (f'(u) × f''(u)) / | (f'(u) × f''(u)) |
• Thus, the binormal vector b(u) is
perpendicular to both f'(u) and f''(u) and
hence perpendicular to the osculating plane.
The line f(u)+tb(u) is the binormal line at f(u).
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• The normal vector is the vector perpendicular
to both tangent and binormal vectors with its
direction determined by the right-handed
system. That is, the unit-length normal
vector n(u) is defined to be
n(u) = ( b(u) × f'(u) ) / | b(u) × f'(u) |
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• The line f(u)+tn(u) is the normal line at f(u). Therefore,
tangent vector f'(u), normal vector n(u) and binormal
vector b(u) form a coordinate system with origin f(u).
The tangent line, binormal line and normal line are the
three coordinate axes with positive directions given by
the tangent vector, binormal vector and normal vector,
respectively. These three vectors are usually referred to
as the moving triad or triad at point f(u). Moving triad
is also calledmoving trihedron. The following figure
shows their relationship. Note that the tangent, normal
and f''(u) vectors are on the same plane.
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Example
• Let us compute the tangent, binormal and normal
vectors of the circular helix curve:
f(u) = ( acos(u), asin(u), bu )
• The first and second derivatives are as follows:
f'(u) = ( -asin(u), acos(u), b )
f''(u) = ( -acos(u), -asin(u), 0 )
• The non-unit-length binormal vector is the crossproduct
of f'(u) and f''(u), in this order:
b(u) = f'(u) × f''(u) = ( absin(u), -abcos(u), a2 )
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• The non-unit-length normal vector is the cross-product
of the binormal vector and the tangent vector, in this
order:
n(u) = b(u) × f'(u) = ( -a(a2 + b2 )cos(u), -a(a2 + b2 )sin(u), 0 )
• If you compare n(u) and f''(u), you will see that these
vectors are parallel to each other (i.e., coefficients are
proportional) and are both parallel to the xy-plane. As
a result, after normalizing all involving vectors, the
normal and the second derivative vectors are identical.
This is shown by the following figure. It is computed
at u = 1.
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Curvature
• As we all know, the tangent vector measures the change of
"distance" and thus gives the speed of a moving point. The
speed change, or acceleration, is measured by the derivative
of the tangent vector, which is the second derivative. In fact,
there is one more interesting interpretation. Take X as a fixed
point and P and Q two moving points. As long as not all of
these three points lie on a line, they uniquely determine a
circle. As both P and Q moves toward X, the circle they
determine approaches a limiting position as shown with dotline
below. This limiting circle is called the circle of
curvature at X and its center and radius, O and r, are
the center and radius of circle of curvature, respectively. More
importantly, 1/r is the curvature at X. Therefore, the larger the
circle of curvature, the smaller the curvature.
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• From the definition of osculating plane, we know that this circle of
curvature must be on the osculating plane. Since the circle of
curvature is tangent to the curve and hence the tangent line, the
center of curvature lies on the normal line.
• The curvature at u, k(u), can be computed as follows:
k(u) = | f'(u) × f''(u) | / | f'(u) | 3
• Let us continue with the previous example of the circular helix
curve. From the computations of f'(u) and f''(u), we have
f'(u) = ( -asin(u), acos(u), b )
f'(u) × f''(u) = ( absin(u), -abcos(u), a2 )
| f'(u) | = SQRT(a2 + b2 )
| f'(u) × f''(u) | = aSQRT(a2 + b2 )
k(u) = a / (a2 + b2 )
• Therefore, the curvature at any point on the curve is a
constant a/(a2 + b2 ).
• Since the radius of the circle of curvature is 1/k, we see that the
center of circle of curvature is located at a distance of
(a2 + b2 )/a from f(u) in the normal direction n(u). 20

Examples
• Consider a straight line:
• f(u) = ( a + up, b + uq, c + ur )
• We have the following:
f'(u) = ( p, q, r )
| f'(u) | = SQRT(p2 + q2 + r2 )
f''(u) = ( 0, 0, 0 )
f'(u) × f''(u) = ( 0, 0, 0 )
k(u) = 0
• Therefore, the curvature of a straight line is zero
everywhere. Note that none of the binormal and
normal vectors is well-defined because f''(u) is a zero
vector.
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• Consider a circle on the xy-plane:
• f(u) = ( rcos(u) + p, rsin(u) + q, 0 )
• Since it is on the xy-plane, the third coordinate function is
always 0. From the given circle's equation we have the
following:
f'(u) = ( -rsin(u), rcos(u), 0 )
f''(u) = ( -rcos(u), -rsin(u), 0 )
f'(u) × f''(u) = ( 0, 0, r2 )
| f'(u) | = r
| f'(u) × f''(u) | = r2 b(u) = (f'(u) × f''(u)) / | f'(u) × f''(u) | = ( 0, 0, 1 )
n(u) = (b(u) × f'(u)) / | b(u) × f'(u) | = (-cos(u),-sin(u),0 )
k(u) = 1/r
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• Thus, the unit-length tangent vector is (-sin(u),
cos(u), 0), the binormal vector is ( 0, 0, 1), and
the normal vector is ( -cos(u), sin(u), 0). The
binormal vector is always perpendicular to
the xy-plane while both the tangent and
normal vectors lie on the xy-plane. The
curvature of a circle is a constant 1/r. As a
result, the radius of the circle of curvature
is r and the circle of curvature is the given
circle itself.
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• Consider the space cubic defines as follows:
• f(u) = ( u, u2 , u3 )
• The following are the computation of
curvature k(u):
f'(u) = ( 1, 2u, 3u2 )
| f'(u) | = SQRT(1 + 4u2 + 9u4 )
f''(u) = ( 0, 2, 6u)
f'(u) × f''(u) = ( 6u2 , -6u, 2 )
| f'(u) × f''(u) | = 2SQRT(1 + 9u2 + 9u4 )
k(u) = 2SQRT(1 + 9u2 + 9u4 ) / (SQRT(1 + 4u2 +
9u4 )) 3
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Why Is the Moving Triad Important?
• The moving triad is not only a mathematical concept. It also
provides vital information of the characteristics of a moving
object. For example, if you are on a airplane or a rollercoaster,
as you are "flying" you are following a curve.
Therefore, you are moving in the direction of the tangent
vector, your "up" vector is in the direction of the binormal
vector and the rate of turning and turning direction are
given by the curvature and the direction of the normal
vector, respectively. That is, while you are rolling or
tumbling, this triad always provides you with the forward,
up and turning directions. Therefore, to correctly animate a
moving object, you need to know the geometric
characteristics of the moving triad.
25

s
N
T
�
B
dα
dα
= ( − sinα i + cosα
j) = n = N
ds ds
ds
ds
Bir önceki slide’dan ödünç alınan bağıntıdaki s, şekildeki T nin dengidir. s birim
vektörünün eğri boyunca alınan uzay türevi N normal vektörünü tanımlar. N
vektörünün uzay türevi de B binormal vektörü tanımlar.
26

in astronomy/rpekunlu/MHD II/FrenetSerretDifGeoCurves.pdf
27

Frenet – Serret Vektör Tabanındaki birim vektörleri
T,N,B olarak gösterelim.
B = T × N
dT
ds
dT
ds
=
N
d
=
ds
r
T
Frenet – Serret Formülleri
dT
= κ N
ds
dN
= − κT + τ B
ds
dB
κ - Eğrilik = −τ
N
ds
τ - Burulma 29

Frenet-Serret Formülleri matris gösterimle aşağıdaki gibidir
⎡T ′ ⎤ ⎡ 0 κ 0⎤
⎡T ⎤
⎢
′
⎥ ⎢
κ 0
⎥ ⎢ ⎥
⎢
N
⎥
=
⎢
− τ
⎥ ⎢
N
⎥
⎢ ⎣ B′ ⎥ ⎦ ⎢ ⎣ 0 −τ
0⎥
⎦ ⎢ ⎣ B ⎥ ⎦
Serret-Frenet Equations
κ (s ) N (s )
�
=
d �
T
ds
+ τ (s ) (s )
B �
= - κ (s ) T (s )
�
dN ds
�
= - τ (s ) N (s )
�
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dB ds
�