Computer Aided Engineering Design

DIFFERENTIAL GEOMETRY OF CURVES 79

From Eq. (3.23), T

r

=

ds()

ds

so that

d

du

ds

du

r

= T using chain rule. Differentiating further gives

d

du

d

du

ds

du

ds

du

2

2

2

= + 2

r T T

Implementing the above result in Eq. (3.29) yields

QP QW T

T

T

T

TT

= +

= =

2

2

3

2

3

3

3

××

⎛

⎝⎜

⎞

⎠⎟

⎡

⎣

⎢

⎤

⎦

⎥

× ⎛

⎝

⎞

⎠

⎡

⎣

⎢

⎢

⎤

⎦

⎥

⎥

× ⎛

⎝

⎞

⎠

⎡

⎣

⎢

⎢

⎤

⎦

⎥

⎥

ds

du

d

du

ds

du

ds

du

u

d

du

ds

du

u

d

ds

ds

du

u

Δ

ΔΔT

Using Eqs. (3.26) and (3.27), we get

QP QW = T N = B

3

3

3

××⎛^3

⎝

⎞

⎠

⎡

⎣

⎢

⎢

⎤

⎦

⎥

⎥

⎛

⎝

⎞

⎠

⎡

⎣

⎢

⎢

⎤

⎦

⎥

⎥

κκ

ds

du

u

ds

du

ΔΔu (3.30a)

From Eqs. (3.29) and (3.30a)

d

du

d

du

ds

du

r r

= B

2

2

3

×

⎡

⎣

⎢

⎤

⎦

⎥

⎛

⎝

⎞

⎠

⎡

⎣

⎢

⎢

⎤

⎦

⎥

⎥

κ (3.30b)

From Eq. (3.30a), QP×QW is parallel to B implying that if a circle is drawn through P,Q and W,
the normal to the plane containing the circle would be parallel to B, that is, the circle would be
contained in the osculating plane for which reason it is termed as the osculating circle (Figure 3.8).
From vector algebra, the radius of curvature ρ at P is given as

ρ =

| || || – |

2| |

=

| || || – |

2 | ( – ) |

=

| || || – |

2| |

WP WQ WP WQ

WP WQ

WP WQ WP WQ

WQ PQ WQ

WP WQ WP WQ

×××PQ WQ

or ρ =

+^1

2

+... 2 +... –^1

2

+...

2

2

2

2

2

2

2

2

2

3

ΔΔΔΔΔ

Δ

u

d

du

d

du

uu

d

du

u

d

du

d

du

u

d

du

d

du

u

r r rrr

r r

×

=

3

3

2

2

3

Δu

d

du

d

du

d

du

u

r

rr

× Δ

or using Eq. (3.30b)