Appendix: Curvature Tensor of a Developable Surface
Here, we use the notation introduced in Sect. 2.1 to prove the geometrical identities announced in Sects. 2.3 to 2.5, relevant to developable surfaces. We consider a general ruled surface, enforce the condition of developability, and derive the expression of the curvature tensor at an arbitrary point on the surface. By doing so, we extend the expressions obtained by Wunderlich [28] and by Starostin and van der Heijden [25] to account for the geodesic curvature κ
g of the center-line.
Let us first calculate the tangent vectors at an arbitrary point y(S,V) of the surface:
$$\begin{aligned} \mathbf{y}_{,S}(S,V) & = \mathbf{d}_{3}(S) + V \mathbf{q}'(S), \end{aligned}$$
(35a)
$$\begin{aligned} \mathbf{y}_{,V}(S,V) & = \mathbf{q}(S) , \end{aligned}$$
(35b)
where q′(S) denotes the total derivative of q defined in Eq. (4) as q=η
d
3+d
1.
Using Eq. (6), we write the following equation
$$ \mathbf{q}' = \eta \omega_{2} \mathbf{d}_{1} + (\omega_{3} - \eta \omega_{1}) \mathbf{d}_{2} + \bigl(\eta'-\omega_{2}\bigr) \mathbf{d}_{3} , $$
(36)
which implies
$$ \mathbf{q} \times \mathbf{q}' = (\eta \omega_{1}- \omega_{3}) \mathbf{q}^\perp + \frac{1}{V_{\mathrm{c}}} \mathbf{d}_{2}, $$
(37)
where V
c is the quantity defined by Eq. (13*), and q
⊥ is the vector
$$ \mathbf{q}^\perp = \mathbf{d}_{2} \times \mathbf{q} = - \mathbf{d}_{3} + \eta \mathbf{d}_{1} . $$
(38)
Later on, we shall show that q
⊥ is a vector perpendicular to q lying in the plane tangent to the surface; hence the notation.
The classical condition for a ruled surface to be developable [23, Sect. 3.II] is that the following three vectors are linearly dependent: the tangent d
3 to the center-line (called the directrix in the context of the geometry of surfaces), the vector q spanning the generatrices, and its derivative with respect to the arc-length along the center-line. This is expressed by (q×q′)⋅d
3=0. In view of Eq. (37), this yields ηω
1=ω
3, which is the constraint of developability announced in Eq. (11a*).
As a result, q′⋅d
2=0, and so y
,S
⋅d
2=0. On the other hand, Eq. (35b) shows that y
,V
⋅d
2=0. The director d
2(S) is orthogonal to both tangents: d
2 is a unit normal at any point of the developable surface.
The element of area on the surface reads
$$ \mathrm{d} a = \vert \mathbf{y}_{,S}\times \mathbf{y}_{,V} \vert \, \mathrm{d}S \,\mathrm{d}V = \bigl\vert \mathbf{d}_{3}\times \mathbf{q} + V \mathbf{q}'\times \mathbf{q}\bigr\vert \,\mathrm{d}S\, \mathrm{d}V = \biggl\vert \biggl(1-\frac{V}{V_{\mathrm{c}}} \biggr) \mathbf{d}_{2} \biggr\vert \,\mathrm{d}S \,\mathrm{d}V. $$
(39)
Noting that 1−V/V
c>0 by the inequality (14b), we arrive at the result announced in Eq. (12).
To compute the curvature tensor K(S,V), we note that the direction of the generatrix is a principal direction of zero curvature, since the surface is developable. Therefore, there exists some scalar field k(S,V) such that
$$ \mathbf{K} = k \mathbf{q}^\perp \otimes \mathbf{q}^\perp . $$
(40)
The quantity k in equation above can be found by contracting with y
,S
on both sides of the equation to give:
$$ \mathbf{y}_{,S}\cdot \mathbf{K}\cdot \mathbf{y}_{,S} =k \bigl(\mathbf{q}^\perp \cdot \bigl(\mathbf{d}_{3}+V \mathbf{q}' \bigr) \bigr)^2 = k \biggl(-1+ \frac{V}{V_{\mathrm{c}}} \biggr)^2 . $$
(41)
By the definition of the curvature tensor (second fundamental form) [23], the left-hand side of the resulting identity is the normal projection of the second derivative y
,SS
:
$$ \mathbf{y}_{,S}\cdot \mathbf{K}\cdot \mathbf{y}_{,S} = \mathbf{y}_{,SS}\cdot \mathbf{d}_{2} = \bigl( \mathbf{d}_{3}' + V \mathbf{q}'' \bigr)\cdot \mathbf{d}_{2} = -\omega_{1} +V \biggl( \frac{\mathrm{d}( \mathbf{q}'\cdot \mathbf{d}_{2})}{\mathrm{d}S} - \mathbf{q'}\cdot \mathbf{d}_{2}' \biggr). $$
(42)
In this equation, q′⋅d
2=ω
3−ηω
1=0 by the developability condition, and \(\mathbf{q}'\cdot \mathbf{d}_{2}' = \mathbf{q}'\cdot (\boldsymbol{\omega}\times \mathbf{d}_{2}) = \mathbf{q}'\cdot (\omega_{1} \mathbf{q}\times \mathbf{d}_{2}) = - \omega_{1} \mathbf{d}_{2}\cdot(\mathbf{q}\times \mathbf{q}') = -\omega_{1} / V_{\mathrm{c}}\). Therefore,
$$ \mathbf{y}_{,S}\cdot \mathbf{K}\cdot \mathbf{y}_{,S} = - \omega_{1} \biggl(1-\frac{V}{V_{\mathrm{c}}} \biggr). $$
(43)
From Eqs. (41) and (43), we can solve for k, giving k=−ω
1/(1−V/V
c). Inserting this result into Eq. (40) yields the expression of curvature tensor announced in Eqs. (15) and (16*). The curvature tensor keeps the same form as in the case of zero geodesic curvature [25, 28] provided that the proper definition of V
c in terms of κ
g is used; see Eq. (13*).