# Tutorial 2: Framed Discrete Curves

A discrete curve $$\gamma$$ in a space $$\mathrm M$$ is just a finite sequence of points in $$\mathrm M$$, $$\gamma = (\gamma_0,…,\gamma_{n-1})$$. In case we have discrete curve $$\gamma$$ in $$\mathbb R^3$$ we can simply draw $$\gamma$$ by joining the points by straight edge segments. The vector that moves a point $$\gamma_i$$ to the next point $$\gamma_{i+1}$$ is called the discrete tangent vector $\mathbf e_i := \gamma_{i+1}-\gamma_i.$Further, the normalized tangent vector will be denoted by$T_i := \frac{\mathbf e_i}{\Vert\mathbf e_i\Vert}.$It is even more convenient to draw the points as small spheres and the lines as tubes that align with the corresponding edge. Actually there is 1-parameter freedom to place the tube around the edge – we could rotate the tube around the edge – which is hidden in the tube’s rotational symmetry.

If we instead would like replace the edges by a less symmetric shape (as e.g. in Albert’s post) or if we want to really mesh a tubular surface around the curve this freedom matters. Therefore we must additionally specify a direction in the normal space of the edge. A discrete normal vector field of $$\gamma$$ is a sequence $$N= (N_0,…,N_{n-2})$$ such that$N_i\perp T_i, \textit{ for all }i=0,…,n-2.$For convenience, $$\mathcal N_i:=T_i^\perp$$. Given such a discrete normal vector field $$N$$ we automatically obtain a second normal vector field from it:$B := T\times N.$The triple $$\sigma= (T,N,B)$$ then yields at each vertex an oriented orthonormal basis of $$\mathbb R^3$$, where the first vector aligns with the tangent direction. This kind of object are well-known in differential geometry and in analogy to the smooth case we will call it a discrete frame. Like in the smooth case,  there are distinguished frames among all the possible ones. A quite old-fashioned but still famous notion is the Frenet frame (see e.g. this wikipedia entry). We will instead work with parallel frames here:

The discrete parallel transport is map that assigns to each point $$i$$ a rotation $$P_i\in \mathrm{SO}(3)$$ that describes how the normal spaces $$\mathcal N_{i-1}$$ and $$\mathcal N_i$$ are connected: If $$T_{i-1}\times T_i = 0$$  we set $$P_i=\mathrm{id}$$, else $$P_i$$ is the unique rotation around $$T_{i-1}\times T_i$$ that maps $$T_{i-1}$$ to $$T_i$$, i.e.$P_i(T_{i-1}) = T_i, \quad P_i(T_{i-1}\times T_i) = T_{i-1}\times T_i.$Now, let $$N$$ be a discrete normal vector field of the discrete curve $$\gamma$$. We say $$N$$ is parallel, if $N_i = P_i(N_{i-1}), \quad i =1,…,n-2.$A parallel discrete frame is then a discrete frame $$(T,N,B)$$, where $$N$$ (and consequently $$B$$) is a parallel discrete normal vector field.

In Houdini rotations are stored as quaternions. So it is convenient to describe the parallel transport in quaternions, i.e. we want to find $$q_i\in \mathbb H$$, $$\Vert q_i\Vert=1$$ such that $P_i(X)= q_i\,X\,\overline{ q_i}, \quad X \in \mathcal N_{i-1}.$Here we identified $$\mathbb R^3$$ with the imaginary quaternions (compare with Albert’s post):$\mathbb R^3 \ni (x,y,y) \longleftrightarrow x\,\mathbf i + y\,\mathbf j + z\,\mathbf k \in\mathrm{Im}(\mathbb H).$Now let us find a formula for $$q_i$$. First, recall that if $$\mathbf v\in \mathbb R^3$$, $$\Vert \mathbf v\Vert=1$$, then the rotation around $$\mathbf v$$ by the angle $$\alpha \in\mathbb R$$ is given by the quaternion$\cos\bigl(\frac{\alpha}{2}\bigr) + \sin\bigl(\frac{\alpha}{2}\bigr)\mathbf v .$

The first condition on $$P_i$$ tell us that $$q_i = a_i + \mathbf v_i$$ where $$\mathbf v_i \perp T_{i-1},T_i$$. In particular, $$T_{i-1}\overline{q}_i = q_i\, T_{i-1}$$. The second condition then yields that $T_i = q_i\, T_{i-1}\, \overline{q}_i =q_i^2\, T_{i-1}$and hence$\bigl(a_i^2 – \Vert\mathbf v_i\Vert^2\bigr) +2a_i \mathbf v_i = q_i^2 = -T_i T_{i-1} = \langle T_i,T_{i-1}\rangle – \mathrm{Im}\bigl(T_iT_{i-1}\bigr).$Here we used that $$T_{i-1}^{-1}= -T_{i-1}$$. Further, since $$\Vert q_i\Vert =1$$, we obtain $$1 – a_i^2= \Vert \mathbf v_i\Vert$$, which we can substitute into the equation above. Comparing real and imaginary part then yields two equations$a_i^2 =\frac{1+\langle T_i,T_{i-1}\rangle}{2}, \quad \mathbf v_i = -\frac{1}{2a_i}\mathrm{Im}\bigl(T_iT_{i-1}\bigr).$

Note that the equation above determines $$q_i$$ only up to sign. Therefore we can assume that $$a_i\geq 0$$. Hence we end up with the following formula for $$a_i$$ and $$\mathbf v_i$$:$a_i = \sqrt{\frac{1+\langle T_i,T_{i-1}\rangle}{2}},\quad \mathbf v_i = \frac{-1}{\sqrt{2+2\langle T_i,T_{i-1}\rangle}}\mathrm{Im}\bigl(T_iT_{i-1}\bigr).$

Now given the discrete parallel transports we can easily construct a parallel normal field $$N$$ as follows: Choose some vector $$N_0$$ normal to the first edge. Then propagate this vector to the whole curve using the transport,$N_i := P_i\circ\cdots\circ P_1(N_0) = \underbrace{q_i\cdots q_1}_{=:\mathfrak q_{0,i}} \,N_0\, \overline{q}_1\cdots \overline{q}_i =\mathfrak q_{0,i}\, N_0 \, \overline{\mathfrak q_{0,i}}.$

Actually that is what is used by Albert to orient the chain links: The chain link is a fixed geometry in its own coordinate system. The copy node makes for each point a copy which is then translated to the current point’s position and stretched by the attribute @pscale. If the node finds an attribute named @orient (a quaternion store as vector4), it additionally performs the corresponding rotation. Hence if the chain was aligned with the $$x$$-axis (i.e. with the quaternion $$\mathbf i$$) we need to specify quaternions that rotate the vector $$\mathbf i$$ to the corresponding tangent vectors $$T$$. Therefore we can choose any quaternion, say $$\psi_0$$, to get the rotation from $$\mathbf i$$ to the first edge $$T_0$$,$T_0 = \psi_o\mathbf i\,\overline{\psi_o},$before we use the parallel transports $$q_i$$ to get the other rotations:$\psi_i := q_i\cdots q_1\psi_0.$

Homework (due 10/12 May): Rebuild the chain rendering presented in the previous post. Write two networks: one using the closed arclength-parametrized curves from Ulrich’s post, and a second one that uses the curve node to build up a generic closed curve as input. (Use the resample node to make it arclength-parametrized).

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