Helix
Suppose the trajectory of a moving particle can be described by the vector valued function \[
\mathbf{r} = \begin{bmatrix}
b t \\
a \cos{\left(t \right)} \\
a \sin{\left(t \right)} \\
\end{bmatrix}
\] where \(a\) and \(b\) are positive constants. The curve that is the image of this function is called a helix (the plural form of helix is helices). A helix is a spiral-like curve, where a curve moves around a central axis at a constant distance. It can be found in nature, such as in DNA structures, or in artificial objects like a coiled spring. The following figure shows two helices with different choices for \(a\) and \(b\).
Here, the parameter \(a\) controls the distance of the points to its central axis, and \(b\) controls the rate at which it moves upward (in the \(\mathbf{k}\) direction).
The TNB frame
The TNB frame, also known as the Frenet-Serret frame, consists of three unit vectors that are orthogonal to one another:
- the unit tangent vector \(\mathbf{T}\)
- the principal unit normal vector \(\mathbf{N}\)
- the (unit) binormal vector \(\mathbf{B}\)
Together, these vectors form a natural coordinate system (indeed a orthonormal basis spanning the 3-dimensional space) that is particularly convenient for describing the motion of a point-like moving object.
Compute the unit tangent vector \(\mathbf{T}\).
The unit tangent vector is given by \[
\mathbf{T} =
\frac{
\frac{d\mathbf{r}}{dt}
}{
\left\| \frac{d\mathbf{r}}{dt} \right\|
}
=
\begin{bmatrix}
\frac{b}{\sqrt{a^{2} + b^{2}}} \\
- \frac{a \sin{\left(t \right)}}{\sqrt{a^{2} + b^{2}}} \\
\frac{a \cos{\left(t \right)}}{\sqrt{a^{2} + b^{2}}} \\
\end{bmatrix}
\] This vector gives us the direction in which this particle moving.
Compute the principal unit normal vector \(\mathbf{N}\).
The principal unit normal vector is given by \[
\mathbf{N} =
\frac{
\frac{d\mathbf{T}}{dt}
}{
\| \frac{d\mathbf{T}}{dt} \|
}
=
\begin{bmatrix}
0 \\
- \cos{\left(t \right)} \\
- \sin{\left(t \right)} \\
\end{bmatrix}
\]
Compute the binormal vector \(\mathbf{B}\).
The binormal vector is simply the cross product between \(\mathbf{T}\) and \(\mathbf{N}\), i.e., \[
\mathbf{B} = \mathbf{T} \times \mathbf{N} =
\begin{bmatrix} \frac{b}{\sqrt{a^{2} + b^{2}}} \\ - \frac{a \sin{\left(t \right)}}{\sqrt{a^{2} + b^{2}}} \\ \frac{a \cos{\left(t \right)}}{\sqrt{a^{2} + b^{2}}}\end{bmatrix} \times
\begin{bmatrix} 0 \\ - \cos{\left(t \right)} \\ - \sin{\left(t \right)}\end{bmatrix}
=
\begin{bmatrix} \frac{a}{\sqrt{a^{2} + b^{2}}} \\ \frac{b \sin{\left(t \right)}}{\sqrt{a^{2} + b^{2}}} \\ - \frac{b \cos{\left(t \right)}}{\sqrt{a^{2} + b^{2}}}\end{bmatrix}
\]
Arc-length and arc-length parametrization
Compute the arc-length function \(s(t) = \int_0^t \| \mathbf{r}'(u) \| du\).
\[
\begin{aligned}
s(t) &=
\int_0^t \left\| \frac{d\mathbf{r}}{dt}(u) \right\| du =
\int_0^t \left\|
\begin{bmatrix} b \\ - a \sin{\left(u \right)} \\ a \cos{\left(u \right)}\end{bmatrix}
\right\| du
\\ &=
\int_0^t \left\|
\sqrt{a^{2} + b^{2}}
\right\| du
\\ &=
t \sqrt{a^{2} + b^{2}}
\end{aligned}
\]
Find an arc-length parametrization for \(\mathbf{r}\).
In the above, we have computed \[
s = t \sqrt{a^{2} + b^{2}}
\] Solve if this equation for \(t\) and plug in the result into the original definition of \(\mathbf{r}\), we get the arc-length parametrization \[
\mathbf{r}(s) =
\begin{bmatrix} \frac{b s}{\sqrt{a^{2} + b^{2}}} \\ a \cos{\left(\frac{s}{\sqrt{a^{2} + b^{2}}} \right)} \\ a \sin{\left(\frac{s}{\sqrt{a^{2} + b^{2}}} \right)}\end{bmatrix}
\]
Curvature and Torsion
Curvature and torsion are key characteristics that define a curve in three-dimensional space. Curvature quantifies the amount by which a curve bends at any given point. On the other hand, torsion measures the rate at which a curve is twisting out of the osculating plane.
Compute the curvature of this curve.
Using the arc-lengh parametrization, we can compute that \[
\mathbf{T}(s) =
\begin{bmatrix} \frac{b}{\sqrt{a^{2} + b^{2}}} \\ - \frac{a \sin{\left(\frac{s}{\sqrt{a^{2} + b^{2}}} \right)}}{\sqrt{a^{2} + b^{2}}} \\ \frac{a \cos{\left(\frac{s}{\sqrt{a^{2} + b^{2}}} \right)}}{\sqrt{a^{2} + b^{2}}}\end{bmatrix}
\] Therefore, the curvature can be computed as \[
\kappa =
\left\| \frac{d\mathbf{T}}{ds} \right\| =
\left\|
\begin{bmatrix} 0 \\ - \frac{a \cos{\left(\frac{s}{\sqrt{a^{2} + b^{2}}} \right)}}{a^{2} + b^{2}} \\ - \frac{a \sin{\left(\frac{s}{\sqrt{a^{2} + b^{2}}} \right)}}{a^{2} + b^{2}}\end{bmatrix}
\right\|
=
\frac{a}{a^{2} + b^{2}}
\] Notice that the curvature is a constant that is only dependent and \(a\) and \(b\) (but remain the same value anywhere on the helix). It is important to note that this answer assumes \(a > 0\). Otherwise we have to replace \(a\) in the numerator by \(|a|\).
Compute the torsion of this curve.
Again, using the arc-length parametrization, we can compute the binormal and the principle unit normal vectors as \[
\mathbf{N}(s) =
\begin{bmatrix} 0 \\ - \cos{\left(\frac{s}{\sqrt{a^{2} + b^{2}}} \right)} \\ - \sin{\left(\frac{s}{\sqrt{a^{2} + b^{2}}} \right)}\end{bmatrix}
\quad\text{and}\quad
\mathbf{B}(s) =
\begin{bmatrix} \frac{a}{\sqrt{a^{2} + b^{2}}} \\ \frac{b \sin{\left(\frac{s}{\sqrt{a^{2} + b^{2}}} \right)}}{\sqrt{a^{2} + b^{2}}} \\ - \frac{b \cos{\left(\frac{s}{\sqrt{a^{2} + b^{2}}} \right)}}{\sqrt{a^{2} + b^{2}}}\end{bmatrix}
\quad\text{and}\quad
\frac{d\mathbf{B}}{ds} =
\begin{bmatrix} 0 \\ \frac{b \cos{\left(\frac{s}{\sqrt{a^{2} + b^{2}}} \right)}}{a^{2} + b^{2}} \\ \frac{b \sin{\left(\frac{s}{\sqrt{a^{2} + b^{2}}} \right)}}{a^{2} + b^{2}}\end{bmatrix}
\] Therefore, the torsion can be computed as \[
\tau = -\frac{d\mathbf{B}}{ds} \cdot \mathbf{N}(s) =
\frac{b}{a^{2} + b^{2}}
\] Notice that the torsion is also a constant that is only dependent and \(a\) and \(b\) (but remain the same value anywhere on the helix).
Decomposition of acceleration
Let \(\mathbf{a} = \frac{d^2 \mathbf{r}}{dt^2}\) be the acceleration vector. It can be decomposed into its tangential and normal components. That is, \[
\mathbf{a} = a_T \mathbf{T} + a_N \mathbf{N}
\] for some scalars \(a_T\) (describing the tangential component) and \(a_N\) (describing the normal component). They are not to be confused with the parameter \(a\) we used to describe the shape of the helix.
Compute \(a_T\) and \(a_N\).
First, we can compute \[
\mathbf{a} =
\frac{d^2 \mathbf{r}}{dt^2} =
\frac{d}{dt} \left(
\frac{d}{dt} \begin{bmatrix} b t \\ a \cos{\left(t \right)} \\ a \sin{\left(t \right)}\end{bmatrix}
\right)
=
\frac{d}{dt} \begin{bmatrix} b \\ - a \sin{\left(t \right)} \\ a \cos{\left(t \right)}\end{bmatrix}
=
\begin{bmatrix} 0 \\ - a \cos{\left(t \right)} \\ - a \sin{\left(t \right)}\end{bmatrix}
\] The tangent component \(a_T\) is simply the project of the acceleration vector \(\mathbf{a}\) in the direction of \(\mathbf{T}\). That is, \[
a_T = \mathbf{a} \cdot \mathbf{T} =
\begin{bmatrix} 0 \\ - a \cos{\left(t \right)} \\ - a \sin{\left(t \right)}\end{bmatrix}
\;\cdot\;
\begin{bmatrix} \frac{b}{\sqrt{a^{2} + b^{2}}} \\ - \frac{a \sin{\left(t \right)}}{\sqrt{a^{2} + b^{2}}} \\ \frac{a \cos{\left(t \right)}}{\sqrt{a^{2} + b^{2}}}\end{bmatrix}
=
0
\] This answer agrees with our intuition since we know that the scalar speed remains constant.
Similarly, \(a_N\) is the projection of \(\mathbf{a}\) in the direction of \(\mathbf{N}\). That is,
\[
a_N = \mathbf{a} \cdot \mathbf{N} =
\begin{bmatrix} 0 \\ - a \cos{\left(t \right)} \\ - a \sin{\left(t \right)}\end{bmatrix}
\;\cdot\;
\begin{bmatrix} 0 \\ - \cos{\left(t \right)} \\ - \sin{\left(t \right)}\end{bmatrix}
=
a
\]
This give us the magnitude of the normal component of the acceleration, which is responsibie for turning.
Recall that scalar speed (i.e., \(\| \mathbf{r}'(t) \|\), which is equivalent to \(\frac{ds}{dt}\)) is given by \[
\frac{ds}{dt} = \frac{d}{dt} s =
\frac{d}{dt} \left(
t \sqrt{a^{2} + b^{2}}
\right)
=
\sqrt{a^{2} + b^{2}}.
\] The magnitude of the tangential component \(a_T\) is also given by the derivative of this scalar speed. Therefore, \[
a_T = \frac{d}{dt} \left( \sqrt{a^{2} + b^{2}} \right) =
0
\]
The normal component \(a_N\) can becomputed using the scalar speed and the curvature: \[
a_N = \kappa \, \left( \frac{ds}{dt} \right)^2
=
\frac{a}{a^{2} + b^{2}}
\left( \sqrt{a^{2} + b^{2}} \right)^2
=
a
\]
Actually, once we know \(a_T = 0\), we can recognize that
\[
\mathbf{a} = 0 \, \mathbf{T} + a_N \, \mathbf{N} = a_N \, \mathbf{N}
\]
Therefore, \(a_N\) must be \(\| \mathbf{a} \|\), i.e.,
\[
a_N = \| \mathbf{a} \| =
\left\|
\begin{bmatrix} 0 \\ - a \cos{\left(t \right)} \\ - a \sin{\left(t \right)}\end{bmatrix}
\right\|
=
a
\]
This agrees with answers obtained through the other two methods.
