# [SOLVED]Curvature and torsion on a helix

#### dwsmith

##### Well-known member
Consider the case of a right circular helical curve with parameterization $$x(t) = R\cos(\omega t)$$, $$y(t) = R\sin(\omega t)$$, and $$z(t) = v_0t$$. Find the curvature and torsion curve.

We can then parameterize the helix
\begin{align*}
x(t) &= R\cos(\omega t)\\
y(t) &= R\sin(\omega t)\\
z(t) &= v_0t
\end{align*}
We have that
\begin{align*}
\frac{d\mathbf{r}}{dt} &= v\hat{\mathbf{u}}\\
\hat{\mathbf{u}} &= \frac{1}{v}\frac{d\mathbf{r}}{dt}\\
v\hat{\mathbf{u}} &= \frac{d\mathbf{r}}{dt}\\
\lvert v\hat{\mathbf{u}}\rvert &= \left\lvert\frac{d\mathbf{r}}{dt}\right\rvert\\
v &= \left\lvert\frac{d\mathbf{r}}{dt}\right\rvert
\end{align*}
From our parameterization, we have that $$\mathbf{r}(t) = R\cos(\omega t)\hat{\mathbf{i}} + R\sin(\omega t)\hat{\mathbf{j}} + v_0t\hat{\mathbf{k}}$$.
Therefore,
\begin{align*}
\frac{d\mathbf{r}}{dt} &= -R\omega\sin(\omega t)\hat{\mathbf{i}} +
R\omega\cos(\omega t)\hat{\mathbf{j}} + v_0\hat{\mathbf{k}}\\
\left\lvert\frac{d\mathbf{r}}{dt}\right\rvert
&= \sqrt{R^2\omega^2 + v_0^2}\\
v &= \sqrt{R^2\omega^2 + v_0^2}
\end{align*}
So our unit vector $$\hat{\mathbf{u}}$$ can be written as
\begin{align*}
\hat{\mathbf{u}} &= \frac{1}{\sqrt{R^2\omega^2 + v_0^2}}
\langle -R\omega\sin(\omega t), R\omega\cos(\omega t), v_0\rangle.
\end{align*}
Since $$\frac{d\hat{\mathbf{u}}}{ds} = \frac{1}{\rho}\hat{\mathbf{n}}$$,
$$\left\lvert\frac{d\hat{\mathbf{u}}}{ds} \right\rvert = \frac{1}{\rho}$$.
Using the fact that $$\frac{d\hat{\mathbf{u}}}{ds} = \frac{1}{v}\frac{d\hat{\mathbf{u}}}{dt}$$, we can now write
\begin{align*}
\frac{1}{\rho} &= \left\lvert\frac{1}{v}\frac{d\hat{\mathbf{u}}}{dt}\right\rvert
\end{align*}
Let's take the time derivative of $$\hat{\mathbf{u}}$$.
\begin{alignat*}{2}
\frac{d\hat{\mathbf{u}}}{dt} &= \frac{R\omega^2}{\sqrt{R^2\omega^2 + v_0^2}}\langle
-\cos(\omega t), -\sin(\omega t), 0\rangle\\
\frac{1}{\rho} &= \frac{R\omega^2}{R^2\omega^2 + v_0^2} &&
\left(\text{curvature}\right)
\end{alignat*}
Now, let's look at $$\frac{d\hat{\mathbf{b}}}{ds} = -\frac{1}{\tau}\hat{\mathbf{n}}$$.
Then $$\left\lvert \frac{d\hat{\mathbf{b}}}{ds} \right\rvert = \frac{1}{\tau}$$.

I am not sure what I can say about $$\frac{d\hat{\mathbf{b}}}{ds}$$

Last edited:

#### Ackbach

##### Indicium Physicus
Staff member
Well, you can write that
$$\tau=-\mathbf{n} \cdot \frac{d \mathbf{b}}{ds},$$
where $\mathbf{b} := \mathbf{t} \times \mathbf{n}$. So here, I'm using $\mathbf{n}$ as the principal normal vector, $\mathbf{t}$ as the unit tangent vector, and $\mathbf{b}$ as the binormal vector. Can you compute $\mathbf{t}, \mathbf{n},$ and $\mathbf{b}$ in terms of arc length?