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[SOLVED] Curvature and torsion on a helix

dwsmith

Well-known member
Feb 1, 2012
1,673
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
Jan 26, 2012
4,192
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?