Curvature and torsion on a helix

In summary, this conversation discusses the case of a right circular helical curve with a given parameterization and finding the curvature and torsion curve for it. The curvature is given by $\frac{R\omega^2}{R^2\omega^2 + v_0^2}$ and the torsion can be computed using the principal normal vector and the unit tangent and binormal vectors in terms of arc length.
  • #1
Dustinsfl
2,281
5
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.
http://img30.imageshack.us/img30/7828/gwi.png

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:
Mathematics news on Phys.org
  • #2
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?
 

FAQ: Curvature and torsion on a helix

What is the difference between curvature and torsion on a helix?

Curvature refers to the amount of bending or change in direction of a curve, while torsion refers to the amount of twisting or rotation of a curve. On a helix, the curvature is constant, while the torsion increases as the helix winds tighter.

How are curvature and torsion calculated on a helix?

The curvature of a helix can be calculated by dividing the distance between two points on the helix by the angle between the tangents at those points. Torsion can be calculated by dividing the change in direction of the tangent vector by the change in arc length.

How do curvature and torsion affect the shape of a helix?

The curvature of a helix determines its overall shape, while the torsion affects the tightness of the helix and how much it twists. Together, they determine the 3D shape of the helix.

Are curvature and torsion important in real-life applications?

Yes, curvature and torsion play important roles in various fields such as physics, engineering, and biology. They are used to describe the shape and movement of objects, as well as the properties of materials.

Can curvature and torsion be negative on a helix?

Yes, curvature and torsion can have negative values on a helix. Negative curvature would result in a helix with a concave shape, while negative torsion would cause the helix to twist in the opposite direction.

Similar threads

Replies
1
Views
1K
Replies
2
Views
2K
Replies
2
Views
882
Replies
2
Views
1K
Replies
1
Views
386
Replies
10
Views
805
Back
Top