multivariable-calculusparametrictrigonometry
I would like to model a braided hose like shown here: https://grabcad.com/tutorials/tutorial-how-to-model-a-braided-hose-in-solidworks
I found these equations for sine along a helix: Parametric Equation of sine wave helically wrapped around a cylinder
I was wondering how the equations provided here would change if the sine wave has to follow a helical path, but in my case the wave is perpendicular to the cylinder surface instead of being in the axial direction.
Really appreciate the help.
Try the following, where $t$ represents the main parameter, $\theta$ is the tilting angle as the sine wave wraps around the cylinder, $k$ is the frequency of the sine wave, and $a$ is the amplitude:
$$\begin{align*} x(t) &= \cos (t \cos \theta - a \sin \theta \sin (k t)) \\ y(t) &= \sin (t \cos \theta - a \sin \theta \sin (k t)), \\ z(t) &= t \sin \theta + a \cos \theta \sin (k t). \end{align*}$$
Fancy animated picture! 
Given the functions $\overline{f},f : [0,T] \to \mathbb{R}$ of law:
$$ \overline{f}(t) = a\sin(\overline{\omega}\,t+\phi), \quad \quad \quad f(t) = a\sin(\omega\,t+\phi) $$
and given the curves of parametric equations:
$$ \begin{aligned} & \mathbf{r}_1(t) = \left(t,\;\overline{f}(t),\;0\right); \\ & \mathbf{r}_2(t) = \left(r\cos t + f(t)\sin\theta\sin t,\;r\sin t - f(t)\sin\theta\cos t,\;r\,t\tan\theta + f(t)\cos\theta\,\right); \\ \end{aligned} $$
the local maximum points are located by:
$$ t^* = \frac{\pi/2+2\,n\,\pi-\phi}{\omega^*}, \quad \quad \quad n \in \mathbb{N}\,. $$
So, all that remains is to impose the equality of the radii of curvature at such points:
$$ \rho_1(t^*) = \rho_2(t^*) \quad \quad \Leftrightarrow \quad \quad \frac{||\mathbf{r}_1'(t^*)||^3}{||\mathbf{r}_1'(t^*) \times \mathbf{r}_1''(t^*)||} = \frac{||\mathbf{r}_2'(t^*)||^3}{||\mathbf{r}_2'(t^*) \times \mathbf{r}_2''(t^*)||} $$
that is, by performing the calculations:
$$ \frac{1}{a\,\overline{\omega}^2} = \frac{a^2\left(1-\cos(4\theta)\right)+8\,r^2}{(2\cos\theta)^2\sqrt{a^2\left(4\,\omega^4+4\,\omega^2+2\right)-2\,a^2\left(2\,\omega^2+1\right)\cos(2\theta)+4\,r^2}} $$
from which:
$$ \boxed{\overline{\omega} = \frac{2\cos\theta\,\sqrt[4]{a^2\left(4\,\omega^4+4\,\omega^2+2\right)-2\,a^2\left(2\,\omega^2+1\right)\cos(2\theta)+4\,r^2}}{\sqrt{a}\,\sqrt{a^2\left(1-\cos(4\theta)\right)+8\,r^2}}} $$
or:
$$ \boxed{\omega = \sqrt{\frac{\sqrt{\left(a^2\left(1-\cos(4\theta)\right)+8\,r^2\right)\left(a^2\,\overline{\omega}^4\left(a^2\left(1-\cos(4\theta)\right)+8\,r^2\right)-8\,(\cos\theta)^4\right)}}{8\,a\,(\cos\theta)^2}-(\sin\theta)^2}}\,. $$
Note: to perform this comparison in Mathematica use AspectRatio -> Automatic.
Best Answer
The projection onto horizontal plane $z=0$ of the 3D curve you want to obtain can be given the following polar representation :
$$r(\theta)=R+r \cos(k \theta) \ \iff \ \begin{cases}x(\theta)&=&(R+r\cos(k \theta))\cos \theta\\y(\theta)&=&(R+r\cos(k \theta))\sin \theta\end{cases}\tag{1}$$
(Consider different values $k=5,10,20...$).
Retrieving the 3D curve out of this 2D curve if obtained through a natural "lifting", i.e., by adding to the 2 equations in (1) the third equation:
$$z(\theta)=a \theta$$
for a certain constant $a$.
Remark: a shift parameter $\phi$ can be added to the equations in (1) by taking $r(\theta)=(R+r \cos(k(\theta+ \phi)))$.