I know for the helix, the equation can be written:
$$x=R\cos(t)$$
$$y=R\sin(t)$$
$$z=ht$$
this is the helix curve, and there are two parameters: outer radius $R$ and the pitch length $2\pi h$. However, I would like to generate the 3D helix with another minor radius $r$. This is not the helix curve, but a 3D object something like spring. I don't know exactly the name of such structure, but when I search helix equation, they usually give the equations for helix curve but not for the 3D helix object (spring). Does anyone know the equation of such object? Thank you so much for any help and suggestions.
ps. the shape looks like the following way:
Best Answer
We can use a local orthonormal basis of a parametrized curve to get a surface of the desired type.
A helix running around the $x$-axis has a parametrization like $$ \vec{r}(t)=(ht,R\cos t, R\sin t). $$ Its tangent vector can be gotten by differentiating $$ \vec{t}=\frac{d\vec{r}(t)}{dt}=(h,-R\sin t,R\cos t). $$ We note that this has constant length $\sqrt{h^2+R^2}$. With a more general curve this is not necessarily the case, and we would normalize this to unit length, and switch to using the natural parameter $s=$ the arc length. This time $ds/dt=\sqrt{h^2+R^2}$, and we can keep using $t$ as long as we remember to normalize.
We get a (local) normal $\vec{n}(t)$ vector by differentiating the (normalized) tangent $$ \vec{n}(t)= \frac{\frac{d\vec{t}}{dt}}{\left\Vert\frac{d\vec{t}}{dt}\right\Vert}=(0,-\cos t,-\sin t). $$ As the name suggests this is orthogonal to the tangent vector (in the direction of change of the tangent). The third basis vector is the binormal $$ \vec{b}(t)=\frac1{\Vert\vec{t}\Vert}\vec{t}\times\vec{n}=\frac{1}{\sqrt{R^2+h^2}}(R,h\sin t,-h\cos t). $$ This is, of course, orthogonal to both $\vec{t}$ and $\vec{n}$.
The key is that we get the desired surface by drawing (3D-)circles with axis direction determined by the direction of the curve, i.e. the tangent. Equivalently, we draw a circle of radius $a$ in the plane spanned by $\vec{n}$ and $\vec{b}$. Hence we get the entire surface $S$ parametrized as $$ S(t,u)=\vec{r}(t)+a\vec{n}(t)\cos u+ a\vec{b}(t)\sin u $$ with $t$ ranging over as many loops as you wish, and $u$ ranging over the interval $[0,2\pi]$. In terms of individual coordinates this reads (barring mistakes due to lack of coffee): $$ \begin{aligned} x(t,u)&=ht+\frac{Ra\sin u}{\sqrt{R^2+h^2}},\\ y(t,u)&=R\cos t-a\cos t\cos u+\frac{ha\sin t\sin u}{\sqrt{R^2+h^2}},\\ z(t,u)&=R\sin t-a\sin t\cos u-\frac{ha\cos t\sin u}{\sqrt{R^2+h^2}}. \end{aligned} $$
Here's what Mathematica-output looks like with parameters $h=1$, $R=3$, $a=0.4$:
Here's the effect of the change to $a=1.0$. The lines on the surface correspond to constant values of $t$ and $u$. These are now more clearly defined.