Two parallel, same circles, and many rods linked one to another. The angle between every rod and the below circle surface is the same. How to prove the envelope of this thing a hyperboloid?
I don't have whether math or physical method to solve this problem.
On
One method is to set up the parametric equations for each line (you know the starting and ending points) and then do algebra to get it into the form of the equation of a hyperboloid.
Another is to start with a circle and a slanted line intersecting it, and prove that the surface swept out is a hyperboloid.
Conversely, one can take the equation of the hyperboloid and factor it, seeing that the left and right hand side each turn into the equation of a plane, concluding that if they intersect they have to do it along a line, and hence the surface is ruled by lines.
If one is familiar with fancier math one can perhaps do it backwards by starting out by asking which surfaces have two sets of ruling lines (doubly ruled surfaces) and then finding that it is just the hyperboloid and hyperbolic paraboloid.
From a physics standpoint it is less clear how to show it.
You can approach this problem as the rotation of a tilted rod around the axis determined by the centers of the two parallel circles on the picture. Indeed, Let $O_1$ and $O_2$ be the centers of the two circles $c_1$ and $c_2$. Denote by $O$ the midpoint of the segment $O_1O_2$. Fix an orthonormal coordinate system $O \, \vec{e}_1 \, \vec{e}_2 \, \vec{e}_3 $, where $\vec{e}_3$ is aligned with $O_1O_2$ and the plane spanned by $O \, \vec{e}_1 \, \vec{e}_2 $ is parallel to the planes of the two circles and is at equal distance from each of them. Then, if we assume that the circles have radii $R$, their equations in the coordinate system $O \, \vec{e}_1 \, \vec{e}_2 \, \vec{e}_3 $ can be written as $$c_1 \, : \,\,\,\, x^2 + y^2 = R^2, \,\,\, z = - \, z_0$$ $$c_2 \, : \,\,\,\, x^2 + y^2 = R^2, \,\,\, z = z_0$$ where $$z_0 \, =\, \frac{|O_1O_2|}{2}$$ Let the angle between the rod and the horizontal plane $z = - \, z_0$ of circle $c_1$ is $\alpha$. As stated in the problem, $\alpha$ is constant, no matter which rod we take. However, by construction, the vertical axis $O \vec{e}_3$ is perpendicular to $z = -\, z_0$. This means that the angle between the rod and $O \vec{e}_3$ is $\frac{\pi}{2} - \alpha$ and is therefore also constant.
In the given coordinate frame $O \, \vec{e}_1 \, \vec{e}_2 \, \vec{e}_3 $ the plane spanned by $O \, \vec{e}_1 \, \vec{e}_2 $ is in fact the plane given by the equation $z = 0$, i.e. the horizontal coordinate plane. Then, $z = 0$ intersects each rod in its midpoint, because the circles $c_1$ and $c_2$ on which the endpoints of the rod rest are at equal distance up and down from $z = 0$. Let us take an arbitrary rod from the picture and let $M$ be its midpoint. Let the distance between $O$ and $M$ be $r$, i.e. $|OM| = r$. Then the position vector of the point $M$ can be written as $$\vec{OM} \, =\, r\, \cos(\theta)\,\vec{e}_1 \, + \, r\, \sin(\theta) \, \vec{e}_2$$ where $\theta$ is the counter-clock-wise angle measured from the $O\vec{e}_1$ axis to the position vector $\vec{OM}$. Moreover, since the angle between the rod and the plane $z = -z_0$ is constant $\alpha$, then the angle between the rod and the plane $z = 0$, parallel to $z = -z_0$, is also $\alpha$.
For all of the above reasons, the family of rods on the picture can be described as one rod, rotated around $O \vec{e}_3$ uniformly at an angle $\theta$. Then, the midpoint $M$ of the rod traverses a horizontal circle $c_0$ of some radius $r$ centerd at the origin $O$, i.e. this is the smaller middle circle drawn on the picture. Its equations are $$c_0 : \,\,\,\,\, x^2 \, + \, y^2 \, = \, r^2, \,\,\,\,\, z= 0$$ or in parametrized form \begin{align} &x = r \, \cos(\theta)\\ &y = r \, \sin(\theta)\\ &z = 0 \end{align} Let us introduce an orthonormal frame $M \, \vec{E}_1 \, \vec{E}_2 \, \vec{E}_3$ attached to the point $M$ and adapted to the circle $c_0$ as follows:
the unit tangent vector to the circle $c_0\,$ : $$ \,\,\,\, \vec{E}_1 \, =\, -\, \sin(\theta) \, \vec{e}_1 \, + \, \cos(\theta) \, \vec{e}_2$$
the unit vector perpendicular to the circle $c_0$, in the circle's plane $z = 0$: $$\,\,\,\, \vec{E}_2 \, =\, \cos(\theta) \, \vec{e}_1 \, + \, \sin(\theta) \, \vec{e}_2 \, = \, \frac{1}{r}\, \vec{OM}$$ the unit vector perpendicular to the circle's plane $z = 0\,$: $$\,\,\,\, \vec{E}_3 \, =\, \vec{e}_3 $$
Now, observe that by construction, as the parameter $\theta$ changes uniformly, the frame $M \, \vec{E}_1 \, \vec{E}_2 \, \vec{E}_3$ rotates uniformly around the vertical direction $O\,\vec{e}_3$ with respect to the fixed frame $O \, \vec{e}_1 \, \vec{e}_2 \, \vec{e}_3$ . On top of this, the origin $M$ of the rotating frame $M \, \vec{E}_1 \, \vec{E}_2 \, \vec{E}_3$ is shifted by $$\vec{OM} \, =\, r\,\cos(\theta) \, \vec{e}_1 \, + \, r\,\sin(\theta) \, \vec{e}_2$$ Since the distances between the origin $O$ and the endpoints of the rod are both equal to the same number $\sqrt{R^2 + z_0^2}$ implies that $O$ is equidistant from the endpoints of the rod and therefore lies on the orthogonal bisector plane of the rod. $M$ also lies on that plane. Therefore, the segment $OM$ is an orthogonal bisector of the rod and in particular $OM$ is perpendicular to the rod. Therefore, the orthogonal projection of the rod on $\vec{OM}$ is zero. But $\vec{OM} = r \, \vec{E}_2$, which means that the orthogonal projection of the rod on $\vec{E}_2$ is zero.
The fact that at each position of the midpoint $M$, the rod's orthogonal projection in the radial direction $\vec{OM} = r \, \vec{E}_3$ is always zero and that the rod forms a constant angle $\frac{\pi}{2} - \alpha$ with the vertical vector $\vec{E}_3 = \vec{e}_3$ implies that the angle between the rod and the vector $\vec{E}_1$ tangent to the circle $c_0$ at $M$ is $\alpha$. Consequently, the unit vector aligned with the rod, pointing from $M$ to the upper circle $c_2$, can be decomposed as $$\cos(\alpha) \, \vec{E}_1 \, + \, \sin(\alpha) \, \vec{E}_3$$ in the rotating coordinate frame $M \, \vec{E}_1 \, \vec{E}_2 \, \vec{E}_3$ and hence, every point $P$ on the rod can be parametrized as $$P \, =\, t\,\cos(\alpha) \, \vec{E}_1 \, + \,t \, \sin(\alpha) \, \vec{E}_3$$
Now, the transformation from the rotating $M \, \vec{E}_1 \, \vec{E}_2 \, \vec{E}_3$ coordinates to the fixed $O \, \vec{e}_1 \, \vec{e}_2 \, \vec{e}_3$ coordinates is \begin{align} \begin{bmatrix}x \\ y \\ z \end{bmatrix} \, =\, \begin{bmatrix}r\,\cos(\theta) \\ r\,\sin(\theta) \\ 0 \end{bmatrix} \, +\, \begin{bmatrix}-\,\sin(\theta) & \cos(\theta) & 0\\ \ \,\,\,\, \cos(\theta) & \sin(\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}X \\ Y \\ Z \end{bmatrix} \end{align} Specifically, for the arbitrary point $P$ on the rod with coordinates in the rotating frame $M \, \vec{E}_1 \, \vec{E}_2 \, \vec{E}_3$ \begin{align} &X \, =\, t \, \cos(\alpha)\\ &Y \, =\, 0\\ &Z \, =\, t \, \sin(\alpha)\\ \end{align} its coordinates in the fixed frame $O \, \vec{e}_1 \, \vec{e}_2 \, \vec{e}_3$ are \begin{align} \begin{bmatrix}x \\ y \\ z \end{bmatrix} \, =\, \begin{bmatrix}r\,\cos(\theta) \\ r\,\sin(\theta) \\ 0 \end{bmatrix} \, +\, \begin{bmatrix}-\,\sin(\theta) & \cos(\theta) & 0\\ \ \,\,\,\, \cos(\theta) & \sin(\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}t \, \cos(\alpha) \\ 0 \\ t \, \sin(\alpha) \end{bmatrix} \end{align} When you perform all the matrix operations, you get in terms of components \begin{align} &x \, =\, r \, \cos(\theta) \, - \, t\, \cos(\alpha)\sin(\theta) \\ &y \, =\, r \, \sin(\theta) \, + \, t\, \cos(\alpha)\cos(\theta) \\ &z \, =\, t \, \sin(\alpha)\\ \end{align} Now, calculate \begin{align} x^2 \, + \, y^2 \, =& \, \Big(\, r \, \cos(\theta) \, - \, t\, \cos(\alpha)\sin(\theta) \Big)^2 \, + \, \Big(\,r \, \sin(\theta) \, + \, t\, \cos(\alpha)\cos(\theta)\,\Big)^2\\ =& \, r^2 \cos^2(\theta) \, - \, 2\, r \, t\, \cos(\alpha)\cos(\theta)\sin(\theta) \,+ \, t^2 \cos^2(\alpha)\sin^2(\theta)\\ &+ \, r^2 \sin^2(\theta) \, + \, 2\, r \, t\, \cos(\alpha)\cos(\theta)\sin(\theta) \,+ \, t^2 \cos^2(\alpha)\cos^2(\theta)\\ =& \, r^2 \cos^2(\theta) \, + r^2 \sin^2(\theta) \, + \, t^2 \cos^2(\alpha)\sin^2(\theta) \, +\, t^2 \cos^2(\alpha)\cos^2(\theta)\\ =&\, r^2 \, +\, t^2 \cos^2(\alpha)\\ \end{align} From $z = t\,\sin(\alpha)$ express $$t \, =\, \frac{z}{\sin(\alpha)}$$ and plug in the previous equality \begin{align} x^2 \, + \, y^2 \, =& \, r^2 \, +\, t^2 \cos^2(\alpha)\\ =& \, r^2 \, + \, \left( \frac{z}{\sin(\alpha)}\right)^2\cos^2(\alpha)\\ \end{align} which is the same as $$x^2 \, + \, y^2 \, =\, r^2 \, + \,\tan^2(\alpha) z^2 $$ which can also be written as $$\frac{x^2}{\tan^2(\alpha)} \, +\, \frac{y^2}{\tan^2(\alpha)} \, - \, z^2 \, = \, \frac{r^2}{\tan^2{\alpha}}$$ which is the equation of a one sheeted rotational hyperboloid.