Defining the process of twisting a prism : Twisting the top face of a prism with no walls.

The prism can show these two behaviors while getting twisted:

  1. An ideal prism (side length not constant) will simply have it's face twisted with no other change,

    like this

  2. A real scenario where the side length is constant and hence there is a slight compression perpendicular to the top face,

    like this

For this post I am concerned about the second point i.e when the side length is constant.

Some more examples I constructed:

square prism

triangular prism

planer prism?!

I am providing the link of a google drive folder where I have uploaded the Geogebra files so you guys can experiment with them.

As I was constructing these, I noticed that the length all the figures were getting compressed was equal (the polygons had equal radii (circum-radii), and the side length was also equal). I did it only till pentagon $>1$.

  1. I hypothesize that it will be equal for every regular polygon given the radius and the side length are equal. Is my hypothesis correct? If yes how to prove it?

I noticed another thing—every $180^\circ$ rotation resulted in the first intersection for every polygon prism not depending on the radius/side length. I tried thinking a lot about it but wasn't able to visualize it.

  1. Why does the first intersection happens after rotating $180^\circ$?

My last but not the least question:

  1. How can we find the relation between the angle by which the top face gets twisted and the changing angle between the polygon side and the side length i.e.


In the process of construction, I found out the locus of the vertices : taking the example of a square prism the vertex $\text{B}_1$ follows : $$x=\sqrt{l^2 - (r\cos (\phi + \pi /2)-h)^2 - (r\sin (\phi + \pi /2)-k)^2}-m \\ y=r\cos(\phi +\pi /2) \\ z=r\sin(\phi + \pi /2) \\ \text{the prism is along x axis}\\ \text{ $(m,h,k)$ are the $x$-, $y$-, and $z$-coordinates of $\text{A}_1$ respectively} \\ \text{$\phi$ is the angle by which the top face is getting rotated.} \\ \text{ $r,l$ are the radius and length of the prism respectively.}$$ Note that I have added a '$+\pi /2$' in the angle to denote the initial coordinate of the vertex.

  • $\begingroup$ For clarity's sake: when you say 'equal radii', you mean equal radii of the polygons' circumcircles? In that case, the result is easy: the distance can be broken down via Pythagoras into the squared xy distance between points on the two polygons, and the z distance between those polygons, and the former will always be maximized when the points are antipodal on the circumcircle — in which case they'll always be exactly a diameter apart. $\endgroup$ Jun 22, 2020 at 20:03
  • $\begingroup$ @Steven yes, radius of a polygon here is referred as the radius of the circumcircle. Also I could not properly understand what you exactly said, can you explain it in more detail or maybe add an answer? $\endgroup$ Jun 22, 2020 at 20:07
  • 2
    $\begingroup$ It is trivial that after a rotation of 180^0 you get the side crossing itself. What is it exactly that puzzles you? $\endgroup$
    – Moti
    Jun 23, 2020 at 0:58
  • $\begingroup$ @Moti can you explain the reason for it being trivial? $\endgroup$ Jun 23, 2020 at 2:00
  • 1
    $\begingroup$ Draw the cylinder representing the inscribed prism and pick one point on the top circumference and let it rotate $180^0$ and you will see how the right angle triangle is created. By using two points you get the twist you demonstrated. $\endgroup$
    – Moti
    Jun 23, 2020 at 4:11

2 Answers 2


First, some observations:

  • If we consider the cylinder containing all the vertices of the prism (including the circumcircles of the polygons), polygon $B$ is moving back and forth in this cylinder like a rotating piston.

  • Given a radius $r$ and a side length $l$, consider the point $B_1$.
    It is always at a distance $l$ from $A_1$.
    So it is always on the sphere of radius $l$ centered at $A_1$.

  • So the point $B_1$ moves along the path which is the intersection of the cylinder and the sphere. So this path depends only on $r$ and $l$.

Now, some answers:

  1. Your hypothesis is correct. You can see that the path of $B_1$ does not depend on how many points are in the polygon. It depends only on $r$ and $l$. The same is true for the motion of B towards and away from A.

  2. This question is really about visualizing a hyperboloid of one sheet. The best way to visualize it, however, is just to make one yourself. Cut two circles of cardboard, make little cuts around their edges to hold the string, and thread some string between the two. Then you can see it in 3D in your own hands.
    After you play with it, it will be obvious why the strings touch only at 180°.

  3. If you are talking about the angle $\angle A_2 A_1 B_1$, then this question does not have a simple answer. Unlike the previous two questions, this angle will depend on $A_2$, whose position relative to $A_1$ depends on the number of sides of the polygon, as well as on $r$ and $l$. You can extend your equation for the location of $B_1$ with some trigonometry to calculate this angle, but unfortunately the formula will just be a big mess.


What you have given agrees with my understanding of the twist or torsion shortening between two circular endrings by twisting one endring relative to the other.

(1) Yes, the same happens for all polygon prisms. the number of sides on end rings plays no role. The number of sides of regular polygon can be between $ ( n=2,\infty )$. Your hypotheses are correct.

(2) After 180 degree shift when strings cross, imagine a parallel thread is sewn for the short region of contact, original thread cut and twist rotation freely continues after the threads run concurrently at the vertex of cone that is a central point between end disc centers. You can also imagine a "ghost" thread that walks freely through the tensioned obstructing thread. So twist going back to 360 deg is has full meaning.

Height /twist relation

Vertical distance between string ends $2h$ String length $ 2 L$ End Ring radius $R$ Coordinates of top and bottom points respectively:

$$ (R \cos ( t + \theta) , \sin (t+ \theta), 2 h ),\; (R \cos t , R Sin t, 0)$$

Distance between two twisted points

$$ R^2 ( \cos ( t + \theta) - \cos ( t )^2+R^2 ( \sin ( t + \theta) - \sin ( t )^2 + 4h^2 = 4 L^2$$


$$ 2 (1-\cos \theta) = \dfrac{4 (L^2-h^2)}{R^2}$$

$$ h= \sqrt{L^2-({R \sin (\theta/2))}^2}$$

$$ r_{min}= R \cos \dfrac{\theta}{2}$$

Shortened distance $h$ and minimum waist radius $ r_{min}$ are sine/cosine trig relations as a functions of $\theta$ .

tallies for

$$ \theta= (0, \pi/2 ,\pi) $$

where we respectively have the cylinder, hyperboloid of one sheet and cone.

Important rotation configurations are

When cylinder full height

$$\theta = (0,\; 2 \pi), h=L $$

In between progressively narrow waisted hyperboloids of one sheet.

When cone $$ \theta = \pi, h^2= L^2- R^2 $$

If we take $ L=5,\; R=3 $ reduced height would be $\sqrt{5^2-3^2}= 4 $ shown in graph for cone.

If $ ( L, R, \theta_{max}) $ are given, then

$$ r_{min}= R \cos \dfrac{\theta_{max}}{2}$$

and $$ \tan \alpha= \dfrac{R \cos \dfrac{\theta}{2}}{\sqrt{L^2-({R \sin (\theta/2))}^2}}$$

The parametric equation of hyperboloid is $$ (x,y,z)= r_{min} (\theta, \theta \cot \alpha, 1)$$

Using these relations we can animate/morph successive deformations as functions of $ \theta $ relations given above and used in plotting 3D and height reduction graph.

The pair of generators diametrically opposite are called asymptotic, as normal curvature vanishes on it for a continuous surface of revolution.

I would suggest (afterwards in view of independence from $n$) you to upload another geogebra dynamic demo like nice present ones with $n=20$ or so. It will show the changing moving hyperbola envelopes beautifully. It would be superset of what you did so far.

There could be typos.

Height Redn with twist Height Redn schematic

  • $\begingroup$ Thanks alot, I am working on it, will update soon. $\endgroup$ Jun 29, 2020 at 13:43

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