My 1st question has a straightforward answer but I'd appreciate hints on a proof. My 2nd question is open from my point of view.

Q1. Is it the case that the maximum convex volume body inside a torus in $\mathbb{R}^3$ is the intersection with a cylinder, as shown below?


Let $C$ be a smooth curve in $\mathbb{R}^3$, whose maximum curvature at any point $x \in C$ is $\le 1$. Now consider a tubular neighborhood of $C$— (used also in Light rays bouncing in twisted tubes)— width of $r<1$.

Q2. Let the curvature of the smooth $C \in \mathbb{R}^3$ be bound by $\le 1$. What is (a description of) the maximum volume convex shape that could move (via rigid motions) through any such tubular neighborhood radius of $\le r$ and overall central-rib $C$ curvature $\le 1$?

          A smooth curve $C$ with curvature everywhere $\le 1$. Tube of radius $r < 1$.

I presume the optimal shape is convex. I suspect this question has been considered previously...?

Related: Sofa in a snaky 3D corridor.

  • 3
    $\begingroup$ For Q1, the question of finding the convex subset of some region with the maximum volume seems to be called the "potato peeling problem" en.wikipedia.org/wiki/Potato_peeling . Though it seems that everything that's published discusses the 2D case? $\endgroup$
    – j.c.
    Jun 9, 2018 at 1:16
  • $\begingroup$ Q1 -- it would seem you might do better by replacing the cylinder with a half-space tangent to the inner equator; at any rate you'd do no worse because this half-space contains the cylinder (if I'm reading your picture right). $\endgroup$ Jun 9, 2018 at 1:34
  • $\begingroup$ For the torus, intersecting with tangent hyperplanes should help define the maximum solid and aid in a proof of optimality, especially if you know what parts of the torus are parts of the optimal surface. For question 2, you need to guarantee that you can twist the solid so that it faces something (osculating plane?) and navigates turns with high torsion. If you normalize and find that the solid for higher curvature is contained in one with lower curvature, then an intersection should navigate safely. Gerhard "Slow Down For Sharp Curves" Paseman, 2018.06.08. $\endgroup$ Jun 9, 2018 at 1:52
  • 5
    $\begingroup$ Noam, not quite because the half space piece includes arcs on the toroidal surface which "bend the wrong way" for convexity. Several tangent planes (a.k.a. cylinder) ensure convexity. Gerhard "Need Bending The Right Way" Paseman, 2018.06.08. $\endgroup$ Jun 9, 2018 at 1:55
  • $\begingroup$ I guess my intuitional certainty re Q1 was unwarranted. $\endgroup$ Jun 9, 2018 at 1:58

3 Answers 3


Not being a geometer, I have a hazy intuition for a proof of question 2, which hopefully someone else can formalize.

I interpret the problem as finding a solid which can pass (using rigid motions) through a constraining tube T, where constraints are that T looks like it "bends no more sharply" than a torus of similar radius and curvature. Indeed, if we have such a curve on a plane, create the associated tube, and then consider the intersection of "enough pieces" of the tube, that intersection should pass through T. Indeed, if T is made of toroidal arcs of a given curvature joined by cylinders, an intersection should look like Joseph's first shape giving intersection of a torus and a cylinder.

(Pause for a proof hint the Joseph's shape is optimal: given a circle C cutting the torus into a shape of genus 0, consider the transverse circles which indicate concavity of the torus. At most one point of the transverse circle can lie in a convex body, and so that point lies on C. Now use tangent hyper planes to C to get Joseph's shape.)

To strengthen the intuition, imagine the containing cylinder being slowly bent around Joseph's shape, resembling tori of sharper curvature, until the limit is reached. The intersection of all those tori is also Joseph's shape, or the intersection of two them, since the curve can bend to the left or to the right in the plane. (We still have our original curve in a plane.) Part of my intuition here which needs professional geometric verification is that for this case, a tubular neighborhood at a point is well approximated by joining two pieces of tori together, possibly of two different curvatures.

Now to mildly generalize out of the plane. Instead of approximating a piece of T at a point by gluing two pieces of tori together "with zero torsion", twist one of the pieces slightly to create a neighborhood for a curve that rises out of the plain. Now consider intersecting a lot of these twisted neighborhoods together. The intersection will be strictly contained in Joseph's shape. How do we show this intersection is convex?

I believe we can do it by using the tangent planes of the tube. As the shape slides through all possible tubes, this means it belongs to all possible neighborhood shapes appearing, and so to their intersection. It is my belief that the shape is actually contained in the intersection of all tori of optimal curvature that share the cutting circle C, so itself a solid of revolution. Hopefully someone who knows geometry can fill in the gaps.

Gerhard "Holes Not Just In Tori" Paseman, 2018.06.08.

  • 1
    $\begingroup$ My dad sold machines for cleaning the tube side of heat exchangers. The basic mechanism was a mildly flexible lance with a nozzle tip of relatively inflexible material. (high pressure water came out the nozzle to clean the inside of the tube.). I remember calculations for an optimal nozzle length given a certain radius of curvature, and proving to him that it suffices to consider the cross section and that looking at a rectangle in an annulus worked. He sold more equipment for more jobs with this applied mathematics. Gerhard "Is Remembering American Powerlance Fondly" Paseman, 2018.06.08. $\endgroup$ Jun 9, 2018 at 2:55
  • $\begingroup$ "imagine the containing cylinder being slowly bent around Joseph's shape" Here's a similar idea. Fix the intersection body illustrated. Now spin the torus around the axis of the cylinder. Intersect all those tori, effectively shaving down the intersection shape. And yes, it should be a solid of revolution. $\endgroup$ Jun 9, 2018 at 13:55

I want to say an intuition on question 1 :

Def :

$$ f(t,s):=\bigg((R +r\cos\ t)\cos\ s,(R+r\cos\ t)\sin\ s,r\sin\ t\bigg) $$

Define a solid torus $T:={\rm conv}\ f([0,2\pi]^2)$ and $d$ to be a Euclidean distance in $\mathbb{E}^3$.

Def : Define $$S_t=f\bigg( t,[0,2\pi)\bigg) ,\ \frac{\pi}{2} < t < \frac{3\pi}{2} $$

Property of Convex Set : Assume that $C$ is a closed convex subset in $T$. Then $$ \bigg|\ S_t\bigcap C \ \bigg|\leq 1 $$

EXE : Let $p\in S_t$. Note that tangent a plane $T_p \partial T$ cuts $ T $ into two components $T_i$. Then one component $\overline{T}_1\bigcap B^d(p,\epsilon)$ is star-shaped at $p$ in $ \mathbb{E}^3$.

In further, if a plane at $p$ is close to $T_p\partial T$ s.t. it has such property, then it is the tangent plane.

Proof : By Gaussian curvature condition, any geodesic in $\partial T_1$ at $p$ has a positive curvature.

EXE : Fix $p_0\in S_\pi$. Then $T_{p_0} \partial T$ cuts $T$ into two components. Clearly, if $C$ is a closed convex set, then it can be assumed in $T_1$.

So now we suffice to rule out points in $\partial T_1$ of negative curvature. If $c(t)\in S_t,\ 0<\frac{\pi}{2}<\frac{3\pi}{2}$ is a curve, then we cut $T_1$ with a surface $M_{c,v} \ :\ x(t,s)=c(t)+sv(t)$ where $v(t)$ is a unit field in $T_{c(t)}\partial T_1$.

Here, $M_{c,v}$ is a ruled surface so that it is nonpositively curved. Hence when it is flat, the resulting body has a maximal volume.


I want to say some for the question 2 :

(1) Construction of tubular surface of curve of const curvature $1$ :

If $e_i$ is a canonical basis in $\mathbb{R}^3$, then we define \begin{align*} c(t)& =c_0(t) + \frac{3}{2}e_3,\\ c_0&=\bigg(\cos (-t),0,\sin(-t) \bigg) \\ &\\ N(t)&=-c_0(t),\ b= (0,1,0)\\ \end{align*}

Hence $T=c'(t),\ N, \ b$ is Frenet-Serre frame so that we define \begin{align*} X(t,\theta )&= c_0(t) + \frac{1}{2} (-c_0\cos\ \theta +b\sin\ \theta ) + \frac{3}{2}e_3\\&=\bigg( \cos\ t\cdot (1-\frac{1}{2}\cos\ \theta ) , \frac{1}{2}\sin\ \theta,-\sin\ t\cdot (1-\frac{1}{2}\cos\ \theta ) \bigg) + \frac{3}{2} e_3\end{align*}

(2) Notation and Definition : If $X$ parametrize a surface $\Sigma$, then we define

$$ \Sigma_{right}=\Sigma\cap \{y\geq 0 \} ,\ \Sigma_{left} = \Sigma\cap \{y\leq 0\},\ \Sigma_0=\Sigma\cap \{ z<\frac{1}{2}\} $$ and a line $l:=\{ (t,0,\frac{1}{2} )|t\in \mathbb{R} \}$

(3) If $S_{rev}$ is a revolution surface of the curve $ X(t,\pi)$ along the line, then $\Sigma_0$ is a graph surface and then we can show that $S_{rev}$ is contained in $\Sigma$, by a direct computation.

(4) If $S$ is a sliding which can be obtained from the intersection between $\Sigma$ and cylinder of radius $\frac{1}{2}$, and if $S$ contains a circle $X(\frac{\pi}{2},\theta)$, then we rotate $S$ along this circle. That is, $(0,0,1)$ goes through $\Sigma_{right}$. If we define $S_{right},\ S_{left}$ wrt $y>0,\ y<0$, and if we do $\pi$-rotation, then $S_{right}$ can be cut as much as possible, but $S_{left}$ not. Hence the resulting $S$ is larger than $S_{rev}$.

(5) If $\alpha$ is a union of piecewise smooth space curves $\alpha_i$ whose curvatures are less than $1$, and if $\alpha$ has a continuous unit tangent vector, then the resulting sliding can be $S_{rev}$

Proof : If $N_i^s,\ N_i^e$ are normal vectors at starting, end points of $\alpha_i$, then we define $\theta_{i(i+1)}=\angle (N_i^e,N_{i+1}^s)$.

When a sliding in tubular surface related to $\alpha_i$ goes to tubular surface related to $\alpha_{i+1}$, if the sliding is cut through $\theta_{i(i+1)}$-rotation, then the sliding goes back and forth freely on two surfaces without further volume change.

Assume that $\theta_{12}=\pi$ and $N_i^s,\ N_i^e$ are orthogonal to a fixed unit vector for all $i$. Hence we map $N_i^s,\ N_i^e$ into $N_i$ in unit circle s.t. distance between $N_i$ and $ N_{i+1}$ is $\theta_{i(i+1)}$.

If sliding is in second tubular surface, then $N_1$ goes to $N_2$ by using half of the circle. Hence if we take $N_3$ in the not used part of the circle, then $\pi$-rotation is not sufficient, which complete the proof.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.