Distance from a point on a torus to the origin

Question

This question may be slightly subtle, but I'd like to better understand why my first attempts to solve the following problem failed. I am defining a point on a torus with minor and major radii $0<r<R<1$ using the parameterization $$ x = (R+r\cos\theta)\cos\phi,\quad y = (R+r\cos\theta)\sin\phi,\quad z = r\sin\theta $$ and would like to determine the bound on $r$ and $\theta$ such that a point is at most unit distance away from the origin, given $R$. This seems like a paradox and perhaps I should have tried to

$$ 1 >= x^2 + y^2 + z^2 = R^2 + 2Rr\cos\theta + r^2 $$

This seemed straightforward, so I wrote the quadratic in $a r^2 + b r + c \geq 0$ as having the discriminant $4(R^2\cos^2\theta - R^2 - 1)$, which must be nonnegative to establish a real bound on $\theta$. I am wondering if this is a conceptually flawed approach, since if I require $$ \cos^2\theta >= \frac{R^2-1}{R^2} $$ This isn't particularly helpful as the left-hand-side is nonnegative and the right-hand-side is strictly negative and thus satisfied for all $\theta$. It does not seem intuitively correct that a point could remain within the unit sphere of the origin independent of $\theta$.

Next I tried the factoring the quadratic in $r$ $$ (r-\lambda_1)(r-\lambda_2) \leq 0 $$ with $$ \lambda_{1,2} = -R\cos\theta\pm\sqrt{1-R^2\sin^2\theta}$$ Looking at the conditions where one term in the product is negative essentially recovers the original quadratic inequality.

Finally, I thought to exploit symmetry and consider the problem in cylindrical coordinates. If $R+r<1$, then the torus lies inside the sphere and all $\theta\in[0,2\pi)$ and $0<r<1-R$ satisfy the bounds. Otherwise we can think of this as an intersection of circles in the $\rho z$ plane. Using the law of cosines

$$ \rho_i = \frac{R^2+1-r^2}{2R},\quad z_i = \sqrt{1-\rho_i^2}$$ and

the unit distance equality is attained when

$$ \theta_i = \tan^{-1}\frac{z}{\rho}=\tan^{-1}\frac{\sqrt{4R^2-(R^2-r^2+1)^2}}{R^2-r^2+1} $$

Of course this angle is from the origin and is not the same as the $\theta$ in the torus parameterization, which is the angle from the $xy$ plane. I think that the angle I actually want is

$$\theta_\text{torus} \geq \tan^{-1}\frac{z_i}{r}=\tan^{-1}\frac{\sqrt{4R^2-(R^2-r^2+1)^2}}{2Rr}$$

Considering the $z\geq 0$ case. So the question is... what was wrong with my first approach? I know it's advantageous use symmetry, but was it essential here?

Edit: My wording may have been unclear. I am trying to determine the set of $r,\theta$ values that for a given $R$ that will lie within the unit sphere.

Answer 1

If $r$ is allowed to vary, then you are considering a family of concentric tori with common major radius $R$. If $R < 1$, then for any $0 < r \le 1-R$, the entire torus will be contained within the unit sphere, since the maximum distance of any point on the torus from the origin is $R + r$.

When $r > 1-R$, then only some portion of the torus will be enclosed in the unit sphere. To understand which points, it is first worth noting that the geometry is radially symmetric about the $z$-axis, so we can assume $y = 0$ (or equivalently $\phi = 0$), which yields the cross-section in the half-plane

$$x = R + r \cos \theta, \quad z = r \sin \theta.$$ Equivalently, this describes a circle with center at $(R,0)$ and radius $r$: $$(x-R)^2 + z^2 = r^2.$$ Then the intersection of this circle with the unit circle $x^2 + z^2 = 1$ is given by

$$x = \frac{1 - r^2 + R^2}{2R}, \quad z = \pm \frac{\sqrt{(1+r+R)(1+r-R)(1-r+R)(-1+r+R)}}{2R}.$$

Consequently, the range of angles $\theta$ for which points on the torus are contained within the unit sphere is given by $$\arccos \frac{1-r^2-R^2}{2Rr} \le \theta \le 2\pi - \arccos \frac{1-r^2-R^2}{2Rr}$$ when $r > 1-R$. Again, for a fixed $R$ but variable $r$, the region of interest is parametrized by $\theta$ and $r$, corresponding to a family of tori. See the following example below, for $R = 0.8$. The surfaces plotted correspond to $r \in \{0.1, 0.2, \ldots, 0.8\}$:

Answer 2

In fact, due to circular symmetry, this 3D issue can be turned into 2D in a vertical cross section plane as described in the figure below (in the case where the sphere $S(0,1)$ intersects the torus ; see remark below) :

Applying cosine law to triangle $OPM_0$ gives :

$$1=R^2+r^2-2rR \ \underbrace{\cos(\pi-\alpha)}_{-\cos(\alpha)}$$

Otherwise said, the limit angle is :

$$\alpha = \cos^{-1}\left(\frac{R^2+r^2-1}{2rR}\right).$$

Any point point $M$ situated on the interior circular arc $M_0M_1$, i.e. having its $\theta$ coordinate not in $(-\alpha,\alpha)$ will be at a distance less than $1$ from the center $O$ of the torus.

Remark : of course, if the sphere $S(O,1)$ doesn't intersect the torus, there are two evident answers : either no point $M$ satisfies the condition, or all points satisfy it.