Parametrizing A Torus And Locating Its Coordinate Curves

Introduction

A torus is the familiar ring-shaped surface formed by revolving a circle around an axis in the same plane as the circle, but not intersecting it. Describing it with two angles makes its geometry especially clear: one angle selects a position around the central hole, and the other selects a position around the circular cross-section.

The parametrization also gives direct access to the surface’s coordinate curves and tangent directions. We will use those derivatives to find a normal vector and establish exactly when this standard parametrization is regular.

The Ring Torus and Its Two Angles

Let $R$ be the major radius, the distance from the axis of revolution to the center of the generating circle, and let $r$ be the minor radius, the radius of that circle. For a ring torus, assume $R>r>0$. Take the axis of revolution to be the $z$-axis, and place the center of the generating circle at $(R,0,0)$ before revolving it.

Use $u$ for the angle around the $z$-axis and $v$ for the angle around the generating circle. Both angles are periodic with period $2\pi$; one convenient parameter domain is $0\leq u<2\pi$, $0\leq v<2\pi$.

Definition: A standard parametrization of the ring torus with major radius $R$ and minor radius $r$, where $R>r>0$, is $$ \mathbf{X}(u,v) = \bigl((R+r\cos v)\cos u,\ (R+r\cos v)\sin u,\ r\sin v\bigr). $$

To see the geometry behind the formula, first consider the circle in the vertical plane $y=0$. Its points can be written as $(R+r\cos v,0,r\sin v)$. Rotating such a point around the $z$-axis through angle $u$ gives the displayed parametrization. The horizontal distance from the $z$-axis is $R+r\cos v$, and the height is $r\sin v$.

The Two Families of Coordinate Curves

A coordinate curve is obtained by fixing one parameter and varying the other. The choice of which parameter is fixed determines which of the two circle families is traced.

Fix $v=v_0$ and let $u$ vary. The height $r\sin v_0$ stays constant, and the distance from the $z$-axis is the constant $R+r\cos v_0$. This is a horizontal circle centered on the $z$-axis. These circles go around the torus’s central hole; they are often called parallels.

Fix $u=u_0$ and let $v$ vary. The point stays in the vertical plane through the $z$-axis with azimuth $u_0$. Its circular cross-section is centered at $(R\cos u_0,R\sin u_0,0)$ and has radius $r$. These are the meridian circles of the torus.

Worked Example: Identifying the Two Circle Families

Take $R=4$ and $r=1$. First fix $v=0$. Since $\cos 0=1$ and $\sin 0=0$,

\mathbf{X}(u,0) = \bigl((4+1)\cos u,(4+1)\sin u,0\bigr) = (5\cos u,5\sin u,0).

This is the horizontal circle of radius $5$ in the plane $z=0$, centered at $(0,0,0)$. It is the outermost parallel of the torus.

Now fix $u=0$. Then $\cos 0=1$ and $\sin 0=0$, so

\mathbf{X}(0,v) = (4+\cos v,0,\sin v).

The first and third coordinates satisfy $(x-4)^2+z^2=\cos^2 v+\sin^2 v=1$, while $y=0$. Thus this is a radius-one circle centered at $(4,0,0)$ in the plane $y=0$, a meridian circle. In particular, at $v=\pi/2$, the point is $\mathbf{X}(0,\pi/2)=(4,0,1)$, which lies on that circle.

Tangent Vectors from the Parameters

At a parameter value $(u,v)$, differentiating $\mathbf{X}$ with respect to each parameter gives the two coordinate tangent vectors. For clarity, write $A=R+r\cos v$. The derivatives are

\mathbf{X}_u = (-A\sin u,\ A\cos u,\ 0), \qquad \mathbf{X}_v = (-r\sin v\cos u,\ -r\sin v\sin u,\ r\cos v).

The vector $\mathbf{X}_u$ is tangent to the parallel because it describes the change in position as $u$ varies with $v$ fixed. Similarly, $\mathbf{X}_v$ is tangent to the meridian because it describes the change as $v$ varies with $u$ fixed. The two vectors are not just labels: their values at a point give the actual tangent directions there.

Worked Example: Tangents and a Normal at the Outer Equator

Let $R=5$, $r=2$, $u=\pi/3$, and $v=0$. Here $A=R+r\cos v=5+2=7$, and $\cos(\pi/3)=1/2$, $\sin(\pi/3)=\sqrt{3}/2$. The point on the torus is

\mathbf{X}\left(\frac{\pi}{3},0\right) = \left(7\cdot\frac12,\ 7\cdot\frac{\sqrt{3}}2,\ 0\right) = \left(\frac72,\frac{7\sqrt{3}}2,0\right).

Substituting the same angles into the derivative formulas gives

\mathbf{X}_u\left(\frac{\pi}{3},0\right) = \left(-\frac{7\sqrt{3}}2,\frac72,0\right), \qquad \mathbf{X}_v\left(\frac{\pi}{3},0\right) = (0,0,2).

In the parameter order $(u,v)$, their cross product is

\mathbf{X}_u\times\mathbf{X}_v = \left(7,7\sqrt{3},0\right).

Its length is $\sqrt{7^2+(7\sqrt{3})^2}=\sqrt{49+147}=14$. Thus the corresponding unit normal is $(1/2,\sqrt{3}/2,0)$. To check it is perpendicular to both tangent vectors, compute

\left(-\frac{7\sqrt{3}}2,\frac72,0\right) \cdot \left(\frac12,\frac{\sqrt{3}}2,0\right) = -\frac{7\sqrt{3}}4+\frac{7\sqrt{3}}4=0, \qquad (0,0,2)\cdot\left(\frac12,\frac{\sqrt{3}}2,0\right)=0.

Both dot products vanish, as required for a normal vector.

The Cross Product and Regularity

The earlier material on normal vectors uses the cross product of the two tangent vectors. In this setting, it can be computed explicitly at every parameter value. With $A=R+r\cos v$, the component calculation is

\begin{aligned} \mathbf{X}_u\times\mathbf{X}_v &= \bigl(A\cos u\cdot r\cos v,\ A\sin u\cdot r\cos v,\ A r\sin v(\sin^2u+\cos^2u)\bigr)\\ &= rA(\cos v\cos u,\ \cos v\sin u,\ \sin v). \end{aligned}

The simplification of the third component uses $\sin^2u+\cos^2u=1$. The vector in parentheses has squared length $\cos^2v\cos^2u+\cos^2v\sin^2u+\sin^2v=\cos^2v+\sin^2v=1$. Hence it has length $1$, and the cross product has length $rA$.

Theorem: For the ring torus parametrization with $R>r>0$, the two tangent vectors are linearly independent at every parameter value. In the parameter order $(u,v)$, their cross product is $r(R+r\cos v)(\cos v\cos u,\cos v\sin u,\sin v)$, with length $r(R+r\cos v)>0$. Its associated unit normal is $(\cos v\cos u,\cos v\sin u,\sin v)$.

Here is why the positivity condition holds. Since $-1\leq\cos v\leq1$, we have $R+r\cos v\geq R-r>0$. Also $r>0$. Therefore $r(R+r\cos v)>0$ for every $u,v$. A cross product is nonzero exactly when its two vectors are linearly independent, so the tangent vectors are independent everywhere. In the rank condition for a parametrized surface, this means the parametrization is regular throughout its parameter domain.

Worked Example: A Point at the Top of a Meridian

Let $R=4$, $r=1$, $u=\pi/4$, and $v=\pi/2$. Since $\cos(\pi/2)=0$, $\sin(\pi/2)=1$, and $\cos(\pi/4)=\sin(\pi/4)=\sqrt{2}/2$, the point is

\mathbf{X}\left(\frac{\pi}{4},\frac{\pi}{2}\right) = \left(4\frac{\sqrt{2}}2,\ 4\frac{\sqrt{2}}2,\ 1\right) = (2\sqrt{2},2\sqrt{2},1).

Here $A=4$. The tangent vectors are

\mathbf{X}_u=(-2\sqrt{2},2\sqrt{2},0), \qquad \mathbf{X}_v=\left(-\frac{\sqrt{2}}2,-\frac{\sqrt{2}}2,0\right).

Their cross product, calculated directly, is

\begin{aligned} \mathbf{X}_u\times\mathbf{X}_v &= \left(0,\ 0,\ (-2\sqrt{2})\left(-\frac{\sqrt{2}}2\right) -(2\sqrt{2})\left(-\frac{\sqrt{2}}2\right)\right)\\ &=(0,0,2-(-2))=(0,0,4). \end{aligned}

The general formula gives the same result: $rA=4$, and $(\cos v\cos u,\cos v\sin u,\sin v)=(0,0,1)$, so the cross product is $(0,0,4)$. Its length is $4$, and its unit normal is $(0,0,1)$. The tangent vectors are both horizontal, so their dot products with this normal are $0$, confirming the result.

Why Parameter Order Matters

The cross product is sensitive to the order of its inputs: $\mathbf{X}_v\times\mathbf{X}_u=-(\mathbf{X}_u\times\mathbf{X}_v)$. Thus reversing the parameter order does not change the tangent plane or the line perpendicular to the surface, but it reverses the direction of the normal. For the standard $(u,v)$ order above, the unit normal points from the center of the generating circle toward the surface point.

Common pitfall: At the point in the outer-equator example, reversing the cross product gives $\mathbf{X}_v\times\mathbf{X}_u=(-7,-7\sqrt{3},0)$. This is still perpendicular to both tangents, but it points opposite to the $(u,v)$-order normal $(7,7\sqrt{3},0)$. Calling the reversed vector the normal for that parameter order without noting the reversal causes an orientation error. To correct it, use $\mathbf{X}_u\times\mathbf{X}_v$ when the specified order is $(u,v)$.

The condition $R>r$ matters as well. It ensures that the surface is a ring torus and that the radial factor $R+r\cos v$ never vanishes. If that factor could vanish, the cross product could be zero at some parameter values, and the rank argument just given would no longer establish regularity there. The ring-torus assumption rules out that case.

Using the Parametrization to Read the Geometry

The two angles answer different geometric questions. The angle $u$ records where the point lies around the central axis, while $v$ records where it lies around the circular tube. Keeping those roles distinct makes it easier to identify which coordinate curve is being followed and which derivative supplies its tangent direction.

For example, a fixed-$v$ curve changes $u$, so its tangent is $\mathbf{X}_u$; a fixed-$u$ curve changes $v$, so its tangent is $\mathbf{X}_v$. Their cross product then supplies a normal direction, with the order determining its sign. These facts make the parametrization useful in later surface calculations, while the regularity check ensures that its two parameter directions really do span a two-dimensional tangent space everywhere on the ring torus.

Check Your Understanding

Use the parametrization and results in this tutorial to answer the following questions.

For fixed $v=v_0$, describe the center, radius, and height of the coordinate circle.
For fixed $u=u_0$, state the plane, center, and radius of the meridian circle.
Write $\mathbf{X}_u$ and $\mathbf{X}_v$ for the ring torus parametrization.
Why does $R>r>0$ guarantee that $\mathbf{X}_u\times\mathbf{X}_v$ is nonzero everywhere?
What changes, and what does not change, when the cross-product order is reversed?