# Archytas curve

The Archytas curve may be constructed as the intersection of a (right circular) cylinder and a horn torus. The center of the torus should be one of the points of the cylinder. The directrix of the torus should be in a plane parallel to the plane of a base of the cylinder; it should have a radius equal to the radius of the base of the cylinder. (The generatrix of the torus should have such a radius as well, since it is a horn torus.)

## Calculations

Cylinder:
    $x^{2}+y^{2}=r^{2}$, for any z.
Horn torus: With center at (r,0), parametrize a circular directrix of radius r with parameter φ:
    $(r,0,0)+r\cdot (\cos(\phi ),\sin(\phi ),0)=(1+r\cos(\phi ),r\sin(\phi ),0)$
Now parametrize a circular generatrix of radius r around that directrix with parameter θ:
    $(r+r\cos(\phi ),r\sin(\phi ),0)+r\cos(\theta )\cdot (\cos(\phi ),\sin(\phi ),0)+r\sin(\theta )\cdot (0,0,1)$
    $=(r+r\cos(\phi )+r\cos(\theta )\cos(\phi ),r\sin(\phi )+r\cos(\theta )\sin(\phi ),r\sin(\theta ))$
How to describe the torus non-parametrically?
Given (x, y, z) —a point on the torus— its horizontal distance to (r, 0) is ${\sqrt {(x-r)^{2}+y^{2}}}$; but the center of a generatrix is at distance r from (r, 0), in either direction:
    ${\Big{\sqrt {(x-r)^{2}+y^{2}}}-r{\Big )}^{2}+z^{2}=r^{2}$
    $z^{2}=r^{2}-{\Big{\sqrt {(x-r)^{2}+y^{2}}}-r{\Big )}^{2}$
    $z^{2}=r^{2}-{\Big(x-r)^{2}+y^{2}+r^{2}-2r{\sqrt {(x-r)^{2}+y^{2}}}{\Big )}$
    $z^{2}=r^{2}-(x-r)^{2}-y^{2}-r^{2}+2r{\sqrt {(x-r)^{2}+y^{2}}}$
    $z^{2}=2r{\sqrt {(x-r)^{2}+y^{2}}}-(x-r)^{2}-y^{2}$
    $z^{2}=2r{\sqrt {(x-r)^{2}+y^{2}}}-(x^{2}+r^{2}-2xr)-y^{2}$
    $z^{2}=2r{\sqrt {(x-r)^{2}+y^{2}}}-x^{2}-y^{2}-r^{2}+2xr$
    $z^{2}=2r{\sqrt {(x-r)^{2}+y^{2}}}-2r^{2}+2xr$
    $z^{2}=2r{\sqrt {x^{2}+r^{2}-2xr+y^{2}}}-2r^{2}+2xr$
    $z^{2}=2r{\sqrt {2r^{2}-2xr}}-2r^{2}+2xr$
* $z^{2}=2r{\sqrt {(x-r)^{2}+y^{2}}}-(x-r)^{2}-y^{2}$;
this works for arbitrary x and y, defining the torus. Now constrain x and y parametrically:
    $x=r\cos(\phi )$,
    $y=r\sin(\phi )$.
The added constraint selects the points on the torus which are also on the cylinder.
    $z^{2}=2r{\sqrt {(r\cos(\phi )-r)^{2}+r^{2}\sin ^{2}(\phi )}}-(r\cos(\phi )-r)^{2}-r^{2}\sin ^{2}(\phi )$
    $z^{2}=2r{\sqrt {r^{2}*(\cos(\phi )-1)^{2}+r^{2}\sin ^{2}(\phi )}}-r^{2}(\cos(\phi )-1)^{2}-r^{2}\sin ^{2}(\phi )$
    $z^{2}=2r^{2}{\sqrt {[(\cos(\phi )-1)^{2}+\sin ^{2}(\phi )}}-r^{2}(\cos(\phi )-1)^{2}-r^{2}\sin ^{2}(\phi )$
    $z^{2}=r^{2}{\Big2{\sqrt {(\cos(\phi )-1)^{2}+\sin ^{2}(\phi )}}-(\cos(\phi )-1)^{2}-\sin ^{2}(\phi ){\Big )}$
                                     ... $-(\cos ^{2}(\phi )+1-2\cos(\phi ))-\sin ^{2}(\phi )$
                                     ... $-\cos ^{2}(\phi )-1+2\cos(\phi )-\sin ^{2}(\phi )$
                                     ... $-2+2\cos(\phi )$
    $z^{2}=r^{2}{\Big2{\sqrt {2-2\cos(\phi )}}-2+2\cos(\phi ){\Big )}$
Cartesian parametrization:
* $z=r{\sqrt {2{\sqrt {2-2\cos(\phi )}}-2+2\cos(\phi )}}$
* $x=r\cos(\phi )$
* $y=r\sin(\phi )$
    $x^{2}+y^{2}=r^{2}\cos ^{2}(\phi )+r^{2}\sin ^{2}(\phi )=r^{2}$
    $z^{2}=r^{2}{\Big2{\sqrt {2-2\cos(\phi )}}-2+2\cos(\phi ){\Big )}$
Let $u=(x-r)^{2}+y^{2}=(r\cos(\phi )-r)^{2}+r^{2}\sin ^{2}(\phi )$
                      $=r^{2}(\cos(\phi )-1)^{2}+r^{2}\sin ^{2}(\phi )$
                      $=r^{2}{\Big(\cos(\phi )-1)^{2}+\sin ^{2}(\phi ){\Big )}$
                      $=r^{2}{\Big\cos ^{2}(\phi )+1-2\cos(\phi )+\sin ^{2}(\phi ){\Big )}$
                      $=r^{2}(2-2\cos(\phi ))$
       $z^{2}=r^{2}{\Big2{\sqrt {u/r^{2}}}-u/r^{2}{\Big )}$
       $z^{2}=2r^{2}{\sqrt {u/r^{2}}}-u$
       $z^{2}=2r{\sqrt {u}}-u$
    $(z^{2}+(x-r)^{2}+y^{2})=2r{\sqrt {(x-r)^{2}+y^{2}}}$
    $(z^{2}+(x-r)^{2}+y^{2})^{2}=4r^{2}((x-r)^{2}+y^{2})$
Replace x − r with x’:
    $(z^{2}+x'^{2}+y^{2})^{2}=4r^{2}(x'^{2}+y^{2})$
Substituting x’ → x and r → a/2, this equation becomes:
    $(x^{2}+y^{2}+z^{2})^{2}=a^{2}(x^{2}+y^{2})$
which is the first equation in the reference:, right under "system of Cartesian equations".
    $(z^{2}+(x-r)^{2}+y^{2})^{2}=4r^{2}((x-r)^{2}+y^{2})$
    $x^{2}+y^{2}=r^{2}$
Replace x with r + x. This would correspond to a shift of the entire curve along the x direction.
    $(z^{2}+x^{2}+y^{2})^{2}=4r^{2}(x^{2}+y^{2})$
    $(x+r)^{2}+y^{2}=r^{2}$
    $x^{2}+r^{2}+2xr+y^{2}=r^{2}$
    $x^{2}+2xr+y^{2}=0$
    $x^{2}+y^{2}=-2rx$
Let a = −2 r, then
* $x^{2}+y^{2}=ax$
    $a^{2}=4r^{2}$, so
* $(x^{2}+y^{2}+z^{2})^{2}=a^{2}(x^{2}+y^{2})$;
these match the system of cartesian equations given in the reference.

## Revision of parametric equations

The parametric equations as given above only show the $z\geq 0$ part of the curve. The entire curve has a reflective symmetry about the $z=0$ plane. The parametrization can fixed so that it yields a closed (periodic) smooth loop with no cusps; it has a tacnode and a crunode.
The fix involves two changes: replacing φ with a triangular wave:
    $\theta (x)=2\arcsin(\sin({\tfrac {1}{2}}(x-\pi )))+\pi$
and prepending a square wave:
    $\sigma (x)=\operatorname {sgn}(\sin(x))$
although for purposes of graphing it may be better to use a smoother σ such as a sigmoid function:
    $\sigma (x)={2 \over \pi }\arctan(1024\sin(x))$,
where the 1024 may be replaced by a higher number; as it goes to ∞ the sigmoid would approach the sign function as a limit.
Then
* $x(\phi )=r\cos(\theta (\phi ))$
* $y(\phi )=r\sin(\theta (\phi ))$
* $z(\phi )=\sigma (\phi )\ r{\sqrt {2{\sqrt {2-2\cos(\theta (\phi ))}}-2+2\cos(\theta (\phi ))}}$
The parametrization may have the entire set of real numbers as domain, in which case it is periodic with period 4π.

## Application of the Archytas curve to finding cube roots

To use the Archytas curve to find cube roots, it is necessary to intersect it with another surface, namely a cone. Such a cone should have its apex coinciding with the center of the torus; thus the calculations are simplified if the center of the torus and the apex of the cone are placed at the origin; then the axis of the cylinder will no longer pass through the origin.
First we will derive the equation of the horn torus. Let the diameter of the generatrix (and also the directrix) be a. Consider a pair of perpendicular axes defining the coordinate system of the plane, with the vertical axis being z, and the horizontal axis being u instead of x, where
    $u={\sqrt {x^{2}+y^{2}}}$
so the uz-plane is any plane of the generatrix, which is being revolved around the z-axis.
There is a circle centered not at the origin but at a/2; it is the generatrix. Its equation is not
    $u^{2}+z^{2}=(a/2)^{2}$
but
    $(u-a/2)^{2}+z^{2}=(a/2)^{2}$
Expand the first square on the LHS:
    $u^{2}+a^{2}/4-ua+z^{2}=a^{2}/4$
Cancel out the $a^{2}/4$:
    $u^{2}+z^{2}-ua=0$
Transpose u a to the RHS and replace u with ${\sqrt {x^{2}+y^{2}}}$:
    $x^{2}+y^{2}+z^{2}=a{\sqrt {x^{2}+y^{2}}}$
Square both sides:
* $(x^{2}+y^{2}+z^{2})^{2}=a^{2}(x^{2}+y^{2})$
This is the (implicit) equation for the horn torus (centered at the origin, with the z-axis as its axis of revolution).
Consider a cylinder whose base is a circle in the xy-plane centered at (a/2,0), with diameter a, and whose axis is parallel to the z-axis.
    $(x-a/2)^{2}+y^{2}=(a/2)^{2}$
    $x^{2}+(a/2)^{2}-ax+y^{2}=(a/2)^{2}$
    $x^{2}+y^{2}-ax=0$
* $x^{2}+y^{2}=ax$
This is the cylinder's equation (for any z).
Consider a cone whose apex is at the origin, whose axis is the x-axis. Let the axes of the plane's coordinate system be x and u, where x is horizontal and u is vertical, $u={\sqrt {y^{2}+z^{2}}}$; the u-axis revolves around the x-axis. Draw a diagonal line passing through the origin. Let its slope be denoted as s; it is the cone's "apical slope" (an ad hoc term).
    $u=sx$
    $sx={\sqrt {y^{2}+z^{2}}}$
* $s^{2}x^{2}=y^{2}+z^{2}$.
This is the cone's equation.
Use the cylinder's equation to perform a substitution in the RHS of the torus's equation
    $(x^{2}+y^{2}+z^{2})^{2}=a^{3}x$
Use the cone's equation to perform a substitution in the LHS of the torus's equation:
    $(x^{2}+s^{2}x^{2})^{2}=a^{3}x$
    $((s^{2}+1)x^{2})^{2}=a^{3}x$
    $(s^{2}+1)^{2}x^{4}=a^{3}x$
Assuming that $x\neq 0$:
    $(s^{2}+1)^{3}=a^{3}$
    $x_{1}^{3}={a^{3} \over (s^{2}+1)^{2}}$
where the subscript 1 means that this is the value of a solution; it is the x-coordinate of the intersection of the three surfaces.
* $x_{1}={a \over {\sqrt[{3}]{(s^{2}+1)^{2}}}}$.
Given $x=x_{1}$, solve for y; use the cylinder's equation.
    $x^{2}+y^{2}=ax$
    $y^{2}=ax-x^{2}$
    $y={\sqrt {ax-x^{2}}}$
so
* $y_{1}={\sqrt {ax_{1}-x_{1}^{2}}}$.
Given x$_1$, y$_1$, solve for z; use the cone's equation.
    $s^{2}x^{2}=y^{2}+z^{2}$
    $z^{2}=s^{2}x^{2}-y^{2}$
* $z_{1}={\sqrt {s^{2}x_{1}^{2}-y_{1}^{2}}}$
Let P denote the intersection of the three surfaces (i.e., of the Archytas curve and the cone). Let O denote the origin.
    $OP={\sqrt {x_{1}^{2}+y_{1}^{2}+z_{1}^{2}}}$
Now substitute the values of y$_1$ and z$_1$:
    $OP={\sqrt {x_{1}^{2}+ax_{1}-x_{1}^{2}+s^{2}x_{1}^{2}-y_{1}^{2}}}$
Cancel out the opposing $x_{1}^{2}$'s:
    $OP={\sqrt {ax_{1}+s^{2}x_{1}^{2}-y_{1}^{2}}}$
Now substitute the value of y$_1$:
    $OP={\sqrt {ax_{1}+s^{2}x_{1}^{2}-ax_{1}+x_{1}^{2}}}$
Cancel out the opposing a x$_1$'s:
    $OP={\sqrt {s^{2}x_{1}^{2}+x_{1}^{2}}}$
    $OP={\sqrt {(s^{2}+1)x_{1}^{2}}}$
* $OP={\sqrt {s^{2}+1}}\ x_{1}$.
Let N be the vertical projection of P onto the xy-plane. Then
    $ON={\sqrt {x_{1}^{2}+y_{1}^{2}}}$
       $={\sqrt {x_{1}^{2}+ax_{1}-x_{1}^{2}}}$
* $ON={\sqrt {ax_{1}}}$
Let B be the intersection of the diagonal line in the slice of the cone in the $z=0$ plane —the word "slice" is being used loosely; it means cross-section— with the circle in the slice of the cylinder in the same plane. (This circular slice may be ascribed as the base of the cylinder.) Circle OBA has its center at $(a/2,0,0)$ and radius a/2. Line OB has slope s with respect to the x-axis.
Circle OBA’s equation is
    $(x-{\tfrac {a}{2}})^{2}+y^{2}=({\tfrac {a}{2}})^{2}$
    $x^{2}+{\cancel {({\tfrac {a}{2}})^{2}}}-ax+y^{2}={\cancel {({\tfrac {a}{2}})^{2}}}$
    $x^{2}+y^{2}=ax$.
Line OB’s equation is
    $y=sx$
Use the line's equation in the circle's equation:
    $x^{2}+s^{2}x^{2}=ax$
    $(s^{2}+1)x^{2}=ax$
Assuming that x is not zero,
    $(s^{2}+1)x=a$,
    $x_{0}={a \over s^{2}+1}$
where x$_0$ means the abscissa of point B. Then
    $y_{0}=sx_{0}$.
Let b denote the length of line segment OB:
    $b={\sqrt {x_{0}^{2}+y_{0}^{2}}}$
    $b={\sqrt {x_{0}^{2}+s^{2}x_{0}^{2}}}$
* $b=x_{0}{\sqrt {s^{2}+1}}$.
Claim: ${a \over OP}={OP \over ON}={ON \over b}$.
Firstly, cross-multiplying the first equation:
    $aON{\overset {?}{=}}OP^{2}$
    $a{\sqrt {ax_{1}}}{\overset {?}{=}}(s^{2}+1)x_{1}^{2}$
    ${\sqrt {ax_{1}}}{\overset {?}{=}}{(s^{2}+1)x_{1}^{2} \over a}$
    $ax_{1}{\overset {?}{=}}{(s^{2}+1)^{2}x_{1}^{4} \over a^{2}}$
Assuming that $x_{1}\neq 0$,
    $a{\overset {?}{=}}{(s^{2}+1)^{2}x_{1}^{3} \over a^{2}}$
    $a^{3}{\overset {?}{=}}(s^{2}+1)^{2}x_{1}^{3}$
    $x_{1}^{3}{\overset {?}{=}}{a^{3} \over (s^{2}+1)^{2}}$
is true; so the first equation is true.
Secondly, cross-multiplying the second equation:
    $bOP{\overset {?}{=}}ON^{2}$
    $bx_{1}{\sqrt {s^{2}+1}}{\overset {?}{=}}ax_{1}$
Assuming that x$_1$ is not zero,
    $b{\sqrt {s^{2}+1}}{\overset {?}{=}}a$
    $b{\overset {?}{=}}{a \over {\sqrt {s^{2}+1}}}$
True ∴ the second equation is true.
Then
    ${\Big{a \over OP}{\Big )}{\Big{OP \over ON}{\Big )}{\Big{ON \over b}{\Big )}$
       $={\Big{a \over {\cancel {OP}}}{\Big )}{\Big{{\cancel {OP}} \over ON}{\Big )}{\Big{ON \over b}{\Big )}$
       $={\Big{a \over {\cancel {ON}}}{\Big )}{\Big{{\cancel {ON}} \over b}{\Big )}$
    ∴${\Big{a \over OP}{\Big )}{\Big{OP \over ON}{\Big )}{\Big{ON \over b}{\Big )}={a \over b}$.
Also,
    ${\Big{a \over OP}{\Big )}{\Big{OP \over ON}{\Big )}{\Big{ON \over b}{\Big )}={\Big{a \over OP}{\Big )}^{3}$
    ∴${\Big{a \over OP}{\Big )}^{3}={a \over b}$,
* ${a \over OP}={\sqrt[{3}]{a \over b}}$
which shows that the Archytas curve (with the aid of a cone of adjustable aperture) serves as a nomogram for finding cube roots.
The point B can be chosen to be any point on the semicircle OBA, so that b can be chosen to be arbitrarily smaller than a, so that any number above 1 may be set equal to $a/b$.
    ${a \over b}={\sqrt {s^{2}+1}}$
Solve for the apical slope s:
    ${a^{2} \over b^{2}}=s^{2}+1$
    ${a^{2} \over b^{2}}-1=s^{2}$
    $s={\sqrt {{a^{2} \over b^{2}}-1}}$
(Euclid's Elements gives a general method for constructing square roots, so finding an s that will suit a given a/b is not a problem.)
Setting $s={\sqrt {3}}$ (i.e., the "apical angle" to 60° degrees) constructs a solution to the Delian cube problem, which is equivalent to finding the cube root of two.
Now give credit where credit is due.