Involute

This section is based on.[3]

There are generically two types of cusps in involutes. The first type is at the point where the involute touches the curve itself. This is a cusp of order 3/2. The second type is at the point where the curve has an inflection point. This is a cusp of order 5/2.

This can be visually seen by constructing a map ${\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} ^{3}}$ defined by

$${\displaystyle (s,t)\mapsto (x(s)+t\cos(\theta ),y(s)+t\sin(\theta ),t)}$$

where ${\displaystyle (x(s),y(s))}$ is the arclength parametrization of the curve, and ${\displaystyle \theta }$ is the slope-angle of the curve at the point ${\displaystyle (x(s),y(s))}$. This maps the 2D plane into a surface in 3D space. For example, this maps the circle into the hyperboloid of one sheet. By this map, the involutes are obtained in a three-step process: map ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {R} ^{2}}$, then to the surface in ${\displaystyle \mathbb {R} ^{3}}$, then project it down to ${\displaystyle \mathbb {R} ^{2}}$ by removing the z-axis:

$${\displaystyle s\mapsto (s,l-s)\mapsto f(s,l-s)\mapsto (f(s,l-s)_{x},f(s,l-s)_{y})}$$

where ${\displaystyle l}$ is any real constant.

Since the mapping ${\displaystyle s\mapsto f(s,l-s)}$ has nonzero derivative at all ${\displaystyle s\in \mathbb {R} }$, cusps of the involute can only occur where the derivative of ${\displaystyle s\mapsto f(s,l-s)}$ is vertical (parallel to the z-axis), which can only occur where the surface in ${\displaystyle \mathbb {R} ^{3}}$ has a vertical tangent plane.

Generically, the surface has vertical tangent planes at only two cases: where the surface touches the curve, and where the curve has an inflection point.

For the first type, one can start by the involute of a circle, with equation

$${\displaystyle {\begin{aligned}X(t)&=r(\cos t+(t-a)\sin t)\\Y(t)&=r(\sin t-(t-a)\cos t)\end{aligned}}}$$

then set ${\displaystyle a=0}$, and expand for small ${\displaystyle t}$, to obtain

$${\displaystyle {\begin{aligned}X(t)&=r+rt^{2}/2+O(t^{4})\\Y(t)&=rt^{3}/3+O(t^{5})\end{aligned}}}$$

thus giving the order 3/2 curve ${\displaystyle Y^{2}-{\frac {8}{9r}}(X-r)^{3}+O(Y^{8/3})=0}$, a semicubical parabola.

Tangents and involutes of the cubic curve ${\displaystyle y=x^{3}}$. The cusps of order 3/2 are on the cubic curve, while the cusps of order 5/2 are on the x-axis (the tangent line at the inflection point).

For the second type, consider the curve ${\displaystyle y=x^{3}}$. The arc from ${\displaystyle x=0}$ to ${\displaystyle x=s}$ is of length ${\displaystyle \int _{0}^{s}{\sqrt {1+(3t^{2})^{2}}}dt=s+{\frac {9}{10}}s^{5}-{\frac {9}{8}}s^{9}+O(s^{13})}$, and the tangent at ${\displaystyle x=s}$ has angle ${\displaystyle \theta =\arctan(3s^{2})}$. Thus, the involute starting from ${\displaystyle x=0}$ at distance ${\displaystyle L}$ has parametric formula

$${\displaystyle {\begin{cases}x(s)=s+(L-s-{\frac {9}{10}}s^{5}+\cdots )\cos \theta \\y(s)=s^{3}+(L-s-{\frac {9}{10}}s^{5}+\cdots )\sin \theta \end{cases}}}$$

Expand it up to order ${\displaystyle s^{5}}$, we obtain

$${\displaystyle {\begin{cases}x(s)=L-{\frac {9}{2}}Ls^{4}+({\frac {9}{2}}L-{\frac {9}{10}})s^{5}+O(s^{6})\\y(s)=3Ls^{2}-2s^{3}+O(s^{6})\end{cases}}}$$

which is a cusp of order 5/2. Explicitly, one may solve for the polynomial expansion satisfied by ${\displaystyle x,y}$:

$${\displaystyle \left(x-L+{\frac {y^{2}}{2L}}\right)^{2}-\left({\frac {9}{2}}L+{\frac {51}{10}}\right)^{2}\left({\frac {y}{3L}}\right)^{5}+O(s^{11})=0}$$

or

$${\displaystyle x=L-{\frac {y^{2}}{2L}}\pm \left({\frac {9}{2}}L+{\frac {51}{10}}\right)\left({\frac {y}{3L}}\right)^{2.5}+O(y^{2.75}),\quad \quad y\geq 0}$$

which clearly shows the cusp shape. Setting ${\displaystyle L=0}$, we obtain the involute passing the origin. It is special as it contains no cusp. By serial expansion, it has parametric equation

$${\displaystyle {\begin{cases}x(s)={\frac {18}{5}}s^{5}-{\frac {126}{5}}s^{9}+O(s^{13})\\y(s)=-2s^{3}+{\frac {54}{5}}s^{7}-{\frac {318}{5}}s^{11}+O(s^{15})\end{cases}}}$$

or ${\displaystyle x=-{\frac {18}{5\cdot 2^{1/3}}}y^{5/3}+O(y^{3})}$

Involutes of a circle

For a circle with parametric representation ${\displaystyle (r\cos(t),r\sin(t))}$, one has ${\displaystyle {\vec {c}}'(t)=(-r\sin t,r\cos t)}$. Hence ${\displaystyle |{\vec {c}}'(t)|=r}$, and the path length is ${\displaystyle r(t-a)}$.

Evaluating the above given equation of the involute, one gets

${\displaystyle {\begin{aligned}X(t)&=r(\cos(t+a)+t\sin(t+a))\\Y(t)&=r(\sin(t+a)-t\cos(t+a))\end{aligned}}}$

for the parametric equation of the involute of the circle.

The ${\displaystyle a}$ term is optional; it serves to set the start location of the curve on the circle. The figure shows involutes for ${\displaystyle a=-0.5}$ (green), ${\displaystyle a=0}$ (red), ${\displaystyle a=0.5}$ (purple) and ${\displaystyle a=1}$ (light blue). The involutes look like Archimedean spirals, but they are actually not.

The arc length for ${\displaystyle a=0}$ and ${\displaystyle 0\leq t\leq t_{2}}$ of the involute is

${\displaystyle L={\frac {r}{2}}t_{2}^{2}.}$

Involutes of a semicubic parabola (blue). Only the red curve is a parabola. Notice how the involutes and tangents make up an orthogonal coordinate system. This is a general fact.

The parametric equation ${\displaystyle {\vec {c}}(t)=({\tfrac {t^{3}}{3}},{\tfrac {t^{2}}{2}})}$ describes a semicubical parabola. From ${\displaystyle {\vec {c}}'(t)=(t^{2},t)}$ one gets ${\displaystyle |{\vec {c}}'(t)|=t{\sqrt {t^{2}+1}}}$ and ${\displaystyle \int _{0}^{t}w{\sqrt {w^{2}+1}}\,dw={\frac {1}{3}}{\sqrt {t^{2}+1}}^{3}-{\frac {1}{3}}}$. Extending the string by ${\displaystyle l_{0}={1 \over 3}}$ extensively simplifies further calculation, and one gets

${\displaystyle {\begin{aligned}X(t)&=-{\frac {t}{3}}\\Y(t)&={\frac {t^{2}}{6}}-{\frac {1}{3}}.\end{aligned}}}$

Eliminating t yields ${\displaystyle Y={\frac {3}{2}}X^{2}-{\frac {1}{3}},}$ showing that this involute is a parabola.

The other involutes are thus parallel curves of a parabola, and are not parabolas, as they are curves of degree six (See Parallel curve § Further examples).

The red involute of a catenary (blue) is a tractrix.

For the catenary ${\displaystyle (t,\cosh t)}$, the tangent vector is ${\displaystyle {\vec {c}}'(t)=(1,\sinh t)}$, and, as ${\displaystyle 1+\sinh ^{2}t=\cosh ^{2}t,}$ its length is ${\displaystyle |{\vec {c}}'(t)|=\cosh t}$. Thus the arc length from the point (0, 1) is ${\displaystyle \textstyle \int _{0}^{t}\cosh w\,dw=\sinh t.}$

Hence the involute starting from (0, 1) is parametrized by

${\displaystyle (t-\tanh t,1/\cosh t),}$

and is thus a tractrix.

The other involutes are not tractrices, as they are parallel curves of a tractrix.

Involutes of a cycloid (blue): Only the red curve is another cycloid

The parametric representation ${\displaystyle {\vec {c}}(t)=(t-\sin t,1-\cos t)}$ describes a cycloid. From ${\displaystyle {\vec {c}}'(t)=(1-\cos t,\sin t)}$, one gets (after having used some trigonometric formulas)

${\displaystyle |{\vec {c}}'(t)|=2\sin {\frac {t}{2}},}$

and

${\displaystyle \int _{\pi }^{t}2\sin {\frac {w}{2}}\,dw=-4\cos {\frac {t}{2}}.}$

Hence the equations of the corresponding involute are

${\displaystyle X(t)=t+\sin t,}$

${\displaystyle Y(t)=3+\cos t,}$

which describe the shifted red cycloid of the diagram. Hence

• The involutes of the cycloid ${\displaystyle (t-\sin t,1-\cos t)}$ are parallel curves of the cycloid

${\displaystyle (t+\sin t,3+\cos t).}$

(Parallel curves of a cycloid are not cycloids.)