Involute

Properties of involutes

In order to derive properties of a regular curve it is advantageous to suppose the arc length ${\displaystyle s}$ to be the parameter of the given curve, which lead to the following simplifications: ${\displaystyle |{\overrightarrow{c}}^{\prime }(s)|=1}$ and ${\displaystyle {\overrightarrow{c}}^{″}(s)=\kappa (s)\overrightarrow{n}(s)}$, with ${\displaystyle \kappa }$ the curvature and ${\displaystyle \overrightarrow{n}}$ the unit normal. One gets for the involute:

${\displaystyle {\overrightarrow{C}}_a(s)=\overrightarrow{c}(s)-{\overrightarrow{c}}^{\prime }(s)(s-a)}$ and

${\displaystyle {\overrightarrow{C}}_a^{\prime }(s)=-{\overrightarrow{c}}^{″}(s)(s-a)=-\kappa (s)\overrightarrow{n}(s)(s-a)}$

and the statement:

• At point ${\displaystyle {\overrightarrow{C}}_a(a)}$ the involute is not regular (because ${\displaystyle |{\overrightarrow{C}}_a^{\prime }(a)|=0}$ ),

and from ${\displaystyle {\overrightarrow{C}}_a^{\prime }(s)\cdot {\overrightarrow{c}}^{\prime }(s)=0}$ follows:

• The normal of the involute at point ${\displaystyle {\overrightarrow{C}}_a(s)}$ is the tangent of the given curve at point ${\displaystyle \overrightarrow{c}(s)}$.
• The involutes are parallel curves, because of ${\displaystyle {\overrightarrow{C}}_a(s)={\overrightarrow{C}}_0(s)+a{\overrightarrow{c}}^{\prime }(s)}$ and the fact, that ${\displaystyle {\overrightarrow{c}}^{\prime }(s)}$ is the unit normal at ${\displaystyle {\overrightarrow{C}}_0(s)}$.

The family of involutes and the family of tangents to the original curve makes up an orthogonal coordinate system. Consequently, one may construct involutes graphically. First, draw the family of tangent lines. Then, an involute can be constructed by always staying orthogonal to the tangent line passing the point.

Cusps

This section is based on work by Huygens and Barrow.

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.

cusp of order 3/2

For the first type, one can start by the involute of a circle, with equation$${\displaystyle \begin{array}{cc}X(t)& =r(\cos t+(t-a)\sin t)\\ Y(t)& =r(\sin t-(t-a)\cos t)\end{array}}$$then set ${\displaystyle a=0}$, and expand for small ${\displaystyle t}$, to obtain$${\displaystyle \begin{array}{cc}X(t)& =r+r{t}^2/2+O(t^4)\\ Y(t)& =r{t}^3/3+O(t^5)\end{array}}$$thus giving the order 3/2 curve ${\displaystyle Y^2-\frac{8}{9r}(X-r{)}^3+O(Y^{8/3})=0}$, a semicubical parabola.

cusp of order 5/2

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$${\displaystyle \{\begin{array}{l}x(s)=s+(L-s-\frac{9}{10}{s}^5+\cdots )\cos \theta \\ y(s)=s^3+(L-s-\frac{9}{10}{s}^5+\cdots )\sin \theta \end{array}}$$Expand it up to order ${\displaystyle s^5}$, we obtain$${\displaystyle \{\begin{array}{l}x(s)=L-\frac{9}{2}L{s}^4+(\frac{9}{2}L-\frac{9}{10})s^5+O(s^6)\\ y(s)=3L{s}^2-2s^3+O(s^6)\end{array}}$$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 $${\displaystyle x=L-\frac{y^2}{2L}\pm \left(\frac{9}{2}L+\frac{51}{10}\right){\left(\frac{y}{3L}\right)}^{2.5}+O(y^{2.75}),y\ge 0}$$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$${\displaystyle \{\begin{array}{l}x(s)=\frac{18}{5}{s}^5-\frac{126}{5}{s}^9+O(s^{13})\\ y(s)=-2s^3+\frac{54}{5}{s}^7-\frac{318}{5}{s}^{11}+O(s^{15})\end{array}}$$or ${\displaystyle x=-\frac{18}{5\cdot 2^{1/3}}{y}^{5/3}+O(y^3)}$

Examples

Involutes of a circle

For a circle with parametric representation ${\displaystyle (r\cos (t),r\sin (t))}$, one has ${\displaystyle {\overrightarrow{c}}^{\prime }(t)=(-r\sin t,r\cos t)}$. Hence ${\displaystyle |{\overrightarrow{c}}^{\prime }(t)|=r}$, and the path length is ${\displaystyle r(t-a)}$.

Evaluating the above given equation of the involute, one gets

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

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 ${\displaystyle 0\le t\le t_2}$ of the involute is

${\displaystyle L=\frac{r}{2}{t}_2^2.}$

Involutes of a semicubic parabola

The parametric equation ${\displaystyle \overrightarrow{c}(t)=(\frac{t^3}{3},\frac{t^2}{2})}$ describes a semicubical parabola. From ${\displaystyle {\overrightarrow{c}}^{\prime }(t)=(t^2,t)}$ one gets ${\displaystyle |{\overrightarrow{c}}^{\prime }(t)|=t\sqrt{t^2+1}}$ 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 ${\displaystyle l_0=\frac{1}{3}}$ extensively simplifies further calculation, and one gets

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

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).

Involutes of a catenary

For the catenary ${\displaystyle (t,\cosh t)}$, the tangent vector is ${\displaystyle {\overrightarrow{c}}^{\prime }(t)=(1,\sinh t)}$, and, as ${\displaystyle 1+{\sinh }^{2}t={\cosh }^{2}t,}$ its length is ${\displaystyle |{\overrightarrow{c}}^{\prime }(t)|=\cosh t}$. Thus the arc length from the point (0, 1) is ${\displaystyle {\int }_0^t\cosh wdw=\sinh t.}$

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

The parametric representation ${\displaystyle \overrightarrow{c}(t)=(t-\sin t,1-\cos t)}$ describes a cycloid. From ${\displaystyle {\overrightarrow{c}}^{\prime }(t)=(1-\cos t,\sin t)}$, one gets (after having used some trigonometric formulas)

${\displaystyle |{\overrightarrow{c}}^{\prime }(t)|=2\sin \frac{t}{2},}$

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.)

Involute and evolute

The evolute of a given curve ${\displaystyle c_0}$ consists of the curvature centers of ${\displaystyle c_0}$. Between involutes and evolutes the following statement holds:

A curve is the evolute of any of its involutes.

Involute and evolute

Application

The most common profiles of modern gear teeth are involutes of a circle. In an involute gear system, the teeth of two meshing gears contact at a single instantaneous point that follows along a single straight line of action. The forces the contacting teeth exert on each other also follow this line and are normal to the teeth. The involute gear system maintaining these conditions follows the fundamental law of gearing: the ratio of angular velocities between the two gears must remain constant throughout.

With teeth of other shapes, the relative speeds and forces rise and fall as successive teeth engage, resulting in vibration, noise, and excessive wear. For this reason, nearly all modern planar gear systems are either involute or the related cycloidal gear system.

The involute of a circle is also an important shape in gas compressing, as a scroll compressor can be built based on this shape. Scroll compressors make less sound than conventional compressors and have proven to be quite efficient.

The High Flux Isotope Reactor uses involute-shaped fuel elements, since these allow a constant-width channel between them for coolant.