# Involute of a Circle

Involute of a Circle
Field: Geometry
Image Created By: Wyatt S.C.

Involute of a Circle

The involute of a circle is a curve formed by an imaginary string attached at fix point pulled taut either unwinding or winding around a circle.

# A More Mathematical Explanation

Note: understanding of this explanation requires: *Alegbra 2, Geometry, Pre-Calculus

When deriving the equation to graph the involute of a circle, it actually has to do with measuring ri [...]

When deriving the equation to graph the involute of a circle, it actually has to do with measuring right triangles.

If you take a point on the involute of a circle with radius 2, where the imaginary string is unwinding and starts at point (2,0), and the string is parallel to the x axis for the first time, that length would be $\pi$. This is because the imaginary string would have unwound a quarter of the circle's circumfrence. So $\frac{4\pi}{4}$ is $\pi$

The radius is 2, so using those two measurements we can find $r$ using the pythagorean theorem(or the distance from the origin to the point on the involute curve. So $r$ would equal$\sqrt{2^2 + \pi^2}$. Then, to find $\theta$ one would have to subtract the angle DAC from $\frac{\pi}{2}$. In this case it would be $\tan^{-1}\left(\frac{\pi}{2}\right)$

Now to form an equation, we just use variables. We name a as the radius, d as the length of the tangent generating out point, r as the distance from the origin to our point (also the hypotenuse of our triangle), $\theta$ is the angle of our point, and $\theta '$ is the angle to the base of the triangle.

So if $r=\sqrt{2^2 + \pi^2}$ then $\pi^2=\sqrt{r^2 + 2^2}$ and using variables $d=\sqrt{r^2-a^2}$

Another equation we can make is that$d=a\theta '$, we know this because of the first thing that we did. We said that when the string would be parallel to the x axis, it would have a length of $\pi$ when the string is parallel to the x axis $\theta '=\frac{\pi}{2}$ or what would be 90 degrees. Then if $a=2$, ${\frac{\pi}{2}}2=\pi$. So $d=a\theta '$

Now we must find what $\theta$ is in terms of $\theta '$.

Because $\theta$ is the angle from the x axis to the hypotenuse of the right triangle and $\theta '$ is the angle from the x axis to the base of the right triangle, $\theta '-\theta$ is the internal angle measure of the triangle. Which can also be represented as $\cos^{-1}\left( \frac{a}{r}\right)$.

Put this all together and $\theta '=\theta+\cos^{-1}\left( \frac{a}{r}\right)$

Simplify this so that $\theta$ is by its self,

$\theta=\theta '-\cos^{-1}\left(\frac{a}{r}\right)$

Then from before, we know that $d=a\theta '$ so we can also say that $\theta '=\frac{d}{a}$

Now we need to get the $d$ out of that equation. All we have to do is substitute what we already found out from before, $d=\sqrt{r^2-a^2}$.

So $\theta '=\frac{\sqrt{r^2-a^2}}{a}$

Substitute this into $\theta '=\theta+\cos^{-1}\left( \frac{a}{r}\right)$

We now have $\theta=\frac{\sqrt{r^2-a^2}}{a}-\cos^{-1}\left(\frac{a}{r}\right)$.

Just subtract from both sides to get $\theta$ on one side and the final equation is $\theta=\frac{\sqrt{r^2-a^2}}{a}-\cos^{-1}\left(\frac{a}{r}\right)$

If you would like to check this, just substitute 2 for $a$ and $[itex]\sqrt{2^2 + \pi^2}$ for $r$ and compare it to $\tan^{-1}\left(\frac{\pi}{2}\right)$

# Why It's Interesting

This is very interesting for many reasons. It is amazing that what looks to be a very complex figure's equation can easily be derived using understanding of just geometry and some pre calculus.

The involute of a circle appears commonly in every day life. Other than the simple tetherball which is more of a model for the involute of a circle. The most commonly used gear system utilizes the involute of a circle. The teeth of the gear are involutes.

This allows the contact of the two interlocking teeth to occur at a single point that moves along the tooth. This allows the transfer of energy to one powered gear to a powerless gear smooth and not require as much energy.