Involute of a Circle

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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.


Contents

Basic Description

The involute of a circle can most easily be represented as a tetherball winding around a pole or unwinding around a pole.

Every 360 degree rotation the ball makes the string shortens or lengthens however many inches thick that the pole is.

The involute of a circle (or anything for that matter) is a type of roulette, to find out more about this topic see Roulette.

See Involute for a short description on what the involute of a circle is if you still are unsure.

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.

See diagram to right for reference, points C,E,G, and I are all points on the Involute but we will be focusing on C. Also, excuse the circle not being radius 2, use your imagination so that it is

Diagram of a few points on the involute of a circle.
Diagram of a few points on the involute of a circle.

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 internal angle of the right triangle from \frac{\pi}{2}. In this case it would be \tan^{-1}\left(\frac{\pi}{2}\right) so

\theta=\frac{\pi}{2}-\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 thatd=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 \sqrt{2^2 + \pi^2} for r and compare it to \tan^{-1}\left(\frac{\pi}{2}\right), they should both come out to \theta=5.669115049


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.

Image:Involute_wheel.gif


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References

http://mathforum.org/mathimages/index.php/Roulette

http://en.wikipedia.org/wiki/Involute#Involute_of_a_circle

http://en.wikipedia.org/wiki/Involute_gear





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