Involute of a Circle

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Involute of a Circle
Involute of a circle.gif
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 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 d=a\theta is in terms of d=a\theta '. Because d=a\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 measure of the angle in between a and r. 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)


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.

Involute wheel.gif


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References

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





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