Difference between revisions of "Involute of a Circle"

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Another equation we can make is that<math>d=a\theta '</math>, 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 <math>\pi</math> when the string is parallel to the x axis <math>\theta '=\frac{\pi}{2}</math> or what would be 90 degrees. Then if <math>a=2</math>,  <math>{\frac{\pi}{2}}2=\pi</math>. So <math>d=a\theta '</math>
 
Another equation we can make is that<math>d=a\theta '</math>, 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 <math>\pi</math> when the string is parallel to the x axis <math>\theta '=\frac{\pi}{2}</math> or what would be 90 degrees. Then if <math>a=2</math>,  <math>{\frac{\pi}{2}}2=\pi</math>. So <math>d=a\theta '</math>
  
Now we must find what <math>d=a\theta</math> is in terms of <math>d=a\theta '</math>. Because <math>d=a\theta</math> is the angle from the x axis to the hypotenuse of the right triangle and <math>\theta '</math> is the angle from the x axis to the base of the right triangle, <math>\theta '-\theta</math> is the measure of the angle in between <math>a</math> and <math>r</math>. Which can also be represented as <math>\cos^{-1}\left( \frac{a}{r}\right)</math>.  
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Now we must find what <math>\theta</math> is in terms of <math>\theta '</math>.
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Because <math>\theta</math> is the angle from the x axis to the hypotenuse of the right triangle and <math>\theta '</math> is the angle from the x axis to the base of the right triangle, <math>\theta '-\theta</math> is the internal angle measure of the triangle. Which can also be represented as <math>\cos^{-1}\left( \frac{a}{r}\right)</math>.  
  
 
Put this all together and <math>\theta '=\theta+\cos^{-1}\left( \frac{a}{r}\right)</math>
 
Put this all together and <math>\theta '=\theta+\cos^{-1}\left( \frac{a}{r}\right)</math>
  
Simplify this so that <math>\theta</math> is by its self, <math>\theta=\theta '-\cos^{-1}\left(\frac{a}{r}\right)</math>
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Simplify this so that <math>\theta</math> is by its self,  
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<math>\theta=\theta '-\cos^{-1}\left(\frac{a}{r}\right)</math>
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Then from before, we know that <math>d=a\theta '</math> so we can also say that <math>\theta '=\frac{d}{a}</math>
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Now we need to get the <math>d</math> out of that equation. All we have to do is substitute what we already found out from before, <math>d=\sqrt{r^2-a^2}</math>.
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So <math>\theta '=\frac{\sqrt{r^2-a^2}}{a}</math>
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Substitute this into <math>\theta '=\theta+\cos^{-1}\left( \frac{a}{r}\right)</math>
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We now have <math>\theta=\frac{\sqrt{r^2-a^2}}{a}-\cos^{-1}\left(\frac{a}{r}\right)</math>.
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Just subtract from both sides to get <math>\theta</math> on one side and the final equation is
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<math>\theta=\frac{\sqrt{r^2-a^2}}{a}-\cos^{-1}\left(\frac{a}{r}\right)</math>
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If you would like to check this, just substitute 2 for <math>a</math> and <math><math>\sqrt{2^2 + \pi^2}</math> for <math>r</math> and compare it to <math>\tan^{-1}\left(\frac{\pi}{2}\right)</math>
 
|other=Alegbra 2, Geometry, Pre-Calculus
 
|other=Alegbra 2, Geometry, Pre-Calculus
 
|AuthorName=Wyatt S.C.
 
|AuthorName=Wyatt S.C.

Revision as of 17:34, 21 April 2012

Inprogress.png
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 \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 <math>\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.

Involute wheel.gif


Teaching Materials

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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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