Difference between revisions of "Involute of a Circle"

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{{Image Description
 
{{Image Description
 
|ImageName=Involute of a Circle
 
|ImageName=Involute of a Circle
|Image=Involute of a circle.gif
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|Image=Involute_of_a_circle.gif‎
 
|ImageIntro=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.
 
|ImageIntro=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.
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|ImageDescElem=The involute of a circle can most easily be represented as a tetherball winding around a pole or unwinding around a pole.
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Every 360 degree rotation the ball makes the string shortens or lengthens however many inches thick that the pole is.
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The involute of a circle (or anything for that matter) is a type of roulette, to find out more about this topic see [[Roulette]].
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See [[Involute]] for a short description on what the involute of a circle is if you still are unsure.
 
|ImageDesc=When deriving the equation to graph the involute of a circle, it actually has to do with measuring right triangles.
 
|ImageDesc=When deriving the equation to graph the involute of a circle, it actually has to do with measuring right triangles.
  
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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
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[[Image:Involute_Trianglesedit.jpg|right|thumb|350px|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 <math>\pi</math>. This is because the imaginary string would have unwound a quarter of the circle's circumfrence. So <math>\frac{4\pi}{4}</math> is <math>\pi</math>
 
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 <math>\pi</math>. This is because the imaginary string would have unwound a quarter of the circle's circumfrence. So <math>\frac{4\pi}{4}</math> is <math>\pi</math>
  
The radius is 2, so using those two measurements we can find  <math>r</math> using the pythagorean theorem(or the distance from the origin to the point on the involute curve. So  <math>r</math> would equal<math>\sqrt{2^2 + \pi^2}</math>. Then, to find  <math>\theta</math> one would have to subtract the angle DAC from  <math>\frac{\pi}{2}</math>. In this case it would be <math>\tan^{-1}\left(\frac{\pi}{2}\right)</math>
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The radius is 2, so using those two measurements we can find  <math>r</math> using the pythagorean theorem(or the distance from the origin to the point on the involute curve. So  <math>r</math> would equal<math>\sqrt{2^2 + \pi^2}</math>. Then, to find  <math>\theta</math> one would have to subtract the internal angle of the right triangle from  <math>\frac{\pi}{2}</math>. In this case it would be <math>\tan^{-1}\left(\frac{\pi}{2}\right)</math> so
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<math>\theta=\frac{\pi}{2}-\tan^{-1}\left(\frac{\pi}{2}\right)</math>
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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), <math>\theta</math> is the angle of our point, and <math>\theta '</math> is the angle to the base of the triangle.
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So if <math>r=\sqrt{2^2 + \pi^2}</math> then <math>\pi^2=\sqrt{r^2 + 2^2}</math>  and using variables
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<math>d=\sqrt{r^2-a^2}</math>
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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>
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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>.
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Put this all together and <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>.
  
Now we can put these into variables, instead <math>\pi</math> for the length of the string from the point tangent
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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>
  
So if you call the radius of the circle <math>a</math> (2 in this case) and
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If you would like to check this, just substitute 2 for <math>a</math> and <math>\sqrt{2^2 + \pi^2}</math> for <math>r</math> and compare it to <math>\tan^{-1}\left(\frac{\pi}{2}\right)</math>, they should both come out to <math>\theta=5.669115049</math>
 
|other=Alegbra 2, Geometry, Pre-Calculus
 
|other=Alegbra 2, Geometry, Pre-Calculus
 
|AuthorName=Wyatt S.C.
 
|AuthorName=Wyatt S.C.
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[[Image:Involute_wheel.gif]]
 
[[Image:Involute_wheel.gif]]
|References=http://en.wikipedia.org/wiki/Involute#Involute_of_a_circle
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|References=http://mathforum.org/mathimages/index.php/Roulette
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http://en.wikipedia.org/wiki/Involute#Involute_of_a_circle
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http://en.wikipedia.org/wiki/Involute_gear
 
http://en.wikipedia.org/wiki/Involute_gear
|InProgress=Yes
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|InProgress=No
 
}}
 
}}

Latest revision as of 18:10, 21 April 2012


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.


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.

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.

Involute wheel.gif


Teaching Materials

There are currently no teaching materials for this page. Add teaching materials.




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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