- #1
saviourmachine
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The idea
An epicycloid is a superimposed circle on another circle. http://www.math.dartmouth.edu/~dlittle/java/SpiroGraph/ can you find a java applet to show you. These epicycloids are tied to each other at their circumferences. But, what does change when using a slightly easier method, and superimpose the second circle (the referent) using it's centre!?
A point on the first circle moves along its circumference, and because it is the centre of the second circle, the whole second circle does move with along it. Now comes the clue: do the same with the second circle. Take a point at the circumference of the second circle and move in the opposite direction (with a negative frequency). Like you can see will this point move along the horizontal axis. Not much have to be imagined to realize that this traject will be equal to [tex]2 \cos{\omega}[/tex]. Of course is this equals the sum of [tex]\exp{j \omega t}[/tex] and [tex]\exp{-j \omega t}[/tex], but it's cool that with rotating in the other direction a meaning is assigned to the concept "negative frequency".
http://www.annevanrossum.nl/pictures/science/Epicycloid.gif
Does anyone know of a clarification of Euler's relations using these drawing techniques?
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Edit: Changed math to Latex.
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Observation: Odd, that the image can't be displayed. That's for paying members?
An epicycloid is a superimposed circle on another circle. http://www.math.dartmouth.edu/~dlittle/java/SpiroGraph/ can you find a java applet to show you. These epicycloids are tied to each other at their circumferences. But, what does change when using a slightly easier method, and superimpose the second circle (the referent) using it's centre!?
A point on the first circle moves along its circumference, and because it is the centre of the second circle, the whole second circle does move with along it. Now comes the clue: do the same with the second circle. Take a point at the circumference of the second circle and move in the opposite direction (with a negative frequency). Like you can see will this point move along the horizontal axis. Not much have to be imagined to realize that this traject will be equal to [tex]2 \cos{\omega}[/tex]. Of course is this equals the sum of [tex]\exp{j \omega t}[/tex] and [tex]\exp{-j \omega t}[/tex], but it's cool that with rotating in the other direction a meaning is assigned to the concept "negative frequency".
http://www.annevanrossum.nl/pictures/science/Epicycloid.gif
Does anyone know of a clarification of Euler's relations using these drawing techniques?
___
Edit: Changed math to Latex.
__
Observation: Odd, that the image can't be displayed. That's for paying members?
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