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Parametrization of a torus

  1. Jun 1, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the parametrization of torus given by:

    x=x(ø,ß)=(3+cos(ø))cos(ß)
    y=y(ø,ß)=(3+cos(ø))sin(ß)
    z=z(ø,ß)=sin(ø),

    for 0≤ø,ß≤2π

    What is the radius of the circle that runs through the center of the tube, and what is the radius of the tube, measured from the central circle?

    2. Relevant equations

    ?

    3. The attempt at a solution

    Hmmmm..... I don't get it. The distance from a point on the torus to the center would be √(x2+y2+z2), I think. But the center of the tube? Not sure how that would work. H

    Help is appreciated. And after that, I may ask some more questions. :smile: I know you folks are ready, willing, and able.
     
  2. jcsd
  3. Jun 1, 2010 #2

    LCKurtz

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    Hint: It might help you to write it like this:

    [tex]\langle x,y,z\rangle=\langle 3\cos\beta,3\sin\beta,0\rangle+
    \langle\cos\phi\cos\beta,\cos\phi\sin\beta,\sin\phi\rangle[/tex]

    and look at the lengths of those vectors.
     
  4. Jun 1, 2010 #3
    But what does the length of those vectors have to do with anything?
     
  5. Jun 1, 2010 #4

    HallsofIvy

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    If [itex]\beta= 0[/itex], [itex]x= 3+ cos(\phi)[/itex], [itex]y= 0[/itex], and [itex]z= sin(\phi)[/itex] so that [itex](x-3)^2+ z^2= 1[/itex]. That's a circle in the xz-plane with center at (3, 0, 0) and radius 1. Similarly, if [itex]\beta= \pi/2[/itex], [itex]x= 0[/itex], [itex]y= 3+ cos(\phi)[/itex], and [itex]z= sin(\phi)[/itex] so that [itex](y-3)^2+ z^2= 1[/itex]. That's a circle in the yz-plane with center at (0, 3, 0) and radius 1. With [itex]\beta= \pi[/itex] and [itex]\beta= 3\pi/2[/itex] we get circles with radii 1 and centers at (-3, 0, 0) and (0, -3, 0), respectively. In general, as [itex]\beta[/itex] ranges from 0 to [itex]2\pi[/itex] we get circles with radii 1 and centers sweeping out a circle with center at (0, 0, 0) and radius 3. That's the "center of the tube".

     
  6. Jun 1, 2010 #5
    Oh, okay. It just seems like if you had donut floating in space that if defined by some parametrization x(∂,ß), y(∂,ß), z(∂,ß), values of ∂,ß would map to a point on the surface. Like I know if you have x(t), y(t), you can regions in the x-y plane into cool things like cartoids, that could not be made with just x, y=f(x). I'm just confused about how exactly the equation for the radius of a sphere in x-y-z space gets turned into the a formula for the distance between the origin and the center of a donut. :confused:
     
  7. Jun 1, 2010 #6

    LCKurtz

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    Here's a picture that may help you understand the parameterization:

    torus.jpg

    Edit: The φ in the picture should be α or vice - versa. Too much trouble to change it now :grumpy:
     
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