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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1568.0. "A Knotty problem" by CIVAGE::LYNN (Lynn Yarbrough @WNP DTN 427-5663) Wed Feb 19 1992 16:34

I am trying to put together a really nice demo of the capabilites of MAPLE
and Mathematica in overcoming the inadequacy of intuition in math. One idea
that I have seen elsewhere and am trying to implement is to show that cold
equations fail to reveal a knotted surface, specifically a trefoil knot, in
3-space. Right now I am going crazy trying to find the parameteric polar
coordinates for the 2-D projection of the center curve of the knot. What I
am seeking looks something like: 

		-
	      /   \
	      |    |
	     -|------
	    /  \  /  \
	   |     /    |
	    \___/ \__/


Anyone have any ideas?
T.RTitleUserPersonal
Name
DateLines
1568.1an exampleALLVAX::JROTHI know he moves along the piersThu Feb 20 1992 02:3422
    An example (hacked together) that shows a rotating trefoil knot
    can be found in 3D::USER$02:[ROTH.PUBLIC].

    The following type of parameterization seems obvious to try:

	r(t) = 1 + a*cos(3*t/2)
	x(t) = r(t)*cos(t)
	y(t) = r(t)*sin(t)
	z(t) = h*sin(3*t/2)

	0 <= t < 4*PI, a & h = some adhoc shape params

    I recommend sweeping a frame along this curve and reorthogonalizing
    it from point to point against the new tangent vector (rather than
    using the Frenet-Serret frames, since these give bad behaviour if
    the curvature gets very small along the curve) if you want to
    show a "thick" curve instead of just a line.

    Torus knots look pretty nice too, and you can intertwine several of them
    for an impressive display.

    - Jim
1568.2Another solution and where it came from.CADSYS::COOPERTopher CooperThu Feb 20 1992 14:1645
    Here's another solution.  It's conceivable that this is equivalent
    to Jim's, but it isn't obvious that it is.

	x(t) = (1-s)*R*sin(t) - s*R*sin(t/2)
	y(t) = (1-s)*R*cos(t) + s*R*cos(t/2)

    R is a size parameter.  If your were to circumscribe a circle around
    the x-y projection of the trefoil it would have radius R.  "s" is a
    shape parameter which should be somewhere between 0 and 2/3.  s=1/3
    seems to be about right.

    Using the same z as Jim and setting its extremes to R we get:

	z(t) = R*sin(3*t/2)

    Here's where these came from:

    My first thought was that the projection was like the figures produced
    by my daughter's spirograph.  The spirograph works by having a pen
    placed through a hole in the interior of a circular gear, which is
    rolled around the circular interior of a larger (internally toothed)
    ring-shaped gear.

    I thought I knew that the name for the figure produced by the
    spirograph is "hypocycloid" -- though it turns out I was slightly wrong
    on that.  Anyway I looked up the hypocycloid's formula and discovered
    that that name applies only when the "interior" point is on the edge
    of the smaller circle.  The correct term is hypotrochoid.  (If the
    moving circle is on the exterior of the fixed circle the result is an
    epicycloid or an epitrochoid, by the way).

    In order to get three lobes the radius of the moving circle had to be
    either 1/3 or 2/3 the radius of the fixed circle.  In the former case
    the result would be a loop-the-loop around the central point, something
    like the familiar pictures showing the apparent retrograde motions of
    the outer planets, so I could set the smaller radius to 2/3 the larger
    one.  I set the distance from the center of the moving circle to the
    tracing point to be a proportion, s, of the radius of the final figure,
    R.  Finally I solved for the radius of the fixed circle in terms of R
    and s and substituted.  When I plotted the result I found that it was
    sideways, so I swapped x and y.  I tried s=1/2 and that didn't work
    (everything crossed at the center point) so I tried s=1/3 and that
    looked pretty good.  That's it.

				    Topher
1568.3Thanks!CIVAGE::LYNNLynn Yarbrough @WNP DTN 427-5663Thu Feb 20 1992 15:551
Thannks to all. Topher, that was exactly what I was looking for.