[math-fun] line segment passing through a hyperbola-shaped slit
Is there a video on the web of a line segment passing through a hyperbola-shaped slit, as in the Eames Mathematica exhibit (and probably other math museum exhibits as well)? I want to show this to my calculus students tomorrow, and was astonished that I couldn't find it on the web even after several minutes of searching ("... Eames ruled surface hyperbola ...". Thanks, Jim Propp
The Exploratorium has this exhibit but I was only able to find still photos of it in my searches. They call it "Hyperbolic Slot". Then I found http://www.youtube.com/watch?v=OJGkdO9gcdY which isn't gerat but should suffice. --Joshua Zucker On Tue, Nov 22, 2011 at 1:58 PM, James Propp <jamespropp@gmail.com> wrote:
Is there a video on the web of a line segment passing through a hyperbola-shaped slit, as in the Eames Mathematica exhibit (and probably other math museum exhibits as well)?
I want to show this to my calculus students tomorrow, and was astonished that I couldn't find it on the web even after several minutes of searching ("... Eames ruled surface hyperbola ...".
Thanks,
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Thanks! It occurs to me to ask: Is there a parametrization of the one-sheeted hyperboloid that makes it manifest that it's a ruled surface? And: Is there a parametrization that makes it manifest that it's a doubly-ruled surface? Jim On Tue, Nov 22, 2011 at 5:11 PM, Joshua Zucker <joshua.zucker@gmail.com>wrote:
The Exploratorium has this exhibit but I was only able to find still photos of it in my searches. They call it "Hyperbolic Slot". Then I found http://www.youtube.com/watch?v=OJGkdO9gcdY which isn't gerat but should suffice.
--Joshua Zucker
On Tue, Nov 22, 2011 at 1:58 PM, James Propp <jamespropp@gmail.com> wrote:
Is there a video on the web of a line segment passing through a hyperbola-shaped slit, as in the Eames Mathematica exhibit (and probably other math museum exhibits as well)?
I want to show this to my calculus students tomorrow, and was astonished that I couldn't find it on the web even after several minutes of searching ("... Eames ruled surface hyperbola ...".
Thanks,
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
In principle, an answer is that the lines of a regulus have Pluecker coordinate 6-vectors which are linear combinations of any 3 distinct fixed members. Since these coordinates are projective / homogeneous, and a line must also satisfy the (quadratic) Grassmann constraint L ^ L = 0, there is only one degree of freedom left over to generate the real regulus. Given a pair of conjugate reguli, the points can be now generated as intersections of pairs of lines, one from each regulus. Each point is specified by two pairs of (inhomogenous) parameters, each pair subject to a quadratic constraint: I think it must be straightforward to parameterise the solutions, though I don't know the answer offhand. Fred Lunnon On 11/23/11, James Propp <jamespropp@gmail.com> wrote:
Thanks!
It occurs to me to ask: Is there a parametrization of the one-sheeted hyperboloid that makes it manifest that it's a ruled surface?
And: Is there a parametrization that makes it manifest that it's a doubly-ruled surface?
Jim
On Tue, Nov 22, 2011 at 5:11 PM, Joshua Zucker <joshua.zucker@gmail.com>wrote:
The Exploratorium has this exhibit but I was only able to find still photos of it in my searches. They call it "Hyperbolic Slot". Then I found http://www.youtube.com/watch?v=OJGkdO9gcdY which isn't gerat but should suffice.
--Joshua Zucker
On Tue, Nov 22, 2011 at 1:58 PM, James Propp <jamespropp@gmail.com> wrote:
Is there a video on the web of a line segment passing through a hyperbola-shaped slit, as in the Eames Mathematica exhibit (and probably other math museum exhibits as well)?
I want to show this to my calculus students tomorrow, and was astonished that I couldn't find it on the web even after several minutes of searching ("... Eames ruled surface hyperbola ...".
Thanks,
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
A pencil of planes in 3-space is the set meeting in a common line. Given any two distinct members E,F of the pencil, the (4-vector homogeneous coordinate representing the equation of the) general member plane of a pencil is expressible as a linear combination a E + b F in this basis. A (linear) correspondence between two pencils with given bases E,F and G,H maps a E + b F to a G + b H . Given two skew lines L,M in 3-space, and a linear correspondence between them defined via pairs of basis planes, the set of lines in which corresponding pairs of planes meet {N} = {(a E + b F) cap (a G + b H)} as a/b varies form a regulus. [This must be well-known, though I don't recall seeing it stated. Writing out the Pluecker coordinate 6-vector of the general line N in terms of E,F,G,H; a,b, it should be immediately apparent that the rank is 4 as a,b vary, yielding an algebraic proof.] It is easy to see that L,M meet every line of the first regulus, and so must lie on its second, conjugate regulus: the latter could be generated dually from any two lines L',M' of the first. But I can't at the moment see how to construct a third line of the second regulus, which would inter alia yield an alternative synthetic proof. Once we have both reguli, the surface points are parameterised immediately as intersections of (any 3 of) planes a E + b F, a G + b H, a' E' + b' F', a' G' + b' H', one pair from each correspondence, in terms of two inhomogeneous parameters a/b, a'/b' [incidentally avoiding the awkward calculation of the intersection point of two lines in 3-space.] Notice that the freedom of a (quadric) regulus with two chosen (axis) lines equals 9+2 = 11, matching the freedom of a correspondence between pencils 11 = 2.4 (for two axis lines) + 2 (for base planes in second pencil) + 1 (for homogeneous weight of any one line), where missing planes and weights can be absorbed into the parameters. As a practical technique for plotting a one-sheet hyperboloid, this method requires some form of adaptive step-control in order to ensure even distribution. I have previously implemented its analogue in 2-space, plotting a conic as the locus of the intersection point of pairs of lines, from two corresponding pencils of lines through common points. Line geometry is discussed in Helmut Pottmann's fine book, and in a series of online articles by Renatus Ziegler. There is a brief summary of Pluecker coordinates in the appendix to my paper on bicycle spokes at https://docs.google.com/leaf?id=0B6QR93hqu1AhZThjZDBkYTQtNWMxNS00OWM3LWI5YWI... (no line-breaks) Fred Lunnon On 11/23/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
In principle, an answer is that the lines of a regulus have Pluecker coordinate 6-vectors which are linear combinations of any 3 distinct fixed members. Since these coordinates are projective / homogeneous, and a line must also satisfy the (quadratic) Grassmann constraint L ^ L = 0, there is only one degree of freedom left over to generate the real regulus.
Given a pair of conjugate reguli, the points can be now generated as intersections of pairs of lines, one from each regulus. Each point is specified by two pairs of (inhomogenous) parameters, each pair subject to a quadratic constraint: I think it must be straightforward to parameterise the solutions, though I don't know the answer offhand.
Fred Lunnon
On 11/23/11, James Propp <jamespropp@gmail.com> wrote:
Thanks!
It occurs to me to ask: Is there a parametrization of the one-sheeted hyperboloid that makes it manifest that it's a ruled surface?
And: Is there a parametrization that makes it manifest that it's a doubly-ruled surface?
Jim
On Tue, Nov 22, 2011 at 5:11 PM, Joshua Zucker <joshua.zucker@gmail.com>wrote:
The Exploratorium has this exhibit but I was only able to find still photos of it in my searches. They call it "Hyperbolic Slot". Then I found http://www.youtube.com/watch?v=OJGkdO9gcdY which isn't gerat but should suffice.
--Joshua Zucker
On Tue, Nov 22, 2011 at 1:58 PM, James Propp <jamespropp@gmail.com> wrote:
Is there a video on the web of a line segment passing through a hyperbola-shaped slit, as in the Eames Mathematica exhibit (and probably other math museum exhibits as well)?
I want to show this to my calculus students tomorrow, and was astonished that I couldn't find it on the web even after several minutes of searching ("... Eames ruled surface hyperbola ...".
Thanks,
Jim Propp
Thanks!
It occurs to me to ask: Is there a parametrization of the one-sheeted hyperboloid that makes it manifest that it's a ruled surface?
I believe the following is a valid parametrisation, for much the same reason as the hyperbolic slot demonstration works: x = p cos(theta) + sin(theta), y = p sin(theta) - cos(theta), z = p, where p is a real number; theta is in the interval (-pi,pi]. When theta is constant, varying p corresponds to motion along a straight line. Substituting these expressions into x^2 + y^2 = z^2 + 1 confirms that this parametrisation does indeed describe the hyperboloid.
And: Is there a parametrization that makes it manifest that it's a doubly-ruled surface?
Yes, if there are twelve constants, {a,b,c,d,e,f,g,h,i,j,k,l}, such that: x = apq + bp + cq + d, y = epq + fp + gq + h, z = ipq + jp + kq + l, x^2 + y^2 = z^2 + 1, for all values of {p,q} real. Equating parts, these constraints collapse into nine quadratic equations in twelve variables, which (when solved) will give a valid parametrisation. (Proof: if either p or q is varied, x, y and z will all change linearly.) Sincerely, Adam P. Goucher
Implementing the method I sketched earlier for a general one-sheet hyperboloid in canonical position x^2/a^2 + y^2/b^2 - z^2/c^2 = w^2 makes it easy to locate lines missing from reguli using symmetry. Four lines from each regulus are face diagonals of the cuboid with corners [+/- a, +/- b, +/- c], and base planes for the correspondence between pencils through two of them can be selected from corner-slices +/- x/a +/- y/b +/- z/c = 1 . The result is gratifyingly elegant: a general point P(s, t) on the surface has homogeneous coordinate vector [w,x,y,z] = [s*t+1, -a*(s+t), b*(s*t-1), c*(-s+t)] . It is trivial to check that P satisfies the (homogeneous) quadric equation; and being linear in s and t separately, P describes a line when either parameter is fixed. A missing subset of measure zero is reached by letting s -> oo or t -> oo . Showing their line intersection sweeping out the surface as the two moving planes rotate about their pencil axes in synchrony might make a worthwhile demonstration? Fred Lunnon On 11/23/11, James Propp <jamespropp@gmail.com> wrote:
Thanks!
It occurs to me to ask: Is there a parametrization of the one-sheeted hyperboloid that makes it manifest that it's a ruled surface?
And: Is there a parametrization that makes it manifest that it's a doubly-ruled surface?
Jim
Postscript: a wrinkle concerning planar orientation. Four of the eight cuboid corner-slice planes comprising one tetrahedron form the two pairs of base planes required to specify each pencil correspondence. If these are oriented consistently (positive constant component 1), the correspondence must map s E + F -> s G - H to generate the required one-sheet hyperboloid. Ellipsoids are generated by reguli of lines in complex space: setting z -> z \i, s,t -> u +/- v \i in the hyperboloid gives ellipsoid x^2/a^2 + y^2/b^2 + z^2/c^2 = w^2 parameterised by Q(u, v) = [w,x,y,z] = [u^2 + v^2 + 1, 2*a*u, b*(u^2 + v^2 - 1), 2*c*v] . Fred Lunnon On 11/25/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
Implementing the method I sketched earlier for a general one-sheet hyperboloid in canonical position x^2/a^2 + y^2/b^2 - z^2/c^2 = w^2 makes it easy to locate lines missing from reguli using symmetry. Four lines from each regulus are face diagonals of the cuboid with corners [+/- a, +/- b, +/- c], and base planes for the correspondence between pencils through two of them can be selected from corner-slices +/- x/a +/- y/b +/- z/c = 1 .
The result is gratifyingly elegant: a general point P(s, t) on the surface has homogeneous coordinate vector
[w,x,y,z] = [s*t+1, -a*(s+t), b*(s*t-1), c*(-s+t)] .
It is trivial to check that P satisfies the (homogeneous) quadric equation; and being linear in s and t separately, P describes a line when either parameter is fixed. A missing subset of measure zero is reached by letting s -> oo or t -> oo .
Showing their line intersection sweeping out the surface as the two moving planes rotate about their pencil axes in synchrony might make a worthwhile demonstration?
Fred Lunnon
On 11/23/11, James Propp <jamespropp@gmail.com> wrote:
Thanks!
It occurs to me to ask: Is there a parametrization of the one-sheeted hyperboloid that makes it manifest that it's a ruled surface?
And: Is there a parametrization that makes it manifest that it's a doubly-ruled surface?
Jim
participants (4)
-
Adam P. Goucher -
Fred lunnon -
James Propp -
Joshua Zucker