A Piecewise Cubic PostScript Trefoil — F.E.J. Linton Math

A Piecewise Cubic PostScript Trefoil — F.E.J. LintonMath/CS Emeritus, Wesleyan Univ., Middletown, CT, USA

8/5/2014 3:40 AM file:///D:/Docs/WesStuff/FRG-2014-15/ICM-SeoulKR/TalkSlides/Talk-001.html

PostScript Trefoil • F.E.J. Linton 1 of 1


TrefoilParameterizations















































Begin with a fragmentof the graph ofy=|x|/√3 ,with origin, a verticaly-axis, andperpendiculars at(x0, y0)=(-3, √3)and(x3, y3)=(1.5, ½√3) ,all as marked.



Sketched in red, a partof the graph of afunction y = p(x)satisfying p(-3)= √3, p'(-3)= √3, p(1.5)= ½√3,and p'(1.5)= -√3.For ...



For if we rotate thesketched arc not justonce through 2π/3radians, ...



For if we rotate thesketched arc not justonce through 2π/3radians, but twice, andthen ...



For if we rotate thesketched arc not justonce through 2π/3radians, but twice, andthen superimpose themirror image of thatresult, ...



For if we rotate thesketched arc not justonce through 2π/3radians, but twice, andthen superimpose themirror image of thatresult, we’d have thesort of trefoil we’reafter.



Now to find coefficientsa, b, c, d for a cubicpolynomial p(x)= ax3+bx2+cx+d ,satisfying p(-3)= √3, p'(-3)= √3, p(1.5)= ½√3, and p'(1.5)= -√3is standard LinearAlgebra, as is ...



... rotating andreflecting its graph inthe plane.



... rotating andreflecting its graph inthe plane.Fortunately, PostScriptlanguage can spareyou many matrixcalculations. For here’show PostScript drawscubic polynomials(crash course part 1):





In our case, we have ax=0=bx , cx=4.5 , x0=-3 ,y(t)= p(x(t)) , and y0=√3 .



In our case, we have ax=0=bx , cx=4.5 , x0=-3 ,y(t)= p(x(t)) , and y0=√3 .The PostScript Bézier curve drawing process usesxn=x(nt/3) and yn=y(nt/3) , for n=0,1,2,3 . The middletwo (with n=1,2) are the so-called control points.



In our case, we have ax=0=bx , cx=4.5 , x0=-3 ,y(t)= p(x(t)) , and y0=√3 .The PostScript Bézier curve drawing process usesxn=x(nt/3) and yn=y(nt/3) , for n=0,1,2,3 . The curvedeparts from (x0, y0) in the direction of (x1, y1) , andarrives at (x3, y3) from the direction of (x2, y2) .



In our case, we have ax=0=bx , cx=4.5 , x0=-3 ,y(t)= p(x(t)) , and y0=√3 .The PostScript Bézier curve drawing process usesxn=x(nt/3) and yn=y(nt/3) , for n=0,1,2,3 . The curvedeparts from (x0, y0) in the direction of (x1, y1) , andarrives at (x3, y3) from the direction of (x2, y2) .The “control points” we need are easy to construct:



We already have thegraph of y=|x|/√3 , theorigin, the start and endpts (x0, y0)=(-3, √3) ,(x3, y3)=(1.5, ½√3) ,a vertical y-axis, andorthogonals at(x0, y0)=(-3, √3) and(x3, y3)=(1.5, ½√3) ,as marked.



We construct theauxiliary line segmentsthat will meet at thecontrol points overx1=-1.5 and x2=0 , thecontrol points becoming(x1, y1)=(-1.5, 2.5√3)and (x2, y2)=(0, 2√3) ,respectively.



We construct theauxiliary line segmentsthat will meet at thecontrol points overx1=-1.5 and x2=0 , thecontrol points becoming(x1, y1)=(-1.5, 2.5√3)and (x2, y2)=(0, 2√3) ,respectively. Now,crash course part 2:



If PostScript’s drawing finger is pointed at (x0, y0) ,here’s how to draw the Bézier curve to (x3, y3) usingcontrol points (x1, y1) and (x2, y2) —



If PostScript’s drawing finger is pointed at (x0, y0) ,here’s how to draw the Bézier curve to (x3, y3) usingcontrol points (x1, y1) and (x2, y2) — use curveto :



If PostScript’s drawing finger is pointed at (x0, y0) ,here’s how to draw the Bézier curve to (x3, y3) usingcontrol points (x1, y1) and (x2, y2) — use curveto :



So we must movePostScript’s writingfinger to (-3, √3) andthen perform theRPN-based PSincantation-1.5 2.5 3 sqrt mul0 2 3 sqrt mul1.5 0.5 3 sqrt mulcurvetoThe result:



So we must movePostScript’s writingfinger to (-3, √3) andthen perform theRPN-based PSincantation-1.5 2.5 3 sqrt mul0 2 3 sqrt mul1.5 0.5 3 sqrt mulcurvetoThe result: (that waseasy).



We do the same forthe analogous pointsbetween 0.6�(-3, √3)and 0.1�(1.5, ½√3) .



We do the same forthe analogous pointsbetween 0.6�(-3, √3)and 0.1�(1.5, ½√3) .



We do the same forthe analogous pointsbetween 0.6�(-3, √3)and 0.1�(1.5, ½√3) .And then we’ll rotate ...once ...



We do the same forthe analogous pointsbetween 0.6�(-3, √3)and 0.1�(1.5, ½√3) .And then we’ll rotate ...once ... and twice ...



We do the same forthe analogous pointsbetween 0.6�(-3, √3)and 0.1�(1.5, ½√3) .And then we’ll rotate ...once ... and twice ...and then reflect acrossthe y-axis.



We do the same forthe analogous pointsbetween 0.6�(-3, √3)and 0.1�(1.5, ½√3) .And then we’ll rotate ...once ... and twice ...and then reflect acrossthe y-axis. Now let'sclean up a little ...



We do the same forthe analogous pointsbetween 0.6�(-3, √3)and 0.1�(1.5, ½√3) .And then we’ll rotate ...once ... and twice ...and then reflect acrossthe y-axis. Now let'sclean up a little ...There! Only ...



We had to “fill” thezones between outerand inner Béziercurves with (opaque)white “paint” in orderthat the unreflectedarms of the trefoil notshine through thereflected ones. Now it’smore satisfying.



Or, the same thing inrainbow colors ... .




References• http://en.wikipedia.org/wiki/Trefoil_knot• http://math.stackexchange.com/questions/148916/trefoil-knot-as-an-algebraic-curve• http://home.adelphi.edu/~stemkoski/knotgallery/• http://katlas.math.toronto.edu/wiki/File:TriquetraCaixaGeral.png• http://katlas.math.toronto.edu/w/images/a/af/TriquetraCaixaGeral.png• Adobe Systems Inc. PostScript Language Reference Manual (the RedBook) (2nd ed.), Addison-Wesley, Reading (MA), 1990, ISBN-13:978-0-201-18127-2 (cf. esp. p. 393)• V. Pratt <[email protected]>, http://boole.stanford.edu/Trefoil.jpg



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A Piecewise Cubic PostScript Trefoil — F.E.J. Linton Math