From - Thu Dec 6 12:33:31 2001 Return-Path: Received: from coxeter.math.toronto.edu (coxeter.math.toronto.edu [128.100.68.3]) by nef.ens.fr (8.10.1/1.01.28121999) with ESMTP id fB6BXMh03510 for ; Thu, 6 Dec 2001 12:33:22 +0100 (CET) Received: (from laurent@localhost) by coxeter.math.toronto.edu (8.11.0/8.11.0/UTMath 1.0) id fB6BX4K62846; Thu, 6 Dec 2001 06:33:04 -0500 Date: Thu, 6 Dec 2001 06:33:04 -0500 From: Larry Siebenmann Message-Id: <200112061133.fB6BX4K62846@coxeter.math.toronto.edu> To: B.Jackowski@GUST.ORG.PL, laurent@math.toronto.edu, luecking@comp.uark.edu, metafont@ens.fr, tjk@ams.org X-Virus-Scanner: AMaVis 0.2.0-pre6 / Virus Scan X-Loop: metafont@nef.ens.fr X-Sequence: 512 Precedence: list Subject: [metafont] Quadratic bezier areas ** Area formula for quadratic bezier spline curves ** Hi all, Back in the spring of 2000 I discussed with Jacko and Dan Leucking the area entrapped by *cubic* bezier curves in the plane. We arrived at various formulae -- some for thinking, some for cooking. One that seems easy to remember is as follows. A planar bezier curve B(t), 0<=t<=1, with control points B(0)=a,b,c,d=B(1) entraps the same area as the rational combination of piecewise linear paths from a to d: (***) .1 (a--d) + .3 (a--b--d) + .3 (a--c--d) + .3 (a--b--c--d) Notice that the coefficient sum is 1; so I called this a "wooly polygon". This provides a formula for the area of any closed path gotten by chaining together several bezier cubic spline curves. Indeed it is the weighted sum .1 A1 + .3 A2 + .3 A3 + .3 A4 of the areas of four polygons gotten by replacing each bezier by a linear approximation in the four ways indicated. The area of any polygon is easily calculated in terms of vector products. (Dan indicated tricks to make vector product calculations particularly efficient.) I asked then whether any explicit area formula for bezier curves in terms of control points is known/published. There is no definitive answer, but one quotation from a research article I found on the web mentions area of cubic beziers as being difficult to calculate: http://nr.stic.gov.tw/ejournal/ProceedingA/v24n5/373-381.pdf Quadratic B-spline for Curve Fitting WU-CHIH HU * AND HSIN-TENG SHEU ** Proc. Natl. Sci. Counc. ROC(A) Vol. 24, No. 5, 2000. pp. 373-381 (Short Communication) << Further, it is not easy to calculate the curve length, the enclosed area of a closed curve or the intersection points of two cubic B-splines as is necessary in most CAD/CAM applications. For the quadratic B-spline function, such computation is much simpler. Further, since a quadratic B-spline curve is closer to its control polygon than a cubic one, it is somewhat easier to manipulate the curve using its control polygon. Hence, the quadratic B-spline function is considered in this paper. >> Here is the direct analog for quadratic splines of the above area formula (***). Let B(t), 0<=t<=1, with control points B(0)=a,b,c=B(1) be a quadratic bezier spline curve. The rational combination (**) (1/3) (a--c) + (2/3) (a--b--c) is a "wooly piecewise linear path" that entraps the same area as the curve B(t). What is more, there is an exceptionally easy proof of (**) as follows. The case when a,b,c are collinear is trivial, so suppose they are not. Then we can and do normalize a,b,c by an affine linear transformation sending a --> (-1,1); b --> (0,-1 ); c --> (1,1). The transformed curve is still quadratic bezier and control points are the assigned images of a,b,c. The transformed curve is the standard parabola y=x^2 (check this by chasing definitions). For the normalized a,b,c, (**) amounts to the assertion that (2/3) of the area (=2) of the triangle abc lies inside the parabola. The simplest integral formula \int x^2 = (1/3) x^3 easily proves this. (See approximate figure.) a = (-1,0) c = (1,0) o--------------------o--------------------o \ | / \ | /| \ | / | \ | / | \ | / | \ | / | \ | / ........Area = (1/3) \ | / | \ | / | ---------------------o---------------------- |(0,0) | | | | | | | | | ---------------------o---------------------- b=(0,-1) By transforming back one deduces (**) for the originally given bezier. --------------- Exercise: The same normalization method proves two more basic facts: (i) Every quadratic bezier curve describes a segment of a parabola. (ii) If a,b,c are the control points for a quadratic bezier curve from a to c, then the axis of the containing parabola is the oriented line from b through the midpoint of the segment a--c. --------------- Does (**) make area calculations for font characters of True Type fonts much easier than our area calculations for Adobe Type 1 fonts (which use cubics)? This remains a moot point -- inasmuch as True Type fonts use many more bezier curves (incidentally, has anyone noticed the proportions, say as measured by a well-done native Times font of each format??). Cheers Laurent S. From - Thu Dec 6 15:34:08 2001 Return-Path: Received: from coxeter.math.toronto.edu (coxeter.math.toronto.edu [128.100.68.3]) by nef.ens.fr (8.10.1/1.01.28121999) with ESMTP id fB6EXth34689 for ; Thu, 6 Dec 2001 15:33:57 +0100 (CET) Received: (from laurent@localhost) by coxeter.math.toronto.edu (8.11.0/8.11.0/UTMath 1.0) id fB6EXoB58494; Thu, 6 Dec 2001 09:33:50 -0500 Date: Thu, 6 Dec 2001 09:33:50 -0500 From: Larry Siebenmann Message-Id: <200112061433.fB6EXoB58494@coxeter.math.toronto.edu> To: B.Jackowski@GUST.ORG.PL, laurent@math.toronto.edu, luecking@comp.uark.edu, metafont@ens.fr, tjk@ams.org X-Virus-Scanner: AMaVis 0.2.0-pre6 / Virus Scan X-Loop: metafont@nef.ens.fr X-Sequence: 513 Precedence: list Subject: [metafont] # Area of quadratic bezier Correction: In exercise (ii) of what I just posted, please change "axis of the containing parabola" ==> "axis direction of the containing parabola" A definition of "axis direction" that behaves naturally under affine linear transformations is:- the unique direction in the plane in which points on the parabola are seen to vanish as they go to infinity. Exercise (iii) Give a further algorithm to find the (parallel!) line that is the parabola axis you met in school. LS From - Tue Dec 11 11:01:28 2001 Return-Path: Received: from mothers.student.bth.se (mothers.student.bth.se [194.47.133.17]) by nef.ens.fr (8.10.1/1.01.28121999) with ESMTP id fBBA1Eh31389 for ; Tue, 11 Dec 2001 11:01:18 +0100 (CET) Received: from MarcusEjnarsson.student.bth.se (itslaptop1.its.hk-r.se [194.47.136.151]) by mothers.student.bth.se (8.10.2+Sun/8.10.2) with ESMTP id fBBA1DQ19148 for ; Tue, 11 Dec 2001 11:01:13 +0100 (MET) Message-Id: <5.1.0.14.2.20011211105619.03859a10@mail.student.bth.se> X-Sender: di97mej@mail.student.bth.se X-Mailer: QUALCOMM Windows Eudora Version 5.1 Date: Tue, 11 Dec 2001 11:03:32 +0100 To: metafont@ens.fr From: Marcus Ejnarsson Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii"; format=flowed X-Virus-Scanner: AMaVis 0.2.0-pre6 / Virus Scan X-Loop: metafont@nef.ens.fr X-Sequence: 514 Precedence: list Subject: [metafont] MetaPost - vbox/hbox - Centering? Hello I have a question about MetaPost and how to center text in vbox/hbox. In my .mp-file I write like this to create a box contatining some text. Anyone who know how I can get the text centered in the box? boxit.c(btex \vbox{\hbox{Voice} \hbox{Activation} \hbox{Detection}} etex); I also wounder if someone know if there is a easy way to get boxes of the same height and width (still centered text) Thankful for any help .... /Marcus Ejnarsson MSc Student at Blekinge Institute of Technology From - Tue Dec 11 11:27:13 2001 Return-Path: Received: from lluna.uib.es (lluna.uib.es [130.206.33.10]) by nef.ens.fr (8.10.1/1.01.28121999) with ESMTP id fBBAQ9h38034 for ; Tue, 11 Dec 2001 11:27:01 +0100 (CET) Received: from ipc4.uib.es (pcignasi.uib.es [130.206.134.178]) by lluna.uib.es (8.10.0/8.10.0) with ESMTP id fBBAFoX13855 for ; Tue, 11 Dec 2001 11:15:50 +0100 (MET) Message-Id: <3C15DEEC.5BE96924@ipc4.uib.es> Date: Tue, 11 Dec 2001 11:24:44 +0100 From: Ignasi =?iso-8859-1?Q?Furi=F3?= Caldentey X-Mailer: Mozilla 4.7 [en] (Win98; I) X-Accept-Language: en,pdf Mime-Version: 1.0 To: metafont@ens.fr Subject: Re: [metafont] MetaPost - vbox/hbox - Centering? References: <5.1.0.14.2.20011211105619.03859a10@mail.student.bth.se> Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: 8bit X-Virus-Scanner: AMaVis 0.2.0-pre6 / Virus Scan X-Loop: metafont@nef.ens.fr X-Sequence: 515 Precedence: list Marcus Ejnarsson wrote: > > Hello > > I have a question about MetaPost and how to center text in vbox/hbox. > > In my .mp-file I write like this to create a box contatining some text. > Anyone who know how I can get the text centered in the box? > > boxit.c(btex \vbox{\hbox{Voice} \hbox{Activation} \hbox{Detection}} etex); > > I also wounder if someone know if there is a easy way to get boxes of the > same height and width (still centered text) > I've been using this: verbatimtex \documentclass{article} \newcommand{\stk}[1]{\parbox[c][2cm][c]{3cm}{\centerline{\begin{tabular}{c}#1\end{tabular}}}} \newcommand{\stkb}[1]{\parbox[c][4.5cm][c]{3cm}{\centerline{\begin{tabular}{c}#1\end{tabular}}}} \begin{document} etex ... boxit.c(btex \stk{Line 1\\Line2} etex, (x,y)); If anybody knows a better solution, I also would like to know it. Ignasi -- ------------------------------------------------------- Ignasi Furió Caldentey ignasi@ipc4.uib.es Dpt. Matemątiques i informątica http://dmi.uib.es Universitat de les Illes Balears http://www.uib.es Fax: +34 971 173003 Tel.: +34 971 173196 ------------------------------------------------------- From - Tue Dec 11 12:41:27 2001 Return-Path: Received: from smtpzilla3.xs4all.nl (smtpzilla3.xs4all.nl [194.109.127.139]) by nef.ens.fr (8.10.1/1.01.28121999) with ESMTP id fBBBfEh51477 for ; Tue, 11 Dec 2001 12:41:14 +0100 (CET) Received: from server-1.pragma-ade.nl (s340-isdn1251.dial.xs4all.nl [194.109.184.227]) by smtpzilla3.xs4all.nl (8.12.0/8.12.0) with ESMTP id fBBBfDNe006897; Tue, 11 Dec 2001 12:41:13 +0100 (CET) Received: from laptop-2.wxs.nl (laptop-1.pragma-ade.nl [200.1.1.26]) by server-1.pragma-ade.nl (8.11.6/8.11.6) with ESMTP id fBBBIOa19119; Tue, 11 Dec 2001 12:18:24 +0100 Message-Id: <5.1.0.14.1.20011211121348.03181d38@server-1> X-Sender: hagen@server-1 X-Mailer: QUALCOMM Windows Eudora Version 5.1 Date: Tue, 11 Dec 2001 12:17:32 +0100 To: Marcus Ejnarsson From: Hans Hagen Subject: Re: [metafont] MetaPost - vbox/hbox - Centering? Cc: metafont@ens.fr In-Reply-To: <5.1.0.14.2.20011211105619.03859a10@mail.student.bth.se> Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii"; format=flowed X-Virus-Scanner: AMaVis 0.2.0-pre6 / Virus Scan X-Loop: metafont@nef.ens.fr X-Sequence: 516 Precedence: list At 11:03 AM 12/11/2001 +0100, Marcus Ejnarsson wrote: >Hello > >I have a question about MetaPost and how to center text in vbox/hbox. > >In my .mp-file I write like this to create a box contatining some text. >Anyone who know how I can get the text centered in the box? > >boxit.c(btex \vbox{\hbox{Voice} \hbox{Activation} \hbox{Detection}} etex); > >I also wounder if someone know if there is a easy way to get boxes of the >same height and width (still centered text) > >Thankful for any help .... To get a proper alignment you need to make sure that 'lines' have the same height and depth which is accomplished by using struts. In principle you should be able to use the high level macros that the package that you use provides. When including the dvi result from btex/etec, text and rules are passed to mp, so you can let tex place the rules, but in most cases the results are worse than using mp to do that. In context, the principle can be demonstrated as follows: \setupcolors[state=start] \starttext \startMPpage picture p, q ; p := image(boxit.c (btex \framed[frame=off]{Voice} etex)) ; q := image(boxit.c (btex \framed[frame=off]{Detection} etex)) ; draw p withcolor transparent (1,.5,blue) ; draw q withcolor transparent (1,.5,red) ; \stopMPpage \stoptext [there are more efficient ways]. Hans ------------------------------------------------------------------------- Hans Hagen | PRAGMA ADE | pragma@wxs.nl Ridderstraat 27 | 8061 GH Hasselt | The Netherlands tel: +31 (0)38 477 53 69 | fax: +31 (0)38 477 53 74 | www.pragma-ade.com ------------------------------------------------------------------------- From - Sat Dec 15 21:15:03 2001 Return-Path: Received: from coxeter.math.toronto.edu (coxeter.math.toronto.edu [128.100.68.3]) by nef.ens.fr (8.10.1/1.01.28121999) with ESMTP id fBFKEth71751 for ; Sat, 15 Dec 2001 21:14:55 +0100 (CET) Received: (from laurent@localhost) by coxeter.math.toronto.edu (8.11.0/8.11.0/UTMath 1.0) id fBFKEsq84898 for metafont@ens.fr; Sat, 15 Dec 2001 15:14:54 -0500 Date: Sat, 15 Dec 2001 15:14:54 -0500 From: Larry Siebenmann Message-Id: <200112152014.fBFKEsq84898@coxeter.math.toronto.edu> To: metafont@ens.fr X-Virus-Scanner: AMaVis 0.2.0-pre6 / Virus Scan X-Loop: metafont@nef.ens.fr X-Sequence: 517 Precedence: list Subject: [metafont] quadratic bezier curves in metafont ** quadratic bezier curves in metafont ** Hi all, My excursion last week into quadratic bezier curves may seem to some of you of interest only to TrueType font buffs. Not so! Let me assure you that metafont can draw an arbitrary quadratic bezier curve. This is well known to experts, but the proof I'll give may not be the most common one. The salient fact is that the restriction (*) below on the cubic bezier curve F(t), 0<=t<=1, with control points F(0)=a,b,c,d=F(1) in the plane R^2 is necessary and sufficient that the curve be quadratic bezier, or equivalently a (possibly degenerate) parabola. (*) d - a = 3 b - c a o ------------------------------------- o d \ / \ / \ / \ / \ / \ / \ / b o ------------- o c \ / \ / \ / \ / o e Furthermore, the quadratic bezier has control points a,e,d where e is the intersection point of the lines ab and dc. (What is more, the quadratic and cubic parametrizations are identical; so the cubic is degenerate.) To prove this I like to consider the coefficient of t^3 in the expansion in powers of t for F(t) = s^3*a + 3s^2*t*b + 3s*t^2*c + t^3*d, where s=(1-t) This coefficient of t^3 is: C = - a + 3 b + -3 c + d and its vanishing is the condition (*)! Now the vanishing is equally the condition that F(t) be quadratic or less, ie a (possibly degenerate) parabola. But for a parabolic arc a to d with tangent lines ab and dc the quadratic control points are well known to be a,e,d. (The degenerate case where a,b,c,d are colinear needs a bit of extra discussion. Exercise!) QED. At this point you can see how to trivially convert any TrueType font outline to a Type1 font outline:- just regard the quadratic bezier segments as special cubic bezier segments by the above recipe. Cheers Laurent S. PS. There is also a *lossy* conversion Type1 to TrueType. Does anyone know the details?? From - Sat Dec 15 21:17:24 2001 Return-Path: Received: from coxeter.math.toronto.edu (coxeter.math.toronto.edu [128.100.68.3]) by nef.ens.fr (8.10.1/1.01.28121999) with ESMTP id fBFKHFh71999 for ; Sat, 15 Dec 2001 21:17:15 +0100 (CET) Received: (from laurent@localhost) by coxeter.math.toronto.edu (8.11.0/8.11.0/UTMath 1.0) id fBFKHE3138962 for metafont@ens.fr; Sat, 15 Dec 2001 15:17:14 -0500 Date: Sat, 15 Dec 2001 15:17:14 -0500 From: Larry Siebenmann Message-Id: <200112152017.fBFKHE3138962@coxeter.math.toronto.edu> To: metafont@ens.fr X-Virus-Scanner: AMaVis 0.2.0-pre6 / Virus Scan X-Loop: metafont@nef.ens.fr X-Sequence: 518 Precedence: list Subject: [metafont] cubic (but non-quadratic) bezier curves ** cubic (but non-quadratic) bezier curves ** Hi all, Recall the "cubic axis" direction C = - a + 3 b + -3 c + d of the bezier cubic curve F(t) = s^3*a + 3s^2*t*b + 3s*t^2*c + t^3*d, where s=(1-t) with control points a,b,c,d. It is just the coefficient of t^3. I like this "cubic axis" direction C because it gives quick insight into the global behavior of F(t) for t \in R. Since bezier curve is a notion that behaves naturally under affine linear transformations of R^2, we demand the same for properties and constructions. (Euclidian notions like angle are banned.) Assuming the vector C is nonzero, F(t) is definitely not quadratic, and the curve F(t) goes to infinity in direction C as t --> \infty. Likewise it goes to infinity in direction -C as t --> -\infty. We need to view the lines in R^2 parallel to C as being the points of an affine line L_C such that one has a natural affine linear projection map p : R^2 --> L_C One can informally identify L_C with any line transversal to direction C and p then sends all points on a line parallel to C to its intersection with L_C. The one problem is that there is no natural choice of this L_C; but since any two choices are naturally related by projection parallel to C, the problem is academic ... we can accept *all* choices. One can distinguish three mutually exclusive global behaviors of F(t) with respect to direction C. -- (non-degenerate) The composition p o F : R --> R is nondegenerate quadratic in t. Then the closed convex hull of the complete curve F(t), t \in R, is a closed half plane with F(t) tangent at a single point to the bounding line (parallel to C). -- (singly-degenerate) The composition p o F is linear nondegenerate. Equivalently, the four control point images under p o F, say a',b',c',d', are in linear progression on L_C. Then the convex hull of the complete curve F(t), t \in R, is the whole plane. Further, F(t) is the graph of a cubic function on any line L_C transverse to the axis C. So F(t) has an unique inflection point I. Choose I as origin. Choose the tangent line to F(t) at I as x-axis (it is a privileged choice of L_C). Choose the line through I in direction C as y-axis. This naturally re-coordinatizes the plane so that F(t) is (t,t^3) after rescaling the axes. -- (doubly-degenerate) The composition p o F is constant. Equivalently, the four control points map under p o F to one point in L_C. Then the complete curve F(t), t \in R, is confined to one line parallel to C. It is the first (non-degenerate) case that is of most geometric interest since one always finds a pair of inflection points, or else one double point, or else (exceptionally) a cusp. No time to explain this further trichotomy today! Cheers Laurent S. From - Tue Dec 18 12:34:29 2001 Return-Path: Received: from coxeter.math.toronto.edu (coxeter.math.toronto.edu [128.100.68.3]) by nef.ens.fr (8.10.1/1.01.28121999) with ESMTP id fBIBYHh78207 for ; Tue, 18 Dec 2001 12:34:18 +0100 (CET) Received: (from laurent@localhost) by coxeter.math.toronto.edu (8.11.0/8.11.0/UTMath 1.0) id fBIBYGL65068 for metafont@ens.fr; Tue, 18 Dec 2001 06:34:16 -0500 Date: Tue, 18 Dec 2001 06:34:16 -0500 From: Larry Siebenmann Message-Id: <200112181134.fBIBYGL65068@coxeter.math.toronto.edu> To: metafont@ens.fr X-Virus-Scanner: AMaVis 0.2.0-pre6 / Virus Scan X-Loop: metafont@nef.ens.fr X-Sequence: 519 Precedence: list Subject: [metafont] # quadratic bezier curves in metafont # Correction to: "quadratic bezier curves in metafont" # Hi all In my recent note "quadratic bezier curves in metafont" the equation (*) for degeneration of a cubic bezier to a quadratic bezier should have been (*) d - a = 3 (c - b) Happily, the figure was for the correct equation (*), and the proof too! Please tell me if there were other slipups! Laurent S.