From:Paul Earwicker <pgearwicker@dra.hmg.gb>To:eppstein@ics.uci.eduSubject:Maximum volume of a tea bagDate:Wed, 7 May 1997 14:30:28 +0100

You may like this problem, We buy tea in square paper bags. These are sealed around all four edges. The question is what is the maximum volume that can be contained?

Good luck

Paul Earwicker

From:eppstein@ICS.UCI.EDUTo:pgearwicker@dra.hmg.gbSubject:Re: Maximum volume of a tea bagDate:Wed, 07 May 1997 11:34:47 -0700

Cute problem -- one that forces me to question my usual assumptions. I've also seen Mylar baloons made like this. I'm not sure there is a smooth surface formed by two squares sealed at the edges, containing a nonzero volume, but of course you didn't say it had to be smooth. You can fill each of the four corners out into a circular cone, but then it seems like you need some kind of crumpling in the middle of each square to get it to match up with the cones, and I don't understand what that part should look like. Do you have a solution?

From:Paul Earwicker <pgearwicker@dra.hmg.gb>To:eppstein@ICS.UCI.EDUSubject:FW: Maximum volume of a tea bagDate:Thu, 8 May 1997 09:56:35 +0100

I do not have a solution but I think crumpling is unavoidable. There is the possibility that the deformation can not be carried out at all. If you get any ideas I would be interested to hear them. The problem came from my wife who is a school teacher. As a project for pupils they decided to investigate a claim by a tea bag maufacturer that his new tetrahedral tea bags gave the tea twice as much space to brew. In a simple minded way she got them to work out the volume of a tetrahedron and then turned to the "easy" case of the square bags....disaster.

Wine boxes have a rectangular Mylar bag.

I am trying to get a reasonably sharp upper bound. I.e. better than the sphere.

I liked your pages.

There is a lot of geometry to discover yet!

Good luck

Paul Earwicker

From:eppstein@ICS.UCI.EDU[SMTP:eppstein@ICS.UCI.EDU]To:pgearwicker@dra.hmg.gbSubject:Re: FW: Maximum volume of a tea bag

I spent the morning playing with origami models -- it turns out that you can fold the two squares to surround a nonzero volume, without any bending or crumpling away from the few well-defined folds.

For instance, by folding each square along vertical and horizontal lines 1/4 and 3/4 of the way across, you get a tic-tac-toe-board like pattern, which with some additional diagonal folds near each corner can be made to surround a 1/2 by 1/2 by 1/2 cube (total volume: 1/8), with some triangular flaps hanging off four of the edges.

Dan Hirschberg pointed out to me that instead folding on lines 1/6 and 5/6 of the way across, you get a rectangular box with slightly bigger volume, 4/27.

Neither of these likely resembles the true optimum solution, but they do at least show that some solution is actually possible. Actually, my intuition is that there may not be an optimum solution, rather that some sequence of suboptimal folded solutions converges to an optimum that can't actually be reached without an infinite amount of folding. Maybe this is what you mean by "the deformation can not be carried out"?

To:eppstein@verity.ICS.UCI.EDUSubject:volume problem using 2 attached square sheetsReply-To:Dan Hirschberg <dan@ics.uci.edu>Date:Mon, 12 May 1997 08:28:04 -0700From:Dan Hirschberg <dan@verity.ICS.UCI.EDU>

Your original solution had volume 0.1250 My slight modification had volume 0.1481+ I have built a paper model with volume 0.1751+ I can describe a slight mod with volume 0.1756+

dan

From:eppstein@ICS.UCI.EDUTo:pgearwicker@dra.hmg.gbSubject:Re: FW: Maximum volume of a tea bagDate:Mon, 12 May 1997 10:32:36 -0700

I got another prettier origami-like solution from Hirschberg, who worked it out over the weekend.

Take each square, and crease it along lines running at 22.5 degree angles from each corner. Ugly ASCII graphical rendition:

------------ |. . | . . . | . . . | . . | . | . | .(If you're doing this by hand, you might find it easiest to make a light crease along the square's diagonals first, then you can make the 22.5deg creases by folding each edge onto the diagonal.)

These creases will meet each other at the four corners of a small square, diagonal to the original square. Crease also the edges of these squares (with the crease direction matching that of the 22.5 degree creases -- in origami terms, make all creases so far "mountain folds").

Finally, make short creases perpendicularly from each corner of the small diagonal square to the nearest edge of the large square, the opposite direction ("valley folds").

Join the edges of these creased squares together, and expand them to form a shape that looks like a cubical box with the two small diagonal squares as two opposite sides, and with tall pyramids extending from the remaining four sides to each of the corners of the original squares. So the boundary of this shape is formed by the two small squares and sixteen 45-67.5-67.5 isosceles triangles.

According to Dan, this shape's volume is roughly 0.1751 (compared with 0.1250 and 0.1481 for the rectangular boxes I described earlier). Choosing angles slightly different from 22.5deg produces a slight improvement, to 0.1756.

This is starting to look closer to what the mylar balloons actually end up shaped like...

-d

Date:Fri, 28 Nov 1997 11:41:16 +1100To:eppstein@ics.uci.eduFrom:Andrew Kepert <andrew@frey.newcastle.edu.au>Subject:teabag

Dear David Eppstein (& others via Cc:) A couple of days ago I read the discussion at the Geometry Junkyard on the teabag problem: http://www.ics.uci.edu/~eppstein/junkyard/teabag.html I agree it is a cute problem. I have managed a slight improvement on Dan Hirschberg's solution, giving 0.190205, which can probably be improved. My solution is at http://maths.newcastle.edu.au/~andrew/teabag You are welcome to lift any part of this into the Junkyard, or just throw in a link. (My department's site is fairly stable, but on the wrong side of the pacific for you.) Regards, Andrew p.s. Similar techniques apply to the tetrahedral teabag. [ Dr Andrew Kepert e-mail: andrew@frey.newcastle.edu.au ] [ http://frey.newcastle.edu.au/~andrew ] [ Central Coast Campus, Univ. of Newc., Ourimbah NSW 2258, AUSTRALIA ] [ Phone: 043 484116 Fax: 043 484145 ] [ Mathematics, University of Newcastle, Callaghan NSW 2308 AUSTRALIA ] [ Phone: 049 215190 Fax: 049 215548 ]

Date:Fri, 28 Nov 1997 15:24:09 +0000From:Paul Earwicker <pgearwicker@dera.gov.uk>Subject:RE: teabagTo:"'Andrew Kepert'" <andrew@frey.newcastle.edu.au>

Dear Andrew, Well thank you for your tea bag notes. Very interesting. I still find it odd that origami has become a branch of geometry. At the moment everything in life seems to be geometry. It was all Bayes theorem last month so things are improving. I am interested in the infinitesimal stuff. Transforms with no shear and no stretch are very constrained. Transforms that leave the length around any closed curve invariant are interesting too. This is mathematical netting (string bags with very small mesh) or cloth. There should be a maximum value for stuffing a square string bag too. I expect the shape and volume to be very similar to the tea bag. My email address has changed (or is about to change) and the new one, which works now, is below. If you have any ideas on how to treat these (often rough) surfaces mathematically I would be very interested. Thank you for sharing your thoughts. Paul Paul Earwicker Room 1052B A2 Building, DERA Farnborough, Ively Road, Farnborough, Hampshire GU14 0LX, United Kingdom. e-mail : pgearwicker@dera.gov.uk Tel: +44 (0) 01252 - 393168 Fax: +44 (0) 01252 - 395090

To:eppstein@ICS.UCI.EDUFrom:andrew@frey.newcastle.edu.au (Andrew Kepert)Subject:teabagDate:27 Dec 1998

I have updated the page to include some further musings on the teabag problem: A volume of 0.2055+ achieved. Upper bound of 0.2182+ established. This is probably close enough to call it a day! Andrew

From:Tom Longtin <tlongtin@sover.net>To:eppstein@ics.uci.eduSubject:Maximum Volume of a TeabagDate:Mon, 9 Nov 1998 13:24:08 -0500 (EST)

Hi David - Read with great interest attempts at maximizing the unit square teabag volume. I don't know how much progress has been made on this but it occurs to me that a curved fold might yield some benefit. The attached gif portrays 2 curved fold lines within a unit square. Left and right edges can be folded so that the main area raises convexedly out of the plane with the 2 end areas becoming concave and perpendicular to the plane. This becomes one-half of the teabag and will fit perfectly its other half. In my 1/4 volume cardboard model (one fold) I used a circle-arc extending 1/4 the edged distance into the square. Other curves/distances might prove more optimum such as parabolic, catenary or elliptical. I didn't calculate the volume because it's not clear to me how to determine the resulting curve along the fold line. This may have already been tried - I got the idea from a McDonald's French fry container. Regards, Tom Longtin

Date:Tue, 10 Nov 1998 11:43:47 +1100To:Tom Longtin <tlongtin@sover.net>From:andrew@frey.newcastle.edu.au (Andrew Kepert)Subject:teabag

Hi Tom, > Read with great interest your attempts at maximizing the unit square > teabag volume. I don't know how much progress has been made since April, > 1998 but it occurs to me that a curved fold might yield some benefit. The no progress at all - it is not a particularly serious project for me. Your configuration is a nice one I hadn't thought of - I only considered teabags with 4-fold rotational symmetry. This was guided by the shape of inflated wine cask liners. I seem to have had more exposure to these in my life than McDonald's (being vegetarian). I did, come to the same realisation about curved folds - they are used in my last example in the curved pleats (half the curved triangles that jut out of the quadrant need to be folded back in) Some of my thoughts about your configuration: * It should be quite easy to work out a volume if you model the main part of the curved surface - it will only need z as (say) a function of x on an interval [-a,a]. For the paper not to stretch, the arc length of this function is 1. The volume is then the integral of z(1-z). I would suggest trying a class of functions (e.g. arcs) with a single parameter (e.g. angle), then using 1-variable calculus or numerical methods. * If you have the skills, this would be one where variational method, such as is used to get the catenary as the chain curve. I would expect that the formulae in this calculus of variations calculation would be would neater than for any of the configurations I dreamed up, but this does not mean that it would be easy. * From an intuitive perspective, I guess that having the corners on a square would be optimal. If you can get hold of a wine cask liner and safely dispose of its contents (easier said than done, although they are also used for 5 litre OJ boxes over here) then you might be able to convince yourself of this. * Your configuration can be inflated slightly more by slanting the vertical sides -- this will make them project below the plane, and so another curved fold would be required, making a curved pleat like I used between the cones in my last example. This was my very first step in this problem, converting If I get an idle moment, I may try the variational problem. Let me know if you get anywhere - I could link or include it. Regards, Andrew

From:David Eppstein <eppstein@ics.uci.edu>To:Tom Longtin <tlongtin@sover.net>Subject:Re: Maximum Volume of a TeabagDate:Thu, 12 Nov 1998 17:30:12 -0800 (PST)

Your idea of using curved folds for the teabag problem is fascinating -- would you mind if I added your message to my page? I'm thinking the volume computations might be easier to work through if you specified the eventual cross section of the folded figure, rather than the shape of the curve. All you need is to specify a given convex curve y=f(x) with f(w/2)=f(-w/2)=0 for some w<1, where the length of the curve from (-w/2,0) to (w/2,0) is 1; then it's possible to fold a shape that in cross-section looks like that curve, and viewed from above looks like a 1 by w rectangle, with indentations in the shape of the same curve. E.g. if the cross-section is a semicircle of length 1 (so radius 1/pi), you can form a shape that is a circular cylinder of radius 1/pi and length 1, minus two cylindrical indentations at the two short ends. The flattened shape of the fold is not a circle, but is instead a sine curve. The two indentations can be reassembled to form a shape that is the intersection of two of these cylinders. So the volume of the overall shape is the same as that of the unindented cylinder (1/pi) minus the two-cylinder intersection (16/3pi^3 unless I've miscalculated). 1/pi - 16/3pi^3 ~ 0.1463, rather worse than the current best solution 1.902, but maybe a semicircular cross-section is not best... If two curved folds are good, maybe four curved folds (one on each side of the square) are even better?

From:Tom Longtin <tlongtin@sover.net>To:David Eppstein <eppstein@ics.uci.edu>Subject:Re: Maximum Volume of a TeabagDate:Fri, 13 Nov 1998 13:05:59 -0500 (EST)

Hi David - On Thu, 12 Nov 1998, David Eppstein wrote: > Your idea of using curved folds for the teabag problem is fascinating -- > would you mind if I added your message to my page? Thanks - please feel free to do so and I will work on more illustrations - possibly a VRML model. Andrew Kepert also responded with interest on this which I hope can be part of the dialog. I'll let him know of your intended posting. Meanwhile I'm studying your volumetric formulas. Here's an arc-folding patent from Herbert G. Bennett: http://patent.womplex.ibm.com/patlist?icnt=US&patent_number=4871080 Tom Longtin

Date:Fri, 13 Nov 1998 13:47:45 -0500 (EST)From:Tom Longtin <tlongtin@sover.net>Subject:Re: Teabag ProblemTo:Andrew Kepert <andrew@frey.newcastle.edu.au>

Hi Andrew - Thanks for your excellent suggestions. David Eppstein will be adding my contribution to his teabag.html which encourages me to work on more illustrations - possibly a VRML model. I hope you will be interested to continue adding your ideas to this dialog. Meanwhile I'm studying your volumetric formulas. Slanting the vertical side and adding another fold of larger radius with the same chord seemed not to work for me. The paper has a tendency to toggle between one fold or the other but not both - if I understand your concept. Here's an arc-folding patent from Herbert G. Bennett: http://patent.womplex.ibm.com/patlist?icnt=US&patent_number=4871080 Sincerely, Tom Longtin

Date:Sat, 14 Nov 1998 18:40:29 +1100 (EST)From:Kepert <andrew>To:tlongtin@sover.netSubject:Re: Teabag Problem

HI Tom, Folds along arcs, or at least the type I have considered (of which yours is an example) are quite easy. So long as the fold ends up being all in one plane, you consider this a mirror plane through which one part of the paper is reflected. Then the condition on it being foldable is that the unfolded (i.e. configuration with the mirrored bit un-mirrored) sheet is bendable - usually a cone or cylinder of some cross-section. To get the slanted side of your chip cup (umm *fries*, sorry) to work, say with a circular cross-section of the main (horizontal) bit, you need therefore to have a circular cylinder in the near-vertical section, with the curve of intersection being an arc of an ellipse (in some slanted plane) This means that the fold-line (when unwrapped) will be an arc of a sine curve. By similar reasoning, the curve where the near-vertical side cuts through the horizontal base-plane will again be an ellipse, but appear as a sine curve before folding. Same wavelength, smaller amplitude, at a guess. On the other hand, if you start out with an arc of a circle as your fold-line, my guess is that an arc of an ellipse would give you the other fold. Probably one where the major axis is equal to the diameter of the circular arc. However, I have no idea what the profile of the surface would be. Andrew

Date:Tue, 17 Nov 1998 21:15:49 -0500 (EST)From:Tom Longtin <tlongtin@sover.net>Subject:Re: Teabag ProblemTo:Kepert <andrew@frey.newcastle.edu.au>

Hi Andrew - Just to let you know that David Eppstein has posted his and my ruminations concerning the teabag problem - please be included on your page or if you like, with your permission, I will forward your contributions thus far to David: www.ics.uci.edu/~eppstein/junkyard/teabag.html I'm looking at a circular curved fold which extends nearly to the center of the square whereby the concave end can no longer be vertical and maintain all edges in the plane - it needs to be inclined inwards thus reducing the volume. Apparently a shallower curve would allow the concave end to slant outwards. I guess my initial try with the curve extending 1/4 into the square just happened to align vertically with the edges remaining in the plane. Needs work! Tom

From:Andrew Kepert <andrew@frey.newcastle.edu.au>To:Tom Longtin <tlongtin@sover.net>, eppstein@ics.uci.eduSubject:Re: Teabag ProblemDate:Wed, 18 Nov 1998 19:21:56 +1100

Tom & David, I have just spent a bit of time fiddling with the teabag problem. The main result of this is a page describing how to do curved folds in general - this expands on the brief description I sent Tom last Saturday. It is at http://frey.newcastle.edu.au/~andrew/teabag/folding/curvedFold.html Secondly I have a question related to this about foldability/bendability. I am hoping that might be able to be answered through the Junkyard: Is every smooth surface whose Gaussian curvature is everywhere zero bendable from paper? Cheers, Andrew p.s. (to TL) I have sent DE the last few e-mails we have exchanged, to be thrown into the junkyard or onto the junkheap, whichever he prefers.

From:David Eppstein <eppstein@ics.uci.edu>To:Andrew Kepert <andrew@frey.newcastle.edu.au>Subject:Re: Teabag ProblemDate:Wed, 18 Nov 1998 17:46:22 -0800 (PST)

I've updated my teabag page to include the messages you both sent me. It still doesn't compare to the nice graphics at http://frey.newcastle.edu.au/~andrew/teabag/ though... Also, I meant to say this before but, looking through all the old messages, I don't see it: David Huffman at UC Santa Cruz (the inventor of Huffman codes) has done quite a bit of work on shapes formed by curved folds; there's some on the web at http://www.sgi.com:80/grafica/huffman/index.html (and also a nice display included in the origami art show going on now in the lobby of the Xerox Palo Alto Research Center, but that's probably not useful information to people in Vermont and Australia...) -- David Eppstein UC Irvine Dept. of Information & Computer Science eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

From:Tom Longtin <tlongtin@sover.net>To:Andrew Kepert <andrew@frey.newcastle.edu.au>Subject:A New WrinkleDate:Wed, 2 Dec 1998 13:44:00 -0500 (EST)

Gentlemen - I thought this fold worth considering - haven't worked out the volume tho. Instead of 2 curved folds 4 worked OK with cardboard - one at each edge. In teabag2.gif the inner diamond would remain planar with a ruled surface from each of its edges to the outer corners. As before with 2 curved folds, the concave sides need to be vertical so the 4 outer edges remain in a plane. Regards, Tom Longtin

From:"Cooper, Christopher [255942]" <255942@academic.oldham.ac.uk>To:eppstein@ics.uci.eduDate:Wed, 24 Jan 2001 14:31:16 -0000

Dear Mr Earwicker, After taking a close look at your work I have also taken an interest in formulating the correct account of the efficiency of a teabag. Under much research and testing I found that it is an inaccurate judgement to assume muddle in the almighty powers of the teabag. Surly something, which as evolved form square too circular and finally into pyramid should not be underestimated. I believe that you should stay well away from the powers which be. For more then five centuries now they have lied dormant. If disturbed it will cause sure Disaster upon all of humanity. Whom drink thy foul broth?

From:James Waldby <j-waldby@pat7.com>To:eppstein@ics.uci.eduReply-To:j-waldby@pat7.comSubject:Re: Volume of a pillow shaped pouch?Date:Thu, 21 Mar 2002 16:06:25 -0800

David Eppstein wrote: > > In article <u9imq2fjqg3f43@corp.supernews.com>, > wayne ingalls <ingallsw.junkinhere@frontiernet.net> wrote: > > > I'm looking for help with a calculation. If you start with a > > rectangular pouch (bag?) and want to fill it with a liquid, > > how would you calculate the volume? Assume that the > > pouch is sealed on all four sides after being filled, so the > > resulting shape is like a pillow. Can someone help me with > > a formula to calculate the volume for a rectangular pouch? > > I don't know a formula, but there is a lot of previous discussion at > <http://www.ics.uci.edu/~eppstein/junkyard/teabag.html> ... I ran a couple of experiments with water-filled plastic bags. One formed of two 102mm squares[1] weighed 7.7 oz (about 218 g) and one of two 114mm squares weighed 9.5 oz (about 269 g) which (assuming one g per ml of water) leads to volume ratios of .2057 and .182, respectively, which is fairly large experimental variation but still indicates that the "0.2055+ achieved" ratio mentioned in the webpage is pretty good. [1] http://www.pat7.com/jp/sq/102.jpg shown after some leakage

From:jmb184@frontiernet.netTo:eppstein@ics.uci.eduSubject:Re: Volume of a pillow shaped pouch?Date:Thu, 21 Mar 2002 16:07:13 -0800

On Thu, 21 Mar 2002 04:02:10 -0000, wayne ingalls <ingallsw.junkinhere@frontiernet.net> wrote: >I'm looking for help with a calculation. If you start with a >rectangular pouch (bag?) and want to fill it with a liquid, >how would you calculate the volume? Assume that the >pouch is sealed on all four sides after being filled, so the >resulting shape is like a pillow. Can someone help me with >a formula to calculate the volume for a rectangular pouch? If you get an answer you are satisfied with, let me know. Last July there was an extended exchange, some in the news group, other messages privately exchanged. I share them because while I don't feel like revisiting the problem, the ideas may be helpful. http://groups.google.com/groups?as_umsgid=%3C3b430a6c.15210760%40news.eas ynet.co.uk%3E&hl=en >On Wed, 04 Jul 2001 12:36:24 GMT, library.treasures@saqnet.co.uk >wrote: > >>Calculate, that is, not just simply experimentally find out, but >>mathematically describe the approximate maximum volume of a common, >>brown paper bag. >> >>When empty and flat, the non-stretch, strong paper bag is 21 cm by 28 >>cm, with the opening at the shorter 21 cm side of the 3 to 4 ratio >>rectangle. > >For a open bag: >Volume = (k*h*a^2)/3 where h= sqrt(b^2 - a^2/2) and 1<k<4/pi >For the dimensions given, Volume = 3466 cc. maximum for an open bag. Retraction. (this was the last posting in the thread) An further reflection and a little playing with paper cutouts, a flat bag of the kind described can certainly enclose a parallelopiped which is a/2 x a/2 x (b - a/4) This accounts for a basic volume of (28 - 5.25)*(10.5^2) or 2508.1875 and leaves flat triangular tabs. If the parallelopiped were stretched into a cylinder, the volume would be 3193.58. The most additional volume possible would be if the two triangular tabs formed in order to create the parallelopiped could be filled. These are each a units in circumference at the base and a/4 units high. If filled, each could account for the volume contained in a cone a/2 units high and a units in circumference at the base. This works out to 61.418 cc per tab. Thus the upper bound on total volume is 3316.4078 cc. John

Date:Mon, 30 Sep 2002 10:16:29 +1000From:Dr Andrew Kepert <andrew.kepert@newcastle.edu.au>Subject:Re: ZIPLOC CONJECTURE AND TEA BAG PROBLEMTo:"Narasimham G.L." <mathma18@hotmail.com>, eppstein@ics.uci.edu, pgearwicker@dra.hmg.gb, dan@ics.uci.edu, andrew@frey.newcastle.edu.au, tlongtin@sover.net, 255942@academic.oldham.ac.uk

G'day Teabaggers, If I understand the note from Narasimham G.L., my "second solution" http://maths.newcastle.edu.au/~andrew/teabag/#TWO includes the configuration discussed below as the "degenerate case" where the triangular facets have zero size. I worked out the volume as pi * 2^(1/2) * (3^(1/2) - 1)/16 which is 2.03276003. My best solution is around 0.2055. 2/pi^2=3D2.02642367 would have been a neat solution, but we have already passed it. On the other hand, no-one has verified my calculations ... My 2.055 is not optimal either -- the profile of the cone can be perturbed from being circular for some gain. A rough variational argument convinced me that the property of the optimal profile curve is that the radius of curvature is proportional to the length of the side of the cone before it hits the axis planes. (See the image of 1/8 of the teabag on the above page for what I mean by this distance.) Note that "radius of curvature" for a cone needs to be defined properly -- it can be done in terms of the tan of the slant angle, or equivalently, a section through the cone at distance 1. Getting a profile curve from this invariant property is another matter, though. A finite-element simulation may be the only way. Cheers, Andrew At 7:51 PM +0000 28/9/02, Narasimham G.L. wrote: > > >To > >David Eppstein, Paul Earwicker, Tom Longtin, Andrew Kepert and >others who participated in the Tea Bag. > > > >Dear Tea Bag problem enthusiasts, > > > >SUB : Tea Bag volume constant could be 2/p2 > > > >With the above problem Ziploc Conjecture recently in sci.math , the >following was brought to notice by David Eppstein > ><http://www.ics.uci.edu/~eppstein/junkyard/teabag.html>http://www.ics.uci.edu/~eppstein/junkyard/teabag.html > > > >The problem has absorbing interest, that it was hitherto not addressed. > > > >I am trying to prove my own Conjecture, but the following are >observations for time being. > >The Ziploc conjecture and Tea Bag problem essentially address the >same problem, i.e., what max volume can be contained by a given >wetted area of water, or a powdery substance, inside a square bag. >It turns out the Ziploc problem is the guessed or conjectured >solution to the Tea Bag, given below. > > > >In the corner of the Tea Bag, one obtains cones filled with water. >This is expected and explainable. The Tea/water bags continue to be >developable surfaces with zero Gauss curvature everywhere. The >regular shapes discernable are cones and cylinder in the middle of >the bag. Other transition pieces also are intrinsically flat, having >an irregular shape, and hitherto are not given a name, presumably >due to their irregular geometry. However, it is sure that one >direction is straight, at any point of the bag. As swept polar angle >around corner containing cone apex is 2*pi*sin(alpha)=2*pi/2=pi, by >virtue of local development, we must have sin(alpha)=1/2 or the >cones must have semi-vertical angle 30 degrees. This is in fact the >case; one finds these right circular (not oblique, as before >thought) cones. In the middle of the bag, peri > > > >It now remains for some one to prove the Ziploc Conjecture !!---I >would appreciate remarks from you all. > > > >Regards, > >Narasimham G.L. > > > >( Attachment same as this text) -- __r-,!\ Dr Andrew Kepert. Lecturer in Maths / Deputy Head _/ `' \ Ph: +61-2-4348 4116 Fax: 4348 4145 / \ School of Applied Sciences \ ! Ourimbah Campus, University of Newcastle )_,--., o<<<< PO Box 127, Ourimbah NSW 2258. 33o21'S 151o22'E \__/ andrew@maths.newcastle.edu.au V http://maths.newcastle.edu.au/~andrew/

From:"Paul Earwicker" <pgearwicker@dstl.gov.uk>To:"Narasimham G.L." <mathma18@hotmail.com>, <eppstein@ics.uci.edu>, <pgearwicker@dra.hmg.gb>, <dan@ics.uci.edu>, <andrew@frey.newcastle.edu.au>, <tlongtin@sover.net>, <255942@academic.oldham.ac.uk>Subject:Re: ZIPLOC CONJECTURE AND TEA BAG PROBLEMDate:Mon, 30 Sep 2002 14:11:56 +0100

Dear All Many thanks for keeping me informed of your efforts on the Tea Bag = problem. Soon my (very)old email address will no longer work and I would be = pleased if you could use the following one. pgearwicker@dstl.gov.uk Thank you very much Paul Paul Earwicker DSTL