PDA

View Full Version : PATENT PENDING! New Polyhedron discovered by Warren Platts!

Warren Platts
2008-Jul-11, 01:46 AM
There are lots of so-called "equal area" truncated polyhedron maps out there, like this one (http://www.csiss.org/map-projections/Polyhedral_Globes/Truncated_Icosahedron.pdf) at csiss.org, but the faces of an ordinary truncated icosahedron (the soccerball shape) aren't of an equal area--the hexagons have about 1.5 times more area than do the pentagons.

So I modified the hexagons so that they are exactly equal to the area of the pentagons. The exact formula is a trade secret for now, but I will say that the if the sides of the pentagons, and hence the pentagon-hexagon boundaries are equal to 1, then the hexagon-hexagon boundaries will be approximately ~0.64 times the pentagon-hexagon boundaries. That should be accurate enough so you can convince yourself that the hexagons and pentagons are of equal area.

Buckminster Fuller took out a patent (#2393676) on a similar polyhedral globe.

A global map that used my polyhedron would be improved over the ordinary truncated icosahedron globe because there is less distortion because the modified hexagons take up less of the surface of the globe than the regular hexagons of the regular truncated icosahedron.

I've scoured the internet for the last week--I can't find a similar construction anywhere. I doubt that even 10010110 will be able to find one! :)

In fact, I'll pay \$20.00 USD to anyone who can find a description of an identical equal area faced truncated polyhedron. :lol:

http://www.bautforum.com/attachment.php?attachmentid=8295&d=1215738755

If you want to try this at home, it's best to use an xacto knife or razor knife rather than scissors. For best folding results, lightly score the outside edges that need folded--it works a lot better.

You saw it here first, folks!

(again)

On bautforum.com

:D

Occam
2008-Jul-11, 02:27 AM
Well, for what it's worth, I think that's pretty cool!

Warren Platts
2008-Jul-11, 02:47 AM
Well, for what it's worth, I think that's pretty cool!Why thank you sir! :cool:

BigDon
2008-Jul-11, 06:33 AM
Warren, nice one! Much better than your eulogy for George Carlin!

So THIS is what you really do. :)

Sure as hell couldn't have been working for Hallmark.

Jeff Root
2008-Jul-11, 08:13 AM
Warren,

In the drawing, some of the horizontal edges are a pixel longer than others,
resulting in some angled edges having very slightly different angles. Is that
an unavoidable result of digitization, or a drawing error? If it is an error in
drawing, I can repair it.

-- Jeff, in Minneapolis

Ivan Viehoff
2008-Jul-11, 09:14 AM
The exact formula is a trade secret for now
Hardly a secret. An equal area truncated icosahedron defines precisely what it is, and the ratios of the irregular hexagons can calculated by anyone with sufficient skill in trigonometry.

Ken G
2008-Jul-11, 09:36 AM
Hardly a secret. An equal area truncated icosahedron defines precisely what it is, and the ratios of the irregular hexagons can calculated by anyone with sufficient skill in trigonometry.That was my reaction also. One only needs a formula for the area of a distorted hexagon with two different length alternating sides, as one may then set its area equal to that of a pentagon with regular sides equal to the longer of the sides of the distorted hexagon. It's all done by inscribing triangles from the center of the polygons, and is straightforward trigonometry.

hhEb09'1
2008-Jul-11, 11:21 AM
There are lots of so-called "equal area" truncated polyhedron maps out there, like this one (http://www.csiss.org/map-projections/Polyhedral_Globes/Truncated_Icosahedron.pdf) at csiss.org, but the faces of an ordinary truncated icosahedron (the soccerball shape) aren't of an equal area--the hexagons have about 1.5 times more area than do the pentagons."So-called"? Equal area maps are not called equal area because the individual polygonal pieces are equal area, it has to do with the projection of the map onto the flat surface. Often, other projections, such as gnomic, are used instead of equal area projections.
So I modified the hexagons so that they are exactly equal to the area of the pentagons. The exact formula is a trade secret for now, but I will say that the if the sides of the pentagons, and hence the pentagon-hexagon boundaries are equal to 1, then the hexagon-hexagon boundaries will be approximately ~0.64 times the pentagon-hexagon boundaries. That should be accurate enough so you can convince yourself that the hexagons and pentagons are of equal area.A better approximation is 0.64070, right? :)

A global map that used my polyhedron would be improved over the ordinary truncated icosahedron globe because there is less distortion because the modified hexagons take up less of the surface of the globe than the regular hexagons of the regular truncated icosahedron. The distortion depends upon the projection. It's often more of a function of the greatest distance between points, rather than the area of the pieces, but there's obviously a relation between the two.

PS: 0.64070 results from microsoft excel =SQRT(3+SQRT((25+10*SQRT(5))/3))-2

Warren Platts
2008-Jul-11, 04:11 PM
Warren, nice one! Much better than your eulogy for George Carlin!

So THIS is what you really do. :)

Sure as hell couldn't have been working for Hallmark.Now that I think about it, there is a certain type of geek who might not mind such a card. :think: Thanks for the idea!

In the drawing, some of the horizontal edges are a pixel longer than others, resulting in some angled edges having very slightly different angles. Is that an unavoidable result of digitization, or a drawing error? If it is an error in drawing, I can repair it.Sorry, the attached file above was chosen because it looks better on screen; the .png file attached below is better for printing if you want to try construct the 3D model; it has 2000 X 1250 pixels instead of 800 X 500; I would upload a bigger file, but the server is only letting me upload 30KB files today for some reason (it says that 146.5 is the maximum .png allowed! :mad:); but I just printed it, and it is crisper than the 800 X 500 (but not quite as crisp as the 8000 X 5000 version!); still, there might be some intrinsic errors--the BASIC program I used to create the file is only accurate to 15 significant digits! ;)

Hardly a secret. An equal area truncated icosahedron defines precisely what it is, and the ratios of the irregular hexagons can calculated by anyone with sufficient skill in trigonometry.Hey, it took me 3-4 days of nonstop obsessing that included solving messy quadratic equations before I could figure it out!

That was my reaction also. One only needs a formula for the area of a distorted hexagon with two different length alternating sides, as one may then set its area equal to that of a pentagon with regular sides equal to the longer of the sides of the distorted hexagon. It's all done by inscribing triangles from the center of the polygons, and is straightforward trigonometry.If you want a more challenging problem, try coming up with a straight-edge and compass construction of a (regular) pentagon with an adjacent equal area hexagon. :D (Hint: the circumradius of the hexagon is slightly smaller than the circumradius of the pentagon.)

There are lots of so-called "equal area" truncated polyhedron maps out there, like this one at csiss.org, but the faces of an ordinary truncated icosahedron (the soccerball shape) aren't of an equal area--the hexagons have about 1.5 times more area than do the pentagons.
"So-called"? Equal area maps are not called equal area because the individual polygonal pieces are equal area, it has to do with the projection of the map onto the flat surface. Often, other projections, such as gnomic, are used instead of equal area projections.
True. Still, it would be nice to have an equal area projection on equal area faces. Besides golf ball dimple patterns (there's been a lot of research into that lately), there would be GIS-related applications for such a map--it allows for 2-dimensional storage of 3D data with a minimum of distortion, which is useful in itself. In addition, there could be practical applications: for example, an ecologist might want a broad-brushed comparison of biodiversity richness based on roughly circular portions of the globe that are also of equal area. The equal area faced truncated icosahedron allows one to do that.

So I modified the hexagons so that they are exactly equal to the area of the pentagons. The exact formula is a trade secret for now, but I will say that the if the sides of the pentagons, and hence the pentagon-hexagon boundaries are equal to 1, then the hexagon-hexagon boundaries will be approximately ~0.64 times the pentagon-hexagon boundaries. That should be accurate enough so you can convince yourself that the hexagons and pentagons are of equal area.
A better approximation is 0.64070, right? Gharg! Not quite as good as 0.640695431407952 though! :)

A global map that used my polyhedron would be improved over the ordinary truncated icosahedron globe because there is less distortion because the modified hexagons take up less of the surface of the globe than the regular hexagons of the regular truncated icosahedron.
The distortion depends upon the projection. It's often more of a function of the greatest distance between points, rather than the area of the pieces, but there's obviously a relation between the two.Your way of putting it is definitely more precise.

PS: 0.64070 results from microsoft excel =SQRT(3+SQRT((25+10*SQRT(5))/3))-2
:clap:Very elegant hhEb09'1! :clap:

That's a lot better than the mess I came up with: =(SQRT(16-4*(1-(SQRT(25+10*SQRT(5)))/SQRT(3))))/2-2

Now all I've got to do is figure out the vertices in 3D Cartesian format, and then convert those to latitude and longitude. . . .

Argos
2008-Jul-11, 05:05 PM
Re the OP, you forgot to say you deserve the Nobel Prize in maths...

Warren Platts
2008-Jul-11, 05:32 PM
Re the OP, you forgot to say you deserve the Nobel Prize in maths...
I figured grant hutchison is jealous of me enough already. :D

. . . . Besides I keep telling you guys I'm going for the Templeton these days!

Warren Platts
2008-Jul-11, 05:42 PM
Question: If =SQRT(3+SQRT((25+10*SQRT(5))/3))-2 and =(SQRT(16-4*(1-(SQRT(25+10*SQRT(5)))/SQRT(3))))/2-2 produce the same answer down to 15 decimal places, does that in itself constitute a proof that the two equations are equivalent? Or could one never be sure that as one added digits, the two equations would not produce different results?

tdvance
2008-Jul-11, 06:09 PM
no--it's not a proof.

After all, (10^1000+1)/(10^1000) and (10^1000+2)/(10^1000) agree on 999 decimal digits, but are unequal.

However, that both formulas were created to solve the same problem and that they both agree to 15 decimal places makes me think that "most likely one can be transformed into the other". That some of the same integers occur between the two (5, 25, 3, 2, 10, as well as some powers of 2) adds to that belief.

Warren Platts
2008-Jul-11, 06:26 PM
Actually, one obvious application of the equal face area truncated icosahedron would be for virtual maps used when rendering celestial bodies in Celestia. An equal face area gnomic projection with 32 faces on a 2D "cut" of the modified soccerball shape would look a lot better than the cylindrical projections they are using now--there would be no "pinched" look at the poles. And that's no lie.

hhEb09'1
2008-Jul-11, 06:35 PM
True. Still, it would be nice to have an equal area projection on equal area faces. Besides golf ball dimple patterns (there's been a lot of research into that lately), there would be GIS-related applications for such a map--it allows for 2-dimensional storage of 3D data with a minimum of distortion, which is useful in itself. In addition, there could be practical applications: for example, an ecologist might want a broad-brushed comparison of biodiversity richness based on roughly circular portions of the globe that are also of equal area. The equal area faced truncated icosahedron allows one to do that.I'm liking this problem Warren :)

Just remember though, it's the polygons that are equal area, not the Earth's surface represented. But approximately, yeah.

:clap:Very elegant hhEb09'1! :clap: Thanks!
Now all I've got to do is figure out the vertices in 3D Cartesian format, and then convert those to latitude and longitude. . . .According to Mathworld (http://mathworld.wolfram.com/Icosahedron.html), the twelve points of the icosahedron are at 1, -1, and the other ten at plus/minus sqrt(5)/5. Those should be easy to translate into lat, and the lon will be just 0, 72, 144, 216, and 288 degrees for one set and 36, 108, 180, 252, and 324 degrees for the other set. Then, you can use your side length to get how far away the points of the pentagons are from those, and use distance/azimuth spherical trig formula for the real points.

Question: If =SQRT(3+SQRT((25+10*SQRT(5))/3))-2 and =(SQRT(16-4*(1-(SQRT(25+10*SQRT(5)))/SQRT(3))))/2-2 produce the same answer down to 15 decimal places, does that in itself constitute a proof that the two equations are equivalent? Or could one never be sure that as one added digits, the two equations would not produce different results?Does 960 = (EXP(PI() *SQRT(43))-744)^(1/3) ? :)

Although mathematicians deal with approximations all the time, equality (in the real numbers) does mean to the end of the decimal digits (1=0.999... notwithstanding :) ). However, those two numbers are exactly equal (take the 2 in the denominator under the radical to a 4, and divide out. Do the subtraction, and put the sqrt(3) in that denominator under the other radical. That should do it.)

Jeff Root
2008-Jul-11, 09:36 PM
Warren,

I didn't see that you have a 2000 x 1250 pixel version until after I finished
the polygons exactly identical, so the whole unfolded polyhedron is a few
pixels wider, and might no longer be correct. I haven't taken the time yet
to print and assemble it to discover if I made any mistakes. Assuming my
printer works.

Cut on the black lines, fold on the blue lines.

I didn't know what name to put on it. It needs a title.

-- Jeff, in Minneapolis

Warren Platts
2008-Jul-11, 10:10 PM
Warren,

I didn't see that you have a 2000 x 1250 pixel version until after I finished
the polygons exactly identical, so the whole unfolded polyhedron is a few
pixels wider, and might no longer be correct. I haven't taken the time yet
to print and assemble it to discover if I made any mistakes. Assuming my
printer works.
Wow Jeff, that pretty sweet!
http://www.bautforum.com/attachments/general-science/8311d1215812129-patent-pending-new-polyhedron-discovered-warren-platts-equalarea3.png

Cut on the black lines, fold on the blue lines.

I didn't know what name to put on it. It needs a title.
I just put one together with tape, so they do go together, but with glue and the tabs, it would probably come out better for sure!

As for titles, I'm vacillating between "equal area faced truncated icosahedron" and "equal face area truncated icosahedron". The latter is a little easier to say and doesn't carry the connotation of being "faced". . . . ;)

Jeff Root
2008-Jul-11, 11:27 PM
Oh, NOW I see that I missed putting in a short line for the edge of one tab.
I'll fix it and replace the image in a few minutes.

Done.

-- Jeff, in Minneapolis

Warren Platts
2008-Jul-11, 11:54 PM
Oh, NOW I see that I missed putting in a short line for the edge of one tab.
I'll fix it and replace the image in a few minutes.

-- Jeff, in MinneapolisDude, relax! :hand: It's Miller Time! You've earned yourself a beer--take a break! :)

Warren Platts
2008-Jul-12, 02:09 PM
Hey Jeff, what software did you use to add the flaps and such?

And here's the code for the BASIC program that makes the bitmap, so you can create a bitmap of any size you want (within reason!).

OPTION ANGLE DEGREES
LET bitmapscale = 1
LET width = 800 * bitmapscale
LET ht = 500 * bitmapscale
SET BITMAP SIZE width, ht
LET left = 7 * (7/8)
LET right = 9 * (7/8)
LET bottomtop = 5 * (7/8)
SET WINDOW -left,right,-bottomtop,bottomtop
SET LINE

REM Tsmall is the length of the short side of the hexagon--Thanks hhEb09'1!
LET Tsmall = SQR(3+SQR((25+10*SQR(5))/3))-2
LET Rhex = SQR((Tsmall^2+Tsmall+1)/3)
LET Rpent = (0.1)*(SQR(50+(10*SQR(5))))

LET pentscale = 1
LET hexscale = pentscale * Rhex / Rpent
LET big = 74.4121373379946
LET little = 45.5878626620054
LET pentht = Rpent * COS(36)
LET shexht = Rhex * COS(big/2)
LET thexht = Rhex * COS(little/2)
LET spacer = 2*thexht*COS(30)
LET scl = 1
LET i = 3/16
LET j = -3/8

REM **************************************
REM north pole pentagon
DRAW pentagon WITH ROTATE(18)*SHIFT(i+spacer,j+2*thexht*SIN(30)+2*the xht+shexht+pentht)*SCALE(scl)
REM south pole pentagon
DRAW pentagon WITH ROTATE(-18)*SHIFT(i,j-2*thexht-shexht-pentht)*SCALE(scl)

DRAW pentup WITH SHIFT(i,j)*SCALE(scl)
FOR a = 1 TO 3 STEP 2
DRAW pentdown WITH SHIFT(i+a*spacer,j) * SCALE(scl)
DRAW pentdown WITH SHIFT(i-a*spacer,j)*SCALE(scl)
DRAW pentup WITH SHIFT(i+(a+1)*spacer,j)*SCALE(scl)
DRAW pentup WITH SHIFT(i-(a+1)*spacer,j)*SCALE(scl)
NEXT a
DRAW pentdown WITH SHIFT(i+5*spacer,j)*SCALE(scl)

160 PICTURE pentagon
161 PLOT 0,0
170 FOR n=0 TO 5
180 PLOT LINES:Rpent*COS(n*360/5),Rpent*(SIN(n*360/5)) ;
190 NEXT n
200 END PICTURE

500 PICTURE hexagon
501 LET big = 74.4121373379946
502 LET little = 45.5878626620054
503 LET x = 0
504 LET y = 0
505 PLOT 0,0
510 FOR n = 1 TO 3
520 PLOT LINES:Rhex*COS(x),Rhex*SIN(y);
525 LET x = x + big
526 LET y = y + big
530 PLOT LINES:Rhex*COS(x),Rhex*SIN(y);
535 LET x = x + little
536 LET y = y + little
537 PLOT LINES:Rhex*COS(x),Rhex*SIN(y);
540 NEXT n
550 END PICTURE

600 PICTURE pentup
610 REM face #1
620 DRAW hexagon WITH ROTATE(-90+(little/2))*SHIFT(0,0)
630 REM face #2
640 DRAW pentagon WITH ROTATE(18)*SHIFT(0,pentht + shexht)
650 REM face #3
660 DRAW hexagon WITH ROTATE(-90-(big/2))*SHIFT(0,-thexht*2)
690 END PICTURE

700 PICTURE pentdown
710 REM face #5
720 DRAW hexagon WITH ROTATE(-90-(big/2))*SHIFT(0,2*thexht*SIN(30))
730 REM face #6
740 DRAW pentagon WITH ROTATE(-18)*SHIFT(0,2*thexht*SIN(30)-shexht-pentht)
750 REM face #7
760 DRAW hexagon WITH ROTATE(90-(big/2))*SHIFT(0,2*thexht*(SIN(30)+1))
790 END PICTURE

1000 END

2008-Jul-12, 05:42 PM
hmmm.. i should show this page to my boss.. he could convert that printout into a CAD drawing and we could cut a piece of thin steel- like, say, 24 gauge- on the flat laser and make a metal ball if he could figure out a way to bend it up in the press brake..
that would be pretty sweet all TIG welded together.

Nowhere Man
2008-Jul-12, 06:02 PM
You'd probably have to cut it into pieces that fit in the brake, and then weld the pieces together.

Fred

Jeff Root
2008-Jul-12, 06:06 PM
I just used Paint Shop Pro to draw the flaps and adjust the edges of two
polygons, then copied and pasted them into your layout.

Don't see line numbers much anymore, but LET statements with the word
LET actually spelled out? So rare I didn't realize what it was for a minute!
What version of BASIC is that? I looked at the QBASIC help, and there is
no SET command. A graphics window would have to be set up differently
in QBASIC, but it has been long enough that I forget how. Does it work
under Windows, DOS, or what? Looks easy enough to convert to QBASIC,
or PowerBASIC which I own but haven't really used yet.

-- Jeff, in Minneapolis

Jeff Root
2008-Jul-12, 06:20 PM
hmmm.. i should show this page to my boss.. he could convert that
printout into a CAD drawing and we could cut a piece of thin steel-
like, say, 24 gauge- on the flat laser and make a metal ball if he could
figure out a way to bend it up in the press brake..
that would be pretty sweet all TIG welded together.
Oh, what a neat idea! I really should have thought of it. For one,
my brother-in-law uses lasers to cut sheet steel, for two, I came up
with an idea for something I wanted him to cut out of sheet steel
with his lasers (although he rejected it as not having enough value
to make a profit... It would probably have to be made in China to be
profitable), and for three, I designed and had someone else shape
the parts for a steel model of a sculpture I want(ed) to build.

-- Jeff, in Minneapolis

tdvance
2008-Jul-12, 06:37 PM
I just used Paint Shop Pro to draw the flaps and adjust the edges of two
polygons, then copied and pasted them into your layout.

Don't see line numbers much anymore, but LET statements with the word
LET actually spelled out? So rare I didn't realize what it was for a minute!
What version of BASIC is that? I looked at the QBASIC help, and there is
no SET command. A graphics window would have to be set up differently
in QBASIC, but it has been long enough that I forget how. Does it work
under Windows, DOS, or what? Looks easy enough to convert to QBASIC,
or PowerBASIC which I own but haven't really used yet.

-- Jeff, in Minneapolis

I remember BASIC-my first (non-natural) language in 8th grade. Of course, once you try C, Perl, Lisp, or almost anything other than maybe COBOL or Fortran or Assembly, you'll never go back to BASIC, compiled or modernized or otherwise.

orionjim
2008-Jul-12, 06:42 PM
.
.
.
Don't see line numbers much anymore, but LET statements with the word
LET actually spelled out? So rare I didn't realize what it was for a minute!
What version of BASIC is that? I looked at the QBASIC help, and there is
no SET command. A graphics window would have to be set up differently
in QBASIC, but it has been long enough that I forget how. Does it work
under Windows, DOS, or what? Looks easy enough to convert to QBASIC,
or PowerBASIC which I own but haven't really used yet.

-- Jeff, in Minneapolis

I've followed some of Warren's work on the Titus/Bode Law and if my memory is correct he's using "Decimal Basic" a freeware program. I did a search in Google and found this site:
http://hp.vector.co.jp/authors/VA008683/english/

Jim

hhEb09'1
2008-Jul-12, 07:02 PM
True. Still, it would be nice to have an equal area projection on equal area faces. Besides golf ball dimple patterns (there's been a lot of research into that lately), I've seen some of the results (Let's see, where did I put that patent file...).

One of the things about the original pattern, which has all edges equal to the same length, the dimple pattern with dimples at each vertex, and at each center of each face, has all the 120 dimples the same distance away from each other, except for the 12 dimples at the centers of the pentagons, which are a slight bit closer to their five mates.

I don't really know if that's good or bad, but someone thinks it is! :)

Warren Platts
2008-Jul-13, 12:40 AM
hmmm.. i should show this page to my boss.. he could convert that printout into a CAD drawing and we could cut a piece of thin steel- like, say, 24 gauge- on the flat laser and make a metal ball if he could figure out a way to bend it up in the press brake..
that would be pretty sweet all TIG welded together.

You'd probably have to cut it into pieces that fit in the brake, and then weld the pieces together.
A stainless steel equal face area truncated icosahedron would be pretty sweet. If we could then do an acid-etching of the world on it, then we would have something worth marketing; a nice floor mounted stainless steel globe would sell for at least \$700 per piece--cf. maps.com (http://www.maps.com/maps.aspx?nav=MS&cid=1,74). Wow, we're getting a nice, little corporation going here. Might as well make some money off this site! :D

Don't see line numbers much anymore, but LET statements with the word LET actually spelled out? So rare I didn't realize what it was for a minute! What version of BASIC is that? I looked at the QBASIC help, and there is no SET command. A graphics window would have to be set up differently in QBASIC, but it has been long enough that I forget how. Does it work under Windows, DOS, or what? Looks easy enough to convert to QBASIC, or PowerBASIC which I own but haven't really used yet.

I've followed some of Warren's work on the Titus/Bode Law and if my memory is correct he's using "Decimal Basic" a freeware program. I did a search in Google and found this site:
http://hp.vector.co.jp/authors/VA008683/english/
That looks right--the help function for the program I have is in Japanese--but you can still figure it out.

I remember BASIC-my first (non-natural) language in 8th grade. Of course, once you try C, Perl, Lisp, or almost anything other than maybe COBOL or Fortran or Assembly, you'll never go back to BASIC, compiled or modernized or otherwise.Lisp! :sick: But yeah, I took the very first computer class our high school taught, with BASIC of course! Visual BASIC still is used a lot.

I've seen some of the results (Let's see, where did I put that patent file...).

One of the things about the original pattern, which has all edges equal to the same length, the dimple pattern with dimples at each vertex, and at each center of each face, has all the 120 dimples the same distance away from each other, except for the 12 dimples at the centers of the pentagons, which are a slight bit closer to their five mates.

I don't really know if that's good or bad, but someone thinks it is!
I don't know much about golf ball engineering, but I'm not sure the equal face area truncated icosahedron (EFATI) will help with golf balls. :think: They're already going with the 120-side regular polyhedron, whatever that's called, and there's no getting rid of the pentagons.

You know a lot about physics, though. What about a new Dungeons and Dragons die? Or a new and improved Magic 8-Ball with 32 possible answers, instead of 20? Since the faces have equal area, the pentagons and hexagons should be equally like to come up.

Jeff Root
2008-Jul-13, 02:05 AM
Since the faces have equal area, the pentagons and hexagons should
be equally like to come up.
Oooo! Intriguing question, at least!

-- Jeff, in Minneapolis

Warren Platts
2008-Jul-13, 03:40 AM
I did fool around a bit with my paper model, and I must say that my impression was that the hexagons and pentagons were coming up about equally. Now of course you can't tell anything from a dozen rolls of a highly imperfect paper model of a polyhedron.

Nevertheless, the following bit of reasoning did appear to my fertile mind: you would figure that as the EFATI rolled around, it would most likely come to a stop when it ran into one of the relatively longer pentagon/hexagon edges compared to the much shorter hexagon/hexagon edges. Now the pentagons have 5 of the long edges, whereas as the hexagons only have 3 such edges--a ratio of 5 to 3.

On the other hand, there are 20 hexagons, but only 12 pentagons--a ratio of 5 to 3.

So the ratios cancel each other out, and so that's how the 50-50 ratio I "observed" could be explained. :neutral:

hhEb09'1
2008-Jul-13, 03:58 AM
Nevertheless, the following bit of reasoning did appear to my fertile mind: you would figure that as the EFATI rolled around, it would most likely come to a stop when it ran into one of the relatively longer pentagon/hexagon edges compared to the much shorter hexagon/hexagon edges. Now the pentagons have 5 of the long edges, whereas as the hexagons only have 3 such edges--a ratio of 5 to 3.I would think it'd be the other way around. Since the short edges on the hexagons are farther from the center, when the die is lying flat on a hexagon, it requires more momentum to flip it over the increased leverarm when rolling over the short side. The pentagon sides are nearly the same distance as the long sides of the hexagons. So your logic might still hold, but inverted.

Roll it again, ten thousand times. :)

Warren Platts
2008-Jul-14, 11:16 AM
Here's an interesting link (http://kiwi.atmos.colostate.edu/BUGS/index.html) on polyhedrons and geodesic grids as applied to climate modeling.

(Still working on the coordinates . . . .)

Warren Platts
2008-Jul-15, 08:00 PM
Below are the latitudes and longitudes. Don't ask me for a proof that they're right; I couldn't explain how I got them even to myself. So feel free to try and prove me wrong.

Also, I was serious about paying a reward (I hereby increase the amount to \$35 USD) to anyone who can find a complete description of the EFATI. Such a description must give the coordinates, or at least the length of the short hex-hex edge (and of course predate my post #1).

The circumradius R for the EFATI polyhedron (with pentagon edge = 1) is 2.160251992878220.

ETA: Cripes! After all that I realized I had the entire planet backwards! It's fixed now. . . . :wall:

1 66.81047400359420 N 36.00000000000000 E
2 66.81047400359420 N 108.00000000000000 E
3 66.81047400359420 N 180.00000000000000
4 66.81047400359420 N 108.00000000000000 W
5 66.81047400359420 N 36.00000000000000 W
6 49.75457717348470 N 36.00000000000000 E
7 49.75457717348470 N 108.00000000000000 E
8 49.75457717348470 N 180.00000000000000
9 49.75457717348470 N 108.00000000000000 W
10 49.75457717348470 N 36.00000000000000 W
11 31.32678094501860 N 9.99719338992966 E
12 31.32678094501860 N 62.00280661007030 E
13 31.32678094501860 N 81.99719338992970 E
14 31.32678094501860 N 134.00280661007000 E
15 31.32678094501860 N 153.99719338993000 E
16 31.32678094501860 N 153.99719338993000 W
17 31.32678094501860 N 134.00280661007000 W
18 31.32678094501860 N 81.99719338992970 W
19 31.32678094501860 N 62.00280661007030 W
20 31.32678094501860 N 9.99719338992966 W
21 7.24685763776526 N 22.50748957297480 E
22 7.24685763776526 N 49.49251042702520 E
23 7.24685763776526 N 94.50748957297480 E
24 7.24685763776526 N 121.49251042702500 E
25 7.24685763776526 N 166.50748957297500 E
26 7.24685763776526 N 166.50748957297500 W
27 7.24685763776526 N 121.49251042702500 W
28 7.24685763776526 N 94.50748957297480 W
29 7.24685763776526 N 49.49251042702520 W
30 7.24685763776526 N 22.50748957297480 W
31 7.24685763776526 S 13.49251042702520 E
32 7.24685763776526 S 58.50748957297480 E
33 7.24685763776526 S 85.49251042702520 E
34 7.24685763776526 S 130.50748957297500 E
35 7.24685763776526 S 157.49251042702500 E
36 7.24685763776526 S 157.49251042702500 W
37 7.24685763776526 S 130.50748957297500 W
38 7.24685763776526 S 85.49251042702520 W
39 7.24685763776526 S 58.50748957297480 W
40 7.24685763776526 S 13.49251042702520 W
41 31.32678094501860 S 26.00280661007030 E
42 31.32678094501860 S 45.99719338992970 E
43 31.32678094501860 S 98.00280661007030 E
44 31.32678094501860 S 117.99719338993000 E
45 31.32678094501860 S 170.00280661007000 E
46 31.32678094501860 S 170.00280661007000 W
47 31.32678094501860 S 117.99719338993000 W
48 31.32678094501860 S 98.00280661007030 W
49 31.32678094501860 S 45.99719338992970 W
50 31.32678094501860 S 26.00280661007030 W
51 49.75457717348470 S 0.00000000000000
52 49.75457717348470 S 72.00000000000000 E
53 49.75457717348470 S 144.00000000000000 E
54 49.75457717348470 S 144.00000000000000 W
55 49.75457717348470 S 72.00000000000000 W
56 66.81047400359420 S 0.00000000000000
57 66.81047400359420 S 72.00000000000000 E
58 66.81047400359420 S 144.00000000000000 E
59 66.81047400359420 S 144.00000000000000 W
60 66.81047400359420 S 72.00000000000000 W

hhEb09'1
2008-Jul-15, 11:39 PM
I used this great circle calculator (http://williams.best.vwh.net/gccalc.htm) to check a couple points.

4 66.81047400359420 N 144.00000000000000 W
5 66.81047400359420 N 72.00000000000000 W
Spherical model, distance in km:
2974.175281237102

4 66.81047400359420 N 144.00000000000000 W
9 49.75457717348470 N 144.00000000000000 W
Spherical model, distance in km:
1895.2512557617694

Their ratio is
0.63723589786996131454406921096288
which is close to the straight line ration of 0.64

Warren Platts
2008-Jul-16, 03:50 AM
I used this great circle calculator (http://williams.best.vwh.net/gccalc.htm) to check a couple points.

4 66.81047400359420 N 144.00000000000000 W
5 66.81047400359420 N 72.00000000000000 W
Spherical model, distance in km:
2974.175281237102

4 66.81047400359420 N 144.00000000000000 W
9 49.75457717348470 N 144.00000000000000 W
Spherical model, distance in km:
1895.2512557617694

Their ratio is
0.63723589786996131454406921096288
which is close to the straight line ration of 0.64
If the great circle distances are supposed to be proportional to the straight line distances, a 0.5% error is too much. I'll check it out.

Warren Platts
2008-Jul-16, 01:13 PM
Thanks for the link hhEb09'1. All the edges I checked (long and short) are the same everywhere. So I tested a pentagon at the poles (e.g., points 1-5 or points 56-60). It's impossible to screw one of those up: all vertices have the same latitude, and the longitudes are separated by exact 72o. Moreover, the diagonal and edge of a regular pentagon should be the golden ratio (1.6180339887498...), yet the corresponding length ratio on the sphere is 1.643424604009040. Close, but not an exact match, yet the pentagon was obviously laid out in a perfectly regular manner; therefore, the discrepancy must have something to do with spherical geometry.

Which leads to a new question: How do I know that the areas when projected on a sphere are equal???

hhEb09'1
2008-Jul-16, 01:16 PM
Well, the relative difference between the spherical distance and the linear distance is greater the longer the line, so we'd expect the ration of the spherical distances to be smaller. I guess the error is in the right direction, I mean.

Let's see, I just checked that website (by finding the distance from pole to pole), and their spherical model seems to be based upon an earth of circumference 40003.2 km. That means the diameter of the sphere is 12733.414 km, So, a spherical distance of 2974.175281237102 is 0.467144990 radians or 2947.20571 km straight line. A spherical distance of 1895.2512557617694 is 0.297681557 radians or 1888.261237 km straight line. The ratio of those two is .64069543, which is what you wanted. Your 0.5% error probably represents a one percent difference between the spherical areas of the figures.

hhEb09'1
2008-Jul-16, 01:18 PM
Which leads to a new question: How do I know that the areas when projected on a sphere are equal???Yes, I just looked at that. My best guess is about a one percent error.

PS: I take that back. I'm going to change my answer to "I don't know" :)

After thinking about it, the error is going to be a lot smaller than that.

PSS: Here's an approach. We can calculate the area of that five sided figure at the pole fairly easily (It's five triangles of the same size, bounded by great circle segments). Multiply that times twelve, and what's left over is divided between the twenty six sided figures.

That's assuming spherical earth still. The oblateness will change things again.

hhEb09'1
2008-Jul-16, 03:12 PM
That means the diameter of the sphere is 12733.414 km, So, a spherical distance of 2974.175281237102 is 0.467144990 radians or 2947.20571 km straight line. The program gives the two angles of each triangle as 56.263288140227864 degrees, or 0.9819796260 radians. Twice that, plus 2pi over 5, minus pi, times five, gives an area of 0.3950183. Times twelve, subtract from 4 pi, and divide by 20 gives 0.39130755 for the area of each six sided figure. So, one percent difference in area, on a spherical earth.

Warren Platts
2008-Jul-16, 03:14 PM
We could calculate the area of one of the interior spherical triangles of the pentagon, and then multiply by 5. The formula for the area of a spherical triangle is:

K = R2 * [(A + B + C) - pi]

where R is the radius of the sphere, and A, B, and C are the interior angles formed by great circles connecting the vertices.

Only problem is, these angles aren't the same as for a flat triangle (the interior angles of a flat triangle in radians add up to pi, and so the formula goes to zero.)

A JAVA applet that could calulate spherical areas given the lats and longs of three points would be nice, but preliminary attempts to google one up are to no avail at this point.

hhEb09'1
2008-Jul-16, 03:16 PM
We could calculate the area of one of the interior spherical triangles of the pentagon, and then multiply by 5. The formula for the area of a spherical triangle is:

K = R * (A + B + C - pi)2Is that square in the right place?

Warren Platts
2008-Jul-16, 03:41 PM
Oops! :doh:

Fixed it.
ETA: Found a formula from Ask Dr. Math (http://mathforum.org/library/drmath/view/65316.html):

My CRC Standard Mathematical Tables contain the basic formula for the
area of a spherical triangle:

Area = pi*R^2*E/180

where

E = spherical excess of triangle, E = A + B + C - 180
A, B, C = angles of spherical triangle in degrees

This is the formula you say isn't helpful because you don't know the
angles, right? Well, the tables also have the following formula for
the spherical excess E:

tan(E/4) = sqrt(tan(s/2)*tan((s-a)/2)*tan((s-b)/2)*tan((s-c)/2))

where

a, b, c = sides of spherical triangle
s = (a + b + c)/2

You can find the sides using either the cosine formula or the
haversine formula, found on the following pages
Will give that a try.

hhEb09'1
2008-Jul-16, 04:12 PM
ETA: Found a formula from Ask Dr. Math (http://mathforum.org/library/drmath/view/65316.html):It's the same one! :)

When I first guessed one percent, my reasoning was that a half percent longer would also mean a half percent wider, and that means a full percent larger area. Which is more or less correct. I got into trouble second-guessing myself when I thought maybe a half percent wider meant half percent shorter--but of course it does not. Next time, I stay with my first impression, until I've worked the numbers and I know for sure. :)

BigDon
2008-Jul-16, 05:07 PM
As a small aside, since we were talking about expensive globes,

I once pulled an "Inspector Clouseau" with a big five footer in my high school library. It was done up in a classic old world design and after I got it spinning put my hand on it to stop it and the friction and weight pulled my hand under the arc that supports the axis. Quite uncomfortable I might add. Managed to unhose myself without drawing attention to the fact I had done it. But it took a bit. since it was a relief globe I was able to get a good grip and push it in the other direction with my free hand and all my weight.

I recommend against that.

Warren Platts
2008-Jul-16, 05:11 PM
It's the same one! :)
By "same one" do you mean the areas of the pentagons and hexagons are equal when projected on the sphere?:confused:

I once pulled an "Inspector Clouseau" with a big five footer in my high school library. It was done up in a classic old world design and after I got it spinning put my hand on it to stop it and the friction and weight pulled my hand under the arc that supports the axis. Quite uncomfortable I might add. Managed to unhose myself without drawing attention to the fact I had done it. But it took a bit. since it was a relief globe I was able to get a good grip and push it in the other direction with my free hand and all my weight.
I hate it when that happens. . . .

Kaptain K
2008-Jul-16, 05:22 PM
Although I'll admit that your equal area polyhedron is elegant, the equal length version is still purtier!

hhEb09'1
2008-Jul-16, 06:48 PM
By "same one" do you mean the areas of the pentagons and hexagons are equal when projected on the sphere?No, I meant it was the same formula as in your previous post, but looking back maybe you meant the formula for calculating excess using tangents? I already calculated the difference in area in this post (http://www.bautforum.com/general-science/76494-patent-pending-new-polyhedron-discovered-warren-platts-2.html#post1283801). A one percent difference.

trinitree88
2008-Jul-17, 05:15 PM
There are lots of so-called "equal area" truncated polyhedron maps out there, like this one (http://www.csiss.org/map-projections/Polyhedral_Globes/Truncated_Icosahedron.pdf) at csiss.org, but the faces of an ordinary truncated icosahedron (the soccerball shape) aren't of an equal area--the hexagons have about 1.5 times more area than do the pentagons.

So I modified the hexagons so that they are exactly equal to the area of the pentagons. The exact formula is a trade secret for now, but I will say that the if the sides of the pentagons, and hence the pentagon-hexagon boundaries are equal to 1, then the hexagon-hexagon boundaries will be approximately ~0.64 times the pentagon-hexagon boundaries. That should be accurate enough so you can convince yourself that the hexagons and pentagons are of equal area.

Buckminster Fuller took out a patent (#2393676) on a similar polyhedral globe.

A global map that used my polyhedron would be improved over the ordinary truncated icosahedron globe because there is less distortion because the modified hexagons take up less of the surface of the globe than the regular hexagons of the regular truncated icosahedron.

I've scoured the internet for the last week--I can't find a similar construction anywhere. I doubt that even 10010110 will be able to find one! :)

In fact, I'll pay \$20.00 USD to anyone who can find a description of an identical equal area faced truncated polyhedron. :lol:

http://www.bautforum.com/attachment.php?attachmentid=8295&d=1215738755

If you want to try this at home, it's best to use an xacto knife or razor knife rather than scissors. For best folding results, lightly score the outside edges that need folded--it works a lot better.

You saw it here first, folks!

(again)

On bautforum.com

:D

Warren. Nice job. Here's an odd thought...in supersymmetry there are 34 slots for particles and antiparticles. The Z and the photon are their own...so if you put in 32...you fill all the boxes...one each?....and like the six blind Hindu's at the elephant, with the spear (tusk), rope(tail), wall(side),snake(trunk),fan(ear), tree(leg)....your box converts particle to particle depending on orientation in Minkowski spacetime. There's a unification theory for you. :shifty::D pete

edit: It fits an N=8 supersymmetry, with automatic inclusion of a graviton...hmmm?...see:http://en.wikipedia.org/wiki/Supersymmetry

Kaptain K
2008-Jul-17, 09:05 PM
Come on, Pete! Warren's ego is already so big he has to haul it around on a wagon! j/k :)

Warren Platts
2008-Jul-17, 09:40 PM
Ugh!

hhEb09'1, I finally got my spreadsheet to agree with yours. Old Dr. Math neglected to mention that the great circle lengths of the sides of the spherical triangle should be in radians and that the radius of the sphere you're interested in doesn't come in to the picture until very last step.

Warren's formula for spherical triangle areas:

Area = E * R * R

where R is the radius of the sphere

and E = 4 * atan(sqrt(tan(s/2)*tan((s-a)/2)*tan((s-b)/2)*tan((s-c)/2)))

and a, b, c are the great circle side lengths of a
congruent spherical triangle on a sphere with a unit radius,

and s is the semiperimeter (a + b + c) * 0.5

With this formula in hand, I was able to explicitly calculate the areas of hexagons as well. The good news is that when all the hexagons and pentagons are added together, they equal the calculated sphere surface area down to 15 significant digits. The bad news is that hexagons are 0.35% too small, and the pentagons are 0.59% too big compared to their ideal size.

Thus, while the EFATI might have some some small intrinsic interest in itself and might find some use in Dungeons and Dragons and Magic 8-Balls, for geographical applications, it's the spherical equal face area truncated icosahedron (SEFATI ???) that we're interested in. For example, if one wanted to divide the Moon into 32 equal areas, the pentagons would have 11,210 more square kilometers (~2.77 million acres) than the hexagons--wars have been fought over less territory.

So, it's back to the drawing board--yet again!

Which is going to change EVERYTHING!

:boohoo: :cry:

Warren Platts
2008-Jul-17, 09:44 PM
Come on, Pete! Warren's ego is already so big he has to haul it around on a wagon! j/k :)
In a wagon is more like it. . . . :sad:

Besides the supersymmetry application doesn't interest me--it involves too much metaphysics! ;)

Then again, maybe there could be a million dollar prize to be found in there somewhere. . . . :think:

Kaptain K
2008-Jul-17, 10:33 PM
I'm just glad you took it in the manner it was intended! ;)

Warren Platts
2008-Jul-19, 02:05 AM
OK I wrote a program that automatically calculates vertex coordinates as a function of the shorter hex-hex edge, and then set it to figure out the short edge length that will result in equal spherical face areas--I'm through messing with golden ratios! Here are the coordinates:

1. 66.8777394132000 N 36.0000000000000 E
2. 66.8777394132000 N 108.0000000000000 E
3. 66.8777394132000 N 180.0000000000000
4. 66.8777394132000 N 108.0000000000000 W
5. 66.8777394132000 N 36.0000000000000 W
6. 49.6873117638799 N 36.0000000000000 E
7. 49.6873117638799 N 108.0000000000000 E
8. 49.6873117638799 N 180.0000000000000
9. 49.6873117638799 N 108.0000000000000 W
10. 49.6873117638799 N 36.0000000000000 W
11. 31.3206180110803 N 10.0756048262850 E
12. 31.3206180110803 N 61.9243951737150 E
13. 31.3206180110803 N 82.0756048262850 E
14. 31.3206180110803 N 133.9243951737150 E
15. 31.3206180110803 N 154.0756048262850 E
16. 31.3206180110803 N 154.0756048262850 W
17. 31.3206180110803 N 133.9243951737150 W
18. 31.3206180110803 N 82.0756048262850 W
19. 31.3206180110803 N 61.9243951737150 W
20. 31.3206180110803 N 10.0756048262850 W
21. 7.3038986232102 N 22.5434294725859 E
22. 7.3038986232102 N 49.4565705274141 E
23. 7.3038986232102 N 94.5434294725859 E
24. 7.3038986232102 N 121.4565705274140 E
25. 7.3038986232102 N 166.5434294725860 E
26. 7.3038986232102 N 166.5434294725860 W
27. 7.3038986232102 N 121.4565705274140 W
28. 7.3038986232102 N 94.5434294725859 W
29. 7.3038986232102 N 49.4565705274141 W
30. 7.3038986232102 N 22.5434294725859 W
31. 7.3038986232102 S 13.4565705274141 E
32. 7.3038986232102 S 58.5434294725859 E
33. 7.3038986232102 S 85.4565705274141 E
34. 7.3038986232102 S 130.5434294725860 E
35. 7.3038986232102 S 157.4565705274140 E
36. 7.3038986232102 S 157.4565705274140 W
37. 7.3038986232102 S 130.5434294725860 W
38. 7.3038986232102 S 85.4565705274141 W
39. 7.3038986232102 S 58.5434294725859 W
40. 7.3038986232102 S 13.4565705274141 W
41. 31.3206180110803 S 25.9243951737150 E
42. 31.3206180110803 S 46.0756048262850 E
43. 31.3206180110803 S 97.9243951737150 E
44. 31.3206180110803 S 118.0756048262850 E
45. 31.3206180110803 S 169.9243951737150 E
46. 31.3206180110803 S 169.9243951737150 W
47. 31.3206180110803 S 118.0756048262850 W
48. 31.3206180110803 S 97.9243951737150 W
49. 31.3206180110803 S 46.0756048262850 W
50. 31.3206180110803 S 25.9243951737150 W
51. 49.6873117638799 S 0.0000000000000
52. 49.6873117638799 S 72.0000000000000 E
53. 49.6873117638799 S 144.0000000000000 E
54. 49.6873117638799 S 144.0000000000000 W
55. 49.6873117638799 S 72.0000000000000 W
56. 66.8777394132000 S 0.0000000000000
57. 66.8777394132000 S 72.0000000000000 E
58. 66.8777394132000 S 144.0000000000000 E
59. 66.8777394132000 S 144.0000000000000 W
60. 66.8777394132000 S 72.0000000000000 W

For a pentagon edge length of 1, the short edge is 0.647486097519402 (compared to 0.640695431407952 for the flat faced version).

The radius of the sphere would then be 2.1661900091144 (compared to 2.160251992878220
for the flat faced version).

Sphere area = 27.2211464756536
Pentagon area = 1.84269298538123
Hexagon area = 1.84269298538123
Sphere area/32 = 1.84269298538127

:whistle:

hhEb09'1
2008-Jul-19, 09:45 AM
OK I wrote a program that automatically calculates vertex coordinates as a function of the shorter hex-hex edge, and then set it to figure out the short edge length that will result in equal spherical face areas--I'm through messing with golden ratios! Nice. :)

So, that is a configuration where the 32 centers of each face (which are the same as the centers of the face of the usual "soccer ball" truncated icosahedron) are centered in equal areas that equipartition out the surface of the sphere. Somehow, I'd expect that property to translate into some sort of C32 bucky ball, but I don't find such an animal on the list (http://en.wikipedia.org/wiki/Fullerene).

Warren Platts
2008-Jul-19, 12:09 PM
I'm sure there are C32 fullerenes (got a spare \$31.50? (http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TGT-4FR4BV2-2&_user=10&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=266ec41d1129007ffa8a75319ad11c51))--but they can't be centered on the faces of a truncated icosahedron because they would have 5 or 6 neighbors which violates the ordinary rules for carbon bonding, which in graphite clusters is limited to three neighbors. But perhaps there are other elements that would allow such bonding?

hhEb09'1
2008-Jul-20, 04:14 PM
It appears, from the little I gleaned from the discussions, that forms of Cn procede from C60 by twos, maybe collapsing a hexagon by identifying opposite pairs with each other. Starting with 20 hexagons and 12 pentagons, you could procede that way down to the 12 pentagon dodecahedron, but there's a comment there that forms below 32 are unstable. C32 would have 6 hexagons and 12 pentagons, right?

Warren Platts
2008-Jul-21, 03:58 PM
It appears, from the little I gleaned from the discussions, that forms of Cn procede from C60 by twos, maybe collapsing a hexagon by identifying opposite pairs with each other. Starting with 20 hexagons and 12 pentagons, you could procede that way down to the 12 pentagon dodecahedron, but there's a comment there that forms below 32 are unstable. C32 would have 6 hexagons and 12 pentagons, right?You would be correct, sir. But apparently, there's more than one way to arrange the hexagons: in two cluster of three
http://www.bautforum.com/attachment.php?attachmentid=8371&stc=1&d=1216655081

or as 3 pairs:
http://www.bautforum.com/attachment.php?attachmentid=8372&stc=1&d=1216655210

But other forms are also available, like this one that has squares and hexagons, but no pentagons at all that I can see.
http://www.bautforum.com/attachment.php?attachmentid=8373&stc=1&d=1216655469

Source: Index of Fullerenes (http://www.chem.qmul.ac.uk/iupac/fullerene2/list.html)

publiusr
2008-Jul-21, 05:25 PM
I wonder if the facets might hold some types of lenses to focus light toward the center. I wonder what this design might be used for...

Any virus shaped like it?

Warren Platts
2008-Jul-21, 05:46 PM
I wonder if the facets might hold some types of lenses to focus light toward the center. I wonder what this design might be used for...

Any virus shaped like it?The original "Fat Man (http://en.wikipedia.org/wiki/Fat_Man#Physics_package)" atomic bomb used explosive lenses in the truncated icosahedron shape.

Warren Platts
2009-Jul-17, 04:42 PM
I've been encouraging my daugher, mandy101, to do some math over the summer so she said she wanted to construct a big cardboard EFATI. So I sat down with her, but couldn't for the life of me remember how to do a simple pencil and compass construction of a regular pentagon. So I told her to google how to draw a pentagon, and she came up with a particularly elegant construction (http://mathcentral.uregina.ca/QQ/database/QQ.09.02/mary1.html) that apparently wasn't discovered until the 1980's by a George Odum. It was based on triangles, yet had the necessary elements of a pentagon. So I figured the necessary ingredients of the corresponding equal area hexagon might be contained therein as well.

And sure enough there was, :clap:, though I have yet to work out the algebra to prove it. But it works empirically within the accuracy of my compass and ruler. In retrospect, the solution is so obvious and elegant, I can't believe that none of us saw it earlier:

Draw a (big) circle.
Inscribe an equilateral triangle pointed up.
Label the peak A.
Find the midpoints of the three lines.
Lable the bottom midpoint B, and the upper left midpoint C, and the right one D.
Now draw a line from B through C and extend it to the edge of the circle, and label the endpoint E. (At this point, you have the necessary ingredients for a pentagon: if EC = 1, then CB = Φ.)
Draw a line from B through D and extend it to the edge of the circle and label the endpoint F.
Draw a line from E to F.
Label the intersection of EF and AC G; Label the intersection of AD and EF H.
At this point you should wind up with a sort of hybrid figure that looks like a cross between a pagan pentacle and a Jewish star of David. It is my claim that AG and AH have a length of 0.6406975.... If fact, the figure EGA (and AHF) forms one third of an EFATI hexagon. Drawing the hexagon is then a trivial exercise, and much more satisfying than measuring the number of millimeters and multiplying by 0.64 (that's a form of cheating when doing compass and ruler constructions--measuring is not allowed!)

Actually, there is even a simpler way to make the construction. While fooling around, I noticed that the longest of the diagonals of the hexagon has the same length as the diagonal in the corresponding pentagon: i.e. Φ! So here is an alternate construction.

Follow steps 1 thru 6 above to obtain your unit length of1, and the diagnonal of length Φ.
Make an equilateral triangle pointed up with side lengths of 1.
Label the top point A, the right, bottom point B, and the left point C.
Now extend the compass to length Φ, and put your compass at point B and draw an arc on the other side of A.
Now put your compass on point C and repeat the above step.
Draw a line from point B through point A and extend to the arc, labeling the endpoint D.
Draw a line from point C through point A and extend to the arc, labeling the endpoint E.
Connect D and E with a line.
Make your compass length 1, and putting the compass at point A, make an arc 90o to the left, and another one to the right.
Put the compass at point C and make an arc that intersects the leftmost arc, and label the intersection F.
Repeat from point B and intersect the rightmost arc, and label the intersection G.
Connect points F and C; connect points B and G.
Now make sure your compass is length 1, and then putting the compass at point D, draw an arc that intersects FC.
Repeat at point E and intersect line BG.
Label the left intersection H, and the right intersection I.
Connect H and D; connect E and I.
If you did everything correctly, then the figure DEIBCH will be a hexagon with an area equal to a regular pentagon with a length of 1.

Now for the algebra....:think:

Cougar
2009-Jul-17, 07:59 PM
Draw a (big) circle.
Inscribe an equilateral triangle pointed up.
Label the peak A.
Find the midpoints of the three lines.
Lable the bottom midpoint B, and the upper left midpoint E, and the right one F.
Now draw a line from B through C...

Uh, what C?

Warren Platts
2009-Jul-17, 08:38 PM
Uh, what C?

Oops. Thanks for pointing that out Cougar. It should read:

Draw a (big) circle.
Inscribe an equilateral triangle pointed up.
Label the peak A.
Find the midpoints of the three lines.
Lable the bottom midpoint B, and the upper left midpoint C, and the right one D.
But see below:

I got hhEb09'1's formula above in terms of Φ: =SQRT(3+SQRT(5+(20/3)*Φ))-2

But if you look at the figure I described (if properly labeled), the line EB = 1 + Φ, and that has to be the same as the length of EF, and that is 2 + GH. Thus solving for GH, then GH = Φ - 1. Needless to say, 0.618033989... is not the same as 0.640695431... a difference of ~2%.

:o Oh well. Back to the drawing board. Literally . . .

Stroller
2009-Jul-18, 06:01 AM
:o Oh well. Back to the drawing board. Literally . . .

Great thread, well done Warren. Not sure if it's a dumb question, but will all the vertices touch a circumscribing sphere? Or would the ends of the hex short sides 'stick out a bit'?

In the UK, you can only patent a 'secret' idea. Once you put it in the public domain, or even share it with others privately, it's no longer patentable. Don't know whether the states has similar rules.

Love the geometric construction. Many moons ago, my logic professor set us the task of finding a geometric construction for the trisection of an arbitrary angle. I worked out an elegant one which seemed pretty accurate.

Next class I came armed with big compass and straightedge for the blackboard. He told us we were all wrong and he could show logically why a geometric solution is impossible. I did my construction on the blackboard and got into an argument with him about the appropriateness of the mathematician or logician telling the geometer what was right and wrong about the way he does his job.

Finally I rubbed out my construction and drew another arbitrary angle on the board. "If logic trumps geometry, logically trisect that more accurately than I could with compass and edge" I told him.

"Easy" he said, "Each trisected angle is a third of the original."

I nearly whacked him with the ruler.

hhEb09'1
2009-Jul-18, 07:27 AM
Oh well. Back to the drawing board. Literally . . .I think its going to take me a while to figure out what we're talking about, again. :)

Warren Platts
2009-Jul-18, 01:33 PM
Great thread, well done Warren. Not sure if it's a dumb question, but will all the vertices touch a circumscribing sphere? Or would the ends of the hex short sides 'stick out a bit'?
Hi Stroller. THanks for the kind words. They all have the same circumradius: ~2.16 for the flat-faced version, and ~2.17 for the spherical version (cf. posts #33 and # 53 above) where the pentagonal side lengths are equal to 1.

In the UK, you can only patent a 'secret' idea. Once you put it in the public domain, or even share it with others privately, it's no longer patentable. Don't know whether the states has similar rules.I was just kidding about patenting the thing. I wouldn't even know where to begin. Besides, once you get your patent, you need mega-\$€ŁĄ to sue everyone who tries to steal it! But I thought it might have a practical application in cartography, and as a fair way to carve up territory on spherical celestial bodies.

Love the geometric construction. Many moons ago, my logic professor set us the task of finding a geometric construction for the trisection of an arbitrary angle. I worked out an elegant one which seemed pretty accurate.

Next class I came armed with big compass and straightedge for the blackboard. He told us we were all wrong and he could show logically why a geometric solution is impossible. I did my construction on the blackboard and got into an argument with him about the appropriateness of the mathematician or logician telling the geometer what was right and wrong about the way he does his job.

Finally I rubbed out my construction and drew another arbitrary angle on the board. "If logic trumps geometry, logically trisect that more accurately than I could with compass and edge" I told him.

"Easy" he said, "Each trisected angle is a third of the original."

I nearly whacked him with the ruler.:lol: I know the feeling!

I think its going to take me a while to figure out what we're talking about, again.I was thinking the same thing! :( I did manage to get your formula in terms of the golden ratio Φ though. :) Since it's such an odd ball, I think it deserves a Greek letter of its own. The next one, Χ, is already taken by chi-squared tests, so I propose Ψ for 0.64069.... :razz:

Stroller
2009-Jul-18, 06:57 PM
I propose Ψ for 0.64069.... :razz:

Seems an appropriate symbol for something you can construct with arcs and straight lines. Almost has a 'square and compass' feel about it. ;)

Warren Platts
2009-Jul-18, 07:19 PM
Seems an appropriate symbol for something you can construct with arcs and straight lines. Almost has a 'square and compass' feel about it. ;)

Unfortunately, that's the problem: there is no 'square and compass' solution for deriving Ψ! :(

hhEb09'1
2009-Jul-19, 03:49 AM
so I propose Ψ for 0.64069.... :razz:I knew you were going to say that!

Stroller
2009-Jul-20, 11:21 AM
Unfortunately, that's the problem: there is no 'square and compass' solution for deriving Ψ! :(

Doh!.

But wait, don't despair. After my dust up with the logic prof, I wrote a little program on my psion pocket computer (long lost) which calculated the error of my geometrical method of trisecting an arbitrary angle for all angles 0-90 degrees and plotted it.

It was a sine wave.

I toyed with the idea of working out a projection to add a correction from the sine wave to the trisection, but never got round to it. Because geometers and engineers are just lazy that way. When asked what two times two is, the engineer tells you it's "about 4", and the geometer passes you a deck of floor tiles to play with.

Yachtsman
2011-Sep-17, 07:43 AM
If your main goal is to provide an effective (accurate and efficient) way to store 3D spherical data for 2D map rendering the best way to do this is to pre-calculate all XY coordinates and store them as XY; nothing would be faster because now it's straight display of the data--it would NOT require additional math; whereas your system would require further math to convert from your proprietary coordinate system into XY; I susect the math needed for your system will actually be slower than the simple math used for radian calculations used in spherical mathematics.

I just don't seen an advantage of your system, I tried to, but I can't see it; If speed is of primary importance then store them as XY already pre-converted; otherwise store them in their original radian coordinates for the fastest mathmatical rendering into 2D.

Nowhere Man
2011-Sep-17, 12:06 PM