PATENT PENDING! New Polyhedron discovered by Warren Platts!

[Warren Platts]

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 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.



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

[Occam]

Well, for what it's worth, I think that's pretty cool!

[Warren Platts]

Well, for what it's worth, I think that's pretty cool!

Why thank you sir!

[BigDon]

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]

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]

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]

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]

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.

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]

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. 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! ); 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. (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

Very elegant hhEb09'1!

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]

Re the OP, you forgot to say you deserve the Nobel Prize in maths...

[Warren Platts]

Re the OP, you forgot to say you deserve the Nobel Prize in maths...

I figured grant hutchison is jealous of me enough already.

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

[Warren Platts]

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]

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]

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]

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.

Very elegant hhEb09'1!

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, 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]

Warren,

I didn't see that you have a 2000 x 1250 pixel version until after I finished
modifying your 800 x 500 version. I added tabs for gluing. I also made all
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]

Warren,

I didn't see that you have a 2000 x 1250 pixel version until after I finished
modifying your 800 x 500 version. I added tabs for gluing. I also made all
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!

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]

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]

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 Minneapolis

Dude, relax! It's Miller Time! You've earned yourself a beer--take a break!

[Warren Platts]

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*thexht+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

[novaderrik]

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]

You'd probably have to cut it into pieces that fit in the brake, and then weld the pieces together.

Fred

[Jeff Root]

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]

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]

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]

.
.
.
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]

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]

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. Wow, we're getting a nice, little corporation going here. Might as well make some money off this site!

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! 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. 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]

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]

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.