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Dividing a sphere into twelve "identical" shapes.


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Janus,

 

That is really interesting. Perhaps the four hexagonal planes (and the three square planes) at just those right angles to each other, conspire to equalize the vectors as Mr. Fuller noticed. Perhaps it really is as "magical" (meaning really really neat) a figure as I have sensed.

 

Imagine what you did there. Cut twelve in 1//4s to get 24. Interesting indeed.

 

On a parallel note, I was rethinking the cutting of the diamond into smaller diamonds. Granted it does not work so neatly if you are trying to make faces, but what if you stay on the surface? I remembered on the way home today, that when I drew my 48 and then 192 diamonds on my clay ball, I never broke the surface. The lines I was drawing where always exactly one radius away from the center of the sphere (cause they where on the surface). It is not a matter of figuring was surface area the shape is subtending, because its ON the surface and the area is exactly what the area is. In the case of the 192 diamonds, that would be 1/192nd the surface area of a sphere.

 

Regards, TAR

Cutting the segments up while remaining on the sphere is no problem, you get this:

 

48spheresec.jpg

 

However, going past this step becomes problematic. Each division is now an irregular triangle and won't divide into three similar triangles.,

 

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Vampares,

 

Ah but the strategic pen hole is mightier than the slice of the steak knife (sword).

 

And its the rendering program that is not worth a wit, without a wit to wield it.

 

Regards, TAR

 

Mike,

 

Maybe you could use that white paint to paint some center lines on the Grand Canary mountain roads. From space (Google Earth) it doesn't look like there are any mid lines. Anyway I am considering reopening the case, so that 3meter diamond might still work. Or perhaps we can do something with those odd shapes in Janus's last rendering.

 

In any case, I am rather sure that my original "Janus" sections of sphere are exactly 12th of a sphere, each. Having both 1/2th the surface area of a sphere, and being exactly 12th the solid angle of a sphere, each. If you divide one of these, exactly down the middle, with laser rays, eminating from the exact center of the sphere, the remaining two peices will each be exactly 1/24th the solid angle of a sphere, and , the curved surface area of each peice will be exactly 1/24th the surface area of a unit sphere.

It does not mater, whether draw the lines, exactly in half this way or that, or along this diagonal or that, as long as you are making the cut with a laser from the center and keeping the laser on the same plane, as you make the cut (a straight cut), there is no choice but to split the solid angle, and the surface area in half, with the same cut.

 

Won't result in identical peices, but the peices will each contain the same elements of solid angle, volume and surface area. I would think. How could it be otherwise?

 

Regards, TAR


Janus,

Problematic, indeed. The diamonds on the other hand, staying on surface, stay diamonds when you half them from the sides and not from the point. You always have a diamond to work with exactly 1/4 the size of the parent. And the halfing lines do some nice things, as I mentioned. On the first iteration, the new lines describe six big squares and eight big triangles, exactly in the pattern of cutting the corners off a cube, like we started the whole process with.

Regards, TAR Edited by tar
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Janus,

Wow, I was looking at your triangle ball with the nice complementary colors. I was noticing the eight (twelve rather) circles cutting through...and thought you had just sliced the aphere in half and then half again, and so on, from three different angles...then I saw you had taken the 12 diamonds and split them on the diagnonals.

Interesting figure we are working with here. Certainly dual, if not quad or something.

Interesting indeed.

Regards, TAR

Mike,

 

Once they do another round of satellite photos, I will look for the great white mike smudge on Grand Canary.

 

Regards, TAR

Edited by tar
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  • 2 weeks later...

 

Ratio of long diagonal to short is sqrt(2) (this is for the flat edged polyhedron not for the spheroid). If the short diagonal is 2 units long, the long diagonal is 2.sqrt(2), and very happily the length of the side is sqrt(3). The small angle is 70.53 degrees.

 

The rhombic dodecahedron is the dual of the cubic octahedron (a nice mixture of equitriangles and squares). The rhombic dodecah can be formed in a really neat way; if you arrange 7 cubes with one in the centre and one each of each of its six faces you get a sort of cross in 3d. Draw a set of lines linking the centre of each of the 6 outer cubes and you have the edges of a rhombic dodecah.

 

If you draw lines along all the short diagonals of a rhombic dodecah you get a cube and along the long diagonals you get an octahedron (the cube and the octahedron being duals of each other)

 

Great shape!

Imatfaal,

 

That 70.53 I am thinking is important, the obtuse angle of course is then 109.47, because the four angles of the flat diamond would have to add to 360. But we have already decided that the diamond is not flat, or not flat yet. Those angles are taken on the projection of the rays eminating from the center of the sphere, as to where they cut the plane that is normal to the sphere and touching the center of the diamond. What is interesting to note, is that the obtuse angle, when measured on a plane, normal to the sphere, at the vertex or junction is actually 120 and the 70.53 is actually a 90 degree angle at the vertex. So the function, I was talking about earlier might be the one that takes 120 to 109.47 in the limit, and takes 90 to 70.53.

 

I am thinking this will work, because when you divide the diamond in half side to side and not vertex

to vertex you keep the same diamond proportions on all 4 diamonds you thusly create from the one. And each of these diamonds can similarly be divided in perfect four, retaining the diamond proportion.

 

As the diamond shape gets smaller, it also gets flatter, never getting completely flat, but necessarily approaching the 109.47 to 70.53 angle.

 

Since the curved sphere sees the angles as 120 and 90, and the "flat" side sees the same angles as 109.47 and 70.53, pi should be in there somewhere.

 

Regards, TAR

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............Since the curved sphere sees the angles as 120 and 90, and the "flat" side sees the same angles as 109.47 and 70.53, pi should be in there somewhere.

 

Regards, TAR

I have not looked back through the tread , so you may well have covered this.

 

I designed a fall out shelter for indoors and a Lambing shelter for farmers on fell tops. both were based on " Buckminster Fullers Geodesic Dome" where he had developed Three ratio's to the sides of a triangle 'A' ',B' 'C' if you used his ratio the base of the half globe, or one third of a globe, came out level at the base, , so you could build it as a building. His designs have been used throughout the world to build vast stadiums and Exhibitions.

 

But you may know all this. just a contribution..

 

The farmers thought i was mad,, although one chap wanted to rear Eagles in them , and people never bothered to go much for fall out shelters. however I did make the 'Press' at the time., . Not sure if it was good press or bad press. Oh and pedal chargers for Fall Out Shelters. They did get sold around the world for a little while !

 

post-33514-0-05802300-1398164713_thumb.jpgpost-33514-0-75316800-1398165701_thumb.jpg

Then moved into computer cables.post-33514-0-15703500-1398165781_thumb.jpg

Then teaching physics , then retired. Short history from Geodesic domes for sheep to retirement.

Mike

Edited by Mike Smith Cosmos
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Mike,

 

Thanks for the quick bio. Important for me to remember, when I notice stuff about the world, that its been looked at before, and thought about before.

 

Also important, I think, for the young folk coming up to remember that some of the things in the world that they take for granted came from somebody's ideas and work bringing the ideas into reality.

 

Sure we have fantastic systems and tremendous knowledge at our finger tips, but somebody was programming Commodore 64's and bringing digital copiers into the business world, and designing pedal powered electrical storage and such, that laid some of the groundwork, for what we have now.

 

Regards, TAR

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  • 2 weeks later...

Here is a crudely done rendition of the globe divided into the same twelve "identical" shapes.

 

The globe photographed from twelve vantage points is a copyrighted Replogle Stereo Relief Globe, by RePlogle Globes Inc., Gustav Brueckmann, Cartographer, with polical boundries and names as they existed 50 or 60 years ago. Used without permission.

 

Rendition executed to see what the world would look like, from 12 directions at once, on a two dimensional space, using the segments we have been talking about. There is no projection involved, all continents are "actual" size, to the same scale, as each of the photos is taken of the same 12" globe, from approximately the same distance.

 

Regards, TARpost-15509-0-78317600-1399213642.jpg

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Here is a crudely done rendition of the globe divided into the same twelve "identical" shapes.

 

The globe photographed from twelve vantage points is a copyrighted Replogle Stereo Relief Globe, by RePlogle Globes Inc., Gustav Brueckmann, Cartographer, with polical boundries and names as they existed 50 or 60 years ago. Used without permission.

 

Rendition executed to see what the world would look like, from 12 directions at once, on a two dimensional space, using the segments we have been talking about. There is no projection involved, all continents are "actual" size, to the same scale, as each of the photos is taken of the same 12" globe, from approximately the same distance.

 

Regards, TAR

I saw Fuller mentioned in this thread I seem to recall, but not his Dymaxion map. Forgive me if I simply missed it.

 

source: >> http://en.wikipedia.org/wiki/Dymaxion_map

The Dymaxion map or Fuller map is a projection of a world map onto the surface of an icosahedron, which can be unfolded and flattened to two dimensions. The flat map is heavily interrupted in order to preserve shapes and sizes.

 

The map was created by Buckminster Fuller. The March 1, 1943, edition of Life magazine included a photographic essay titled "Life Presents R. Buckminster Fuller's Dymaxion World". The article included several examples of its use together with a pull-out section that could be assembled as a "three-dimensional approximation of a globe or laid out as a flat map, with which the world may be fitted together and rearranged to illuminate special aspects of its geography."[1] Fuller applied for a patent in the United States in February 1944, the patent application showing a projection onto a cuboctahedron. The patent was issued in January 1946.

The 1954 version published by Fuller, the Airocean World Map, used a modified but mostly regular icosahedron as the base for the projection, which is the version most commonly referred to today. This version depicts the Earth's continents as "one island," or nearly contiguous land masses. more ...

300px-Dymaxion_2003_animation_small1.gif

Edited by Acme
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Acme,

 

No doubt I could have gotten the idea from Mr. Fuller, but I also got it from suggestions on this thread, and I have not copied Mr. Fuller's projection onto the cuboctohedron. I do not even recall ever seeing his projection onto a cuboctohedron. The icosahedron pictured in the link, is not the division into twelve equal segments we are talking about here. We are talking about the one Janus modeled. The cuboctohedron is obviously the figure upon which the diamond segments are built, as that each of the diamond shape's center's are located exactly on a vertex of the cubooctohedron, but the diamond shape, we have already determined is not condusive to making a hedron out of, as the four points of the diamond are not on the same plane.

 

However, retaining the curved surface of the segment, results in no projection onto a flat surface required.

 

The idea is therefore similar to, but not exactly the projection of Fuller. And although it is built on the form of a cubooctohedron, it is NOT a projection of the features of the Earth onto the flat surfaces of a cuboctohedron, and does not use at all the icosahedron, nor does it unfold well, nor is it "flat", except it is a photo on a two D surface.

 

There remains some interesting attributes of the figure that can be explored.

 

Perhaps you can help, or someone can help with establishing the truth or falseness of a conjecture I have related to the figure Janus modeled, and the division of the globe, I have presented here. The globe I used, was a 12" globe. (12" in diameter). I built the sloppy divisions, using peices of tape, 6 inches long. This appeared to me, to work out exactly. Leading me to make the conjecture that the distance between the corners of the diamonds, when measured along the surface of the sphere (as in a peice of tape or a string that could lay flat on the sphere) was "exactly" r. Does anyone know spherical geometry well enough to calculate the length of string or tape it would take to make the edges of the diamonds on the surface of the sphere? It already seems to me that the lengths are identical, but are the lengths equal to r?

 

Regards, TAR

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There is no projection involved, all continents are "actual" size, to the same scale, as each of the photos is taken of the same 12" globe, from approximately the same distance.

It's still curved pieces projected onto a flat image. It is a projection. You can see distortion closer to the edges of the pieces. There is less distortion of size compared to the usual Mercator maps.

 

It can be improved. If you look at the Dymaxion map that Acme posted, you'll see that the surface gets "split" along water. Yours handles Australia really well but cuts up many land masses, making it a poor map along those cuts. You could move where the cuts are, and you can also join the cut pieces differently. For example, instead of joining the main South America piece on a corner over water, if you join it to the rest of the continent then it reduces the extreme distortion of distances between the two pieces. That's an easy fix. Perhaps harder is something that Fuller solved, is getting optimal cuts or whatever. A lot of your pieces are part land, part water. Is it possible to get more pieces that are all water, and others that more closely fit the land, as Fuller's does? This might require more effort than you're willing to spend though. Also it assumes that a land map is what's important, and that distortion and breaks over the ocean are acceptable.

 

 

A tool would make this easier. Perhaps a wire polyhedron that could be rotated over a physical globe. There are also downloadable polyhedrons that can be added to Google Earth. I'm sure it would be possible to create one for a rhombic dodecahedron, and to have it fixed in space so that you could rotate the globe underneath and try out different mappings. While trying to find something like that, I came across this related info:

http://www.progonos.com/furuti/MapProj/Normal/ProjPoly/projPoly.html

Edited by md65536
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Acme,

 

No doubt I could have gotten the idea from Mr. Fuller, but I also got it from suggestions on this thread, and I have not copied Mr. Fuller's projection onto the cuboctohedron. I do not even recall ever seeing his projection onto a cuboctohedron. The icosahedron pictured in the link, is not the division into twelve equal segments we are talking about here. snip...

I was not suggesting you got it from Fuller & I know it's not the same division as you have done. I was merely giving another perspective and reference on dividing a sphere that -dare I say it- you might read up on to broaden your knowledge. My bad for presuming anyone could even remotely be interested. :doh:

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Acme,

 

I am sixty. I probably saw Fuller's projection in the 60s. It most likely was one of the reasons why I was investigating the divisions of a sphere, with my clay balls and such. Because I wanted to better visualize a sphere, and understand solid angles and such.

 

The division into 12 segments I presented at the beginning of the thread is not new to the world, but it was new to me. I learned the names and such of these figures in this thread. But the concepts and the "neatness" of the arrangement, were things I noticed myself. Obviously the characteristics of a sphere, are not my invention. I am just noticing stuff about it, and talking about it on a mathematics board. Perhaps it will interest someone, perhaps it won't.

 

I do not have the understanding of solid geometry and calculus it would take to figure if the length of the diamond's side is exactly r, when measured on the surface of the sphere, but it would be "neat" if it was, and would have implications as to what Pi is. That is, as to "why" Pi is. Just one of the things I care about figuring out. If it has already been figured out fine, I am just looking for the answer, one way or the other.

 

Can anyone help me, in this regard? The figure Janus modeled is already known. Is it known what the length of the arc, describing one side of the "diamond", as measured on the surface of the sphere, is?

 

Regards, TAR

MD65536,

 

Thanks for the link, and the suggestions. You are right, I could pick better continent preserving starting points, but I liked the idea of using the North and South Poles as vertices. And the four lines coming from the poles seemed rather regular and understandable, so I put them on the prime meridian, and the 180 "date line" area, and let everything else fall where ever it fell.

 

My thinking on this is that the Rhombic dodecahedron projected onto the sphere, is more important than the sphere projected onto the Rhombic dodecahedron. This is why I am considering that my rendering is not a projection, but twelve pictures taken of the actual curved surfaces, and laid out in a way that you can see the whole sphere at once.

 

Perhaps it is not very useful as a map, but more as a "visualization", being able to "see" all sides of the globe at once. I always had trouble with the projections. Not that they were not mathematically correct, but that they distorted the place, and I didn't know how to make the corrections. We are used to making the corrections required when we look at a ball, so why not keep it a ball. The twelve divisions I think is a nice balance, sort of vector equalibrium, of the place, that gives one the "idea" of what curvature is about.

 

Was standing out on my deck today, probably looked like a crazy man, drawing lines in the air, dividing the world into these diamonds, with the four "equatorial" diamonds, and the four polar diamonds and the four diamonds below, using my neck/head/eyes as the center point.

 

Seems really neat and clean, how the "90 degree" juntures and the "120 degree" junctures work out, to "form" the sphere. And one can divide the 3D world using these 12 identical diamonds. I think it has possibilities.

 

We could call the center of the Milkyway the "below" 4-point, and the opposing point in the sky the "above" 4-point, choose something that was on the resulting "equator" as a 4-point, and that would determine the other three 4-points. The 3-points would also all be thusly defined. Then the Celestial Sphere would be divided in 12 equal areas and each of these areas could be divided in quarters (from mid edge to mid edge, not point to point), and each of those quarters into quarters, as many times as would be useful. And each subdivision would be exactly defined in both solid angle, and direction. Most celestial objects would keep their designated position for thousands of years, and if we picked a very far away galaxy as our equatorial starting 4-point, we would have a "wireframe" with-in which to measure and chart every object and motion within it, against.

 

Regards, TAR

Edited by tar
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post-15509-0-36174500-1399505398.jpg

 

Here is another arrangement, of the same 12 sections, showing the four Northern "polar" divisions, the four equatorial divisions, and the four Southern "polar" divisions.

 

Still would like to know, if anyone can do the math, if the sides of the diamonds are exactly r or not.

(When measured on the surface of the sphere.)

 

Regards, TAR

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attachicon.gif2world12low.jpg

 

Here is another arrangement, of the same 12 sections, showing the four Northern "polar" divisions, the four equatorial divisions, and the four Southern "polar" divisions.

 

Still would like to know, if anyone can do the math, if the sides of the diamonds are exactly r or not.

(When measured on the surface of the sphere.)

 

Regards, TAR

What math are you looking for?

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Unity+,

 

The way to find the length of the circle segment that corresponds with the sides of the diamonds.

 

Earlier I think it was MD65536 that let me know the angles of the diamonds, but that was the polyhedron, "flat" diamond shape. I am more, at this point interested in the length of the arc from corner to corner, measured along the surface of the sphere, not the straight line that would go "subsurface" between the points.

 

Regards, TAR

Imatfaal,

 

Sorry, I misidentified the provider of the angles. It was you.

"Ratio of long diagonal to short is sqrt(2) (this is for the flat edged polyhedron not for the spheroid). If the short diagonal is 2 units long, the long diagonal is 2.sqrt(2), and very happily the length of the side is sqrt(3). The small angle is 70.53 degrees."

 

Unity+,

 

So anyway, knowing that as a plane the diamond shape has the above angles, and as a sphere section the obtuse angles seem to join at 120 degrees if you think of a plane tangent to the sphere at every 3-point in question, and the acute angles seem to be 90 degrees if you look at the 4-point junctions, from the perspective of a plane tangent to the sphere at a 4-point.

 

I am interested in what makes a 109.47 degree angle a 120 angle and a 70.53 degree angle a 90 degree one.

 

And I am interested in learning how one deals with angles when they are on the surface of a sphere, and not on a plane. Primarily to figure out if the diamonds that you get when you cut a big 1/12th of a sphere diamond into 4 similar diamonds, are the same 70.53 angle when looked at, as if they are transribed onto a plane inside the sphere, or if they retain any of the 90 degree, 4-point type characteristics when being on the surface of the sphere.

 

Regards, TAR

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...And I am interested in learning how one deals with angles when they are on the surface of a sphere, and not on a plane. Primarily to figure out if the diamonds that you get when you cut a big 1/12th of a sphere diamond into 4 similar diamonds, are the same 70.53 angle when looked at, as if they are transribed onto a plane inside the sphere, or if they retain any of the 90 degree, 4-point type characteristics when being on the surface of the sphere.

 

Regards, TAR

Spherical trigonometry @Wiki

 

 

Spherical trigonometry is that branch of spherical geometry which deals with the relationships between trigonometric functions of the sides and angles of the spherical polygons (especially spherical triangles) defined by a number of intersecting great circles on the sphere. Spherical trigonometry is of great importance for calculations in astronomy, geodesy and navigation.

...

Spherical polygons

 

A spherical polygon on the surface of the sphere is defined by a number of great circle arcs which are the intersection of the surface with planes through the centre of the sphere. Such polygons may have any number of sides. Two planes define a lune, also called a "digon" or bi-angle, the two-sided analogue of the triangle: a familiar example is the curved surface of a segment of an orange. Three planes define a spherical triangle, the principal subject of this article. Four planes define a spherical quadrilateral: such a figure, and higher sided polygons, can always be treated as a number of spherical triangles.

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Thank you Acme

 

I read the wiki article, still not sure how to work my problem, when to be in degrees or radians or rads, and I am not sure what angles and distances I "have" already, when I can use an identity, and when I can "figure".

 

But I suppose if trig is what I need, trig is what I need to remember and work with.

 

Got my sister's advanced high school math book out, and will see if I can figure it out. Thank you for the direction though. It is obvious that trig is what I need to be understanding, to do the problem.

 

Have run into sine and cosine and tangent and cotangent before, as I actually did take trig in high school, but I think I was more interested in girls and drinking and smoking, and having fun, than in understanding the implications and uses of the log tables and tables of the trigonimetric functions and the like. I liked them then, and I actually had fun looking at the tables in the back of my sister's book, looking for the patterns.

 

So, as no one else yet has given me a yes or a no as to the exact length of the side of the diamond on the spheroid, I suppose I will have to figure it out myself...using the trig you have pointed me to.

 

Might take a while...but if no one else gets back with the answer, I will eventually.

 

Regards, TAR


But, this being a math forum, I would not mind any help anyone would like to give, in setting up the problem and properly executing the math. My thinking is that I have three angles of the triangle made when you run a great circle through though the center of the diamond from obtuse point to obtuse point. The acute point is already a known 90 degree angle, the obtuse point is already a known 120 degree angle and half that would be 60, so the angles are 60, 60 and 90. From this, according to the wiki article, I think I have 3 out of 6 unknowns know, and think I might be able to get the other three from it. If not, if I need an arc length, I could, I suppose, assume that my two arcs of the triangle, that are also arcs of the diamond, are exactly 1, and thusly "find" the length of arc that goes from obtuse point to obtuse point, as that I would have 5 out of 6 to plug into the identities. Then whatever answer I get for the length of arc from obtuse point to obtuse point, I can plug back into the identities, knowing the three angles, treat the other two identical arcs as unknowns, and see if they turn out to be 1. If they do, then they are 1. If some other number comes up, then they are not 1. That is the plan, I am not confident though, in my ability to execute it, and will gladly accept any help, suggestions, directions, encouragement, or correction.


According to my sister's math book, the sine of 90 derees is .0000 the tangent is .0000 and the cosine is 1.000. The Cotangent seems to be ------, which I am not sure means either infinity or undetermined.

 

60 degrees in the chart shows sine=.5000, Tangent-.5774, Cotangent=1.7321 and Cosine=.8660.

 

Addition information, given by the chart is that a 60 degree angle is equivalent to 1.0472 radians and a 90 degree angle is equivalent to 1.5708 radians.

 

In support of my thought, that Pi must reside in these angles, I just this minute made the interesting observation/calculation that 1.0472 is 1/3 of Pi. (not surprising as that 60 is one sixth of 360 and a radius is 1/2 a diameter,) but interesting, none the less.

 

But still to figure or to understand, for me, is when you can use a radian as a length, and when you can use it as an angle, whether the length of the arc is exactly r, or 1.0472 r, or .9549r...or indeed some other length. It "looks" to be around r somewhere, as that one can tape up a 12" globe with approximately 6 inch pieces of tape, into the diamond arrangement, using the poles and the equator as guides.

 

Which is more likely to be correct, that the arc length is exactly r, the arc length is exactly 1/3 pi, or that the length is some other number?


No doubt I can assume that 1.5708 is likewise 1/4 Pi without doing the math (using the calculator.)

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Thank you Acme

 

I read the wiki article, still not sure how to work my problem, when to be in degrees or radians or rads, and I am not sure what angles and distances I "have" already, when I can use an identity, and when I can "figure".

 

But I suppose if trig is what I need, trig is what I need to remember and work with.

...

You're welcome TAR.

 

My specialty is generalities. While I knew what was called for, I have not studied and have no skill in spherical trig. No shortage here of folks well suited to the task though and with a little luck you may coax them out. :)

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  • 2 weeks later...

post-15509-0-42477900-1400983216_thumb.jpgpost-15509-0-18803200-1400983218_thumb.jpgpost-15509-0-36181100-1400983219_thumb.jpgAcme,

 

Have not gotten the trig down yet. But am still working the problem. Figured out on Thursday how the diamonds relate to the cube.

 

Drew the diamonds onto the cube. The 8 three points are the corners of the cube, and the 6 four points are located right in the center of the faces of the cube. The center of each edge is the location of one of the 12 balls that fit exactly around a center ball of the same diameter. Each midpoint of an edge is the center of a diamond.

 

Relating a cube to the 12 sections of the globe, the top of the cube would have the top half of 4 diamonds, and each of the bottom halfs of those four diamonds would be on one of the sides of the cube, being the triangle you get when you draw an x corner to corner on each side. The diamonds are thusly "folded" in half by the edge. The long axis of the diamond goes from center of top to center of side, and the short axis of the diamond goes from corner to corner.

 

Works out exactly, internal anglewise. I made two clay figures, a sphere and a cube. Sized them, so the distance from center of cube to middle of edge was approximately the same distance as a radius of the sphere. Then I cut the sphere into the 12 sections, and cut the cube into the same 12 sections, with the same internal angles, so that you could replace a peice of the cube with a peice of the sphere and vice a versa. Pretty neat.

 

Looks like this.

 

Regards, TAR

 

 

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  • 1 month later...

post-15509-0-06222100-1404265243_thumb.jpgpost-15509-0-69366200-1404265244_thumb.jpgpost-15509-0-40045400-1404265240_thumb.jpgJanus,

 

Made a string version of shape with the 12 diamonds we are talking about here.

 

Then blew a balloon up, inside it and drew with marker on the balloon, dividing each diamond into 4 to get 48 sections, then divided each of those diamonds in quarters to make the 192 sections. (and then once more to make the 768

 

What was interesting was that the shield shape we got dividing each diamond into four, "stayed in the 120 degree corner. That is a diamond divided into 4 had two shield shapes a diamond divided into 16 had 2 shield shapes and the diamond with 64 sections had only two as well in the 120 degree corner.

 

Led me to think that any continuing division would always have 24 of the sections shield shape, and the rest, exactly the same symetrical diamond shape. With enough divisions the 24 would become tiny areas, eight of them, analogous to the corners of a cube, and with enough divisions, just the imaginary "points" of the corners of cubes.

 

Still seems to me we can make something of this way of dividing the cube, or the sphere. Any amount of pixels in that 12, 48, 192, 768 ... sequence could be used to define solid space around a point where each of the pixels minus the 24 shield shaped ones, were exactly the same size and shape, and in a definite, known position.

 

Regards, TAR

post-15509-0-53011400-1404271056_thumb.jpg

 

Thread recap.

 

Take a cube. Cut off the corners to the midpoints of the edges and you have a cuboctahedron.

 

Put the center point of twelve balls on the 12 vertices of a cuboctahedron and you have a close pack situation with 4 intersecting hexogonal plane orientations and a rather neat situation.

 

Put a dot on a sphere in the exact location of each of the twelve balls an draw lines halfway between the points and you wind up with the twelve sections of the sphere or the dual of the cuboctahedron, the spherical rhombic dodecahedron, that Janus rendered so nicely.

 

Draw the cube on the sphere and the sphere on the cube, using this scheme and you have the twelve sections of the sphere, all identical shapes with internal angles of 90 and 120 degrees.

 

Regards, TAR

Edited by tar
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  • 2 months later...

post-15509-0-40611600-1411310515_thumb.jpgpost-15509-0-52353700-1411310517_thumb.jpgAddendum,

 

Built a Spherical Rhombic Dodecahedron out of 24 pieces of wire of equal length, 6 orange wire nuts and 8 blue wire nuts.

 

The orange wire nuts are on the "four points" analogous to the center of each of the six sides of a cube.

The blue wire nuts are on the "three points" analogous to the 8 corners of a cube.

 

Another image to take is that of a globe with an orange nut at the North Pole, one at the South and the other four equally spaced around the equator. Each of the blue then, four in the northern hemisphere and four in the southern are placed exactly in the middle of the polar orange and two of the equatorial ones.

 

Attached are a picture of the SRD with the 12 sectioned cube pictured earlier in the center, oriented with the center of each face aligned with an orange nut, and each corner aligned with a blue, and a picture of the SRD with the close packed arragement of twelve balls around a center ball (ignore the extras), with each of the 12 balls, aligned with the center of one of the 12 diamond shapes of the Spherical Rhombic Dodecahedron.

 

Regards, TAR

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  • 1 month later...

Another neat observation about the spherical rhombic dodecahedron made with wire and wire nuts.

 

If you join the blue wire nuts each to the three closest other blues with string, you get a perfect cube.

If you join the orange wire nuts each to the four closest other orange with string, you get a perfect octagon.

 

Regards, TAR

 

I have only done it the one way, then the other. I will get some colored yarn and do both and take a picture, later today, after work.

Its pretty neat.


My edit is not working, I meant octahedron, not octagon.

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Another neat observation about the spherical rhombic dodecahedron made with wire and wire nuts.

 

If you join the blue wire nuts each to the three closest other blues with string, you get a perfect cube.

If you join the orange wire nuts each to the four closest other orange with string, you get a perfect octagon.

 

Regards, TAR

 

I have only done it the one way, then the other. I will get some colored yarn and do both and take a picture, later today, after work.

Its pretty neat.

My edit is not working, I meant octahedron, not octagon.

 

Tar - this thread is so long you are forgetting what was discussed six months ago regarding joining diagonals to make cube and octahedron

 

http://www.scienceforums.net/topic/82363-dividing-a-sphere-into-twelve-identical-shapes/?view=findpost&p=799542

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