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John Harby
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Flexable Software, LLC
Flexable Software, LLC

39 followers
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WORKING ON DIAGRAMS

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SPACECRAFT DESIGN 1

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Farsha Beach Sharm El Sheikh Egypt
A stunning beach with some of the best diving available with easy access. A few bars and cafes, and only a couple of quid to hire a lounger. The reef is teaming with fish and other sea critters and drops off to huge depths about 100m or so from the beach. Its a bit of a climb down, and up again, so not for those with bad walking difficulties. There is a sloping road down, less steep than the wooden steps in the video, but even that is very steep. A fantastic beach and just a couple of quid from Sharm town center.

#sharm #sharmelsheikh #egypt #farsha #farshabeach

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The Norwegian Refugee Council has said a lack of food, water, and medicine inside the city are pushing people 'to the bring of desperation.'

#Fallujah

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#Iran  .

Iran's Supreme Leader Ayatollah Ali Khamenei rules out co-operating with the US or the UK to resolve regional issues.

Are we in the US of China now? My rep with Hu Jintao is intact. I didn't lose face.

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Foxiest chick in the world

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Are they going to keep letting the Mexicans gang up in USA and rip off the Americans ideas with their NEVER ending Spanish Interrogations 
4 cubes can be glued trivially face-to-face in R^3 forming a ring, their centers being the vertices of a square. We have seen, however, that only in R^4 can we have a ring of 3 cubes (glued face-to-face), their centers being the vertices of an equil. trig; what was more, we saw that it was possible (in R^4) to have 4 cubes (always glued f.-to-f. through squares) whose centers are the vertices of a reg. tetrahedron.
What happens with reg. tetrahedra? As a matter of fact it is impossible to glue in R^3 reg. tetrahedra face-to-face (through equil. trigs) forming a closed ring; the best we can do is gluing 5 tetrah. face-to-face in R^3 in the said manner without forming a ring [in fact the dihedral angle between the faces of a tetrah. being 70.5 deg, we can never complete 360 deg around an edge of a reg. tetrah.
What happens in R^4? Is it possible to form a ring of tetrahedra that closes in R^4? and how many tetrah. (always glued in the said manner) are needed to form such a closed ring?
The animation herewith shows that such a ring is easy to obtain in R^4 & that it is formed of 4 reg. tetrah. glued face-to-face through equil. trigs.
the hypersolid I present is a convex one in R^4 & is formed by a 4-ring of reg. tetrahedra (in light blue); in the development (or net) I use, there appear to be a gap in the ring, but that is due to the manner the reg tetrah. were unwrapped from R^4 to R^3( such that they all fit in R^3, which is impossible without showing a gap) [the 2 equil. trigs are labled both 1-2-4 & are actually glued in R^4 between themselves (in a ring every tetrah. must be glued simultaneously to 2 other ones (the one which preceeds & the one which follows): this gluing uses 2 faces of each tetrahedra, so that leaves 2 faces free in each tetrah. Those 2 free faces are glued to 2 trirectangular tetrahedra so that the hypersolid is composed of 4 reg. tetrahedra (the 4-ring) + 8 trt's all with a common vertex labeled 7. To put things in another way: the hypersolid of which you see the development in R^3 is the cumulation of 4 tetrahedric 4-pyramids whose bases are reg. tetrahedra & lateral faces are trt tetrah.
The reg tetrah. that form the ring are: 1-2-3-4, 1-2-3-5, 1-2-4-6 & 1-2-5-6 ; the 1st is covered in red & maroon, the 2nd in pink & purple, the 3rd in blue & light green & the 4th in yellow & dark green.
For those who might enjoy having the coordinates of the 7 vertices in the fig., they are: (2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2),(0,0,0,-2),(0,0,-2,0) & (0,0,0,0) the last the origin of coordinates and the common vertex to the 8 trt's.
Please be well, the whole pack!... (lol)
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