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Topic: Visualiing Derivatives with Cubes
Replies: 9   Last Post: Sep 8, 2013 1:53 AM

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frank zubek

Posts: 222
Registered: 5/12/09
Re: Visualiing Derivatives with Cubes
Posted: Sep 2, 2013 3:35 PM
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K.U. wrote;
Zubek keeps repeating a few names but there are more. I don't know what
everyone is up to, don't make it to all the conferences (like the SNEC ones
- -- I was in of the founders of SNEC, more so Russell Chu, and see Chris in
Philadelphia maybe once a year).
- -------------------------------------
well yes I have spoke to Chris, to Chu, Koski, you, and who knows who, but it is irrelevant, the point is, the cube contains all those attributes you somehow assuming, are of attributes of the tetra. the 60 deg. coordinate, there is no difference, you assisting that we have to accept some "SYNERGETIC DEFINITIONS" and even if we do
as I have show you zillions of times, 9/8, 8/9 are fraction proportions os square roots.
1.125 sqr.= 1.060660172
Show you with models, that number of blocks = volume, and that is the purpose of tetrahedral modeling that is the purpose of the concentric hierarchy to show the RELATIVE relation of solids.
Two of your cubes = rh. dodeca vol. two of my cubes also = rh. dodeca vol. show you that we are modeling from same cubes JUST dissected differently. YOUR CUBE IS OF VOLUME 8. The beauty is in the simplicity, have show that I model all your structures from the SAME old cube, even all the barycentric cuts of -A- and -B- modules where they vol. is .1111.... x 72 = 8 your cube is of vol. 8
and (1.125)(72)=81 and 81/729=.1111....

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