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Area enclosed by a tesseract (4th Dimensional Hypercube)
I was reading this book the other day The Time Machine by H.G. Wells, and he described moving through dimensions. I pondered the thought a little bit, and began wondering. If the area of a square is L x W, and the volume of a cube is L x W x H, then what is the area enclosed by a tesseract or a fourth dimensional hypercube. Is it the amount of space it takes up within multiple dimensions? Maybe... What do you think?
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Re: Area enclosed by a tesseract (4th Dimensional Hypercube)
Well, if you take the idea from The Time Machine, then you would take time to be the fourth dimension. In which case, the volume occupied by the tesseract would be L x W x H x T with "T" being the amount of time occupied.
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Re: Area enclosed by a tesseract (4th Dimensional Hypercube)
W*X*Y*Z
W is the extra dimension in the Minkowski space. |
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This web site claims to have some answers for you. http://www.physicsinsights.org/hypercubes_1.html |
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