I had the problem, that I do not see how to construct the angle at the base of the pyramide. However I thought, that things become easier if one simplifies the geometry
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1. This reversed pyramide without bottom was invented for finding cubic roots (filling its volume by liquid - which volume is equal volume of two cubes - value that we seek - is the side of the square on the upper surface of the liquid in the pyramide
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>(you don't describe any method of finding a cubic root).
Of course, I do not. I am basically doing nothing else but using your method. Rewriting it for myself has clarified the things for me. I am now convinced that one can either use an inverted pyramide or the prisma (put on its tip) in rather equivalent ways.
I am basically using your equations. (I am not sure whether my way to write potencies is generally understood: a**3 = a * a * a).
The only other difference is, that I replace h in the volume formula in a different way. Thereby you calculate k = SA/L, while I calculate something I should not have called k, but may be q = h/L.
For the pyramide I get h = 3L, which is consistent with k = (9 + 1/2)**1/2.
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Of course, I do not. I am basically doing nothing else but using your method. Rewriting it for myself has clarified the things for me. I am now convinced that one can either use an inverted pyramide or the prisma (put on its tip) in rather equivalent ways.
I am basically using your equations. (I am not sure whether my way to write potencies is generally understood:
a**3 = a * a * a).
The only other difference is, that I replace h in the volume formula in a different way. Thereby you calculate k = SA/L, while I calculate something I should not have called k, but may be q = h/L.
For the pyramide I get h = 3L, which is consistent with k = (9 + 1/2)**1/2.
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