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). I explain how to use the pyramide for solving the problem of duplicating of the cube. But i don't understand, how can be used for this problem a prisma you suggest (you don't described any method of finding a cubic root). 2. "...because the equation of the circle is only quadratic..." - it's true for plane, but volume of the pyramide in three dimensions is cubic, and it is significantly used in this method. For example, with this pyramide you can't find any root of fourth or fifth degree or upper. 3. I prefer to call this method "direct measurement of the volume and splitting along one coordinate". The pyramide was well known to Greeks, and measuring the volume with displaced fluid was used by Archimedes (remember his famous law) - new here is only combination of these method for solving an old unsolvable problem. 4. I can't decide, is this method dirty or not. Using this method for calculating cubic roots, imho - not practical. For today the only use of this method i can imagine - saying several words about it on geometry lecture for university students.
>(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.
I explain how to use the pyramide for solving the problem of duplicating of the cube.
But i don't understand, how can be used for this problem a prisma you suggest (you don't described any method of finding a cubic root).
2. "...because the equation of the circle is only quadratic..." - it's true for plane, but volume of the pyramide in three dimensions is cubic, and it is significantly used in this method. For example, with this pyramide you can't find any root of fourth or fifth degree or upper.
3. I prefer to call this method "direct measurement of the volume and splitting along one coordinate".
The pyramide was well known to Greeks, and measuring the volume with displaced fluid was used by Archimedes (remember his famous law) - new here is only combination of these method for solving an old unsolvable problem.
4. I can't decide, is this method dirty or not. Using this method for calculating cubic roots, imho - not practical.
For today the only use of this method i can imagine - saying several words about it on geometry lecture for university students.
Reply
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.
Reply
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