298 PYRAMID OF CEPHRENES. - PART V. - CONTENT EQUAL TO CIRCUMFERENCE, CUBE EQUAL TO DISTANCE OF MOON. — THE QUADRANGLE IN WHICH THE PYRAMID STANDS. — SPHERE EQUAL TO CIRCUMFERENCE. — CUBE OF ENTRANCE PASSAGE IS THE RECIPROCAL OF THE PYRAMID. — THE PYRAMIDS OF EGYPT, TEOCALLIS OF MEXICO, AND BURMESE PAGODAS WERE TEMPLES SYMBOLICAL OF THE LAWS OF GRAVITATION AND DEDICATED TO THE CREATOR — EXTERNAL PYRAMID OF MYCERINUS EQUAL TO CIRCUMFERENCE EQUAL TO 19 DEGREES, AND IS THE RECIPROCAL OF ITSELF. — CUBE EQUAL TO CIRCUMFERENCE. — INTERNAL PYRAMID EQUAL TO CIR-CUBE EQUAL TO CIRCUMFERENCE. — THE SIX SMALL PYRAMIDS.—THE PYRAMID OF THE DAUGHTER OF CHEOPS EQUAL TO CIRCUMFERENCE EQUAL TO 2 DEGREES, AND IS THE CUMFERENCE. -- RECIPROCAL OF THE PYRAMID OF CHEOPS. — The Pyramid of Cephrenes. ALL that Herodotus says of the dimensions of the pyramid of Cephrenes is, that they are far inferior to those of Cheops' pyramid, for we measured them. The following are the measurements recently made:— JOMARD. Present height = 138 metres =452-64 feet English Former height =455-64 Jomard supposes this pyramid to have lost about 3 feet from the top, which, if added to the height he has given, 452.64 feet, will make the height to the apex = 455.64 feet, and 4565 feet = 11⁄2 stade. This will make the height to the apex of the pyramid of Cephrenes equal to the height to the platform of the pyramid of Cheops. Let the height to apex = 11⁄2 stade = 4565 feet 394 &c. units and base = 684 by 703 feet then height x base = 592 by 608 units, pyramid = 394 &c. x 592 × 608 = 1⁄2 circumference = = 1 5 of circumference 150 degrees. Thus the height to the apex, 1 stade, will nearly accord with the height assigned by Jomard, Belzoni, and Vyse; and the base, 684 by 703 feet, will somewhat exceed the base of Wilkinson. circumference, 684 x 695 feet, would also very nearly = but then Wilkinson's height to the apex would be much greater than the other measurements, and would make the height to the apex of Cephrenes pyramid = the height to the apex of the pyramid of Cheops. 468 feet 10 plethrons the height to the apex of the pyramid of Cheops; and 466 feet, according to Wilkinson, would the height to the apex of the pyramid of Cephrenes. Thus the dimensions we have assigned to the pyramid of Cephrenes will be that one side of the base = 24 stades =15 plethrons 607-5 units, and the other side = 15 plethrons less 15 units = 592.5 units. Perimeter of base (608+592) × 2 = = 2400 units = 60 plethrons less 30 units height to apex 11⁄2 stade = = 10 plethrons less 10 units. The pyramid of Cephrenes is said by Jomard to rise not from the level of the natural rock, but out of an excavation or deep cut made in the solid rock all round the pyramid. This pyramid, says Greaves, is bounded on the north and west sides by two very stately and elaborate pieces that have not been described by former writers. About 30 feet in depth, and more than 1400 in length, out of a hard rock, these buildings have been cut in perpendicular, and squared by the chisel, as I suppose, for the lodgings of the priests. They run along at a convenient distance, parallel to the two sides of this pyramid, meeting in a right angle. The side of the quadrangle, or one side of half the quadrangle described, exceeds 1400 feet. Five stades = 1405 feet. The side of the quadrangle that enclosed the tower of Belus was twice the side of the base of that pyramid. Supposing the side of the base of the pyramid of Cephrenes to equal half the side of the quadrangle of Greaves, or 2 stades, or 702-5 feet, such a base would nearly agree with 707.9 feet, the former base of Vyse. If each side of rectangle on the north and west sides = 1410 feet 1220 units, = = distance of Mercury from the Sun. Thus the distance of Mercury will 90 cubes, and distance of Belus = 150 times the distance of Mercury 150 x 90 cubes. = The distance of Saturn = 25 times the distance of Mercury 25 x 90 cubes. The assigned base of pyramid=592 × 608 units; 1 (529+ 608)=600. If the side of square base of pyramid = 601 units, and height base = circumference, then 5 times the cube of 5 x 6013-1085409005 units. the side of the base Distance of moon units. 9.55 circumference = 1085730026 Hence 5 times the cube of the side of the base of the pyramid of Cephrenes will 9.55 circumference = distance of the moon from the earth. The cube of Cheops will be to the cube of Cephrenes as 5: 4. Distance of Mercury from the sun will = 150 times the distance of the moon from the earth, 150 × 5=750 cubes of Cephrenes, 150 × 4=600 cubes of Cheops. = Should one side of the base of the pyramid = 610 units, and the other side 592 units, the cube of the greater side will 61032 circumference; the mean of the 2 sides=1 (610+592)=601. The cube of the mean will = 6013 moon. = distance of the A sphere having a diameter = 601 units will = circumference. If the base of pyramid be a square having a side = 601 units, and height = 392, &c., then height x base = 392 &c. x 6012 circumference. Pyramid circumference. = Cube of height cube of side of base :: 3923, &c. : 6013 :: If 12 radii divided the circumference of the earth into 12 equal parts, then pyramid would = 5 of these parts, and the cube of the side of the base would = the 12 radii. The inclined side of the pyramid will = 494 &c. units, distance of the moon. and 4943 &c. = So cube of height: cube of inclined side :: 3923, &c. : 4943, 1 Cube of inclined side cube of side of base ::: distance of the moon Cube of inclined side = cube of side of base. :: 5 : 9. distance of the moon وو = 512 = 4096 Thus 2 cubes of 4 times perimeter of pyramid of Cephrenes = distance of Jupiter = 2 cubes of 4 times perimeter of pyramid of Cheops. = = Sphere having diameter = 601 units side of base of Cephrenes 6013 × 5236 circumference, or sphere of Cephrenes circumference = twice the pyramid of Cheops. Cube of Cephrenes 6013 distance of the moon = = circumference = Cone of Cephrenes pyramid of Cheops Cylinder = circumference height x area of the base of Cheops. = |