Let us assume that this method was to be used for the Aswân obelisk. I think that the largest levers practicable would be 15 metre tree-trunks used with a mechanical advantage of 10 to 1. Not more than 30 levers could be used on each side of the obelisk. The number of men required to raise the obelisk can be found to be about 56 per lever, assuming that they all heave at the end. I hardly see how such a number can be put on a horizontal lever unless we assume that a cross-baulk is attached along the ends of the levers and the whole loaded with stones. The levers would have to be dismounted at each heave and the time taken would be considerable. The method, however, is a possibility, so I include it as an alternative to the embankment. {42} (42) Before leaving this subject, it is as well to ascertain if the obelisk is strong enough to bear the internal strain due to its own weight when it is supported at its centre of gravity. The volume of a truncated cone is given by the formula V = H/3 (A^2 + A⁠_a_ + _a_^2). In the shaft of this obelisk, H is 37.25, A is 4.20 and _a_ is 2.50 metres. Substituting, we have: V = 37.25/3{(4.2)^2 + 4.2 × 2.5 + (2.5)^2} from which we find that the volume of the shaft is 426 cubic metres. Aswân granite weighs about 2.679 tons per cubic metre, which makes the shaft weigh 1143 tons. The weight of the pyramidion is {(base)^2 (height) (unit weight)}/3⁠, or (2.50)^2 (4.50) (2.679)/3 = 25 tons, so that the total weight of the obelisk would have been 1143 + 25 = 1168 tons. The distance of the centre of gravity of a tapering square-sectioned solid from the butt is given by the formula: {¼ H (A^2 + 2⁠A⁠_a_ + 3⁠_a_^2)}/(A^2 + A⁠_a_ + _a_^2). Here H is 37.25 m.; A = 4.2 m.; _a_ = 2.5 m. Substituting we get: (37.25/4) {[(4.20)^2 + 2 (2.50 × 4.20) + 3 (2.50)^2]/[(4.20)^2 + (2.50 × 4.20) + (2.50)^2]}. That is, the distance of the C. G. from the butt, (L⁠N on fig. 11), is 15.35 metres. Taking the pyramidion by itself. Its height is 4.50 metres, so that its C. G. must be one-fourth that distance from the base, which makes 1.12 metres. If _x_ is the distance of the centre of gravity of the whole obelisk from the butt, by taking moments about the butt we have: (Total weight) × _x_ = (weight of pyramidion) × (1.12 + length of shaft) + (weight of shaft) × 15.35, or 1168 _x_ = 25 × 38.37 + 1143 × 15.35, from which _x_ = 15.84. _That is, the distance of the centre of gravity of the whole obelisk from the butt is 15.84 metres._ The breadth of the obelisk at its centre of gravity is 4.2 − (15.84/37.25) × (4.2 − 2.5) or 3.49 metres.