The volume of a truncated cone is given by the formula V = H/3 (A2 + Aa + a2).
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 (A2 + 2Aa + 3a2)}/(A2 + Aa + a2).
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.