(43) Let us assume that the obelisk is balanced at its C. G., and find the stresses due to bending. The weight on each side will be equal. Taking the right hand half, its weight will act at its C. G. Using the formula for the C. G. of a tapering square-sectioned solid, quoted above, we get: (15.84/4) {[(4.20)^2 + 2 (4.20) (3.49) + 3 (3.49)^2]/[(4.20)^2 + (4.20) (3.49) + (3.49)^2]} = 7.43 metres, which means that the centre of gravity of the right-hand half of the obelisk will act at a distance of 7.43 metres from the butt, or 15.84 − 7.43 = 8.41 metres from the balancing point, or C. G. of the whole obelisk. The sum of the moments to the right of the C. G. of the whole obelisk will be half the total weight multiplied by 8.41 = 584 × 8.41. Then, if _s_ is the internal stress set up due to the bending of the obelisk when supported at its C. G., we have: (Section modulus) (stress) = sum of moments on one side of support. The modulus of the square section is one sixth the cube of the depth, so we have: {[(3.49 × 39.37)^3]/6} _s_ = 584 × 8.41 × 39.37 × 2240. From which _s_ = 1001 pounds per square inch (39.37 being the reduction of metres to inches). {43} The modulus of rupture for granite from Aswân is given as 1500 pounds per square inch, so it will be seen that the obelisk, if not converted into a live load (by a jerk, for instance) can be supported at its C. G. without breaking. It is rather difficult to say how far the Egyptians were able to carry their calculations. The erection could well have been rehearsed by means of a scale-model, which could have been further used for obtaining the weight and the position of the centre of gravity. I do not think that they ever troubled about the bending-moment; at any rate, their mathematics were not sufficiently advanced for its determination. It may be that, since in all the obelisks we know of, whose taper does not vary to any great extent, can be supported anywhere, the Egyptians never had a case of such a monument breaking by its own weight. Another interesting point arises in connection with this, and that is, since in obelisks (and all beams) of the same proportion, the bending stress due to their own weight depends on the linear dimension, and therefore the fact that a granite scale-model does not break will be no indication that the monument itself will not break when similarly supported. If the 108 cubit (56.70 metres) obelisk of Ḥatshepsôwet, mentioned by Ṭḥutiy (section 3), does indeed apply to _one_ and not to the _two_ placed butt to butt on the boat shewn in the Dêr el-Baḥari sculpture, then, if the proportions are about the same as the Aswân obelisk, the stress set up when supported at its centre of gravity (see section 40) would be in the nature of 56.70 × 1001/41.75 = 1360 pounds per square inch, which is perilously near the breaking stress of 1500 pounds per square inch.