[9] The check the stress in the levers. Referring to figure 6, (Stress) (Section modulus) = Sum of moments on one side of fulchrum, _i. e._ (s × .0982 × (25)^3) = (1170 × 2240 × 216)/(30 × 21) = 586 pounds per square inch, which is well within the powers of any wood. [10] It will be seen that, if the obelisk lies at too low a level to be rolled _downwards_ to the valley, it can be raised by tilting backwards and forwards by means of levers acting from the north and south trenches alternately, as mentioned in section 21. If the butt were raised even a metre above its present level, it would enormously reduce the quantity of rock to be removed before the obelisk could be rolled out. Then, about Q, the moment of the horizontal force of the ropes round the obelisk to the moment of the weight will be, from the figure, as 9 to 2, so if _n_ be the total number of men required to pull the obelisk over, then (_n_ × 100) = (2 × 1170 × 2240)/9 which gives 5824 men as against the 8000 men which would be required if the levers were not used. It is an enormous number, but I do not see how they could manage with less. A bank of sand just in front of the lower edge of the obelisk would make the second turn an easy matter, and if from thence the obelisk is rolled downwards on soft sand, I think that the 5824 men will still be ample, as the sand can be undercut in front of the edge and so make the rolling approximate to that of a cylinder. (24) As to the size of the ropes required for the rolling out of the obelisk, all we can do is to obtain a very rough idea as to it. If they spread the men out slightly fanwise, I do not see how they could have used more than 40 ropes. The strain per rope will be, as we have seen, (2/9 × 1170/40) = 6.5 tons per rope. The rope used was probably the very best palm-rope, newly made. The safe load which can be put on coir rope, which is of about the same strength, is given by the formula: Load in cwts = (Circumference in inches)^2 divided by 4 (_Military Engineering_, 1913, Part III A, p. 49). Substituting, we have (6.5 × 20 × 4) = C^2 which gives a circumference of 22.8 inches and a diameter of 7 ⁠¼ inches. If such a rope were used it would require handling loops on it.