Returning to our pyramids. We know that the centre of gravity of a pyramid is at one-fourth of its height, or distance from the base, and if we lay one of 3959 miles long (the radius of the earth) over a fulcrum, so that 989-3/4 miles of its length be on one side of it and 2969-1/4 miles on the other, it will be in a state of equilibrium. This does not mean, however, that there are equal masses of matter on each side of the fulcrum, for we know that the mass of the base part must be considerably greater than that of the apex part, and that it must be counterbalanced by the greater leverage of the apex part, due to its greater distance from the point of support. This being so, in the case of a pyramid consisting of gasiform, liquid, or solid matter, the attractive power of the 989-3/4 miles of the base part would be greater than that of the 2969-1/4 miles of the apex part, and the plane of equal attraction of the two parts would be less than 990 miles from the base of the pyramid. This is virtually the same argument we have used before repeated, but it is placed in a simpler and more practical light, and shows that the plane of attraction in a pyramid will not be at its centre of gravity but nearer to its base, and that it must be at or near its centre of volume. Thus the plane of attraction in one of the pyramids we have been considering of 3959 miles in length, and consequently the radial distance of the region of maximum attraction of the earth, would not be at 990 miles from the base or surface, but at some lesser distance.
Now, if we take a pyramid, such as those we have been dealing with, whose base is 1 square and height 3959, its volume would be the square of the base multiplied by one-third of the height, that is 1^{ 2} × 3959/3 = 1319·66, the half of which is 659·83. Again, if we take the plane of division of the volume of the pyramid into two equal parts to be 0·7937 in length on each side, and consequently (from equal triangles) the distance from the plane to the apex to be 0·7937 the total height of 3959, which is 3142·258; then, as we have divided it into a frustum and a now smaller pyramid, if we multiply the square of the base of this new pyramid by one-third of the height we have 0·7937^{ 2} × 3142·258/3, or 0·62996 × 1047·419 = 659·83, which is equal to the half-volume of the whole pyramid as shown above. Thus we get 3959 less 3142·258 = 816·74 miles as the distance from the base of the plane of division of the pyramid into two equal parts, which naturally agrees with the division of the earth into the two equal volumes that we have extracted from the table of calculations, where we have supposed the earth to be made up of the requisite number of such pyramids. So that it would seem that we are justified in considering that the greatest density of the earth must be at the meeting of the two half-volumes, outer and inner, into which we have divided it.
