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. Considering, then, that one-half of the volume and mass of the earth is contained within 817 miles in depth from the surface, this half must have an average density of 5·66 times that of water, the same as the whole is estimated to have. Also, as we have seen already, that, taking its mean diameter at 7918 miles, its mass will be equivalent to 1,471,168,987,476 cubic miles, one-half of this quantity, or 735,584,493,738 cubic miles will represent the half-volume of the earth reduced to the density of water. With these data let us find out what must be the greatest density where the two half-volumes meet, supposing the densities at the surface and for 9 miles down to remain the same as in the calculations we have already made, ending with specific gravity of 3 at 7900 miles in diameter.