NOVEMBEE 10, 1911] 



SCIENCE 



641 



instead of the associate, of the other, thus 

 separating dissimilar elements like copper and 

 potassium. 



The elements are placed in the order of their 

 atomic weights, and with interspaces equal to 

 the average difference of the atomic weights 

 of adjacent elements. These differences are 

 found to increase for the different groups 

 of rings in the simplest arithmetical ratio 

 12 3 4, where, since the atomic weights 

 are rarely whole numbers, the ratio numbers 

 will not be exact digits, and the first term 

 will be some very small number instead of 

 exactly zero. The regularly recurring resem- 

 blances of the elements break them up into 

 groups determining the number of circles of a 

 given size and the number of interspaces and 

 of elements upon each kind of circle; and all 

 these relations are according to the same 

 simplest geometrical ratio, 2 4 8 16 32. The 

 length of each circle is obtained by combining 

 these ratios. 



Starting with the best determined circles, 

 the lithium ring covers a range of atomic 

 weights equal to sixteen (H — Ne = 16) and 

 contains eight elements (an octant), and so the 

 average interspace is put as two. The sodium 

 series is wholly symmetrical with the Li ring, 

 and so is put on a second equal coil of a spiral 

 beside it. The third ring agrees exactly with 

 the other two from potassium to titanium, 

 potassium having valence and sp. gr. about 

 like sodium and so on to titanium, which has 

 valence and sp. gr. about 4, like silicon. 

 Here the resemblance ceases, and vanadium, 

 instead of agreeing with phosphorus, con- 

 tinues downward with valence and sp. gr. 5, 

 and so on to the iron triad group with valence 

 and sp. gr. equal to 8; and then the curve 

 turns ; valence and sp. gr. grow less (copper = 

 7, zinc = 6, etc.), until the last half of this 

 band from Ge to Br is exactly homologous 

 with the rearward half (in Fig. 1) of the 

 preceding circles. Thus is established a sec- 

 ond larger type of circle containing sixteen 

 elements (a double octant), and making an 

 advance in atomic weight of about 48, so that 

 the interweight becomes about 3 and the inter- 

 space in this circle is made 3. This changes 



the spiral into a helix. Four such circles can 

 be constructed. If there are two octant circles 

 with an interspace of 2, and four double 

 octants with interspace 3, there should be 8 

 quadruple octants with interspace 4, but the 

 curve advances only one quadrant of the first 

 circle of this type, and does this with an inter- 

 space of 4, but becomes so complex that it falls 

 asunder spontaneously, giving up atoms of 

 helium whose combining weight is 4, indica- 

 ting that the additions in this last curve have 

 been by fours. 



In the other direction symmetry demands a 

 diminishing of the helix to a single half octant 

 with interspace 1, and hydrogen stands at the 

 beginning and helium (4) at the end of this 

 ring. One must search among the nebulae for 

 elements light enough to fill the two gaps, 

 and we find among the simpler spectra in 

 the simplest nebulse the lines 4,340 t.m. and 

 4,862 t.m. belonging to hydrogen, the next 

 higher line, 4,959 t.m., belonging to nebulium, 

 which has probably greater density than hy- 

 drogen because it is found more concentrated 

 in the center of the nebulse, for which reasons 

 we may assume nebulium to be a dyad and to 

 take the second place in our circle. The next 

 higher line, 5,007 t.m., I have tentatively as- 

 signed to the next element in this half-octant, 

 which, from its position, must be a halogen, 

 and which I have called proto-fluorine. The 

 next higher number, 5,876 t.m., belongs to 

 helium — which completes the circle. 



The helix must close with the half of a 

 quarter-octant — a single element which must 

 have valence = 0, a density much less than 

 hydrogen, and atomic weight much less than 

 unity. Where would one search for such an 

 element better than in the corona? Indeed, 

 coronium is found to be probably lighter than 

 hydrogen, since its lines are found further 

 from the sun than those of hydrogen. It occu- 

 pies also the position of the second element 

 extrapolated by Mendeleeff for which he ob- 

 tained the atomic weight 0.4, and suggested 

 its identity with coronium. I have, by a sim- 

 ilar method, obtained the value 0.3, and the 

 symmetry of the helix would suggest that it 

 should be still smaller. 



