vol. LXl.] PHILOSOPHICAL TRANSACTIONS. 213 



versally, n being the number of rows, the number of the interstices will be 

 {2m — 2a) X {n — \) — 2b X {n —2) — 2c X (n — 3), &c. That is, 2m X 

 (b _ I) —2a X (n — \) — 2b X {n — 2) — 2c X (n — 3), &c. By com- 

 paring this expression with the former expression of the number of the circles, 

 it will appear, that when n, the number of the rows of circles, is infinitely aug- 

 mented, the number of interstices is to the number of circles, ultimately as 2 to 

 1 . For the two expressions always consist of an equal number of terms. The 

 same numerical terms in both are affected with the same signs. The first term 

 of the latter, 2m X (n — l), is ultimately double the first term of the former, 

 mn, when n is infinitely increased, and each succeeding term of the latter is 

 double the corresponding term of the former. Therefore the whole of the latter 

 expression, is ultimately to the whole of the former, as 2 to l. That is, the 

 i.Mmber of interstices is ultimately double the number of circles : whence it fol- 

 lows, as in the former case, that the whole space covered by the circles, is to 

 the whole space occupied by the interstices, as half the area of one circle, to the 

 whole area of one interstice. 



In this demonstration, I have supposed the number of circles on the several 

 lines AG, HP, Qx, &c. to decrease continually. Had I supposed them to decrease 

 by fits, and in any manner imaginable, still the conclusion would have been the 

 same.* Therefore let the figure of the finite space, including the circles thus 

 closely arranged, with their interstices, be what it will, the proportion of the 

 space covered by all the circles, to the space taken up in interstice, is ultimately 

 that of half the area of one circle to the whole area of one interstice. 



Now, that this is the proportion of 39 to 4, very nearly, will appear by com- 

 puting one of the interstitial areas. The method of computing the interstitial 

 area is obvious. Let a, b, h be the centres of the 3 circles, which close the in- 

 terstice T*^. Join AB, AH, BH. The right lines ab, ah, bh, pass through the 



• Suppose the number of circles on the first row to be m, on the 2d, m — a, on the 3d, m — a -\- b, 

 on the 4th, m — a + b — c, on the 5th, m— a + b— c + d, and so on, and each of these num- 

 bers to be infinitely increased. Then, n being the number of rows, the whole number of circles 

 will be nm — a y. (n — 1) + b X (n — 2) — c x (n — 3) + d x (« — 4), &c. Number jthe 

 interstices formed by every 2 contiguous rows, and add them all together, and the whole number 

 of interstices will be found to be 2ffi x (« — ))— 2a x (n — 1) + 26 x (/t — 3) — 2c x (n — 3) 

 + '2d X (n — 5), &c. Now, by comparing these two expressions, it appears that both consist of 

 the same number of terms : that the same numerical terms in order from the first, have the same 

 signs : that the first term of the latter, 2m x (n — 1), is ultimately the double of the first term of 

 the former, when n is infinitely increased : that of the terms following the first, the negative terms 

 of the latter are each double the corresponding negative terms of the former: and each positive term 

 of the latter differs from the double of the corresponding positive term of the former, by a number 

 which vanishes with respect to either of those corresponding terms, when n becomes infinite. There- 

 fore, when n becomes infinite, the whole of the latter expression becomes the double of the whole 

 of the former. Hence the conclusion is as before. — Orig. 



