568 Mr. Playfair’s Account of a 
any column on the plummet at O, estimated in the direction 
of the meridian ON, if b be the breadth of it in the di- 
rection of the radius, d the difference of the sines of the azi- 
muths of the two edges, E the angle which the length of the 
column subtends at O, if its density were = 1, would be bd 
x sin. E. But if the density of the rock be expressed by any 
other number, the attraction just found must be multiplied by 
that number in order to give A the real attraction of the co- 
lumn. Thus if Q denote the density of the granular quartz, 
and M that of the micaceous schistus, we have in the former 
case A = bd Q. sin. E, and in the latter, A = bd M. sin. E. 
In these formulas b = 666.66 feet, and d — by the con- 
struction already explained ; therefore A = (55.55) O sin. E, 
or — (55-55) sin. E. 
The calculation of sin. E is very easy, for the length of 
each column or its depth below O being 1440 feet, and the 
middle of the first ring being 333.33 feet distant from O ; 
of the second 1000, reckoning from O, if n be the number 
of any ring, the distance of its centre from O is — - - x 1000, 
144 ° 3 X »-44 . 
so that tan. E = 2 n — 1 2 « — 1 
— - — - X IOOO 
The sine corresponding to this tangent taken from the 
tables, and multiplied into 55.55, and the product into O or M, 
will give the attraction of the column. Therefore to have the 
attraction of the ring of columns of the order ?i, the quantity 
now obtained must be multiplied by 12, that being the num- 
ber of columns in one ring, having all by hypothesis the 
same altitude, so that the whole attraction of the ring = 
( 666.66 ) Q sin. E, &c. 
