Mr. Knight on the Penetration , &c. 
311 
ing its base. From every point d, on this side of b, draw the 
radius dC to the centre of the hemisphere, and (if the number 
of cylinders is to be 2 ?i*) take the arc bs equal to n times the 
arc bd, draw se perpendicular to Cb, and with the centre C 
and radius Ce describe the arc er cutting Cd in r. Through 
all the points (r) thus found, draw the curve line brC, termi- 
nated at b and C, and it shall be half the base of one of the 
required cylinders. 
It is, in the first place, evident, from the construction, that 
the half cylinder, whose base is beCrb, is contained between 
two planes CabC, CacC, making with each other the angle 
bCc = ; consequently the whole base of the hemisphere 
may be pierced by 2 n such cylinders as this is the half of. 
Let atmb be the intersection of the surfaces of the half cy- 
linder and hemisphere ; amd a great circle passing through a 
and d, and meeting atmb at m. Call the radius of the sphere 
r, Cr is the cosine of the arc bs to the radius r, by construc- 
tion ; it is also the cosine of the arc md to the same radius ; 
therefore md = bs = n x bd. 
Put bd = <p ; md = n x <p ; d*4 = ij/. Moreover, put A for 
the spherical space atmbdcna contained by the arcs anc, cdb 
and the curve atmb ; and let S be the solidity of the portion of 
the hemisphere contained between the quadrant ancC and the 
surface (brCatmb) of the half cylinder. It is easy to see that 
* I do not intend in to represent an even number only, n may be §, or f , or -f, &c. 
and in express any number whatever. 
S S 2 
