The Sphere and least Circumscribing Cone. 229 
ar(eo+rr 
We have seen that the convex surface =2°- , abe 
ar e+r 
and that the base = r mis Sa op 
and since x = 37, the convex surface is three times the base, and 
the base, consequently, one fourth of the entire surface of the 
cone. ‘Therefore the base of the cone is equal to the surface of 
the hemisphere, and the convex surface, to three times the same. 
Hence it is obvious that the ratio of 3 to 1 cannot belong to the 
in those mathematical works to which I have access. 
_ I will notice but one other circumstance relating to these bod- 
les; which is, that just one-third of the spheric surface is cut off 
on the upper side by the circle of contact. For AC: BC:: 
BC: FC; therefore FC is one-third of GC, and GF one-third of 
GD, and hence the surface BGH is one-third of the entire spheric 
surface. It appears, then, that if a sphere be viewed from a point, 
whose distance from its surface equals its diameter, one third of 
the surface is visible. 
As this is one instance of the application of a very simple for~ 
mula, for determining the amount of spheric surface visible at 
any given distance, I will take this opportunity of presenting it. 
If the number of diameters expressing the distance, be made the 
humerator of a fraction, whose denominator is twice the same 
number plus one, this fraction will express the portion of surface 
visible at that distance. Thus, at the distance of 
i 
1 diameter, } is seen, 
2 od ee te 
3 ¢é 3 it 
10, “ 439 “ . and generally, 
s —s te whether x be integral or frac- 
tional ; approaching nearer one-half, as the distance increases, 
Without the possibility of ever equaling it. t emonstra- 
tion of this formula is only a generalization of the example al- 
teady given, I omit it. 
The earth is about 115 lunar diameters from the moon, and 
therefore we can see 318 (=°498) of her entire surface: while, 
tom the moon, only 22 (='491) of the earth’s surface can be 
seen. Of the solar surface, on the other hand, we have about -498 
in view, while -49998 of the earth’s surface is visible from the 
Sun. So an observer, elevated one mile from the earth, sees 5,'55 
> (sss5+1) of its surface, = 7,';5, or about 25,000 square miles. 
