- EGYPT 239
Calculate 9 of 9, since the basket is half of an egg. The result
is 1. Calculate what is left as 8. Calculate 9 of 8. The result is 3 6- Calculate what is left of this 8 after this 3 6 18 is taken away.
The result is 7 9. Calculate 4 2 times with 7 9. The result is 32.
Behold, this is the surface. You have found it correctly.
If we interpret the basket as being a hemisphere, the scribe has first doubled
the diameter of the opening from 4 2 to 9 "because the basket is half of an egg."
(If it had been the whole egg, the diameter would have been quadrupled.) The
procedure used for finding the area here is equivalent to the formula 2d • | • | · d.
Taking (|)^2 as representing ð/4, we find it equal to (nd?)/2, or 2ðÃ^2 , which is
indeed the area of a hemisphere of radius r.
This value is also the lateral area of half of a cylinder of height d and base
diameter d. If the basket is interpreted as half of a cylinder, the opening would
be square and the number 4 2 would be the side of the square. That would mean
also that the "Egyptian ð" (ð/4 = 64/81), used for area problems was also being
applied to the ratio of the circumference to the diameter. The numerical answer is
consistent with this interpretation, but it does seem strange that only the lateral
surface of the cylinder was given. That would indicate that the basket was open
at the sides. It would be strange to describe such a basket as "half of an egg."
The main reason given by van der Waerden (1963, pp. 33-34) for preferring this
interpretation is an apparent inaccuracy in Struve's statement of the problem. Van
der Waerden quotes Ô. E. Peet, who says that the number 4 2 occurs twice in the
statement of the problem, as the opening of the top of the basket and also as its
depth. This interpretation, however, leads to further difficulties. If the surface
is indeed half of a cylinder of base diameter 4 2, its depth is not 4 2; it is 2 4.
Van der Waerden also mentions a conjecture of Neugebauer, that this surface was
intended to be a domelike structure of a sort seen in some Egyptian paintings,
resembling very much the small end of an egg. That interpretation restores the
idea that this problem was the computation of the area of a nonruled surface, and
the approximation just happens to be the area of a hemisphere.
1.2. Volumes. One of the most remarkable achievements of the Egyptians is the
discovery of accurate ways of computing volumes. As in the case of surface areas,
the most remarkable result is found in the Moscow, not the Ahmose, Papyrus. In
Problem 41 of the Ahmose Papyrus we find the correct procedure used for finding
the volume of a cylindrical silo, that is, the area of the circular base is multiplied
by the height. To make the numbers easy, the diameter of the base is given as 9
cubits, as in Problems 48 and 50, so that the area is 64 square cubits. The height is
10 cubits, giving a volume of 640 cubic cubits. However, the standard unit of grain
volume was a khar, which is two-thirds of a cubic cubit, resulting in a volume of 960
khar. In a further twist, to get a smaller answer, the scribe divides this number by
20, getting 48 "hundreds of quadruple hekats." (A khar was 20 hekats.) Problem 42
is the same problem, only with a base of diameter 10 cubits. Apparently, once the
reader has the rule well in hand, it is time to test the limits by making the data
more cumbersome. The answer is computed to be 1185 6 54 khar, again expressed
in hundreds of quadruple hekats. Problems 44-46 calculate the volume of prisms
on a rectangular base by the same procedure.