Problems of Volume

In this section we find the volumes of various objects

Example:

Problem 41: Find the volume of a cylindrical granary of diameter 9 and height 10.

Take away 1/9 of 9, namely, 1; the remainder is 8. Multiply 8 times 8; it makes 64. Multiply 64 times 10; it makes 640 cubed cubits. Add 1/2 of it to it; it makes 960, its contents in khar. Take 1/20 of 960, namely 48. 4800 hekat of grain will go into it.

Method of working out:

1 becomes 8
2 becomes 16
4 becomes 32
\8 becomes 64

1 becomes 64
\10 becomes 640
\1/2 becomes 320
Total: 960
1/10 becomes 96
\1/20 becomes 48

The method by which volume is found is by subtracting from the diameter its 1/9th and then squaring what is left, then multiplying by the height. Thus 640 cubed cubits is the volume. Since a khar is 3/2 of a cubed cubit (although Chace's editionsays it is 2/3rds) we add the 1/2 of 640 to 640 to get 960, which is the volume in khar. THen we divide by 20 to get it in terms of hundreds of quadruple hekat, for which the answer is 48.

changed November 14, 2008