# Problems of Volume

### In this section we find the volumes of various objects

**Example:**

Problem 41:Find the volume of a cylindrical granary of diameter9and height10.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. 4800hekatof grain will go into it.

Method of working out:1 becomes 8

2 becomes 16

4 becomes 32

\8 becomes 641 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.