Seeing that we have the dimensions in connection with Fig. 239, it will, perhaps, be better to explain how to find its volume before passing on. It should be remembered that whilst the calculations that follow apply to the vessel in Fig. 239, the same principle is applicable to all circular articles.
The distance of the centre of gravity of the segment A D E H A from O will equal -
The cube of the chord / 12 times area of segment
= 10.6 X 10.6 X 10.6 / (11.78 X 3.75 - 5.3 X 5.3) X 12 = 6.17 in.
Distance of centre of gravity of segment from centre line equals -
6.17 - 5.3 + 2.5 = 3.37 in.
Then the volume of the vessel equals the volume of the centre cylindrical portion, together with the volume swept out by the segment revolving around the centre line -
Volume = (2.5)2 x 3.1416 x 10.6 „ + 2 x 3.37 x 3.1416 x 16.09 „ = 208.03 + 344.59 „ = 552.62 cubic inches.
To find the number of gallons the above would have to be divided by 277.274 (the number of cubic inches in a gallon). It will thus be seen that the vessel will hold just under two gallons.