Barrel Shaped Vessel 263Barrel Shaped Vessel 264

Fig. 239.

For the benefit of those readers who can manipulate figures, we will now explain how to obtain the radius of the pattern disc by calculation, and then afterwards show how the same result can be found by construction. The arc A D E, in Fig. 239, which is described from the centre, O, is a quarter of a circle, of radius 7 in.; hence its length will be -

15 X 3.1416 / 4 = 11.78 in.

The length of the chord A E will be -

Barrel Shaped Vessel 265 = 10.6 in.

The distance O G, from the rule stated above, will then be -

The distance of the centre of gravity, G, from the centre line C F will equal-

OG - OH + AC.

The diameter A B = 5 in., and in this case O H = H E = 5.3 in.; hence -

K G = 6.75 - 5.3 + 2.5 = 3.95 in.

Having obtained the distance of the centre of gravity from the centre line, the radius of the pattern disc can now be calculated, as in connection with Fig. 238.

Let l = length of arc. " r1 = radius of bottom. " r = distance of centre of gravity from centre line.

Then -

R =Barrel Shaped Vessel 266

= √99.3 = 10 in. (nearly).

Whilst, perhaps, somewhat uninteresting, the calculations, as explained above, are very important, and have wide application in finding the areas of surfaces of this character. They can also be applied to finding the capacity or cubic contents of vessels such as those shown in Figs. 239, 240, and 245.